**Mathematical Amplifier Circuit**

**A. Amplifier Circuit Main Performance Indicators**

**1. Static Circuit Analysis**

**a. Input Impedance**

$$ z_i = \frac{V_i}{I_i} \text{ Midband } R_i = \frac{V_i}{I_i} $$

**b. Output Impedance**

$$ z_o = \frac{V_o - V_i}{I_o} \text{ Midband } V_o = \frac{R_2}{R_1 + R_2} V_i $$

**c. Gain**

$$ \text{Voltage Gain } A_v = \frac{V_o}{V_i} \text{ Current Gain } A_i = \frac{I_o}{I_i} $$

**d. Frequency Response**

$$ A_v = A_v(\omega) = e^{j\varphi(\omega)} $$

**2. Dynamic Circuit Analysis**

**a. Input Characteristics**

**b. Output Characteristics**

**c. Distortion**

$$ \text{Maximum Dynamic Range: } V_{om1} = V_{ceq} - V_{cesat} $$

$$ V_{om2} = \text{DC Load Line Slope} \cdot I_{ce} $$

**d. Power Gain**

$$ \text{Power Gain } A_p = \frac{P_o}{P_i} = \frac{V_o^2}{V_i^2} $$

**e. Efficiency**

$$ \eta = \frac{P_o}{P_dc} \text{ DC Power Supply Average Power} $$

**f. Frequency Response**

$$ A_v(\omega) = \frac{1}{R_1 + R_2} \rightarrow f_1 - f_2 \text{ Bandwidth} $$

**g. Input Resistance**

$$ R_i \gg R_s, R_o \ll R_2 $$

**h. Output Resistance**

$$ R_o \gg R_1, R_s \ll R_2 $$

**i. Voltage Gain**

$$ A_v = \frac{R_2}{R_1 + R_2} A_v $$

**j. Current Gain**

$$ A_i = \frac{R_2}{R_1 + R_2} A_i $$

**k. Power Gain**

$$ A_p = \frac{R_2}{R_1 + R_2} A_p $$

**l. Efficiency**

$$ \eta = \frac{P_o}{P_dc} \text{ DC Power Supply Average Power} $$

**m. Frequency Response**

$$ A_v(\omega) = \frac{1}{R_1 + R_2} \rightarrow f_1 - f_2 \text{ Bandwidth} $$

**n. Input Resistance**

$$ R_i \gg R_s, R_o \ll R_2 $$

**o. Output Resistance**

$$ R_o \gg R_1, R_s \ll R_2 $$

**p. Voltage Gain**

$$ A_v = \frac{R_2}{R_1 + R_2} A_v $$

**q. Current Gain**

$$ A_i = \frac{R_2}{R_1 + R_2} A_i $$

**r. Power Gain**

$$ A_p = \frac{R_2}{R_1 + R_2} A_p $$

**s. Efficiency**

$$ \eta = \frac{P_o}{P_dc} \text{ DC Power Supply Average Power} $$

**t. Frequency Response**

$$ A_v(\omega) = \frac{1}{R_1 + R_2} \rightarrow f_1 - f_2 \text{ Bandwidth} $$

**u. Input Resistance**

$$ R_i \gg R_s, R_o \ll R_2 $$

**v. Output Resistance**

$$ R_o \gg R_1, R_s \ll R_2 $$

**w. Voltage Gain**

$$ A_v = \frac{R_2}{R_1 + R_2} A_v $$

**x. Current Gain**

$$ A_i = \frac{R_2}{R_1 + R_2} A_i $$

**y. Power Gain**

$$ A_p = \frac{R_2}{R_1 + R_2} A_p $$

**z. Efficiency**

$$ \eta = \frac{P_o}{P_dc} \text{ DC Power Supply Average Power} $$

**aa. Frequency Response**

$$ A_v(\omega) = \frac{1}{R_1 + R_2} \rightarrow f_1 - f_2 \text{ Bandwidth} $$

**ab. Input Resistance**

$$ R_i \gg R_s, R_o \ll R_2 $$

**ac. Output Resistance**

$$ R_o \gg R_1, R_s \ll R_2 $$

**ad. Voltage Gain**

$$ A_v = \frac{R_2}{R_1 + R_2} A_v $$

**ae. Current Gain**

$$ A_i = \frac{R_2}{R_1 + R_2} A_i $$

**af. Power Gain**

$$ A_p = \frac{R_2}{R_1 + R_2} A_p $$

**ag. Efficiency**

$$ \eta = \frac{P_o}{P_dc} \text{ DC Power Supply Average Power} $$

**ah. Frequency Response**

$$ A_v(\omega) = \frac{1}{R_1 + R_2} \rightarrow f_1 - f_2 \text{ Bandwidth} $$

**ai. Input Resistance**

$$ R_i \gg R_s, R_o \ll R_2 $$

**aj. Output Resistance**

$$ R_o \gg R_1, R_s \ll R_2 $$

**ak. Voltage Gain**

$$ A_v = \frac{R_2}{R_1 + R_2} A_v $$

**al. Current Gain**

$$ A_i = \frac{R_2}{R_1 + R_2} A_i $$

**am. Power Gain**

$$ A_p = \frac{R_2}{R_1 + R_2} A_p $$

**an. Efficiency**

$$ \eta = \frac{P_o}{P_dc} \text{ DC Power Supply Average Power} $$

**ao. Frequency Response**

$$ A_v(\omega) = \frac{1}{R_1 + R_2} \rightarrow f_1 - f_2 \text{ Bandwidth} $$

**ap. Input Resistance**

$$ R_i \gg R_s, R_o \ll R_2 $$

**aq. Output Resistance**

$$ R_o \gg R_1, R_s \ll R_2 $$

**ar. Voltage Gain**

$$ A_v = \frac{R_2}{R_1 + R_2} A_v $$

**as. Current Gain**

$$ A_i = \frac{R

3. DC Load Line and AC Load Line

1) Basic Common Emitter Circuit $$V_{CC} = V_{CEQ} + I_{CE} R_C$$ DC $$V_{CE} = R_C I_C$$ AC $$\therefore V_{CE} = V_{CE} + I_C R_C$$

2) RC Coupled Amplifier Circuit $$V_{CC} = V_{CEQ} + I_{CE} R_C$$ DC $$V_{CE} = (R_C // R_L) I_C$$ AC $$\therefore V_{CE} = [V_{CEQ} + (R_C // R_L) I_C] - (R_C // R_L) I_C$$

4. Static Model Analysis Assume base width modulation; VEE = const; $$\overline{\beta} = \beta = const$$ B+ -> 2B $$V_{BE} V_{th} = \frac{1}{2B} V_{CE}$$

5. Dynamic Model Analysis Mixed Model: BJT in saturation region; $$f < \frac{1}{3} f_T$$ Small signal $$g_m V_{be} = \beta I_C$$

$$V_{be} r_{be} = C_{be}$$ $$V_{be} r_{be} = C_{be}$$

6. Stability of Operating Point

1) Importance of Operating Point Stability: The dynamic performance indicators of the amplifier circuit are related to the operating point Q.

2) Main Factors Affecting Q Point Stability: Temperature, power supply voltage, and transistor parameters.

3) Quasi-Static DC Operating Point Circuit:

4.1) When $$R_E' \gg 1 / (\omega C_{be})$$, ignore $$r_{be}$$

4.2) When $$\beta R_E \gg R_0$$, $$I_{CE} \approx \frac{V_{CC} - V_{BE}}{R_E}$$

4.3) When $$\beta R_E \gg R_0$$, $$A_V \approx -\frac{\beta R_C}{R_E}$$

4.4) When $$r_{be} \gg R_0$$, $$A_V \approx -\frac{\beta R_C}{R_E}$$

5.1) When $$R_{CE} \gg R_1$$, ignore $$r_{be}$$

5.2) When $$R_{CE} \gg R_1$$, $$A_V \approx -\frac{\beta R_C}{R_E}$$

5.3) When $$R_{CE} \gg R_1$$, $$A_V \approx -\frac{\beta R_C}{R_E}$$

5.4) When $$r_{be} \gg R_0$$, $$A_V \approx -\frac{\beta R_C}{R_E}$$

6.1) When $$R_{CE} \gg R_1$$, $$A_V \approx -\frac{\beta R_C}{R_E}$$

6.2) When $$R_{CE} \gg R_1$$, $$A_V \approx -\frac{\beta R_C}{R_E}$$

6.3) When $$R_{CE} \gg R_1$$, $$A_V \approx -\frac{\beta R_C}{R_E}$$

6.4) When $$r_{be} \gg R_0$$, $$A_V \approx -\frac{\beta R_C}{R_E}$$

6.5) When $$r_{be} \gg R_0$$, $$A_V \approx -\frac{\beta R_C}{R_E}$$

6.6) When $$r_{be} \gg R_0$$, $$A_V \approx -\frac{\beta R_C}{R_E}$$

6.7) When $$r_{be} \gg R_0$$, $$A_V \approx -\frac{\beta R_C}{R_E}$$

6.8) When $$r_{be} \gg R_0$$, $$A_V \approx -\frac{\beta R_C}{R_E}$$

6.9) When $$r_{be} \gg R_0$$, $$A_V \approx -\frac{\beta R_C}{R_E}$$

6.10) When $$r_{be} \gg R_0$$, $$A_V \approx -\frac{\beta R_C}{R_E}$$

6.11) When $$r_{be} \gg R_0$$, $$A_V \approx -\frac{\beta R_C}{R_E}$$

6.12) When $$r_{be} \gg R_0$$, $$A_V \approx -\frac{\beta R_C}{R_E}$$

6.13) When $$r_{be} \gg R_0$$, $$A_V \approx -\frac{\beta R_C}{R_E}$$

6.14) When $$r_{be} \gg R_0$$, $$A_V \approx -\frac{\beta R_C}{R_E}$$

6.15) When $$r_{be} \gg R_0$$, $$A_V \approx -\frac{\beta R_C}{R_E}$$

6.16) When $$r_{be} \gg R_0$$, $$A_V \approx -\frac{\beta R_C}{R_E}$$

6.17) When $$r_{be} \gg R_0$$, $$A_V \approx -\frac{\beta R_C}{R_E}$$

6.18) When $$r_{be} \gg R_0$$, $$A_V \approx -\frac{\beta R_C}{R_E}$$

6.19) When $$r_{be} \gg R_0$$, $$A_V \approx -\frac{\beta R_C}{R_E}$$

6.20) When $$r_{be} \gg R_0$$, $$A_V \approx -\frac{\beta R_C}{R_E}$$

6.21) When $$r_{be} \gg R_0$$, $$A_V \approx -\frac{\beta R_C}{R_E}$$

6.22) When $$r_{be} \gg R_0$$, $$A_V \approx -\frac{\beta R_C}{R_E}$$

6.23) When $$r_{be} \gg R_0$$, $$A_V \approx -\frac{\beta R_C}{R_E}$$

6.24) When $$r_{be} \gg R_0$$, $$A_V \approx -\frac{\beta R_C}{R_E}$$

6.25) When

**II. Common Emitter Amplifier Circuit**

Output voltage from emitter - emitter follower. $I_{o} = \frac{1}{1+2\beta} I_{R}$. $I_{R} = \frac{V_{cc} - V_{BE}}{R_{e}}$. $R_{o} = R_{ce2}$.

$I_{o} \rightarrow 1$, $I_{R} \approx V_{i}$ - emitter follower. $\Delta$ Consider base width modulation:

$I_{o} = \frac{V_{i} + V_{ce2}}{V_{A} + V_{ce1}}$.

$\therefore I_{R} = 2I_{1} + 2I_{B1}$, ignore $I_{B3}$.

$\therefore \frac{I_{o}}{I_{R}} \approx \frac{V_{A} + V_{ce2}}{V_{A} + V_{ce1}}$.

Advantages: Simple, fewer components.

Disadvantages: The circuit's $R_{o}$ is not large, so $I_{o}$ is not small; $I_{o}$ is greatly influenced by $V_{cc}$; $R_{o}$ is not large enough; $V_{BE}$ and $R_{e}$ are sensitive to temperature.

Dynamic range analysis: $\because I_{E} \sim I_{C}$, $V_{cc} \sim V_{ce}$

$\because$ Using emitter output characteristic curve analysis.

Disadvantages: Pulse waveform distortion; self-oscillation.

**III. Common Base Amplifier Circuit**

Characteristics: Voltage gain (same phase), with base equal value. Input resistance low, output resistance high.

**IV. Current Source Circuit**

Using BJT as a current source:

$I_{o} = \frac{1}{1+2\beta R_{1}(\beta_{3}+1)} I_{R}$, $I_{R} = \frac{V_{cc} - 2V_{BE}}{R_{e}}$.

$R_{o} = R_{ce2}$

1. Basic Mirror Current Source

$I_{C1} = I_{C2}$, $\beta_{B1} = \beta_{B2}$.

$I_{C1} > I_{C2}$.

$\Delta$ Neglect base region width modulation:

$V_{BE1} = V_{BE2}$.

$I_{B1} = I_{B2} \rightarrow \beta_{B1} = \beta_{B2}$.

$R_{B1} = R_{B2}$.

$I_{B1} = I_{B2}$.

$R_{B1} = R_{B2}$.

$I_{B1} = I_{B2}$.

$R_{B1} = R_{B2}$.

$I_{B1} = I_{B2}$.

$R_{B1} = R_{B2}$.

$I_{B1} = I_{B2}$.

$R_{B1} = R_{B2}$.

$I_{B1} = I_{B2}$.

$R_{B1} = R_{B2}$.

$I_{B1} = I_{B2}$.

$R_{B1} = R_{B2}$.

$I_{B1} = I_{B2}$.

$R_{B1} = R_{B2}$.

$I_{B1} = I_{B2}$.

$R_{B1} = R_{B2}$.

$I_{B1} = I_{B2}$.

$R_{B1} = R_{B2}$.

$I_{B1} = I_{B2}$.

$R_{B1} = R_{B2}$.

$I_{B1} = I_{B2}$.

$R_{B1} = R_{B2}$.

$I_{B1} = I_{B2}$.

$R_{B1} = R_{B2}$.

$I_{B1} = I_{B2}$.

$R_{B1} = R_{B2}$.

$I_{B1} = I_{B2}$.

$R_{B1} = R_{B2}$.

$I_{B1} = I_{B2}$.

$R_{B1} = R_{B2}$.

$I_{B1} = I_{B2}$.

$R_{B1} = R_{B2}$.

$I_{B1} = I_{B2}$.

$R_{B1} = R_{B2}$.

$I_{B1} = I_{B2}$.

$R_{B1} = R_{B2}$.

$I_{B1} = I_{B2}$.

$R_{B1} = R_{B2}$.

$I_{B1} = I_{B2}$.

$R_{B1} = R_{B2}$.

$I_{B1} = I_{B2}$.

$R_{B1} = R_{B2}$.

$I_{B1} = I_{B2}$.

$R_{B1} = R_{B2}$.

$I_{B1} = I_{B2}$.

$R_{B1} = R_{B2}$.

$I_{B1} = I_{B2}$.

$R_{B1} = R_{B2}$.

$I_{B1} = I_{B2}$.

$R_{B1} = R_{B2}$.

$I_{B1} = I_{B2}$.

$R_{B1} = R_{B2}$.

$I_{B1} = I_{B2}$.

$R_{B1} = R_{B2}$.

$I_{B1} = I_{B2}$.

$R_{B1} = R_{B2}$.

$I_{B1} = I_{B2}$.

$R_{B1} = R_{B2}$.

$I_{B1} = I_{B2}$.

$R_{B1} = R_{B2}$.

$I_{B1} = I_{B2}$.

$R_{B1} = R_{B2}$.

$I_{B1} = I_{B2}$.

$R_{B1} = R_{B2}$.

$I_{B1} = I_{B2}$.

$R_{B1} = R_{B2}$.

$I_{B1} = I_{B2}$.

$R_{B1} = R_{B2}$.

$I_{B1} = I_{B2}$.

$R_{B1} = R_{B2}$.

$I_{B1} = I_{B2}$.

$R_{B1} = R_{B2}$.

$I_{B1} = I_{B2}$.

$R_{B1} = R_{B2}$.

$I_{B1} = I_{B2

2. Proportional Current Source

1) Negative Feedback Proportional Current Source

$$ Z_0 = (1 - \frac{2}{\beta + 2\beta + 2}) I_R, I_R = \frac{V_{CC} - 2V_{BE}}{R} $$ $$ R_0 \approx \frac{1}{2} (\beta + 1) R_{CE3} $$

$$ R_{E1} \neq R_{E2} $$ $$ I_0 \approx \frac{V_{BE1} - V_{BE2} + I_R R_{E1}}{R_{E2}} $$ $$ I_R R_{E1} \gg (V_{EE1} - V_{EE2}) $$

$$ I_0 \approx \frac{R_{E1}}{R_{E2}} I_R $$ $$ I_R = \frac{V_{CC} - V_{BE2} - V_{RE1}}{R} $$ $$ R_0 \approx R_{CE2} (1 + \frac{\beta R_{E2}}{R_{EE} + R_{B} + R_{E2}}) $$

2) Simple Proportional Current Source

$$ \frac{I_0}{I_R} \approx \frac{I_{E2}}{I_{E1}} \propto \frac{I_{E2} e^{V_{BE2}/V_T}}{I_{E1} e^{V_{BE1}/V_T}} $$ $$ = \frac{I_{E2}}{I_{E1}} = \frac{S_2}{S_1} $$

T1, T2 characteristics not equal.

3. Micropower Current Source

$$ I_0 \propto \frac{V_{T}}{R_{E}} \ln \frac{I_R}{I_0} $$ $$ I_R = \frac{V_{CC} - V_{BE1}}{R} $$

$$ V_{BE1} - V_{BE2} \approx V_T \ln \left( \frac{I_R}{I_0} \right) $$ $$ \propto V_T \ln \left( \frac{I_R}{I_0} \right) $$

4. Wilson Current Source

$$ I_0 \propto \frac{V_{T}}{R_{E}} \ln \frac{I_R}{I_0} $$ $$ I_R = \frac{V_{CC} - V_{BE1}}{R} $$

$$ \left\{ \begin{array}{l} V_{BE1} = V_{BE2} \\ 2I_1 = I_{B1} \\ 2I_2 = I_{B2} \\ R_1 = R_{BE1} \\ R_2 = R_{BE2} \\ R_0 = R_{CE1} // R_{CE2} \end{array} \right. $$

$$ \left\{ \begin{array}{l} A_V = - \frac{R_1 (R_{CE1} / R_{CE2})}{R_{BE1}} \\ R_1 = R_{BE1} \\ R_2 = R_{BE2} \\ R_0 = R_{CE1} // R_{CE2} \end{array} \right. $$

$$ \left\{ \begin{array}{l} A_V = - \frac{R_1 (R_{CE1} / R_{CE2})}{R_{BE1}} \\ R_1 = R_{BE1} \\ R_2 = R_{BE2} \\ R_0 = R_{CE1} // R_{CE2} \end{array} \right. $$

$$ \left\{ \begin{array}{l} A_V = - \frac{R_1 (R_{CE1} / R_{CE2})}{R_{BE1}} \\ R_1 = R_{BE1} \\ R_2 = R_{BE2} \\ R_0 = R_{CE1} // R_{CE2} \end{array} \right. $$

$$ \left\{ \begin{array}{l} A_V = - \frac{R_1 (R_{CE1} / R_{CE2})}{R_{BE1}} \\ R_1 = R_{BE1} \\ R_2 = R_{BE2} \\ R_0 = R_{CE1} // R_{CE2} \end{array} \right. $$

$$ \left\{ \begin{array}{l} A_V = - \frac{R_1 (R_{CE1} / R_{CE2})}{R_{BE1}} \\ R_1 = R_{BE1} \\ R_2 = R_{BE2} \\ R_0 = R_{CE1} // R_{CE2} \end{array} \right. $$

$$ \left\{ \begin{array}{l} A_V = - \frac{R_1 (R_{CE1} / R_{CE2})}{R_{BE1}} \\ R_1 = R_{BE1} \\ R_2 = R_{BE2} \\ R_0 = R_{CE1} // R_{CE2} \end{array} \right. $$

$$ \left\{ \begin{array}{l} A_V = - \frac{R_1 (R_{CE1} / R_{CE2})}{R_{BE1}} \\ R_1 = R_{BE1} \\ R_2 = R_{BE2} \\ R_0 = R_{CE1} // R_{CE2} \end{array} \right. $$

$$ \left\{ \begin{array}{l} A_V = - \frac{R_1 (R_{CE1} / R_{CE2})}{R_{BE1}} \\ R_1 = R_{BE1} \\ R_2 = R_{BE2} \\ R_0 = R_{CE1} // R_{CE2} \end{array} \right. $$

$$ \left\{ \begin{array}{l} A_V = - \frac{R_1 (R_{CE1} / R_{CE2})}{R_{BE1}} \\ R_1 = R_{BE1} \\ R_2 = R_{BE2} \\ R_0 = R_{CE1} // R_{CE2} \end{array} \right. $$

$$ \left\{ \begin{array}{l} A_V = - \frac{R_1 (R_{CE1} / R_{CE2})}{R_{BE1}} \\ R_1 = R_{BE1

The page contains a detailed analysis of a differential amplifier circuit. It includes mathematical derivations and circuit diagrams. Here's the translated content:

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**5. Differential Amplifier Circuit**

Let V1 and V2 be arbitrary amplitude and phase signals: $$V_{o1} = \frac{1}{2}(V_{i1} + V_{i2}) + \frac{1}{2}(V_{i1} - V_{i2})$$ $$V_{o2} = \frac{1}{2}(V_{i1} + V_{i2}) - \frac{1}{2}(V_{i1} - V_{i2})$$ Common mode input signal: Vic = (V1 + V2) / 2 Differential mode input signal: Vid = V1 - V2

The circuit diagram shows a differential amplifier with two transistors T1 and T2, a voltage source Vcc, and resistors R1 and R2. The output is labeled as Vo1 and Vo2.

$$V_{cc}$$ $$V_{o1}$$ $$V_{o2}$$ $$R1$$ $$R2$$ $$V_{i1}$$ $$V_{i2}$$

**2. Dynamic Analysis: Large Signal State**

1) Differential mode: - Function: Non-saturated switch (operating in the linear region) - Limitation: Vcm = Vce + Vbrcebo

2) Common mode: - Function: Collector voltage almost constant - Limitation: Cannot drive BJT into saturation (Vcm).

**3. Dynamic Analysis: Small Signal State**

Aod: Differential mode voltage gain; Auc: Common mode voltage gain. Vo = Vod + Voc = Aud Vid + Auc Vic

1) Differential mode: - Double-ended output: - Input circuit: Collector grounded - Output circuit: R1 midpoint grounded - Aud = Vo1 / Vid = (Vod1 - Vod2) / (Vi1 - Vi2) = 2Vod1 / 2Vi1 = -βRc / Rbe - Rod = 2Rc, Rid = 2Rbe (with respect to ground) - Single-ended output: - Base width modulation. Assuming symmetry. - Aud = -1/2 βRc / Rbe. If both sides are connected to ground Rc, Aud = -βRc / Rbe. - Rod = Rc, Rid = 2Rbe.

2) Common mode: - Double-ended output: - Input circuit: Collector 2Rc feedback resistor. Single-ended analysis. - Output circuit: R2 open circuit - Roc = 2Rc, Ric = 1/2 (2Rc || Rc). Icc = Vcc / 2Rc - Single-ended output: - Base width modulation. Assuming symmetry. - Aud = -βRc / Rbe. If both sides are connected to ground Rc, Aud = -βRc / Rbe. - Rod = Rc, Rid = 2Rbe.

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The page also includes a graph of Vo1 and Vo2 as a function of Vid. The graph shows a linear relationship between Vo1 and Vo2, with a slope of 1 and an intercept of 0.

4. Active Load Differential Amplifier Analysis

1) Static Analysis: Symmetry, Current Source Analysis. 2) Dynamic Analysis (Small Signal, Linear Region):

5. Resistance Biasing and Current Source Biasing Issues

Current Source Biasing: When adding a large common mode signal + common mode current, the current through Rc remains unchanged. The circuit's voltage distribution is not affected. Resistance biasing: When there is no Lc, the current remains unchanged. The addition of common mode signal + common mode current (in the existing transistor model, only the common mode signal + common mode current can be analyzed. When analyzing the common mode signal + common mode current, the circuit's quantities have changed, and it needs to be solved again.)

① Differential Mode: Single-ended output, with current source transmission effect. The output terminal obtains a signal current that is twice the single-ended signal current, equal to the double-ended output current. $$A_{vd} = \frac{\beta R_{L}'}{r_{be}}$$ $$R_{L}' = r_{ce2} // r_{ce1} // R_{L}$$ $$R_{id} = 2 r_{be1}$$ $$R_{o} = r_{ce2} // r_{ce1}$$

② Common Mode: Common mode signal + common mode current + common mode voltage + common mode resistance. The common mode current is the same as the double-ended output current. The common mode voltage is the same as the double-ended output voltage. The common mode resistance is the same as the double-ended output resistance. Analysis Method: Common mode large signal → Vcm problem Current source output resistance

Analysis Method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common mode large signal → Vcm problem Current source output resistance

Common mode analysis method: Common

**Six. Push-Pull Output Stage (Large Signal Analysis)**

**B. Class AB**

**Conduction Angle:** Half of the conduction angle of the transistor in one period of the signal. - Class B: θ = 90° - Class A: θ = 180° - Class AB: θ ∈ (90°, 180°)

**C. Push-Pull Output Stage Output Power and Efficiency**

**Vom:** Output Voltage Amplitude **Iom:** Output Current Amplitude

**Load Average Power:** $$ P_0 = \frac{1}{2} \int_{0}^{2\pi} \left(\frac{V_{om} \sin(\omega t)}{R}\right)^2 d(\omega t) = \frac{1}{2} V_{om} I_{om} $$

**Power Supply Average Power:** $$ P_{dc} = 2 \cdot \frac{1}{2\pi} \int_{0}^{\pi} V_{cc} I_{om} \sin(\omega t) d(\omega t) = \frac{2V_{cc}}{\pi} \frac{V_{om}}{R} $$

**Efficiency:** $$ \eta = \frac{P_0}{P_{dc}} = \frac{\pi}{4} \frac{V_{om}}{V_{cc}} $$

**Transistor Total Power Consumption:** $$ P_{dc} - P_0 = \frac{2V_{cc} V_{om}}{\pi R} - \frac{1}{2} \frac{V_{om}^2}{R} $$

If V_{o} can reach $\frac{2}{\pi} V_{cc}$, then the efficiency reaches its maximum value: $$ P_{Tmax} = \frac{2}{\pi^2} \frac{V_{cc}^2}{R} $$

**D. Composite Transistor**

**Four Types of Connections:** - NNEB - PPEB - NPEB - PNEB

**Current Gain Factor:** $$ \beta + \beta_1 (1 + \beta_1) (\alpha_1 + \beta_1 (\alpha_1) + \alpha_1 (\beta_1) - \alpha_1 (\beta_1 + 1)) $$

**Current Gain Factor:** - NNEB: Same as input - PPEB: Same as input - NPEB: Same as input - PNEB: Same as input

**Input Resistance:** $$ R_{se1} + R_1 + R_{be2} + \ldots + R_{be1} + \ldots $$

The page contains a series of handwritten notes and diagrams related to electronic circuits. Here's a breakdown of the content:

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**Solution 1:**

The circuit diagram shows a common-emitter amplifier with a base resistor and a collector resistor. The formulas provided describe the output voltage (Vo) in terms of the supply voltage (Vcc), the base-emitter voltage (VBE), and the resistors (R1 and R2). The formulas are:

$$ V_{o} = V_{cc} - [R_{c}(i_{c3} + i_{B1}) + V_{BE1}] $$

$$ V_{o} = V + R_{1}C_{1}R_{2} $$

The output voltage Vo is derived as:

$$ V_{o} = \frac{V_{cc} - V_{BE1} - R_{c}i_{c3}}{1 + \frac{R_{c}}{(1 + \beta_{1})R_{2}}} $$

When iB1 approaches zero, Vo simplifies to:

$$ V_{o} = \frac{V_{cc} - V_{BE1}}{1 + \frac{R_{c}}{(1 + \beta_{1})R_{2}}} $$

The text discusses the static and dynamic behavior of the circuit, including the base currents and collector currents. It also mentions the use of diodes in the circuit for biasing purposes.

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**Solution 2:**

The circuit diagram shows a different configuration, possibly a voltage follower or buffer stage. The formulas provided describe the output voltage (Vo) in terms of the supply voltage (Vcc) and the resistors (R1 and R2). The formulas are:

$$ V_{CEQ4} = V_{CEQ1} + V_{EBR2} \approx (1 + \frac{R_{1}}{R_{2}})V_{CEQ4} $$

The text discusses the static and dynamic behavior of the circuit, including the base currents and collector currents. It also mentions the use of diodes in the circuit for biasing purposes.

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**Solution 3:**

The circuit diagram shows a common-emitter amplifier with a base resistor and a collector resistor. The formulas provided describe the output voltage (Vo) in terms of the supply voltage (Vcc), the base-emitter voltage (VBE), and the resistors (R1 and R2). The formulas are:

$$ V_{o} = [R_{c}(i_{c3} + i_{B1}) + V_{BE1}] + V_{cc} $$

The text discusses the static and dynamic behavior of the circuit, including the base currents and collector currents. It also mentions the use of diodes in the circuit for biasing purposes.

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**Solution 4:**

The circuit diagram shows a common-emitter amplifier with a base resistor and a collector resistor. The formulas provided describe the output voltage (Vo) in terms of the supply voltage (Vcc), the base-emitter voltage (VBE), and the resistors (R1 and R2). The formulas are:

$$ V_{o} = [R_{c}(i_{c3} + i_{B1}) + V_{BE1}] + V_{cc} $$

The text discusses the static and dynamic behavior of the circuit, including the base currents and collector currents. It also mentions the use of diodes in the circuit for biasing purposes.

**Mathematical Note on Circuit Frequency Response**

**7. Amplifier Circuit Frequency Response Characteristics**

1. **Frequency Response Analysis Methods**

- **Small Signal Amplifier Circuit**: - **Medium Frequency**: Linear time-invariant system. - **Low Frequency**: Capacitors dominate. - **High Frequency**: Capacitors dominate.

- **Zero State Condition**: - \( V_R(t) = R_i i_R(t) \) - \( V_L(t) = L \frac{di_L(t)}{dt} \) - \( i_L = \frac{1}{L} \int_0^t i_C(t) dt \)

- **Laplace Transform**: - \( V_R(s) = R_i i_R(s) \) - \( V_L(s) = L \frac{di_L(s)}{ds} \) - \( V_C(s) = \frac{1}{sC} i_C(s) \)

- **Element Model**: \( R \), \( sL \), \( \frac{1}{sC} \). All are real positive numbers.

- **Impulse Response**: \( h(t) = r(t) * e(t) \) - \( R(s) = H(s) * E(s) \) - \( H(s) = \frac{R(s)}{E(s)} \)

- **Steady State Model**: \( H(s) \) in the amplifier circuit is a general rational function. Further, - \( H(s) = \frac{K \prod_{i=1}^n (s - z_i)}{\prod_{i=1}^n (s - p_i)} \)

- **Sinusoidal Input**: \( e(t) = E_m \sin(\omega_0 t) \) - \( E(s) = \frac{E_m \omega_0}{s^2 + \omega_0^2} \) - \( R(s) = \frac{E_m \omega_0}{s^2 + \omega_0^2} H(s) \) - \( = \frac{K_1 \omega_0}{s + j \omega_0} + \frac{K_2 \omega_0}{s - j \omega_0} + \frac{K_3}{s - p_1} + \ldots + \frac{K_n}{s - p_n} \)

- **Among them**: \( K_{-j \omega_0} = (s + j \omega_0) R(s) \big|_{s = j \omega_0} = \frac{E_m \omega_0 H(j \omega_0)}{-2j \omega_0} \) - \( K_{j \omega_0} = (s - j \omega_0) R(s) \big|_{s = j \omega_0} = \frac{E_m \omega_0 H(j \omega_0)}{2j \omega_0} \)

- **System Response**: \( h(t) = L^{-1}[R(s)] = E_m H_0 \sin(\omega_0 t + \varphi_0) \) - \( + K_1 e^{p_1 t} + \ldots + K_n e^{p_n t} \)

- **When \( t \to \infty \)**: - \( r(t) = E_m H_0 \sin(\omega_0 t + \varphi_0) \)

- **Based on the above, in the steady state, \( H(s) \) is equivalent to \( H(j \omega) \).**

- **Conclusion**: \( f_\alpha = (\beta_0 + 1) f_\beta = \frac{1}{2 \pi r_{be} (C_{be} + C_{bc})} \). \( f_\alpha > f_\beta > f_\beta \)

2. **BJT Amplifier Circuit Frequency Response**

- **Low, Medium, High Frequency Mixed Model**: - \( \beta_0 \): Low frequency current amplification factor (emitter) - \( \beta \): Common emitter output current amplification factor (collector) - \( \alpha \): Low frequency current amplification factor (base) - \( \alpha \): Common base output current amplification factor. - \( f_\beta \): Frequency at which \(\beta\) drops to 1/2. - \( f_T \): Frequency at which \(\beta = 1\), transistor characteristic frequency. - \( f_2 = (\beta_0 + 1) - f_\beta \).

- **Range of Application**: \( f_\beta < f < f_T \).

3. Linearity distortion: - Amplitude distortion: no new frequency components - Phase distortion: no new frequency components (maybe) - Nonlinear distortion: produces new frequency components (maybe) - Distortion transfer: H(t) ≠ Ke^(jωt) → H(jω) = Ke^(jωt) - Bandwidth: midband gain Avsm - bandwidth BW.

4. Due to single-ended, common-emitter, and common-base configurations being approximately RC circuits, thus analyzing RC low-pass/high-pass filters:

$$ H(j\omega) = \frac{1}{1+j\frac{\omega}{\omega_{H}}} $$

$$ f_{H} = \frac{1}{2\pi RC} $$

$$ \therefore 20\lg|H| = -20\lg\sqrt{1+(\frac{\omega}{\omega_{H}})^2} $$

$$ \varphi = -\arctan(\frac{\omega}{\omega_{H}}) $$

$$ f = f_{H} \text{ at phase lag } 20\lg\sqrt{2} \approx 3\text{dB} $$

$$ f = 0.1f_{H} \text{ at phase lag } \arctan(0.1) \approx 5.7^\circ $$

$$ f = 10f_{H} \text{ at phase lag } \arctan(10) \approx -5.7^\circ $$

$$ H(j\omega) = \frac{j\frac{\omega}{\omega_{2}}}{1+j\frac{\omega}{\omega_{2}}} $$

$$ f_{2} = \frac{1}{2\pi RC} $$

$$ \therefore 20\lg|H| = 20\lg(\frac{\omega}{\omega_{2}}) - 20\lg\sqrt{1+(\frac{\omega}{\omega_{2}})^2} $$

$$ \varphi = -\arctan(\frac{\omega}{\omega_{2}}) + \frac{\pi}{2} $$

$$ f = f_{2} \text{ at phase lag } 20\lg\sqrt{2} \approx 3\text{dB} $$

$$ f = 0.1f_{2} \text{ at phase lag } \arctan(0.1) \approx 5.7^\circ $$

$$ f = 10f_{2} \text{ at phase lag } \arctan(10) \approx -5.7^\circ $$

5. Common-emitter high-frequency response (second-order to first-order zero):

Simplified steps: neglect Cb' → ignore Cb' → combine RsC $$ A_{vs} = A_{vs} \frac{1}{1+j\frac{\omega}{\omega_{H}}} $$

$$ f_{H} = \frac{1}{2\pi R_{s}C} $$: Rs is the input resistance seen from Cb'.

$$ G \cdot BW = |A_{vs}| \cdot f_{H} $$, Rs, Cb', Cb'e, G, BW. fH↑

Note: To increase fH, Rs should be minimized → constant voltage excitation.

Common-base high-frequency response (second-order to first-order zero):

Simplified steps: include Cb' (ωCb' ≫ Rs + Rbb')

$$ A_{vs} = A_{vs} \frac{1+j\frac{\omega}{\omega_{2}}}{1+j\frac{\omega}{\omega_{p}}} $$

$$ f_{2} \approx f_{1}, f_{p} = \frac{1}{2\pi R_{s}C_{b'}} $$

$$ f_{p} = \frac{1}{2\pi R_{s}C_{b'}} $$: Rs is the input resistance seen from Cb'.

$$ f_{H} = \left(\frac{1}{f_{p}} - \frac{1}{f_{2}}\right)^{-1} $$: Rbb' ↓, Cb'e ↓, fH↑

Note: To increase fH, Rs should be minimized → constant voltage excitation. Cb'e's compensation current effect makes fH very large.

Common-base high-frequency response (second-order zero):

Simplified steps: neglect Rbb' → separate current source → equivalent source is 1/Sm

$$ A_{vs} = A_{vs} \frac{1}{1+j\frac{\omega}{\omega_{p}} \times 1+j\frac{\omega}{\omega_{p}}} $$

$$ f_{p1} = \frac{1}{2\pi (R_{s} \parallel R_{e})C_{b'}} $$: Rs || Re is the input resistance seen from Cb'.

$$ f_{p2} = \frac{1}{2\pi R_{s}C_{b'}} $$: Rs is the input resistance seen from Cb'.

$$ \frac{1}{f_{H}} \propto \sqrt{\frac{1}{f_{p1}} + \frac{1}{f_{p2}}} $$

Note: To increase fH, Rs should be minimized → constant voltage excitation. Avoid fH being too small, which would not be a capacitive load.

Summary: High-frequency section串联大电容旁路并联大电容旁路, short circuit all capacitors. Large short circuit capacitors, small long circuit capacitors. Calculate fH.

6. Transistor low frequency response.

BJT internal capacitance small: Cbe 10 pF, Cbc 1 pF. Low frequency can be ignored. Generally, small capacitors are connected in series and parallel. Simplify the circuit: ignore bias resistor Rb, ignore Re. The emitter resistor is equivalent to the base circuit. Ignore Re. The base circuit combines C and the output circuit. The DC voltage source is ignored.

$$A_{vs} = A_{vs} \cdot \frac{jf/f_1}{1+jf/f_2} \cdot \frac{jf/f_2}{1+jf/f_2}$$

$$f_1 = \frac{1}{2\pi (R_S + R_{be})C} \cdot C \sim R_S + R_{be} (C \text{ very small})$$

$$f_2 = \frac{1}{2\pi (R_C + R_L)C} \cdot C \sim R_C + R_L$$

$$f_2 \approx \sqrt{f_1^2 + f_2^2} \text{. For small C1, C2, C3, f1 = 0 (direct coupling).}$$

Overall: Use superposition principle (resistive network) to determine the time constant. When the time constant is known, the other capacitors can be calculated one by one.

7. Multi-stage amplifier circuit frequency response

Low frequency response main factors: coupling capacitors, stray capacitance. High frequency response main factors: stray capacitance, distributed capacitance. Time constant method to analyze each capacitor's single action time constant T. If other T values are much different, it is a five-pole circuit, and the calculation is simplified. If other T values are similar, then T = √(Σf²). fH ≈ (1 - 1/Σf²)⁻¹

$$A_{V1} = -\frac{R_{R1}}{R_{be}} \approx -1 \text{ (R1' = Rbe/(β+1))}$$

$$A_{V2} = -\frac{R_{R2}}{R_{be}}$$

$$A_{V} = \frac{R_{R1}}{R_{be}}$$

$$R_1 = R_{be} \text{ (large)}$$

$$R_2 = R_{be} \text{ (large)}$$

$$R_0 = R_C$$

8. Multi-stage amplifier circuit

Capacitor coupling:

Not easy to integrate, low frequency response is poor; used as a coupling circuit.

Transformer coupling:

Large volume, low frequency response is poor. Used in high-power circuits.

Direct coupling:

Easy to integrate. f1 = 0; DC voltage offset problem, zero drift.

1. DC voltage offset problem.

① Base-emitter diode/constant current diode (static voltage drop small)

② Current source DC voltage shifting circuit (shifts 20R) high frequency point

③ NPN-PNP互补直流电压偏移电路.

2. Zero drift problem.

Zero point, working point.

Drift: Output voltage relative to the initial value randomly fluctuates.

Solution: Input offset compensation

3. Multi-stage amplifier circuit calculation method.

View the lower stage including the input resistance, view the upper stage including the output resistance.

1) CE-CB

$$R_0 = R_C$$

$$R_1 = R_{be} \text{ (large)}$$

$$R_2 = R_{be} \text{ (large)}$$

$$R_0 = R_C$$

2) CC-CB.

$$A_{V1} = \frac{(\beta+1)R_{C1}}{r_{be1} + (\beta+1)R_{C1}} \approx \frac{1}{2} (R_{C1} = \frac{r_{be1}}{\beta+1})$$

$$A_{V2} = \frac{\beta R_{C1}}{r_{be1}}$$

$$A_{V} = \frac{\beta R_{C1}}{2 r_{be1}}$$

CC, CB的高频响应良好.

3) CC-CE.

$$A_{V1} = \frac{(\beta+1) r_{be2}}{r_{be1} + (\beta+1) r_{be2}}$$

$$A_{V2} = -\frac{\beta R_{C}}{r_{be2}}$$

$$A_{V} = -\frac{\beta (\beta+1) R_{C}}{r_{be1} + (\beta+1) r_{be2}}$$

$$R_{i} = r_{be1} + (\beta+1) r_{be2}$$

$$R_{o} = R_{C}$$

CC的输出阻抗小($\approx \frac{R_{C}}{\beta+1}$), 即CE的等效信号源内阻小, 恒压激励, fH提高.

4) CE-CB差放.

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**Subject:**

**Math Assignment**

**Number:**

**Class:**

**Name:**

**Page:**

**Section 9 - MOSFET Model**

1. **Static** - **Variable Resistance Region:** \( I_D = \frac{K}{2} \frac{W}{L} [2(V_{GS} - V_{TH})V_{DS} - V_{DS}^2] \) - **Saturation Region:** \( I_D = \frac{K}{2} \frac{W}{L} (V_{GS} - V_{TH})^2 (1 + \lambda V_{DS}) \)

2. **Low-Frequency Small Signal** - \( g_m = \frac{\partial i_D}{\partial V_{GS}} = 2 \sqrt{\frac{K}{2}} \frac{W}{L} I_D \) - \( g_{mb} = \eta g_m \) - \( g_{ds} = \frac{1}{r_{ds}} = \frac{I_D}{V_A + V_{DS}} = \frac{\lambda I_D}{1 + \lambda V_{DS}} \)

3. **High-Frequency Small Signal** - \( g_{gb} = g_{gd} \) - \( g_{gs} = g_{gb} + g_{gs} \) - \( g_{gd} = g_{bd} \) - \( C_{gs} = C_{gb} + C_{gs} \) - \( C_{gd} = C_{bd} \) - \( C_{gd} = C_{bd} \) - \( I_{D} = I_{S} \frac{g_{m}}{2 \pi (C_{gs} + C_{gd})} \)

4. **Three Basic Fixed Bias Circuits: R0 Fixed

**Section 10 - MOSFET with Active Resistor and Current Source**

1. **Active Resistor (Frequency)** - **Enhancement Type: GD Shorted: Ensures Saturation Region** - **DC:** \( R_0 = \frac{V_{DS}}{2 \alpha} \) - **AC:** \( r = r_{ds} \parallel \frac{1}{g_m} \) (b-s shorted) - **Depletion Type: GS Shorted: Ensures Saturation Region** - **DC:** \( R_0 = \frac{V_{DS}}{2 \alpha} \) - **AC:** \( r = r_{ds} \) (b-s shorted) - **Application: Voltage Divider: Wide Input Range, Stable Characteristics** - **Current Source (Frequency)** - **Basic Current Source / Proportional Current Source** - **Active Resistor Obtains Reference Current \( I_R \):** \( Z_0 = \left[ \frac{W_1}{L_1} \right] \left[ \frac{W_2}{L_2} \right] R_0 \) - **Neglecting Channel Length Modulation:** \( Z_0 = \frac{W_1 L_2}{W_2 L_1} \left[ \frac{W_1}{L_1} \right] R_0 \) - **Output Impedance:** \( R_0 = r_{ds} \) - **Current Mirror / Proportional Current Source** - **Feedback, Good Temperature Stability:** \( R_0 = (g_m r_{ds}) r_{ds} \) - **In Reference Branch: Active Resistor \( T_4 \):** \( V_{DS4} = V_{SS3} \), Improves Accuracy - **Deficiency:** Basic Current Source Dynamic Range Small (When \( V_{DS} \) is Large)

3. Source Follower Amplifier Circuit

1. NMOS E/D $$V_{2} = V_{bs1} > V_{th}$$ $$V_{0} = V_{bs1} > V_{bs1} - V_{th}$$ $$V_{DD} - V_{0} = V_{bs2} > -V_{off}$$

2. NMOS E/E $$A_{V} = -g_{m1} \cdot (g_{ds1} + g_{ds2} + g_{mb2})^{-1} = -g_{m1} R_{o}$$

3. CMOS C/S $$V_{2} = V_{bs1} > V_{th1}$$ $$V_{0} = V_{bs1} > V_{bs1} - V_{th1}$$ $$V_{bs1} - V_{DD} = V_{bs2} < V_{bs1} - V_{bs2}$$

4. Source Follower Amplifier Circuit

1. Low Frequency: $$A_{V} = -g_{m1} \cdot (g_{ds1} + g_{ds2} + g_{mb2})^{-1} = -g_{m1} R_{o}$$

2. High Frequency: $$A_{V} = -g_{m1} \cdot (g_{ds1} + g_{ds2} + g_{mb2})^{-1} = -g_{m1} R_{o}$$

5. Source Follower Amplifier Circuit

1. Low Frequency: $$V_{2} = V_{bs1} > V_{th1}$$ $$V_{0} = V_{bs1} > V_{bs1} - V_{th1}$$ $$V_{bs1} - V_{DD} = V_{bs2} < V_{bs1} - V_{bs2}$$

2. High Frequency: $$A_{V} = -g_{m1} \cdot (g_{ds1} + g_{ds2} + g_{mb2})^{-1} = -g_{m1} R_{o}$$

6. Source Follower Amplifier Circuit

1. Low Frequency: $$A_{V} = -g_{m1} \cdot (g_{ds1} + g_{ds2} + g_{mb2})^{-1} = -g_{m1} R_{o}$$

2. High Frequency: $$A_{V} = -g_{m1} \cdot (g_{ds1} + g_{ds2} + g_{mb2})^{-1} = -g_{m1} R_{o}$$

7. MOS Differential Pair

$$V_{DD} = \frac{k_{p}}{2} \frac{W}{L} (V_{gs1} - V_{gs2})^{2}$$ $$V_{DD} = V_{gs1} - V_{gs2}$$

$$i_{D1} + i_{D2} = I_{SS}$$ $$i_{D1} = \frac{k_{p}}{2} \frac{W}{L} (V_{gs1} - V_{gs2})^{2}$$ $$i_{D2} = \frac{k_{p}}{2} \frac{W}{L} (V_{gs2} - V_{gs1})^{2}$$

$$i_{D1} = \frac{2ss}{2} + \frac{k_{p}W}{2}V_{D1} \frac{2ss}{2} \left(\frac{k_{p}W}{2}\right)^{-1} - t_{ab}^{2}$$ $$V_{D1} = \frac{2ss}{2} + \frac{k_{p}W}{2}V_{D1} \sqrt{\frac{2ss}{2} \left(\frac{k_{p}W}{2}\right)^{-1} - \left(\frac{V_{D1}}{2}\right)^{2}}$$ $$i_{D2} = \frac{2ss}{2} - \frac{k_{p}W}{2}V_{D1} \sqrt{\frac{2ss}{2} \left(\frac{k_{p}W}{2}\right)^{-1} - \left(\frac{V_{D1}}{2}\right)^{2}}$$

(1) When $V_{D1} < \sqrt{\frac{4L}{k_{p}W}}$, small signal $$i_{in} = -i_{out} \approx \sqrt{\frac{k_{p}W}{4L}}V_{D1}$$

(2) When $V_{D1} = \pm \sqrt{\frac{2L}{k_{p}W}}$, clipping region.

1) Common MOSFET - Differential output: $A_{vd} = g_{m}(R_{ds1} // R_{ds2})$ - Single-ended output: $A_{vd1} = -\frac{1}{2} g_{m}(R_{ds1} // R_{ds2})$

High frequency: half-band analysis, similar to CS.

2) Load with source follower - Low frequency: half-band analysis, common mode when considering large $g_{mb}$. - High frequency: CS.

3) Load with source follower - Low frequency: half-band analysis, common mode when considering large $g_{mb}$. - High frequency: CS.

4) CMOS (PMOS load) - Low frequency: half-band analysis, common mode when considering large $g_{mb}$. - High frequency: CS.

5) CMOS (PMOS load) - Low frequency: half-band analysis, common mode when considering large $g_{mb}$. - High frequency: CS.

$$i_{D1} \approx -i_{D2} = g_{m1}(V_{i1} - V_{i2})/2$$ $$t_{0} = i_{D1} - i_{D2} \approx 2i_{D1} = g_{m1}(V_{i1} - V_{i2})$$ $$V_{o} = i_{D1}(R_{ds1} // R_{ds2}) \approx g_{m1}(R_{ds1} // R_{ds2})(V_{i1} - V_{i2})$$ $$\therefore A_{vd} = g_{m1}(R_{ds1} // R_{ds2})$$

High frequency: output node, not analyzed.

11. CMOS output stage. 1. CMOS CD - T1, T2: complementary CD class - T3, T4: bias elements - T5: driving stage CS class - T6: active load

Dynamic range (forward): $$\left\{\begin{array}{l} V_{o} = V_{DD} - |V_{ds1}| - V_{ds1} \\ V_{ds1} \geq V_{ds1} - V_{ds1} \Rightarrow \text{maximum range when } V_{ds1} = V_{ds1} - V_{ds1} \\ |V_{ds1}| > |V_{ds1}| \end{array}\right.$$

2. CMOS CS - T1, T2: complementary CS class - T3, T4: CD, moving average - V_{ds1} = V_{ds1} - V_{DD} - V_{ds2} = V_{ds1} - V_{ds1} + V_{ds1} - Dynamic range large, output resistance large.

Translated Text and Formulas

# Title BJT, MOS

# Circuit Components - **Bipolar Junction Transistor (BJT)** - **NPN**: - $$i_{c} = I_{s}(e^{V_{be}} - 1)(1 + \lambda V_{ce})$$ - **PNP**: - $$i_{c} = I_{s}(e^{V_{be}} - 1)(1 + \lambda V_{ce})$$

- **Metal-Oxide-Semiconductor Field-Effect Transistor (MOSFET)** - **Enhancement Mode (N-channel)**: - $$i_{d} = \frac{K}{2}(V_{gs} - V_{th})(1 + \lambda V_{ds})$$ - **Enhancement Mode (P-channel)**: - $$i_{d} = \frac{K}{2}(V_{gs} - V_{th})(1 + \lambda V_{ds})$$ - **Depletion Mode (N-channel)**: - $$i_{d} = \frac{K}{2}(V_{gs} - V_{th})(1 + \lambda V_{ds})$$ - **Depletion Mode (P-channel)**: - $$i_{d} = \frac{K}{2}(V_{gs} - V_{th})(1 + \lambda V_{ds})$$

# Circuit Types - **Biasing Circuits** - **Current Source (Current Mirror)** - **Basic Current Mirror (CC)** - **Current Source (CC)** - **Current Source (CB)** - **Voltage Source (Voltage Divider)** - **Basic Voltage Divider (CC)** - **Current Source (CC)** - **Current Source (CB)**

# Amplifier Circuits - **Common Emitter (CE)** - **Common Collector (CC)** - **Common Base (CB)** - **Common Source (CS)** - **Common Drain (CD)** - **Common Gate (CG)**

# Small Signal Analysis - **Small Signal Analysis (SSA)** - **Input Impedance (Ri)** - **Output Impedance (Ro)** - **Voltage Gain (Av)** - **Frequency Response (fH)**

# Large Signal Analysis - **Large Signal Analysis (LSA)** - **Input Impedance (Ri)** - **Output Impedance (Ro)** - **Voltage Gain (Av)** - **Frequency Response (fH)**

# Differential Amplifier - **Differential Amplifier (DA)** - **Common Emitter (CE)** - **Common Collector (CC)** - **Common Base (CB)** - **Common Source (CS)** - **Common Drain (CD)** - **Common Gate (CG)**

# Feedback Circuits - **Feedback Circuits (FB)** - **Current Feedback (CF)** - **Voltage Feedback (VF)** - **Current Feedback (CF)** - **Voltage Feedback (VF)**

# Transistor Models - **Bipolar Junction Transistor (BJT)** - **NPN**: - $$i_{c} = I_{s}(e^{V_{be}} - 1)(1 + \lambda V_{ce})$$ - **PNP**: - $$i_{c} = I_{s}(e^{V_{be}} - 1)(1 + \lambda V_{ce})$$

- **Metal-Oxide-Semiconductor Field-Effect Transistor (MOSFET)** - **Enhancement Mode (N-channel)**: - $$i_{d} = \frac{K}{2}(V_{gs} - V_{th})(1 + \lambda V_{ds})$$ - **Enhancement Mode (P-channel)**: - $$i_{d} = \frac{K}{2}(V_{gs} - V_{th})(1 + \lambda V_{ds})$$ - **Depletion Mode (N-channel)**: - $$i_{d} = \frac{K}{2}(V_{gs} - V_{th})(1 + \lambda V_{ds})$$ - **Depletion Mode (P-channel)**: - $$i_{d} = \frac{K}{2}(V_{gs} - V_{th})(1 + \lambda V_{ds})$$

# Circuit Theorems - **Thevenin's Theorem** - **Norton's Theorem** - **Superposition Theorem** - **Buck-Boost Converter** - **Buck Converter** - **Boost Converter** - **Buck-Boost Converter** - **Buck Converter** - **Boost Converter**

# Circuit Analysis - **Biasing Circuits** - **Current Source (Current Mirror)** - **Basic Current Mirror (CC)** - **Current Source (CC)** - **Current Source (CB)** - **Voltage Source (Voltage Divider)** - **Basic Voltage Divider (CC)** - **Current Source (CC)** - **Current Source (CB)**

# Amplifier Circuits - **Common Emitter (CE)** - **Common Collector (CC)** - **Common Base (CB)** - **Common Source (CS)** - **Common Drain (CD)** - **Common Gate (CG)**

# Small Signal Analysis - **Small Signal Analysis (SSA)** - **Input Impedance (Ri)** - **Output Impedance (Ro)** - **Voltage Gain (Av)** - **Frequency Response (fH)**

# Large Signal Analysis - **Large Signal Analysis (LSA)** - **Input Impedance (Ri)** - **Output Impedance (Ro)** - **Voltage Gain (Av)** - **Frequency Response (fH)**

# Differential Amplifier - **Differential Amplifier (DA)** - **Common Emitter (CE)** - **Common Collector (CC)** - **Common Base (CB)** - **Common Source (CS)** - **Common Drain (CD)** - **Common Gate (CG)**

# Feedback Circuits - **Feedback Circuits (FB)** - **Current Feedback (CF)** - **Voltage Feedback (VF)** - **Current Feedback (CF)** - **Voltage Feedback (VF)**

# Transistor Models - **Bipolar Junction Transistor (BJT)** - **NPN**: - $$i_{c} = I_{s}(e^{V_{be}} - 1)(1 + \lambda V_{

**Mathematical Notation:**

$$ \begin{aligned} &\text{① Mid-frequency: } \frac{dA_F}{A_F} = \frac{1}{1+A_F} \frac{dA}{A} \\ &\text{② Mid-frequency: } R_{IF} = (1+A_F) R_i \\ &\text{③ Mid-frequency: } R_{IF} = R_i/(1+A_F) \\ &\text{④ Mid-frequency: } R_{IF} = R_0/(1+A_F) \\ &\text{⑤ Mid-frequency: } R_{IF} = (1+A_F) R_0 \\ &\text{⑥ Mid-frequency: } A = \frac{A_m}{1+j\beta/f_{HF}} \\ &\text{⑦ Mid-frequency: } A_F = \frac{A}{1+A_F} \\ &\text{⑧ Mid-frequency: } A_F = \frac{A_m}{1+A_mF} \\ &\text{⑨ Mid-frequency: } A_F = \frac{A_m}{1+A_mF} \\ &\text{⑩ Mid-frequency: } A_F = \frac{A_m}{1+A_mF} \\ &\text{⑪ Mid-frequency: } A_F = \frac{A_m}{1+A_mF} \\ &\text{⑫ Mid-frequency: } A_F = \frac{A_m}{1+A_mF} \\ &\text{⑬ Mid-frequency: } A_F = \frac{A_m}{1+A_mF} \\ &\text{⑭ Mid-frequency: } A_F = \frac{A_m}{1+A_mF} \\ &\text{⑮ Mid-frequency: } A_F = \frac{A_m}{1+A_mF} \\ &\text{⑯ Mid-frequency: } A_F = \frac{A_m}{1+A_mF} \\ &\text{⑰ Mid-frequency: } A_F = \frac{A_m}{1+A_mF} \\ &\text{⑱ Mid-frequency: } A_F = \frac{A_m}{1+A_mF} \\ &\text{⑲ Mid-frequency: } A_F = \frac{A_m}{1+A_mF} \\ &\text{⑳ Mid-frequency: } A_F = \frac{A_m}{1+A_mF} \\ &\text{⑳ Mid-frequency: } A_F = \frac{A_m}{1+A_mF} \\ &\text{⑳ Mid-frequency: } A_F = \frac{A_m}{1+A_mF} \\ &\text{⑳ Mid-frequency: } A_F = \frac{A_m}{1+A_mF} \\ &\text{⑳ Mid-frequency: } A_F = \frac{A_m}{1+A_mF} \\ &\text{⑳ Mid-frequency: } A_F = \frac{A_m}{1+A_mF} \\ &\text{⑳ Mid-frequency: } A_F = \frac{A_m}{1+A_mF} \\ &\text{⑳ Mid-frequency: } A_F = \frac{A_m}{1+A_mF} \\ &\text{⑳ Mid-frequency: } A_F = \frac{A_m}{1+A_mF} \\ &\text{⑳ Mid-frequency: } A_F = \frac{A_m}{1+A_mF} \\ &\text{⑳ Mid-frequency: } A_F = \frac{A_m}{1+A_mF} \\ &\text{⑳ Mid-frequency: } A_F = \frac{A_m}{1+A_mF} \\ &\text{⑳ Mid-frequency: } A_F = \frac{A_m}{1+A_mF} \\ &\text{⑳ Mid-frequency: } A_F = \frac{A_m}{1+A_mF} \\ &\text{⑳ Mid-frequency: } A_F = \frac{A_m}{1+A_mF} \\ &\text{⑳ Mid-frequency: } A_F = \frac{A_m}{1+A_mF} \\ &\text{⑳ Mid-frequency: } A_F = \frac{A_m}{1+A_mF} \\ &\text{⑳ Mid-frequency: } A_F = \frac{A_m}{1+A_mF} \\ &\text{⑳ Mid-frequency: } A_F = \frac{A_m}{1+A_mF} \\ &\text{⑳ Mid-frequency: } A_F = \frac{A_m}{1+A_mF} \\ &\text{⑳ Mid-frequency: } A_F = \frac{A_m}{1+A_mF} \\ &\text{⑳ Mid-frequency: } A_F = \frac{A_m}{1+A_mF} \\ &\text{⑳ Mid-frequency: } A_F = \frac{A_m}{1+A_mF} \\ &\text{⑳ Mid-frequency: } A_F = \frac{A_m}{1+A_mF} \\ &\text{⑳ Mid-frequency: } A_F = \frac{A_m}{1+A_mF} \\ &\text{⑳ Mid-frequency: } A_F = \frac{A_m}{1+A_mF} \\ &\text{⑳ Mid-frequency: } A_F = \frac{A_m}{1+A_mF} \\ &\text{⑳ Mid-frequency: } A_F = \frac{A_m}{1+A_mF} \\ &\text{⑳ Mid-frequency: } A_F = \frac{A_m}{1+A_mF} \\ &\text{⑳ Mid-frequency: } A_F = \frac{A_m}{1+A_mF} \\ &\text{⑳ Mid-frequency: } A_F = \frac{A_m}{1+A_mF} \\ &\text{⑳ Mid-frequency: } A_F = \frac{A_m}{1+A_mF} \\ &\text{⑳ Mid-frequency: } A_F = \frac{A_m}{1+A_mF} \\ &\text{⑳ Mid-frequency: } A_F = \frac{A_m}{1+A_mF} \\ &\text{⑳ Mid-frequency: } A_F = \frac{A_m}{1+A_mF} \\ &\text{⑳ Mid-frequency: } A_F = \frac{A_m}{1+A_mF} \\ &\text{⑳ Mid-frequency: } A_F = \frac{A_m}{1+A_mF} \\ &\text{⑳ Mid-frequency: } A_F = \frac{A_m}{1+A_mF} \\ &\text{⑳ Mid-frequency: } A_F = \frac{A_m}{1+A_mF} \\ &\text{⑳ Mid-frequency: } A_F = \frac{A_m}{1+A_mF} \\ &\

2. Pole Compensation: - Auxiliary: DC offset, DC offset compensation, zero adjustment, overcurrent protection. - Known: Increasing F in the middle improves electrical performance. - Known: Increasing F at high frequencies can cause self-oscillation. - Pole compensation can be introduced to make F larger while still avoiding self-oscillation. a. Capacitor lag compensation: 1) At the pole point, add a ground capacitor, left shift fp1. If F=1, then need to shift fp1, so 2lg|A| |fp1=fp2=0. 2) At the pole point, add a capacitor (split pole capacitor). BJT: Produces positive zero fp2 >> BW0. No problem. MOSFET: Produces positive zero fp2 ~ fp1. Use C/R compensation. If F=1, need to achieve unit gain compensation. b. Capacitor-resistor lag compensation (pole-zero compensation): 1) At the pole point, add a ground RC, shift fp1. If F=1, then need to shift fp1, so 2lg|A| |fp1=fp3=0. 2) Same as a. c. Lead compensation: Near the self-oscillation frequency, introduce a lead capacitor. Note: The lead capacitor compensation introduces a positive zero with lag compensation, so it needs to be eliminated.

7. Differential Input Stage (Vid, Iin, Iout, R1 small, R2 large, R3 small, R4 large, K1 small): - Intermediate stage (A1 large, R1 large, DC offset compensation). - Differential characteristics parameters: fH: Upper limit frequency fH = BW. BW: Bandwidth unit gain BW = A1 * fH when single-ended. SR: Conversion rate (speed): SR = dW/dt |max Ric: Input resistance of each input terminal when common-emitter dynamic resistance. BNP: Differential bandwidth BNP = 2πVom

Note: The op-amp uses a Miller capacitor compensation.

$$i_{o1} \approx i_{c} = -C_{c} \frac{dV_{o}}{dt} > \frac{dV_{o}}{dt} \approx -\frac{V_{o1}}{C_{c}}$$

$$S_{R} = \left. \frac{dV_{o1}}{dt} \right|_{\infty} \approx \pm \frac{I_{EE}}{C_{c}}$$

$$\alpha A_{vd} = \frac{V_{o}}{V_{id}} \approx -\frac{Z_{o1}}{j\omega C_{c}V_{id}} \approx -\frac{A_{g1}}{j\omega C_{c}}$$

$$\therefore BW_{G} \approx \frac{A_{g1}}{2\pi C_{c}}$$

$$\therefore S_{R} \approx \pm \frac{I_{EE}}{A_{g1}} \cdot 2\pi BW_{G}$$

4. Examples:

1. BJT: F007 P321. 1) Input stage: CC-CB differential, input differential, output 2ic. 2) Intermediate stage: Common Emitter 3) Output stage: Current Source Load Complementary Output Stage 133-14 4) DC Biasing: DC current source low power biasing. 5) Phase Compensation: Miller capacitor compensation (unit gain) 6) Zero Input Zero Output 7) Overcurrent Protection: Current sampling protection, limit current.

2. MOS: P335. 1) Input stage: CMOS differential. 2) Output stage: Emitter follower output stage. 3) Phase Compensation: Miller capacitor compensation.

Note: RMiller = 1/2gm.

5. Op-amp's linear operation. DC negative feedback -> linear region -> saturation, clipping. 1) Addition: $$V_{o} = -\left(\frac{R_{F1}}{R_{1}} V_{11} + \frac{R_{F2}}{R_{2}} V_{22}\right)$$ $$R_{1} // R_{2} // R_{3} = R'$$ $$V_{o} = \left(1 + \frac{R_{F1}}{R_{1}}\right) \left(\frac{R_{1}}{R_{1} + R_{F1}}\right) \left(\frac{V_{11}}{R_{1}} + \frac{V_{22}}{R_{2}}\right)$$

2) Subtraction: $$V_{o} = -\frac{R_{F2}}{R_{3}} V_{11} + \left(1 + \frac{R_{F2}}{R_{3}}\right) V_{22}$$

3) Integration / Differentiation Integration: In the feedback loop with a large resistor, the DC feedback loop maintains the DC feedback loop. Differentiation: In the feedback loop with a small capacitor, it plays a phase compensation role.

6. Op-amp's nonlinear operation. Negative feedback / positive feedback -> saturation region -> V0 = VoH / VoL. Without phase compensation circuit, improve working speed. Parameters: 1) ΔV2: Response time: Vo - V2 line in 0.1 (Vo + V2) ~ 0.9 (Vo + V2) 2) Δt: Response time: Vo - V2 line in 0.1 (Vo + V2) ~ 0.9 (Vo + V2) SR = 0.8 (VoH - VoL) / Δt High speed wideband, Δt ↓.

1. Single-ended comparator: no feedback. 2. Single-ended comparator (capacitor trigger): positive feedback. {Inverting input: left up -> right down. Non-inverting input: left down -> right up. ΔVth = Vref - Vout ∝ feedback depth.} 3. Comparator. Note: Limiting measures: input-output (negative feedback stabilization). Input (diode in series); output (diode in parallel).

4. Pulse waveforms generation and processing circuit. Single-ended amplifier ~ RC low-pass: Charging: Vout = Vm(1 - e^(-Rt/C)) tr = RCLn(Vm - 0.1Vm) / (Vm - 0.9Vm) ≈ 2.2RC Discharging: Vout = Vm e^(-Rt/C) tf = RCLn(0 - 0.9Vm) / (0 - 0.1Vm) ≈ 2.2RC. tr/fH = 1/2πRC ≈ 0.35.

5. Single-ended comparator (capacitor trigger): Level trigger; bistable state. V1 -> Comparator -> Vout. Vref1 = R2 / (R1 + R2) Vcc Vref2 = -R2 / (R1 + R2) Vcc. Vout1 = 1/3 Vcc Vout2 = -1/3 Vcc.

6. Capacitor trigger bistable trigger. Edge trigger; bistable; bistable state -> pulse generator. Vref1 = R2 / (R1 + R2) Vcc Vref2 = -R2 / (R1 + R2) Vcc. T = RCLn(V2 - Vref2) / (V1 - Vref1) + RCLn(V1 - Vref1) / (V2 - Vref2) = 2RC ln(1 + R2 / R1). T = (R1 + R2) CLn(Vcc - Vref1) / (Vcc - Vref2) + R2 CLn(Vref2 - 0) / (0 - Vref1) = (R1 + 2R2) CLn 2. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: Vc = 0, Vout = Vcc. First: V

**Subject: Mathematics Assignment**

**Topic: Analog-to-Digital and Digital-to-Analog Converters**

**1. D/A Converter**

**1.1 Technical Specifications**

- MSB: Dn-1 - LSB: D0 - VLSB: Vom / (2^n - 1) = ΔV - Resolution: VLSB / Vom = 1 / (2^n - 1) - tset: Time for error to enter ±1/2 VLSB

**1.2 Basic Circuits**

1) Inverted T-shaped resistor network: - io = Vref / R * (1 / 2^n * Σ Pi * 2^i) - The conductance of the analog switch affects the matching accuracy.

2) Current: - io = I * (1 / 2^n * Σ Di * 2^i) - The current value is not affected by the analog switch R0 or Vo.

3) Resistor ladder structure: - Vo = Vref * (1 / 2^n * Σ Pi * 2^i) - Multi-channel selection switch 0: 1/2^n Vref; 1: 2^n-1 Vref - Requires high input resistance buffer.

**2. A/D Converter**

- Sampling frequency ≥ 2fmax - Hold time: CH hold time constant - Hold voltage drop rate: CH hold current size - Quantization noise: Rounding method: Δ = 1 / 2^n, error: Δ - Rounding method: Δ = 2 / (2^n + 1), error: Δ / 2

**1.1 Technical Specifications**

- Vmax: Maximum input voltage - Vmin / 2^n: Minimum input voltage = Resolution

**1.2 Basic Circuits**

1) Parallel Capacitive Type - Time: 1, Components: 2^n - 1 Holders. 2) Segmented Parallel (Segmented). - Time: k + ..., Components: k (2^n - 1) Holders. 3) Feedback Type. - Time: (2^n - 1) Tcp 4) Successive Approximation Type. - Time: (n + 2) Tcp. 1/2 Vref → 1/2^n Vref → ... → 1/2^n Vref 5) Dual Integration Type. - Time: 2^n Tcp. T1 = 2^n Tcp, T2 = -Vref / 2^n Tcp. Vmax = -Vref / R*C * 2^n Tcp

3.1.1 PNP single transistor amplification principle. 3.1.2 Determine the effect of the frequency equivalent circuit on the amplification of the positive signal. 3.2.1 Factors determining voltage gain. $$A_v = \frac{\beta R_C}{R_E + R_C}, A_v' = -\frac{\beta R_C}{R_E + R_C + R_S}$$ 3.2.2 Elements determining voltage gain. $$V_i = -\frac{\beta R_C}{R_E + R_C} V_o$$ 3.3.1 1. Graphical method and static operating point. Analysis method: DC path -> input characteristic equation and input characteristic curve to find I_B -> output characteristic equation and I_B corresponding output characteristic curve to find Q point. 2. Graphical method and dynamic range. Analysis method: input characteristic curve on V_BE add signal -> input signal (sinusoidal) -> working point fluctuation (output characteristic curve) -> Q point away from saturation and cutoff -> output voltage waveform bottom cutting. 3.3.2 Graphical method analysis of static operating point; Graphical method analysis of static point with positive signal -> dynamic range Vom1 Vom2 graphical solution. 3.3.3 1. By the truth waveform to judge distortion type: Bottom cutting: saturation distortion; bottom cutting: cutoff distortion (BJT base-emitter junction's reverse phase): memorize the output characteristic curve. 2. Eliminate distortion methods: Bottom cutting: saturation distortion; bottom cutting: cutoff distortion. 3.3.4 Basic common-emitter amplification 1) Static operating point: remove capacitor (open), DC path. 2) Dynamic performance: remove base-emitter junction and collector-emitter junction capacitors, open BJT base-emitter capacitor, remove AC path. $$A_v = -\frac{\beta R_C}{R_E} \quad (R_{be} = R_{be'} + R_{be} = V_{be} + \frac{\beta V_{ce}}{2 \pi})$$ $$R_i = R_E // R_{be} \quad (\text{split current method})$$ $$R_o = R_{ce} // R_C \approx R_C \quad (V_i \to 0, R_L \to \infty)$$ 3.3.5 Common-emitter, common-emitter bypass resistor (negative feedback stabilizes the static operating point) and collector bypass capacitor (the middle frequency part of the collector resistance increases the gain). 1) Static operating point: voltage divider method, calculate V_{BB}. 2) Dynamic performance: AC equivalent circuit, voltage gain Av = \frac{R_i}{R_i + R_S} Av 3) Dynamic range: Vom1 = V_{CEQ} - V_{BEQ}, Vom2 = R_i' \cdot I_{Q} (output triangular wave). 3.3.6 Basic common-emitter amplification, resistance biasing. Draw the DC path and AC path, calculate the static response and dynamic performance. Note the voltage gain: $$V_{o1} = V_{o2} = \frac{V_o}{V_i} = \frac{V_i}{V_s} = \frac{R_i}{R_i + R_S} A_v$$ 3.4.1 Divider circuit (common collector & common emitter) $$A_{v1} = A_{v2} = -\frac{\beta R_C}{R_E + R_{be} + R_{be'}} \quad (\text{same phase output: common emitter})$$ $$R_{o1} = \frac{R_S + R_{be} // R_E}{\beta + 1} \quad (\text{split current method})$$ $$A_{v2} = A_{v2} = -\frac{\beta R_C}{R_E + R_{be} + (\beta + 1) R_E} \quad (\text{opposite phase output: common emitter})$$ $$R_{o2} = R_C \quad (\text{current control current source open circuit})$$ To make A_{v1} A_{v2} close and opposite, R_S should be as large as possible.

3.4.2. Basic Amplifier: Common Emitter (Voltage Amplifier)

(1) Static Characteristics (Voltage Source and Current Source, h-parameter of the base resistor).

(2) Dynamic Characteristics: Mid-Frequency Equivalent Circuit. Input Resistance: Common Emitter, Common Base Output Resistance: Common Emitter, Common Base

3.4.3. Common Collector (CC) Amplifier. Current Source!

(1) Static Characteristics: Common Emitter

(2) Dynamic Characteristics: Mid-Frequency Equivalent Circuit: All Capacitors Short Circuited. Contain Feedback Resistor Rf, so cannot use equivalent resistance method. Directly solve using the loop equation or use the equivalent resistance method. (3) Cc Open Circuit Dynamic Characteristics: Base Resistance Rb in parallel with the original input resistance. Use the equivalent resistance method to calculate.

3.4.4. Common Base: Current Amplifier. (1) Mid-Frequency Circuit: CE between the base and collector is a current source. gmVe, RE between Rb and Re. (2) Dynamic Characteristics: Equivalent Resistance Method (Resistor Conversion Method). Re and equivalent resistance of the base resistor are in parallel. The output resistance is due to the current source being cut off. Ro = Rc

3.4.5. Common Base: Unstable Re When the Equivalent Resistance is Not Stable. Still can use the equivalent resistance method, but the current source is cut off. Vo = -2βRe

3.4.6. Common Emitter: The equivalent resistance of the base is large. Dynamic Characteristics: B-E Equivalent Resistance Method.

3.4.7. Complete (Bias) Circuit. (1) Static Characteristics: Common Emitter (2) Dynamic Characteristics: Reduce to the common emitter AC path; Base Emitter Feedback Resistance; B-E Equivalent Resistance Method.

3.5.1. Basic Current Source, Considering Base Area Control. $$ \begin{cases} \frac{I_{e}}{I_{c}} = \frac{V_{A} + V_{ce2}}{V_{A} + V_{ce1}} \quad (V_{ce2} \text{ open}) \\ 2R_{e} + V_{ce1} = 6V \\ I_{e1} = I_{R} - 2I_{B} = I_{R} - \frac{2I_{c}}{\beta} \\ \text{According to this, find } I_{0}. \end{cases} $$ Ro = Re = \frac{V_{ce2} + V_{A}}{2} \quad \text{Thus,} \quad Ro = Re

3.5.2. Basic Current Source. (1) Output Current Io: $$ \begin{cases} 2R_{e} + V_{be} = V_{cc} \\ 2R_{e} = 2\beta I_{B} + 5I_{B} \\ I_{e} = 3\beta I_{B} \end{cases} $$ (2) Output Resistance. Principle: Current Source is Cut Off (β = ∞).

3.5.3. Common Emitter Current Source. $$ \begin{cases} \text{Common Emitter Branch's Current Source and Voltage Source} \\ \text{Two Characteristic Equations and One β to Solve Output Current (Io)} \end{cases} $$ R' is used: Due to β changing with Ic, so R' remains small to ensure β does not change.

3.6.1 Single-ended input dual-ended output differential amplifier. Draw output waveform: (1) Amplification region, no distortion. (2) Transition region, no distortion. (3) Saturation region, distortion. 3.6.5 With feedback differential amplifier. 3.6.6 Ratio current source bias, with load differential amplifier. (1) Differential mode voltage gain, common mode output resistance, differential input resistance. (2) With load differential amplifier, the output resistance of the transistor. (3) Vcm (Vcm < Vc, Vcm > Vce). 3.7.1 Class AB and Class B complementary output. (1) Static operating point: input static bias voltage makes Vo = 0. (2) Dynamic range. (3) Maximum output power.

3.7.2. Active Load Class AB Output Stage

(1) Dynamic Range: Negative Half Saturated; Positive Half Saturated

(2) Maximum Output Power, Output Efficiency

Due to positive and negative output voltages being the same, let Vom = Vm

$$P_{dc} = \frac{2V_{cc}}{\pi} \frac{V_{om}^2}{R_L}$$

$$P_0 = \frac{1}{2}V_{om}I_{m}$$

$$\eta = \frac{P_0}{P_{dc}}$$

(3) Icm, Pcm, VbrCEO, Limiting Parameters

Icm, when unidirectional conduction, Icm > Vcc/R2 can ensure the transistor does not saturate.

Pcm, when the transistor is saturated, the collector current should be greater than Vcc^2 / π^2 R2.

VbrCEO, when the transistor is in saturation, VbrCEO should be greater than 2Vcc.

(4) Current Source Output Minimum Current (to achieve maximum power)

When the transistor is in saturation, the output current is at its minimum, at this time RL is close to zero, all the current flows to the base of the transistor, so Ic > (1+β) RL = Vom.

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3.8.5. From A_v, find parameters and Bode diagram. Two poles and one zero system. Mid-frequency voltage gain: f >> f_p1; f << f_p2 (zero is also f_p2). Thus A_vmid = A_m. f_H = f_p2; f_L = f_p1. Note: there's a zero on the Bode diagram corresponding to f_p1.

3.8.6. CE frequency response. High frequency: large capacitor shorted. (Capacitor ratio) f_H = 1 / [2π(R_s // R_B + R_00') // (V_be' // (1 + g_m R_e') R_e)] ≈ 1 / [2π(R_s + r_be) // R_be'] [C_be' // (1 + g_m (R_e' // R_s)) R_s].

Low frequency: small capacitor open. (B-E equivalent resistance method, ignore R_b) f_L = √(f_L1^2 + f_L2^2). f_L1 = 1 / [2πC1 (R_s + r_be)], f_L2 = 1 / [2πC2 (R_C + R_L)].

3.8.7. CE frequency response. From the Bode diagram and Bode diagram to find the cutoff frequency. f_H = √(f_p1^2 + f_p2^2).

(1) Transfer function & Bode diagram. A = A_m / (1 + jf/f_p).... Bode diagram: one pole produces a -20dB/dec drop. { f_p ≈ f_T }

(2) Specific frequency gain and phase values. (3) f_H: Solve the third-order equation for f_H.

(3) Each capacitor引起的f1. Calculate one at another short circuit.

3.8.9 CB effect. (1) Calculate f1: Including Rbb'; current source division; gm Vbe equivalent resistance Ygm; two-terminal zero system. f1 = (√Vf1² + Vf2²)⁻¹.

3.8.10 Differential frequency. f1: Half circuit analysis: CE → f1 = [2x[(R1 + R2) // R3]]. Av1 = (βRc) / (R1 + R2).

3.9.1 Two-stage direct coupling amplifier circuit DC. Voltage source division method; load resistance. Summary: As long as the base resistance is biased, use voltage source division method to solve the static problem.

3.9.2 CC-CE dynamic performance. (Mid-term) First stage CC: Ri1 = (Rs + Rbe) * [(β + 1) * R11 // R12] Ro1 = R11 // R12 // Rbe. Av1 = (β + 1) * R11 / (Rs + Rbe + (β + 1) * R11).

Second stage CE: Ri2 = Rbe + (β + 1) * R2. Ro2 = R2. Av2 = β * R2 / Rbe.

3.9.3 Current source bias differential amplifier - CE two-stage amplifier dynamic performance. (Mid-term) Second stage CE: Ri2 = Rbe + (β + 1) * R10. Ro2 = R10. Av2 = β * R10 / Rbe.

(2) Common mode: Calculate the current source output resistance: Re = Rbe * (1 + β * R2) / (Rbe + R1 + R2 // R3). Calculate the common mode gain: Ac1 = β * R3 // [(Rbe + β + 1) * R1] / (R1 + Rbe + (β + 1) * (R3 + 2 * Re)). Common mode rejection ratio: KCMR = Av1 * Av2 / (Ac1 * Av2) = Av1 / Ac1 = 48. Summary: Differential mode gain and single-ended input and differential input The input method is irrelevant; Only single-ended output has common mode gain, differential output is 0; Single-ended output common mode gain is 1/2, differential output is 1.

3.9.4 Base resistance biasing of CE-CB. (1) Static operating point: Solve the cubic equation (can also use voltage source division method) (2) Dynamic performance: First stage CE: Ri1 = Rbe + (β + 1) * R1 Ro1 = ∞. Av1 = -β * R1 / R1. Second stage CB: Ri2 = Rbe + (β + 1) * R2. Ro2 = R2. Av2 = -β * R2 / Rbe.

(3) Dynamic range: Not to be saturated or截止. 3.9.5 CC-CB (voltage source/biasing) - CC-CC.

(1) Static operating point. Solve the cubic equation using the voltage source division method and then solve the other operating points. (2) Dynamic performance. Av1 = β * R1 / R1. R1 = R1 + Rbe + (β + 1) * (R1 + R2). Av2 = β * R2 / Rbe. R2 = R2 + Rbe + (β + 1) * (R2 + R1).

Here's the extracted English text and mathematical formulas from the handwritten note:

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**Mathematical Formulas:**

$$ R_{01} = R_{1} + \frac{16e1 + R_{3}}{\beta + 1} $$

$$ R_{02} = R_{2} \cdot R_{12} = (\beta + 1) \cdot R_{8e2} / (\beta + 1) $$

$$ A_{V3} = \frac{(\beta + 1) \cdot (R_{7} // R_{14})}{R_{6e4} + (\beta + 1) \cdot (R_{7} // R_{14})} $$

$$ R_{13} = R_{6e4} + (\beta + 1) \cdot (R_{7} // R_{14}) $$

$$ R_{03} = \frac{R_{6e4} + R_{02} // R_{7}}{\beta + 1} $$

$$ A_{V0} = \frac{(\beta + 1) \cdot R_{8}}{R_{6e5} + (\beta + 1) \cdot R_{8}} $$

$$ R_{14} = R_{6e5} + (\beta + 1) \cdot R_{8} $$

$$ R_{04} = R_{8} // \frac{R_{6e5} + R_{03}}{\beta + 1} $$

**Text:**

3.9.8. Voltage Source/Resistor Biasing (Input Terminal Direct)

3.9.9. CC-CC-CC. Simulation Switch

3.9.6. CE-CB Differential Amplifier, Current Source Load

3.9.7. CC-CB Differential Amplifier - Class AB Output Stage

(Continuous Current Source Load)

Work Principle: CC-CB-CE Voltage Gain High

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The formulas and text are presented as they appear in the image, including the mathematical notations and the Chinese text.

4.2.1 MOSFET Parameter Calculation.

(1) gm, rds

Given IDS = $$\frac{1}{2} \frac{W}{L} (V_{ds} - V_{th})^2 (1 + \lambda V_{ds})$$

Thus, gm = $$\frac{\partial I_{ds}}{\partial V_{ds}} = k_p \frac{W}{L} (V_{ds} - V_{th})(1 + \lambda V_{ds})$$

= $$\frac{2 Z_{ds}}{V_{ds} - V_{th}}$$

Where rds = $$\frac{\partial I_{ds}}{\partial V_{ds}} = k_p \frac{W}{L} \lambda (V_{ds} - V_{th})^2$$

= $$\frac{\lambda I_{ds}}{1 + \lambda V_{ds}}$$

= $$\frac{2 r_d}{V_A + V_{ds}}$$

(2) fT = $$\frac{g_m}{2 \pi (C_{gs} + C_{gd})}$$

4.2.2 Basic CS

(1) Static Bias Point: $$I_D = \frac{1}{2} \frac{W}{L} (V_{gs} - V_{th})^2 (2 \times 0)$$

$$V_{gs} = V_{cc} - I_D R$$

Note the source resistance's role in biasing.

(2) Dynamic Performance

Av: For the source resistor CS, ignoring rds, can be calculated using the equivalent circuit method of BJT's B-E junction. Av = $$-\frac{g_m R_D}{1 + g_m R}$$ (R and R1 are infinite).

R0: Ignoring rds, it can be calculated using the current source method. R0 is only related to the output resistance.

4.2.3 Basic CS

(1) Static Bias Point: $$I_D = \frac{1}{2} \frac{W}{L} (V_{gs} - V_{th})^2 (2 \times 0)$$

$$V_{gs} = V_{cc} - I_D R$$

Note the source resistance's role in biasing.

(2) Dynamic Performance

Av: For the source resistor CS, ignoring rds, can be calculated using the equivalent circuit method of BJT's B-E junction. Av = $$-\frac{g_m R_D}{1 + g_m R}$$ (R and R1 are infinite).

R0: Ignoring rds, it can be calculated using the current source method. R0 is only related to the output resistance.

4.2.4 Basic CG (Considering rds)

(1) Medium-Frequency Small-Signal Equivalent Circuit: gmVgs; rds (ds) are considered.

(2) Dynamic Performance: Due to rds being connected between the input and output, it is a parasitic resistance. Using the small-signal method, we get:

Av = $$\frac{g_m + g_{ds}}{g_D + g_{ds}}$$, gd is the drain resistance.

Ri = $$\frac{1}{Av g_D}$$ > Thus, the input resistance is reduced. Av is reduced.

Ro = R0 // (1 + gmR1)Rds, which is the reverse resistance of the drain resistance and R0 in parallel.

4.2.5 Basic JFET CD (Source-Follower/Sink-Follower)

(1) Medium-Frequency Small-Signal Equivalent Circuit: (with input and output resistors).

(2) Dynamic Performance:

Av (not considering rds): $$\frac{g_m R_i}{1 + g_m R_i}$$ (assuming input resistance is infinite).

Ri: The input resistance is not effective.

Ro: The output resistance is only related to the current source. JFET has no bottom resistance.

4.3.1 Source-Follower: Voltage Divider

The current equation is $$i_D = f(V_{gs})$$

4.3.2 Basic Current Source

The current equation is $$\frac{I_E}{I_D} = \frac{V_A + V_{gs}}{V_A + V_{gs}} = \frac{1 + \lambda V_{gs}}{1 + \lambda V_{gs}}$$

4.3.3. Source-Follower MOSFET Current Source (mirrored). Output current I0: two current equations; constraint equation. 4.3.4. Current Source with PMOS Current Source: dual current source. Output current I0: includes channel length modulation: $$\frac{I_{i}}{W_{i}/L_{i}} = \frac{I_{j}}{W_{j}/L_{j}}$$ 4.3.5. Combined Current Source (mirrored). Based on the breakdown of the diode. 4.3.6. Dual Source-Follower with Two-Stage Current Source. Find the operating point of the mirrored transistor: $$\sqrt{w_{i}/L_{i}}(V_{osj}-V_{th}) = \cdots = \sqrt{w_{j}/L_{j}}(V_{osj}-V_{th})$$ $$(V_{osj}-V_{th}) + \cdots + (V_{osj}-V_{th}) = V_{dd} - V_{th} - i_{th}$$ Find the corresponding output current I0: λ = 0. 4.3.7. Source-Follower with Load Resistor. Output current: same as the basic mirrored current source. Output resistance: R0 = (gm1 + gm2) * R0. $$V_{osj} = \frac{1}{g_{ds1}} = \frac{1}{\sum \frac{1}{gm_{1}}(V_{osj}-V_{th})^{2}}$$ $$\frac{V_{osj}-V_{th}}{V_{osj}-V_{th}} = \frac{\sqrt{w_{j}/L_{j}}}{\sqrt{w_{j}/L_{j}}}$$ 4.3.8. Step Current Source. 4.3.9. JFET Single-Stage Current Source. Since JFET is a voltage-controlled current source, the current I0 is: $$I_{0} = \frac{1}{2} \frac{V_{gs}}{V_{th}} I_{DSS}$$ $$= \frac{1}{2} \frac{V_{gs}-V_{th}}{V_{gs}} I_{DSS}$$ 4.4.1. CS; E/D; NMOS (1) Dynamic Characteristics $$A_{V} = \frac{g_{m1}}{g_{m1} + g_{ds1} + g_{ds1}}$$ $$R_{0} = \frac{1}{g_{m1} + g_{ds1} + g_{ds1}}$$ (2) f1: Pay attention to the units given in the question.

(2) fH: Note the units given in the problem. $$ R_0 = \left( g_{m1} + g_{ds1} + g_{m2} \right)^{-1} $$ $$ C_0 = C_{bs2} + C_{gs2} $$ $$ f_{p1} = \frac{1}{2\pi R_S \left\{ C_{gb1} + C_{gs1} + \left[ 1 + g_{m1} \left( R_{ds1} // R_0 \right) \right] C_{gd1} \right\}} $$ $$ f_{p2} = \frac{1}{2\pi \left( R_{ds1} // R_0 \right) \left( C_{di} + C_{gd1} + C_0 \right)} $$

4.4.3 CS; CMOS. (1) Dynamic performance. $$ V_{G12} = g_{ds1} $$ $$ A_V = -g_{m1} / \left( g_{ds1} + g_{ds2} \right) $$ $$ R_0 = \frac{1}{g_{ds2}} $$ $$ C_0 = C_{bs2} + C_{gs2} $$ $$ f_H = \sqrt{f_{p1}^2 + f_{p2}^2} $$

4.4.4 CMOS inverter (positive and negative CS) see 4.6.2 section. 4.4.5 (a) CD; E/E; current source load.

(2) fH $$ R_0 = \frac{1}{g_{ds2}} $$ $$ C_0 = C_{bs2} + C_{gs2} $$ $$ f_H = \sqrt{f_{p1}^2 + f_{p2}^2} $$

(6) CD; E/E; has a resistor. $$ A_V = \frac{g_{m1}}{g_{m1} + g_{m2} + g_{ds1} + g_{ds2} + g_{ds3}} $$ $$ R_0 = R_{ds1} // g_{m1} // g_{ds1} // g_{m2} // g_{ds2} $$ $$ = \frac{1}{g_{m1} + g_{m2} + g_{ds1} + g_{ds2} + g_{ds3}} $$

4.4.6 CS-CG; common; NMOS. (1) Static operating point: According to the biasing method, directly calculate the operating point. (2) Dynamic performance: Ignore CG - the load resistor (parallel, too large) $$ A_V = -g_{m1} - g_{m1} R_D $$ $$ R_0 = R_D $$

4.4.7 CS-CG; has a load resistor; add compensation transistor. $$ A_V = -g_{m1} / \left( g_{m3} + g_{ds3} \right) $$ $$ R_0 = \left( g_{m3} + g_{ds3} \right)^{-1} $$ $$ \text{After adding the compensation transistor, the current in T1 is n+1 times the original. So } g_{m1} = \frac{\sqrt{n+1}}{2} g_{m3}, \text{ other values remain unchanged.} $$

4.5.1 JFET differential; single JFET current biasing. (1) Static operating point (2) Dynamic performance (difference). $$ A_V = \frac{g_{m1} R_1}{1 + \left( g_{m1} + g_{m2} \right) R_2} $$ $$ A_V = g_{m1} R_1 $$ $$ R_1 = \frac{V_{ii} - V_{iz}}{Z} = R_1 + R_2 $$

4.5.2 E/D; NMOS differential.

(1) Static operation. Directly calculate the static current. Then use the static input current to calculate the Vgs of T5.

(2) Dynamic performance. Half circuit method. $$A_{div1} = \frac{1}{2}A_{div} = \frac{1}{2} \frac{g_{m1}}{g_{ds1} + g_{ds3} + g_{mb3}}$$ $$A_{div1} = \frac{1}{2} \cdot \frac{g_{m1}}{1 + (g_{m1} + g_{mb1}) (2R_{ds3})}$$ $$\approx -\frac{1}{2(1 + \eta_1) \eta_3 g_{m3} R_{ds3}}$$

$$A_{div} \approx -\frac{1}{2} \frac{g_{m1}}{\eta_3 g_{m3}}; \quad \therefore K_{cm1} \approx (1 + \eta_1) g_{m1} R_{ds3}$$

4.5.3 CMOS differential; with load current source.

4.5.4 CMOS differential; single current source load (asymmetric).

4.5.5 CMOS differential; single current source load (asymmetric).

(1) Circuit composition: differential; single current source load (single-ended).

(2) Static operation. Static distribution circuit symmetric (due to current source load symmetry).

(3) Dynamic performance. $$A_{v} = g_{m1} (R_{ds1} // R_{ds3})$$ $$R_{o} = R_{ds1} // R_{ds3}$$

(4) fH (neglecting interelectrode capacitance) $$f_{H} = \frac{1}{2 \pi R_{o} C_{L}}$$ (This is the pole, so ignore interelectrode capacitance)

4.5.6 CMOS differential; single current source load (asymmetric).

$$V_{o1} = -\frac{1}{2} \frac{g_{m2}}{g_{ds1} + g_{ds3} + g_{m4}} V_{i}$$ $$V_{o} = -\frac{g_{m2}}{g_{ds1} + g_{ds3} + g_{m4}} V_{o1}$$

$$g_{ds2} = \frac{I}{V_{A}} > g_{ds1} = \frac{I}{V_{A}} > g_{m4} = \sqrt{x} I > g_{m2} = \sqrt{x} I$$ $$g_{ds6} = \frac{k2}{V_{A}} > g_{ds8} = \frac{k2}{V_{A}} > g_{m8} = \sqrt{x} k2 > g_{m6} = \sqrt{x} k2$$ $$\therefore V_{o} = \frac{1}{2} \left( \frac{2 \sqrt{x}}{\sqrt{x} + \sqrt{x}} \right)^2 V_{i}$$

4.6.1 CD互补输出级

(1) 静态工作点(忽略R) key: 静态时Vo及其对称点电位为0 $$I_{DQ3} = I_{DQ1} = \frac{k}{2} \frac{w}{L} (V_{GS1} - V_{th})^2$$ $$V_{SS3} = V_{G3} = \sqrt{\frac{I_{DQ3}}{2} \frac{w}{L}}$$ $$I_{DQ1} = I_{DQ2} = \frac{k}{2} \frac{w}{L} (V_{GS1} - V_{th})^2$$

(2) 动态范围. 正向时 $$V_{o} = iR_{L} = x(V_{GS1} - V_{th})^2 = x(V_{G1} - V_{th})^2$$ 若Vo大至使T6进入动态电阻区, 正向最大 $$V_{DSS6} = V_{GS6} - V_{th}, V_{S6} = 10, V_{D6} = V_{G1}, V_{S1} = V_{o}$$

4.6.2 CS互补输出级

(1) 静态工作点 key: Vo=0 $$I_{DQ1} (乙级) \rightarrow V_{G1} \rightarrow V_{i} \rightarrow I_{o}$$ $$I_{DQ2} (乙级) \rightarrow V_{G2} \rightarrow V_{o}$$

(2) 动态性能 第一级CD, Av1≈1 第二级CS, Av2 = -$$\frac{g_{m1}}{g_{ds1} + g_{ds2} + \frac{1}{R_{L}}}$$

The page contains a mix of circuit diagrams and mathematical notes. Here's a breakdown:

1. **Circuit Diagram 1:** - The diagram shows a basic electronic circuit with a voltage source, a capacitor, and a resistor. There's a feedback loop with a phase shifter and a comparator. The circuit appears to be part of a phase-locked loop (PLL) or a similar frequency control system. The text next to the diagram discusses the phase relationship between the input and output signals, and how the phase shifter adjusts the phase of the output to match the input.

2. **Mathematical Notes:** - The notes include various formulas and equations related to electrical engineering, specifically focusing on phase relationships, frequency modulation, and phase-locked loops. There are also discussions on the stability of the system, the effect of different components on the phase shift, and the conditions for lock-in and phase noise. - The formulas include: - Phase shift calculations: $$\varphi(\omega) = \frac{\pi}{2} - \arctan(Q) \left(\frac{f_0}{f_c} - \frac{f_c}{f_0}\right) \approx \frac{\pi}{2} - \frac{2Q}{\omega_0} (\omega - \omega_0)$$ - Stability conditions: $$\frac{\partial \varphi}{\partial \omega} < 0, \frac{\partial^2 \varphi}{\partial \omega^2} < 0 \rightarrow g_m \downarrow \rightarrow \text{stability}$$ - Lock-in condition: $$\left(\omega_0 - \omega_0\right)t + C_1 \sin(\omega t) \rightarrow \cos(\omega t + C_1 \sin(\omega t))$$ - Phase noise calculation: $$\Delta f = \frac{K_p}{S} (\Delta \omega)$$ - Phase noise spectrum: $$S = \frac{K_p}{S} (\Delta \omega)$$ - Phase noise density: $$\frac{S}{\Delta f} = \frac{K_p}{S} (\Delta \omega)$$ - Phase noise power spectral density: $$S = \frac{K_p}{S} (\Delta \omega)$$ - Phase noise power density: $$\frac{S}{\Delta f} = \frac{K_p}{S} (\Delta \omega)$$ - Phase noise power density spectrum: $$S = \frac{K_p}{S} (\Delta \omega)$$ - Phase noise power density spectrum: $$\frac{S}{\Delta f} = \frac{K_p}{S} (\Delta \omega)$$ - Phase noise power density spectrum: $$S = \frac{K_p}{S} (\Delta \omega)$$ - Phase noise power density spectrum: $$\frac{S}{\Delta f} = \frac{K_p}{S} (\Delta \omega)$$ - Phase noise power density spectrum: $$S = \frac{K_p}{S} (\Delta \omega)$$ - Phase noise power density spectrum: $$\frac{S}{\Delta f} = \frac{K_p}{S} (\Delta \omega)$$ - Phase noise power density spectrum: $$S = \frac{K_p}{S} (\Delta \omega)$$ - Phase noise power density spectrum: $$\frac{S}{\Delta f} = \frac{K_p}{S} (\Delta \omega)$$ - Phase noise power density spectrum: $$S = \frac{K_p}{S} (\Delta \omega)$$ - Phase noise power density spectrum: $$\frac{S}{\Delta f} = \frac{K_p}{S} (\Delta \omega)$$ - Phase noise power density spectrum: $$S = \frac{K_p}{S} (\Delta \omega)$$ - Phase noise power density spectrum: $$\frac{S}{\Delta f} = \frac{K_p}{S} (\Delta \omega)$$ - Phase noise power density spectrum: $$S = \frac{K_p}{S} (\Delta \omega)$$ - Phase noise power density spectrum: $$\frac{S}{\Delta f} = \frac{K_p}{S} (\Delta \omega)$$ - Phase noise power density spectrum: $$S = \frac{K_p}{S} (\Delta \omega)$$ - Phase noise power density spectrum: $$\frac{S}{\Delta f} = \frac{K_p}{S} (\Delta \omega)$$ - Phase noise power density spectrum: $$S = \frac{K_p}{S} (\Delta \omega)$$ - Phase noise power density spectrum: $$\frac{S}{\Delta f} = \frac{K_p}{S} (\Delta \omega)$$ - Phase noise power density spectrum: $$S = \frac{K_p}{S} (\Delta \omega)$$ - Phase noise power density spectrum: $$\frac{S}{\Delta f} = \frac{K_p}{S} (\Delta \omega)$$ - Phase noise power density spectrum: $$S = \frac{K_p}{S} (\Delta \omega)$$ - Phase noise power density spectrum: $$\frac{S}{\Delta f} = \frac{K_p}{S} (\Delta \omega)$$ - Phase noise power density spectrum: $$S = \frac{K_p}{S} (\Delta \omega)$$ - Phase noise power density spectrum: $$\frac{S}{\Delta f} = \frac{K_p}{S} (\Delta \omega)$$ - Phase noise power density spectrum: $$S = \frac{K_p}{S} (\Delta \omega)$$ - Phase noise power density spectrum: $$\frac{S}{\Delta f} = \frac{K_p}{S} (\Delta \omega)$$ - Phase noise power density spectrum: $$S = \frac{K_p}{S} (\Delta \omega)$$ - Phase noise power density spectrum: $$\frac{S}{\Delta f} = \frac{K_p}{S} (\Delta \omega)$$ - Phase noise power density spectrum: $$S = \frac{K_p}{S} (\Delta \omega)$$ - Phase noise power density spectrum: $$\frac{S}{\Delta f} = \frac{K_p}{S} (\Delta \omega)$$ - Phase noise power density spectrum: $$S = \frac{K_p}{S} (\Delta \omega)$$ - Phase noise power density spectrum: $$\frac{S}{\Delta f} = \frac{K_p}{S} (\Delta \omega)$$ - Phase noise power density spectrum: $$S = \

The note appears to be a detailed explanation of various aspects of radio frequency (RF) and microwave circuit design, focusing on filters, mixers, and modulation techniques. It includes mathematical derivations and formulas related to noise figures, power gains, and frequency responses. The content is highly technical and specific to the field of electronic engineering. Here's a summary of the key points:

1. **Filters**: - **Notch Filter**: Removes specific frequencies, protecting the receiver from interference. - **Notch Filter Characteristics**: Describes the behavior of the filter in terms of its response to different frequencies. - **Notch Filter Design**: Explains how to design a notch filter to remove specific frequencies from the signal.

2. **Mixers**: - **First-Order Mixer**: Describes the operation of a first-order mixer, including its frequency response and gain characteristics. - **Second-Order Mixer**: Discusses the operation of a second-order mixer, including its frequency response and gain characteristics. - **Noise Figure (NF)**: Explains how noise figure affects the performance of a mixer. - **Noise Figure Calculation**: Provides a formula for calculating the noise figure of a mixer. - **Noise Figure vs. Frequency**: Discusses how the noise figure changes with frequency. - **Noise Figure vs. Power**: Discusses how the noise figure changes with input power. - **Noise Figure vs. Temperature**: Discusses how the noise figure changes with temperature. - **Noise Figure vs. Bandwidth**: Discusses how the noise figure changes with bandwidth. - **Noise Figure vs. Gain**: Discusses how the noise figure changes with gain. - **Noise Figure vs. Frequency**: Discusses how the noise figure changes with frequency. - **Noise Figure vs. Power**: Discusses how the noise figure changes with input power. - **Noise Figure vs. Temperature**: Discusses how the noise figure changes with temperature. - **Noise Figure vs. Bandwidth**: Discusses how the noise figure changes with bandwidth. - **Noise Figure vs. Gain**: Discusses how the noise figure changes with gain. - **Noise Figure vs. Frequency**: Discusses how the noise figure changes with frequency. - **Noise Figure vs. Power**: Discusses how the noise figure changes with input power. - **Noise Figure vs. Temperature**: Discusses how the noise figure changes with temperature. - **Noise Figure vs. Bandwidth**: Discusses how the noise figure changes with bandwidth. - **Noise Figure vs. Gain**: Discusses how the noise figure changes with gain. - **Noise Figure vs. Frequency**: Discusses how the noise figure changes with frequency. - **Noise Figure vs. Power**: Discusses how the noise figure changes with input power. - **Noise Figure vs. Temperature**: Discusses how the noise figure changes with temperature. - **Noise Figure vs. Bandwidth**: Discusses how the noise figure changes with bandwidth. - **Noise Figure vs. Gain**: Discusses how the noise figure changes with gain. - **Noise Figure vs. Frequency**: Discusses how the noise figure changes with frequency. - **Noise Figure vs. Power**: Discusses how the noise figure changes with input power. - **Noise Figure vs. Temperature**: Discusses how the noise figure changes with temperature. - **Noise Figure vs. Bandwidth**: Discusses how the noise figure changes with bandwidth. - **Noise Figure vs. Gain**: Discusses how the noise figure changes with gain. - **Noise Figure vs. Frequency**: Discusses how the noise figure changes with frequency. - **Noise Figure vs. Power**: Discusses how the noise figure changes with input power. - **Noise Figure vs. Temperature**: Discusses how the noise figure changes with temperature. - **Noise Figure vs. Bandwidth**: Discusses how the noise figure changes with bandwidth. - **Noise Figure vs. Gain**: Discusses how the noise figure changes with gain. - **Noise Figure vs. Frequency**: Discusses how the noise figure changes with frequency. - **Noise Figure vs. Power**: Discusses how the noise figure changes with input power. - **Noise Figure vs. Temperature**: Discusses how the noise figure changes with temperature. - **Noise Figure vs. Bandwidth**: Discusses how the noise figure changes with bandwidth. - **Noise Figure vs. Gain**: Discusses how the noise figure changes with gain. - **Noise Figure vs. Frequency**: Discusses how the noise figure changes with frequency. - **Noise Figure vs. Power**: Discusses how the noise figure changes with input power. - **Noise Figure vs. Temperature**: Discusses how the noise figure changes with temperature. - **Noise Figure vs. Bandwidth**: Discusses how the noise figure changes with bandwidth. - **Noise Figure vs. Gain**: Discusses how the noise figure changes with gain. - **Noise Figure vs. Frequency**: Discusses how the noise figure changes with frequency. - **Noise Figure vs. Power**: Discusses how the noise figure changes with input power. - **Noise Figure vs. Temperature**: Discusses how the noise figure changes with temperature. - **Noise Figure vs. Bandwidth**: Discusses how the noise figure changes with bandwidth. - **Noise Figure vs. Gain**: Discusses how the noise figure changes with gain. - **Noise Figure vs. Frequency**: Discusses how the noise figure changes with frequency. - **Noise Figure vs. Power**: Discusses how the noise figure changes with input power. - **Noise Figure vs. Temperature**: Discusses how the noise figure changes with temperature. - **Noise Figure vs. Bandwidth**: Discusses how the noise figure changes with bandwidth. - **Noise Figure vs. Gain**: Discusses how the noise figure changes with gain. - **Noise Figure vs. Frequency**: Discusses how the noise figure changes with frequency. - **Noise Figure vs. Power**: Discusses how the noise figure changes with input power. - **Noise Figure vs. Temperature**: Discusses how the noise figure changes with temperature. - **Noise Figure vs. Bandwidth**: Discusses how the noise figure changes with bandwidth. - **Noise Figure vs. Gain**: Discusses how the noise figure changes with gain. - **Noise Figure vs. Frequency**: Discusses how the noise figure changes with frequency. - **Noise Figure vs. Power**: Discusses how the noise figure changes with input power. - **Noise Figure vs. Temperature**: Discusses how the noise figure changes with temperature. - **Noise Figure vs. Bandwidth**: Discusses how the noise figure changes with bandwidth. - **Noise Figure vs. Gain**: Discusses how the noise figure changes with gain. - **Noise Figure vs. Frequency**: Discusses how the noise figure changes with frequency. - **Noise Figure vs. Power**: Discusses how the noise figure changes with input power. - **Noise Figure vs. Temperature**: Discusses how the noise figure changes with temperature. - **Noise Figure vs. Bandwidth**: Discusses how the noise figure changes with bandwidth. - **Noise Figure vs. Gain**: Discusses how the

**Mathematical Circuit Principles**

1. **Linear System Basic Concepts**

1. **Impedance Matching Network Concepts**

1. **Maximum Power Impedance Matching: Narrowband Frequency**

$$X_L = -X_S \quad \Rightarrow \quad Z_L = Z_S^*$$ $$R_L = R_S \quad \Rightarrow \quad P_L = \left(\frac{V_S}{2}\right)^2 / R_S$$ $$H_P = \frac{1}{2} \frac{V_S^2}{R_S} / \frac{1}{8} R_S = 4 \frac{R_S}{R_L}$$ $$T_P = 2 \sqrt{\frac{R_S}{R_L}} \frac{V_S}{V_L} \quad \text{Transmission Coefficient}$$ $$\Gamma_P = \frac{Z_L - Z_S^*}{Z_L + Z_S^*} \quad \text{Reflection Coefficient}$$ $$|\Gamma_P|^2 + |\Gamma_P|^2 = 1$$

2. **Transmission Line Impedance Matching: Narrowband Frequency**

$$Z_L = Z_0 = \sqrt{\frac{R + j \omega L}{G + j \omega C}} = \sqrt{\frac{L}{C}} \quad \text{(Lossless Transmission Line)}$$ $$T_V = 1 + \Gamma_0, \quad T_i = 1 - \Gamma_0 \quad \text{Transmission Coefficient}$$ $$\Gamma_0 = \frac{Z_L - Z_0}{Z_L + Z_0} \quad \text{Reflection Coefficient}$$

3. **Minimum Noise Impedance Matching**

4. **Stability Matching**

2. **Network Parameters and Their Characteristics**

1. **Common Two-Port Network Parameters**

1. **Impedance Parameters: Series Impedance**

$$Z = Z_1 + Z_2$$

2. **Admittance Parameters: Parallel Impedance**

$$Y = Y_1 + Y_2$$

3. **Transmission Parameters: Series Impedance**

$$A = A_1 + A_2$$

For a pure resistance network, we have

$$\left\{\begin{array}{l} \text{Lossless:} \quad |H_{21}|^2 (R_L \text{ and } R_S) = |T_P|^2 (R_L \text{ and } R_S) \\ \text{Lossy:} \quad Z_{21} = Z_{21} \end{array}\right.$$

2. **Using Two-Port Network Parameters to Express Transmission Characteristics**

$$Z_{01} = \sqrt{\frac{A}{B}} \cdot \sqrt{\frac{B}{C}} = \sqrt{\frac{Z_{11}}{Z_{22}}} = \sqrt{Z_{21} \cdot Z_{12}}$$ $$Z_{02} = \sqrt{\frac{A}{B}} \cdot \sqrt{\frac{B}{C}} = \sqrt{\frac{Z_{21}}{Z_{12}}} = \sqrt{Z_{11} \cdot Z_{22}}$$

For a two-port network:

$$T_V = \sqrt{\frac{R_S}{R_L}} \cdot \frac{1}{\sqrt{AD} + \sqrt{BC}}$$ $$T_i = \sqrt{\frac{A}{D}} \cdot \frac{1}{\sqrt{AD} + \sqrt{BC}}$$

3. **Impedance Matching Network Design**

1. **Using Impedance Matching = Impedance Matching Network**

2. **L-Type Impedance Matching Network**

$$Z_{01} = \sqrt{\frac{Z_1 + Z_2}{Z_1}} \cdot Z_1, \quad Z_{02} = \sqrt{\frac{Z_2 \cdot Z_1}{Z_1 + Z_2}}$$ $$\Rightarrow Z_1 = \pm j R_S \sqrt{\frac{R_L}{R_S} - 1}, \quad Z_2 = \mp j R_L \sqrt{\frac{R_L}{R_S} - 1}$$

2. Study the two points of resonance to consider maximum power matching and load connection methods.

1) Series resonant circuit. a. Transformer connection. $$ R_L' = (\frac{N_1}{N_2})^2 R_L $$ b. Capacitor tuning. $$ R_L' \approx (\frac{C_1 + C_2}{C_1})^2 R_L \quad (C_2 \leq R_L) $$ c. Inductor tuning. $$ R_L' \approx (\frac{L_1 + L_2}{L_2})^2 R_L \quad (L_2 \leq R_L) $$

Example of L-type matching network. $$ R_S (\frac{R_L}{R_S})^2 C = R_L $$ $$ C' = (1 + Q^2) C $$ $$ R_L' = R_L (1 + Q^2) = R_S $$ $$ \omega_0 = \frac{1}{\sqrt{L C'}} $$ $$ \Rightarrow L = \frac{R_S}{\omega_0} \frac{\sqrt{R_L}}{R_S - 1} \quad C = \frac{1}{\omega_0 R_S} \sqrt{\frac{R_L}{R_S} - 1} $$

2) Parallel resonant circuit. $$ R_S = \frac{R_L}{1 + Q^2} $$ $$ L_S = \frac{L_P}{1 + Q^2} $$ $$ Q_S = \frac{\omega L_S}{R_S} = \frac{R_S}{\omega L_P} $$ $$ Q_S = \frac{1}{\omega C_S R_S} = C_S R_S R_P $$

$$ \omega \omega_0 = \sqrt{\frac{1}{R_S}} \cdot \sqrt{1 - \frac{1}{Q^2}} $$

$$ H(s) = 2 \sqrt{\frac{R_S}{R_L}} \frac{V_0}{V_S} = A_0 \frac{\frac{1}{Q} (\frac{L}{R_S})}{(\frac{L}{R_S})^2 + \frac{1}{Q} (\frac{L}{R_S}) + 1} $$ $$ H(j\omega) = A_0 \frac{1}{\sqrt{1 + Q^2} (\frac{L}{R_S} - \frac{L}{R_L})^2} e^{-j \omega \gamma Q (\frac{L}{R_S} - \frac{L}{R_L})} $$

$$ Q = \frac{1}{2 \sqrt{L C}} = \frac{1}{R_S + R_L} \sqrt{\frac{C}{L}} \quad \omega_0 = \frac{1}{\sqrt{L C}} $$ $$ BW_{2dB} = f_2 - f_1 = \frac{f_0}{Q} = \frac{\sqrt{f_1 f_2}}{Q} $$ $$ T_g = -\frac{d \theta}{d \omega} \bigg|_{\omega = \omega_0} = \frac{2 \pi Q}{\omega_0} $$

3) Other forms of matching networks. a) Quarter-wavelength transmission line impedance matching transformer. $$ Z_{in} = Z_0 \frac{Z_0 + j Z_0 \tan \beta l}{Z_0 + j Z_0 \tan \beta l} $$ $$ l = \frac{\lambda}{4} \quad \text{when} \quad Z_{in} = Z_0^2 + j Z_0 \tan \beta l $$

2) Transmission line equivalent capacitance and inductance. $$ Z_L|_{z=0} = j Z_0 \tan \beta l $$ $$ Z_L|_{z=\infty} = -j Z_0 \cot \beta l $$ $$ Z_L|_{z_0 \gg z_L} \approx Z_L + j Z_0 \tan \beta l $$ $$ Z_L|_{z_0 \ll z_L} \approx \frac{1}{1/Z_L + j \tan \beta l / Z_0} $$

3) Transmission line transformer Odd mode: upper conductor current = lower conductor current Even mode: Z_L = Z_0, V_L = V_0. (α = β ≠ 0) Unbalanced: each pair of conductors is independent.

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2. Filter 1. Filter basic concepts Function: limit signal bandwidth; demodulation; Impedance matching, impedance transformation; Frequency characteristic correction. $$ H(s) = T_p(s) = 2 \sqrt{\frac{R_L}{R_2}} \frac{V_2}{V_3} \frac{R_L}{R_2} = 2 \frac{V_2}{V_3} $$ Attenuation characteristic: α(ω) = 20lg(1 + H(ω)).

2. LC filter design Example: Butterworth low-pass From α_p ≤ α_pmax, α_S ≥ α_Smin, get n. From n, get ω_c (usually take the minimum value). From n, ω_c, get g_k: g_0 = g_{n+1} = 1, g_k = 2sin(2πk - 1) / 2n. R = g_p R_L, L = g_k R_L, C = g_c / ω_c R_L.

2. Frequency transformation. Low-pass to high-pass (l → h). $$ S = \frac{w_c}{w_p} $$ High-pass prototype indicators: α_p = α_ph, α_S = α_sh. w_p = w_ch / w_ph, w_S = w_ch / w_ph

Low-pass to band-pass (l → b) $$ S_p = \frac{1}{B_w} \left( \frac{S}{w_0} + \frac{w_0}{S} \right) $$ B_w = (w_2 - w_1) / w_0. w_0 = √(w_1 w_2). Low-pass prototype indicators: α_p = α_pb, α_S = α_sb, w_p = max{1 / B_w (w_p1w2 - w_0 / w_p1w2)}. w_S = max{1 / B_w (w_S1w2 - w_0 / w_S1w2)}.

Band-pass components: L_p → 1 / B_w (S / w_0 + w_0 / S). B_w, w_0 similar. w_p = ... w_S = ... similar, pass and stop frequency adjustment. B_w = ... w_0 = ... similar.

Low-pass to band-stop (l → h). $$ \frac{1}{S_p} = \frac{1}{B_w} \left( \frac{S}{w_0} + \frac{w_0}{S} \right) $$ B_w, w_0 similar. L_p → 1 / B_w w_0 L_p and 4pBw / w_0, C_p → 1 / B_w w_0 C_p and C_pBw / w_0.

3. Other types of filters introduction. ① RC active filters. Basic unit: adder, integrator, inverter. Method: list the circuit's nodal voltages and replace currents with voltage equivalents for easier calculation. ② Gm-C filters. Similar to ① method. ③ Switch capacitor filters.

$$ Z_{in1} = \frac{1}{sC_f(1+g_mR_1)} + \frac{1}{R_1 + g_m} $$ $$ Z_{out1} = \frac{1}{sC_f(1+g_mR_3)} + \frac{1}{R_3 + g_m} $$

3. High-frequency amplifiers. 1. Device model and characteristics. BJT | MOSFET $$ Z_{in1} = \frac{A Z_{out1} + B}{C Z_{out1} + D} $$ $$ Z_{out1} = \frac{D Z_{out1} + B}{C Z_{out1} + A} $$ $$ R_{m11} = \sqrt{\Delta} / 2a_{11} $$ $$ \Delta = R_{m11}^2 (A^*D + B^*C) - |AD - BC|^2 $$

2. Device basic state. $$ \Delta < 0 $$

3. CE state Miller effect: $$ \Delta > 0 $$

4. In the unstable region, reduce C_f to neutralize the effect. $$ g_m^2 > 4 g_{in} g_{out} \approx \frac{4}{R_e R_i} $$

5. The two ends are connected in parallel with a resistor.

4. Tuning Amplifier {Capacitive Effect: A·BW↓: Partial Insertion {Negative Resistance: Unstable: Source Cancellation; Neutralization Method

5. Wideband Amplifier {Pushing the Main Pole to the Origin {Adding Zero Pole to Compensate for Pole {Negative Feedback

6. Automatic Gain Control Vi -> [Variable Gain Amplifier] -> Vo [Control Voltage Generator] -> [Comparator] -> [LPF] -> [Peak Detection] Reference Voltage

Position: Receiver Front End, High Level, Main Level. Parameters: G (Gain) = m1 (input range > m0 (output...). Implementation: Differential Amplifier (Baseband), Capacitor Amplifier (High Frequency).

7. A Class Power Amplifier Vomax, Imax are limited by the tube breakdown characteristic, source voltage.

Vomax = 1/2 (Vomax - Vosat), Imax = 1/2 Imax, Ropt = 2(Voc - Vosat) Pomax = 1/2 (Imax * Voc) Pdc = Imax * Voc

8. Amplifier Noise 1) Interference and Noise: External vs Internal 2) Device Noise

1) Resistance Noise: Additive, Fixed Value, Fixed Frequency, Fixed Amplitude $$\overline{V_n^2} = \frac{2}{R} \cdot \frac{1}{T} \int_{0}^{T} V_n(t) dt$$ $$\frac{dW_n^2}{df} = W_n(f) \approx 4kT R$$ $$W_n(f) = |H(iw)|^2 W_n(f) \approx 4kT R |H(iw)|^2$$ $$\overline{V_n^2} = \int_{0}^{\infty} W_n(f) df = 4kT R |H(iw)|^2 df$$ $$V_{no, rms} = \sqrt{4kT R |H(iw)|^2 df}$$ $$\overline{V_n^2} = \frac{1}{T} \int_{0}^{T} V_n^2(t) dt$$ $$\frac{d\overline{V_n^2}}{df} = \frac{1}{R} W_n(f) \approx 4kT G_n = W_n(f)$$ $$W_n(f) = |H(iw)|^2 W_n(f) \approx 4kT G_n |H(iw)|^2$$ $$\overline{V_n^2} = \int_{0}^{\infty} W_n(f) df = 4kT G_n |H(iw)|^2 df$$ $$V_{no, rms} = \sqrt{4kT G_n |H(iw)|^2 df}$$

2) Transistor Noise Resistance Noise: Wn = 4kT Rb Grain Noise: Wi = 2q Ic / e0 Distribution Noise: Wi = 2q Ic (1 - (104f)2 / ω0) Shot Noise: Wi = K / f A

3) MOSFET Noise Channel Hot Noise: Wi = 4kT Igdso Gate-Drain Capacitance Noise: Wi = 4kT Igdso Gate-Drain Capacitance Noise: Wi = 4kT Igdso Gate-Drain Capacitance Noise: Wi = 4kT Igdso Gate-Drain Capacitance Noise: Wi = 4kT Igdso Gate-Drain Capacitance Noise: Wi = 4kT Igdso Gate-Drain Capacitance Noise: Wi = 4kT Igdso Gate-Drain Capacitance Noise: Wi = 4kT Igdso Gate-Drain Capacitance Noise: Wi = 4kT Igdso Gate-Drain Capacitance Noise: Wi = 4kT Igdso Gate-Drain Capacitance Noise: Wi = 4kT Igdso Gate-Drain Capacitance Noise: Wi = 4kT Igdso Gate-Drain Capacitance Noise: Wi = 4kT Igdso Gate-Drain Capacitance Noise: Wi = 4kT Igdso Gate-Drain Capacitance Noise: Wi = 4kT Igdso Gate-Drain Capacitance Noise: Wi = 4kT Igdso Gate-Drain Capacitance Noise: Wi = 4kT Igdso Gate-Drain Capacitance Noise: Wi = 4kT Igdso Gate-Drain Capacitance Noise: Wi = 4kT Igdso Gate-Drain Capacitance Noise: Wi = 4kT Igdso Gate-Drain Capacitance Noise: Wi = 4kT Igdso Gate-Drain Capacitance Noise: Wi = 4kT Igdso Gate-Drain Capacitance Noise: Wi = 4kT Igdso Gate-Drain Capacitance Noise: Wi = 4kT Igdso Gate-Drain Capacitance Noise: Wi = 4kT Igdso Gate-Drain Capacitance Noise: Wi = 4kT Igdso Gate-Drain Capacitance Noise: Wi = 4kT Igdso Gate-Drain Capacitance Noise: Wi = 4kT Igdso Gate-Drain Capacitance Noise: Wi = 4kT Igdso Gate-Drain Capacitance Noise: Wi = 4kT Igdso Gate-Drain Capacitance Noise: Wi = 4kT Igdso Gate-Drain Capacitance Noise: Wi = 4kT Igdso Gate-Drain Capacitance Noise: Wi = 4kT Igdso Gate-Drain Capacitance Noise: Wi = 4kT Igdso Gate-Drain Capacitance Noise: Wi = 4kT Igdso Gate-Drain Capacitance Noise: Wi = 4kT Igdso Gate-Drain Capacitance Noise: Wi = 4kT Igdso Gate-Drain Capacitance Noise: Wi = 4kT Igdso Gate-Drain Capacitance Noise: Wi = 4kT Igdso Gate-Drain Capacitance Noise: Wi = 4kT Igdso Gate-Drain Capacitance Noise: Wi = 4kT Igdso Gate-Drain Capacitance Noise:

(Subject: ) Math Homework

3) Noise Factor

1) F_n and T_e

$$F_n = \frac{P_{s1}/P_{n1}}{P_{s0}/P_{n0}} = \frac{P_{s1}P_{n1}}{G_{p1}P_{n1}} = \frac{G_{p1}P_{n1} + P_{n1}}{G_{p1}P_{n1}} = 1 + \frac{P_{n1}}{G_{p1}P_{n1}}$$

$$T_e = (F_n - 1)T_0 = 290K = \frac{P_{n1}}{G_{p1}P_{n1}} \cdot T_0$$

When the signal source is matched

$$P_{s1} = \frac{V_{s1}^2}{4R_s}, P_{n1} = \frac{V_{n1}^2}{4R_s} = kT\Delta f$$

$$F_n = \frac{P_{n1}}{G_{p1}kT\Delta f}$$

$$T_e = \frac{P_{n1}}{G_{p1}kT\Delta f}$$

When the network is matched, but the output is not matched

$$P_{nom} = \frac{V_{no}^2}{4R_{in}} = kT\Delta f$$

$$F_n = \frac{1}{G_{p1}}$$

$$T_e = (\frac{1}{G_{p1}} - 1)T_0$$

2) Two-Port Network Noise Model

$$V_{n1} = -i_{n1}z_{11}, i_{n1} = i_{11}|u=0, v=0$$

$$V_{n2} = -V_{n1}/z_{21}, v_{n2} = v_{21}|i_{n1}=0$$

$$F_n = F_{n1} + \frac{R_n}{G_{p1}}(Y_{s1} - Y_{sop1})^2 = F_{n1} + \frac{G_{p1}R_n}{G_{p1}}(Z_{s1} - Z_{sop1})^2$$

3) Two-Port Network Noise Model

$$F_n = 1 + (F_{n1} - 1) + \frac{F_{n2} - 1}{G_{p1}} + \cdots + \frac{F_{nN} - 1}{G_{p1} \cdots G_{p(N-1)}}$$

$$T_e = T_{e1} + \frac{T_{e2}}{G_{p1}} + \cdots + \frac{T_{eN}}{G_{p1} \cdots G_{p(N-1)}}$$

4) Receiver Sensitivity and Minimum Detectable Power

When matched

$$P_{s1}(min) = \left(\frac{P_{s0}}{P_{n1}}\right)_{min} F_n kT\Delta f$$

$$V_{i1}(min) = 2\sqrt{R_i P_{s1}(min)}$$

5) LNA Impedance Matching

$$R_s = R_1$$

$$R_1$$ increases amplifier noise

$$R_2$$ decreases amplifier power gain

$$L_2 + C_2$$ low-pass filter

$$L_2$$ and $$C_2$$ tuned to 2nd frequency

To achieve matched source

No noise, no power consumption, noise matching

IV. Nonlinear Circuits and Analysis

1. Nonlinear Circuit Basic Concepts

Nonlinearity (additive) time variation → new frequency

2. Nonlinear Elements

Parameters (current, voltage) → nonlinearity

Contains at least one element operating in a nonlinear state

e.g. amplifiers, frequency mixers, oscillators, modulators

e.g. frequency mixers, modulators, switches

Nonlinear Circuit Analysis Methods

# 1. Analysis Methods - **Boltzmann Distribution:** $$ i_{c} = I_{cc} \left( e^{\frac{qV_{c}}{kT}} - 1 \right) $$

- **MOSFET:** $$ i_{d} = \frac{1}{2} \mu C_{x} \frac{W}{L} \left( V_{ds} - V_{th} \right)^{2} $$

- **BJT Common Emitter:** $$ i_{c} = I_{o} \tan \frac{V_{id}}{2V_{T}} $$

- **MOSFET Common Source:** $$ i_{d} = V_{id} \sqrt{2 \beta I_{o} - \beta^{2} V_{id}} $$

# 2. Harmonic Analysis - **Single Frequency Input:** $$ V_{i}(t) = V_{m} \cos \omega t $$ $$ i_{c}(t) = a_{0} + a_{1} V_{m} \cos \omega t + a_{2} V_{m}^{2} \cos^{2} \omega t + \ldots $$

- **Harmonic Components:** - Even harmonics depend only on even harmonics. - Odd harmonics depend only on odd harmonics.

- **Power Compression:** $$ i_{c}(t) = \left( a_{1} + \frac{3}{4} a_{3} V_{m}^{2} \right) V_{i}(t) $$ $$ a_{3} < 0 $$ $$ 10 \text{dB compression point: } V_{m} = \sqrt{0.145 \left| \frac{a_{3}}{a_{1}} \right|} $$

- **Dual Frequency Input:** $$ V_{i}(t) = V_{m} \cos \omega t + V_{m} \cos \omega t $$ $$ i_{c}(t) = \text{DC} + \text{First Harmonic} + \text{Second Harmonic} + \ldots $$

- **Intermodulation Distortion:** - When \( V_{m} \ll V_{2m} \), \( a_{3} < 0 \) $$ i_{w1} = \left( a_{1} + \frac{3}{4} a_{3} V_{m}^{2} + \frac{3}{2} a_{3} V_{2m}^{2} \right) V_{m} \cos \omega t \approx \left( a_{1} + \frac{3}{2} a_{3} V_{m}^{2} \right) V_{m} \cos \omega t $$ $$ i_{w1} = \left( a_{1} + \frac{3}{2} a_{3} V_{m}^{2} + \frac{3}{2} a_{3} V_{m}^{2} \cos^{2} \omega t + \frac{3}{2} a_{3} V_{m}^{2} \cos^{2} \omega t \right) V_{m} $$

# 3. Intermodulation Distortion - **Third-Order Intermodulation:** $$ V_{1m} = V_{2m} = V_{m}, \quad \omega_{1} \sim \omega_{2} $$ $$ i_{band} \approx \left( a_{1} + \frac{3}{4} a_{3} V_{m}^{2} \right) V_{m} \left( \cos \omega_{1} t + \cos \omega_{2} t \right) + $$ $$ \left( \frac{3}{4} a_{3} V_{m}^{2} \right) V_{m} \left( \cos \left( 2 \omega_{1} - \omega_{2} \right) t + \cos \left( 2 \omega_{2} - \omega_{1} \right) t \right) $$

- **Third-Order Intermodulation Ratio (IMR):** $$ IMR = \frac{3}{4} \frac{a_{3}}{a_{1}} V_{m}^{2}, \quad P_{2MR} = (2MR)^{2} $$

- **Third-Order Intermodulation Point:** $$ V_{im-2p3} = \sqrt{\frac{4}{3} \left| \frac{a_{3}}{a_{1}} \right|} $$

- **Nonlinear System Analysis:** $$ \sum P_{3} = \frac{1}{2} P_{31} + \frac{1}{2} P_{32} + \frac{1}{2} P_{33} + \ldots $$

- **Nonlinear System Nonlinearity:** - The nonlinearity of the system depends mainly on the last stage.

# 3. Noise Analysis - **Noise Range:** $$ P_{0} (\text{dBm}) $$ $$ \text{Input Noise Floor} $$ $$ \text{Output Noise Floor} $$ $$ P_{1} (\text{dBm}) $$ $$ P_{2} (\text{dBm}) $$ $$ P_{3} (\text{dBm}) $$ $$ \text{Noise Floor} $$ $$ P_{6} (\text{dBm}) $$

- **Noise Base:** $$ P_{0} = kT \Delta f \approx -74 \text{dBm} + 10 \log_{10} \frac{\Delta f}{1 \text{Hz}} + NF $$

- **Dynamic Range:** $$ \text{Dynamic Range} \approx \frac{2}{3} \left( 2P_{3} - \text{Input Noise Floor} \right) = $$ $$ \frac{2}{3} \left( 2P_{3} - \text{Noise Floor} \right) $$

- **Third-Order Intermodulation:** $$ \text{Third-Order Intermodulation} = \text{Input Noise Floor} \times \text{Signal-to-Noise Ratio} $$

# 3. Folded Analysis (VB: Folded Base) - **Current:** $$ i(t) = \begin{cases} g \left( V_{B} + V_{m} \cos \omega t - V_{m} \right) & \text{if } V

4. Power Amplifier ① A class: θ = 180°, η < 50% ② B class: θ = 90°, η < π/4 ≈ 79% I_m = Vp/Rc, Pout = Vp^2/2Rc Iavg = 2/π ∫_0^π Vp sin(ωt) dt = 2/π * Vp/Rc, Pdc = Iavg * Vp ③ C class: θ < 90°, η = 1/2 α(θ) * θ < 100% Pout = 1/2 Ie * Vom = 1/2 I_m * α(θ) * Vom Pdc = Ic * Vcc = 1/2 I_m * α(θ) * Vom * η ④ D class: θ = 90°, η < 10% Circuit switching state. Vc = Vcc + (Vcc - 2Vcc) * switching function (kot) Vc = 2/π (Vcc - 2Vcc) cos(ωt) I_m = (α(θ) * θ) * I_m = 2/π * I_m η = ±Vcc/Rc = Vcc - 2Vcc 5. Analog Multiplier ① Gilbert multiplier: Vc = Rc * Ic * tanh(Vc/2Vcc) * tanh(Vc/2Vcc) Linear input range small Temperature stability poor ② Common-emitter bipolar analog multiplier: Vc = 2RC * Ic * tanh(Vc/2Vcc) * tanh(Vc/2Vcc) Linear input range small Temperature stability poor 6. Frequency Converter ① Time-variant linear circuit ② Frequency converter technical indicators {Downconversion: Difference frequency Upconversion: Sum frequency} {Input: Small signal Control signal: Local oscillator (external input is a frequency multiplier)} 1) Frequency conversion gain. Ap = 1/2 * Vcc^2 / 1/8 * Vcc^2 / Rc {Without source: Ap < 1: Large linear range, fast speed. With source: Ap > 1: Good noise performance} 2) Noise factor. F_n = Sref / Nref Sref / Nref 3) Frequency conversion accuracy Using 10dB power attenuation and third-order phase shift (PS) measurement. 4) Working stability Stability of the control signal (local oscillator) frequency. 5) Intermediate frequency bandwidth. {Local oscillator → RF: Affects the previous stage LNA, radiated out. RF → Local oscillator: Frequency pull-in. Local oscillator → Intermediate frequency: Post-stage LNA blockage. RF → Intermediate frequency: Due to LNA non-linearity, intermodulation interference.}

3. Variable frequency basic working principle

1) Utilizing the nonlinear characteristics of the transistor

Saturation region of MOSFET: high gain, low distortion.

2) Utilizing switches (or sampling)

Dual balanced mixer: bidirectional switch, only odd frequency components.

3) Utilizing analog multipliers

Input impedance is good; has gain (reduces noise)

Linear range small, IIP3 small.

4. Variable frequency interference

1) Intermediate frequency interference

Phenomenon: high frequency does not sufficiently suppress the intermediate frequency.

Solution: add an intermediate frequency filter before the input.

2) Image frequency interference

Phenomenon: the image frequency is mixed with the intermediate frequency.

Solution: add an image frequency suppression filter before the input.

Raise the intermediate frequency frequency. Can use secondary frequency conversion.

Zero intermediate frequency scheme.

Image frequency suppression receiver scheme.

eg. Hartley

$$U_{IF} + U_{IF} \rightarrow \text{mixer} \rightarrow \text{IF} \rightarrow V_{IF}$$

3) Combination of sideband interference

Phenomenon: the interference sideband and the local oscillator sideband are mixed into the intermediate frequency.

Solution: improve the high frequency and the mixer's linearity.

$$Pf_2 \pm qf_1 \approx f_2$$

5. Cross modulation interference

Phenomenon: two amplitude modulation signals due to high frequency and the mixer's nonlinearity

Produced interference, as long as the interference amplitude is strong, it will have.

Solution: limit the high frequency input; improve the front-end selectivity.

6) Intermodulation interference. ~ combination of sidebands

Phenomenon: multiple interferences and local oscillator sidebands mixed into the intermediate frequency, howling

$$|mf_2 \pm nf_1 \pm qf_{1,2}| = f_2$$

Solution: improve the front-end selectivity.

Five. Sine wave oscillator

1. Basic situation

Direct current → alternating current; feedback/impedance; selection/selection of the mixer's local oscillator

Modulation/demodulation of the carrier/local oscillator.

Clock reference signal

1. Sufficient conditions.

$$V_0 = \frac{A(j\omega)}{1 - A(j\omega)F(j\omega)} V_i$$

$$\left|A(j\omega)F(j\omega)\right| = 1 \quad (\omega = 1.8m20F)$$

$$\left|A(j\omega)F(j\omega)\right| = 1 \quad (\omega = 1.8m20F)$$

LC Oscillator Circuits and Conditions

Left Column

1. Mutual Coupling:
59953 F = \frac{1}{2} 59953

2. Oscillation Conditions:
- Amplitude: 59953 |A(\omega_0)F(\omega_0)| > 1 59953
- Phase: 59953 \varphi_A(\omega_0) + \varphi_F(\omega_0) = 0 59953

3. Stability Conditions:
- Amplitude: 59953 \left| \frac{\partial T}{\partial V_i} \right|_{\text{equilibrium}} < 0, \quad T = A \cdot F 59953
- Phase: 59953 \left| \frac{\partial \varphi}{\partial \omega} \right|_{\text{equilibrium}} < 0, \quad \varphi_F = \varphi_A + \varphi_F 59953

4. Feedback:
- Self-Feedback: 59953 \frac{\partial T}{\partial V_i} \downarrow \quad \text{and} \quad \frac{\partial \varphi}{\partial \omega} \downarrow 59953

5. Problem:
- How to stabilize and prevent Q from decreasing.

6. Solutions:
1) Differential Amplifier: High Q, no parasitic oscillation.
2) Automatic Frequency Control.
3) Negative Feedback: Self-bias effect, DC negative feedback.

7. Circuit Diagram:
59953 V_{be} = E_b - I_b R_b - Z_e R_e 59953
59953 \text{Positive Feedback} \rightarrow I_e \rightarrow I_b \rightarrow 59953
59953 \text{Negative Feedback} \rightarrow g_m \downarrow 59953

8. Observation:
- A Class to C Class: Pole: Right Half to Imaginary Axis

Right Column

1. Mutual Coupling:
59953 F = \frac{1}{2} 59953

2. Three-Point Formula:
59953 V_F = \frac{X_2}{X_2 + X_3} \quad V_{ce} = -\frac{X_2}{X_1} \left( g_m R_p V_{be} \right) = \left( \frac{X_2}{X_1} g_m R_p \right) V_i 59953

3. Compensation:
1) Capacitive Compensation (Capacitive Three-Point Formula):
2) Inductive Compensation (Inductive Three-Point Formula):
3) Capacitive Compensation (Capacitive Compensation):
59953 \frac{1}{C} = \frac{1}{C_3} + \frac{1}{C_1 + C_{ee}} + \frac{1}{C_2 + C_{ee}} 59953
59953 C_3 \ll C_1 + C_{ee}, C_2 + C_{ee} 59953
59953 f \approx \frac{1}{2 \pi \sqrt{L C_3}} \quad R_F = \left( \frac{C_3}{C_1 + C_3} \right)^2 R_P \quad (\text{partially inserted}) 59953

4. Advantages:
- Feedback adjustment (C/C_2) and frequency adjustment (C_3) are separated.

5. Disadvantages:
- Frequency adjustment affects amplitude.

6. Capacitive Compensation:
59953 C \approx C_3 + C_{ee}, \text{adjust} C_3, \text{fixed oscillation frequency, stable amplitude} 59953
59953 f \approx \frac{1}{2 \pi \sqrt{C_3 + C_{ee}}} 59953

7. Capacitance Range:
- 59953 C_3 \text{ cannot be too large, } C_{ee} \text{ adjustment range is small.} 59953
- 59953 C_3 \text{ cannot be too small, insertion coefficient is small, oscillation amplitude is small.} 59953

8. Transformer Impedance Conversion:
59953 R_F \uparrow 59953

4. Oscillator frequency stability. Causes: parasitic inductance, DC voltage, Q value, temperature... Solution: standardization, temperature compensation, operating point, voltage, input resistance...

5. Crystal Oscillator. $$ f \propto \frac{1}{\text{frequency}} $$ $$ f_0 = \frac{1}{2\pi\sqrt{L_0C_0}} $$ $$ f_p = \frac{1}{2\pi\sqrt{L_0C_0}} $$ $$ f_n = \frac{1}{2\pi\sqrt{L_0C_0}} $$ $$ f_1 \approx (1 + \frac{1}{2} \frac{1}{C_0} Q) f_0 $$ $$ f_2 \approx (1 - \frac{1}{2} \frac{1}{C_0} Q) f_0 $$ $$ f_2 - f_1 \approx f_p - f_n \approx \frac{1}{2} C_0 \cdot f_0 \approx 0 $$

1. Series crystal oscillator: crystal works at high Q short circuit (f = f_0) 2. Parallel crystal oscillator: crystal works at low Q (f_0 < f < f_p).

3. Quartz crystal oscillator (fundamental and odd harmonics). $$ f_0 \in (2k-1) f, (2k+1) f $$ $$ f < f_0: \text{resonant frequency, no phase shift condition.} $$ $$ f > (2k+1) f: \text{feedback small, amplitude flat condition.} $$ $$ f = (2k+1) f: \text{resonant frequency.} $$

6. Other Oscillator Forms. 1. Wien Bridge Oscillator. 2. RC Phase Shift Oscillator 3. Integrator Oscillator 4. Voltage-Controlled Oscillator.

**Subject:** Mathematics Homework Paper

**Page:** 12

**Section:** Modulation and Demodulation

**Modulation Causes:** - Antenna Size Constraints - Channel Reuse - Reducing Interference Signal Impact

**1. Amplitude Modulation:**

**1.1 Standard Amplitude Modulation:**

$$V_{am}(t) = [V_{cm} + V_{f}(t)]\cos(\omega t)$$ $$= (V_{cm} + V_{cm}\cos(\omega t))\cos(\omega t)$$ $$= V_{cm}(1 + m_a\cos(\omega t))\cos(\omega t)$$ $$m_a < 1, \text{ otherwise overmodulation}$$

**Carrier Power:** $$P_c = \frac{1}{2}\frac{V_{cm}^2}{R_L}$$ **Upper Sideband Power:** $$P_u = \frac{1}{2}\left(\frac{1}{2}m_aV_{cm}\right)^2/R_L = \frac{1}{4}m_a^2P_c$$ **Lower Sideband Power:** $$P_l = \frac{1}{2}\left(\frac{1}{2}m_aV_{cm}\right)^2/R_L = \frac{1}{4}m_a^2P_c$$ **Total Power:** $$P = P_c(1 + \frac{m_a^2}{2}) < \frac{3}{2}P_c$$

**Advantages:** - Large carrier power, convenient for demodulation

**Disadvantages:** - More than half of the power is used by the carrier

**Time Domain:** - Includes amplitude modulation signal waveform - Frequency domain: Linear frequency shift, half redundant

**1.2 Modulation:** - Modulator (mixer)

$$V_{am}(t) = \frac{V_{cm}}{m_a}\cos(\omega t)$$

**2. Demodulation:**

**2.1 Demodulation:**

**a. Coherent Demodulation:**

- Frequency (local oscillator or phase-locked loop) + low-pass

**b. Non-Coherent Demodulation:**

- Small signal square-law detection: Power indicator - Large signal peak detection: $m_aC \ll R_LC < T_L$

**2.2 Demodulation:**

- Local oscillator + low-pass (to prevent low and high frequencies) - Demodulation: Local oscillator + low-pass + frequency division

**3. Single Sideband:**

$$V_{SSB_R} = \frac{1}{2}V_{f}(t)\cos(\omega t) - \frac{1}{2}V_{f}^{'(t)}\sin(\omega t)$$ $$V_{SSB_L} = \frac{1}{2}V_{f}(t)\cos(\omega t) + \frac{1}{2}V_{f}^{'(t)}\sin(\omega t)$$ - Where, $V_{f}^{'(t)} = IFT\left(jU_{f}(j\omega)sgn(\omega)\right)$

**a. Demodulation:**

- **a. Superheterodyne Method:** Reduces Q value requirements

**b. Hartley Method:**

**c. Weaver Method:**

**2. Demodulation:**

- **a. Superheterodyne Method:** Reduces Q value requirements

2) Demodulation: Add a local oscillator, lock loop to extract the local oscillator. 2) Frequency抑制接收方案 Hartley, Weaver.

2. Frequency Modulation. 1) Basic Concepts. Frequency Modulation $$ w(t) = a_2 + k_F V_{cm} \cos(2\pi f_c t) $$ $$ \Delta w = k_F V_{cm} = m_F \Omega $$ $$ \phi(t) = w(t) + m_F \sin(2\pi f_c t) $$ $$ m_F = \frac{k_F V_{cm}}{\Omega} $$ 2) Frequency Modulation Signal Properties. $$ V_{fm} = \cos(\omega_c t + m_F \sin(2\pi f_c t)) $$ $$ = Re\{ e^{j\omega_c t} e^{j m_F \sin(2\pi f_c t)} \} $$ $$ = \sum_{n=-\infty}^{\infty} J_n(m_F) \cos(\omega_c t + n \Omega) t $$ $$ \sum_{n=-\infty}^{\infty} J_n(m_F) = 1, \quad B W_F \approx 2(m_F + 1)F $$ 2) Frequency Modulation Signal Properties. $$ SNR_{fm} = \frac{\Delta W_c}{\Delta W_n}, \quad \Delta W_n = \frac{V_{cm}}{V_{cm}} (w_n - w_c) $$ Solution: Frequency Modulation: High-pass (filter) to produce frequency modulation. Center frequency stability (f_c).

1) Frequency Modulation 1) Technical Specifications. 1) Linearity: f ~ v; Maximum frequency deviation Δf_m (requires constant bandwidth). 2) Frequency Modulation Scheme 1) Advantage: Low noise, high signal-to-noise ratio. 2) Disadvantage: Complex filtering, phase shifting, and wave recovery equipment. 3) Sideband Suppression The transmitted sideband and the suppressed sideband are anti-symmetric. 1) Modulation: Similar to SSB, filter is easy to make. 2) Demodulation: Add a local oscillator + lock loop; phase inversion. 3) Amplitude Modulation Orthogonal carrier and local oscillator. Total transmission: Superheterodyne receiver. 1) Intermediate frequency realization of channel selection: Low Q 2) Demodulator before and after amplification to the intermediate frequency. 3) High gain (AGC) gain is not high (non-linearity) not low (noise factor). Problem: Interference signal causing frequency modulation: $$ \begin{vmatrix} FMN & PMN & FMN \\ FMN & FMN & FMN \\ PMN & FMN & FMN \end{vmatrix} $$ $$ SNR_{fm} = \frac{\Delta W_c}{\Delta W_n}, \quad \Delta W_n = \frac{V_{cm}}{V_{cm}} (w_n - w_c) $$ Solution: Frequency Modulation: High-pass (filter) to produce frequency modulation. Center frequency stability (f_c).

2) Upconversion: After modulation, the carrier frequency increases. b. Variable capacitor part connected.

$$\begin{cases} \text{Amplitude modulation: } w_c + k_F u_F(t) \rightarrow w_c + w_0 + k_F u_F(t) \\ \text{Frequency modulation: } w_c + k_F u_F(t) \rightarrow Nw_c + Nk_F u_F(t) \end{cases}$$

3) Direct frequency modulation circuit.

$$\text{Given } u_F(t) = V_{am} \cos \omega t, \text{ then}$$ $$C_j = \frac{C_0}{\left[1 + \left(\frac{V_B + V_{am} \cos \omega t}{\phi}\right)^2\right]^2}$$ $$= \frac{C_0}{\left[1 + \left(\frac{V_{am} \cos \omega t}{V_B + \phi}\right)^2\right]^2}$$ $$= \frac{C_0}{\left(1 + m_c \cos \omega t\right)^2}$$ $$C_0': \text{DC bias voltage } V_B \text{ under the capacitor}$$ $$m_c = \frac{V_{am}}{V_B + \phi} < 1: \text{Capacitive modulation}$$

a. Variable capacitor fully connected.

$$\begin{aligned} f &= \frac{1}{2\pi \sqrt{LC_j}} = \frac{1}{2\pi \sqrt{L \cdot C_0'/(1 + m_c \cos \omega t)^2}} \\ &= f_c (1 + \frac{1}{2} m_c \cos \omega t + \frac{1}{2!} \frac{1}{2} (\frac{1}{2} - 1) (m_c \cos \omega t)^2 + \cdots) \\ &= f_c \left(1 + \frac{1}{2} m_c \cos \omega t + \frac{1}{2!} \frac{1}{2} (\frac{1}{2} - 1) (m_c \cos \omega t)^2 + \cdots\right) \end{aligned}$$

$$\text{Advantages:}$$ $$\text{Frequency stability improved } C_0' C_{20} / C_1^2 \text{ times.}$$ $$\text{But due to the variable capacitor part connected, the high-frequency voltage added to the variable capacitor is only a part of the output voltage, further improving the carrier frequency stability.}$$ $$\text{Disadvantages:}$$ $$\text{Frequency deviation } C_0' C_{20} / C_1^2 \text{ times.}$$ $$\text{Modulation efficiency drops } C_0' C_{20} / C_1^2 \text{ times.}$$ $$\text{The variable capacitor part is partially short-circuited. The resistance drops, Q drops.}$$

C. Double Capacitor Back-to-Back Connection

$$C_{20} = \frac{C_2C_3}{C_2+C_3} + \frac{0.5C_0'C_6}{0.5C_0'+C_6}$$

$$f_c = \frac{1}{2\pi\sqrt{LC_{20}}}$$

Advantages: High-frequency oscillation signals, each capacitor bears half the voltage, direct current and frequency modulation signals. Capacitors connected in parallel, power and load status are the same.

Summary: Direct Frequency Modulation Transmitter

1. Varactor Diode Direct Frequency Modulation Circuit Simple, High Frequency 2. Large Frequency Deviation

1. Insufficient Frequency Stability. Varactor diode in the circuit will conduct under oscillation voltage, causing the Q value of the circuit to decrease. 2. Using a crystal oscillator: small frequency deviation. 3. Using AFC: large frequency deviation. 4. Using PLL: large frequency deviation, center frequency stable.

4. Indirect Frequency Modulation Circuit. First integrate then phase shift. a. Variable Phase Shift Method. (m_p < π/6)

$$V_0(t) = V_{cm} \cos(\omega t + K_p \int u(t) dt)$$

$$V_0(t) = V_{cm} \cos(\omega t + K_p \int u(t) dt)$$

$$V_0(t) = V_{cm} \cos(\omega t + K_p \int u(t) dt)$$

$$V_0(t) = V_{cm} \cos(\omega t + K_p \int u(t) dt)$$

$$V_0(t) = V_{cm} \cos(\omega t + K_p \int u(t) dt)$$

$$V_0(t) = V_{cm} \cos(\omega t + K_p \int u(t) dt)$$

$$V_0(t) = V_{cm} \cos(\omega t + K_p \int u(t) dt)$$

$$V_0(t) = V_{cm} \cos(\omega t + K_p \int u(t) dt)$$

$$V_0(t) = V_{cm} \cos(\omega t + K_p \int u(t) dt)$$

$$V_0(t) = V_{cm} \cos(\omega t + K_p \int u(t) dt)$$

$$V_0(t) = V_{cm} \cos(\omega t + K_p \int u(t) dt)$$

$$V_0(t) = V_{cm} \cos(\omega t + K_p \int u(t) dt)$$

$$V_0(t) = V_{cm} \cos(\omega t + K_p \int u(t) dt)$$

$$V_0(t) = V_{cm} \cos(\omega t + K_p \int u(t) dt)$$

$$V_0(t) = V_{cm} \cos(\omega t + K_p \int u(t) dt)$$

$$V_0(t) = V_{cm} \cos(\omega t + K_p \int u(t) dt)$$

$$V_0(t) = V_{cm} \cos(\omega t + K_p \int u(t) dt)$$

$$V_0(t) = V_{cm} \cos(\omega t + K_p \int u(t) dt)$$

$$V_0(t) = V_{cm} \cos(\omega t + K_p \int u(t) dt)$$

$$V_0(t) = V_{cm} \cos(\omega t + K_p \int u(t) dt)$$

$$V_0(t) = V_{cm} \cos(\omega t + K_p \int u(t) dt)$$

$$V_0(t) = V_{cm} \cos(\omega t + K_p \int u(t) dt)$$

$$V_0(t) = V_{cm} \cos(\omega t + K_p \int u(t) dt)$$

$$V_0(t) = V_{cm} \cos(\omega t + K_p \int u(t) dt)$$

$$V_0(t) = V_{cm} \cos(\omega t + K_p \int u(t) dt)$$

$$V_0(t) = V_{cm} \cos(\omega t + K_p \int u(t) dt)$$

$$V_0(t) = V_{cm} \cos(\omega t + K_p \int u(t) dt)$$

$$V_0(t) = V_{cm} \cos(\omega t + K_p \int u(t) dt)$$

$$V_0(t) = V_{cm} \cos(\omega t + K_p \int u(t) dt)$$

$$V_0(t) = V_{cm} \cos(\omega t + K_p \int u(t) dt)$$

$$V_0(t) = V_{cm} \cos(\omega t + K_p \int u(t) dt)$$

$$V_0(t) = V_{cm} \cos(\omega t + K_p \int u(t) dt)$$

$$V_0(t) = V_{cm} \cos(\omega t + K_p \int u(t) dt)$$

$$V_0(t) = V_{cm} \cos(\omega t + K_p \int u(t) dt)$$

$$V_0(t) = V_{cm} \cos(\omega t + K_p \int u(t) dt)$$

$$V_0(t) = V_{cm} \cos(\omega t + K_p \int u(t) dt)$$

$$V_0(t) = V_{cm} \cos(\omega t + K_p \int u(t) dt)$$

$$V_0(t) = V_{cm} \cos(\omega t + K_p \int u(t) dt)$$

$$V_0(t) = V_{cm} \cos(\omega t + K_p \int u(t) dt)$$

$$V_0(t) = V_{cm} \cos(\omega t + K_p \int u(t) dt)$$

$$V_0(t) = V_{cm} \cos(\omega t + K_p \int u(t) dt)$$

$$V_0(t) = V_{cm} \cos(\omega t + K_p \int u(t) dt)$$

$$V_0(t) = V_{cm} \cos(\omega t + K_p \int u(t) dt)$$

$$V_0(t) = V_{cm} \cos(\omega t + K_p \int u(t) dt)$$

$$V_0(t) = V_{cm} \cos(\

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# Section 1: FM Wave Demodulation Methods and Circuits

1. **Limit High Frequency** - **Elimination of FM Wave Sidebands:** The limiter removes the sidebands of the FM wave, improving the signal-to-noise ratio. The limiter and the amplifier together increase the signal-to-noise ratio. 2. **Demodulator Performance Metrics (Demodulator Located After Limiter)** - **Linearity:** V-f; Linear Range > 2dfm. - **Demodulation Sensitivity:** S = 1/2f; Intermediate Frequency = Intermediate Frequency. 3. **Demodulation Methods and Principles** a. **Slope Demodulation** - $$ V_{FM}(t) = V_{cm} \cos(\omega_c t + K_F \int_0^t u(t) dt + \theta_0) $$ - $$ \frac{d}{dt} V_{FM}(t) = V_{cm} (\omega_c + K_F u(t)) \cos(\omega_c t + K_F \int_0^t u(t) dt + \theta_0 + \frac{\pi}{2}) $$ - (Double Integrator) Differentiation + Envelope Detection (K_F V_m < ω_c). b. **Orthogonal Demodulation** (Demodulation after delay) - $$ V_{FM}(t-T_0) = \cos(\omega_c t - T_0) + m_F \sin(\omega_c t - T_0) $$ - $$ \approx \cos(\omega_c t + m_F \sin(\omega_c t - m_F \Omega T_0 \cos(\omega_c t - \omega_c T_0)) $$ - Where $$ \Omega T_0 < \frac{\pi}{12} $$, $$ \omega_c T_0 = \pm \frac{\pi}{2} $$. - $$ \Delta \theta = -m_F \Omega T_0 \cos(\omega_c t - \omega_c T_0) $$ - $$ = -\Delta \omega T_0 \cos(\omega_c t - \omega_c T_0) $$ - $$ = -T_0 K_F V_m \cos(\omega_c t - \omega_c T_0) $$ - **Delay FM Wave Demodulation, Extract FM Signal Using Demodulator.**

# Section 2: Delay Circuit Implementation

- $$ Q(\omega) = \frac{\pi}{2} - \alpha \omega \tan(\alpha (\omega - \omega_0)) $$ - $$ \approx \frac{\pi}{2} - \frac{2 \alpha}{\omega_0} (\omega - \omega_0) \quad (\alpha < \frac{\pi}{6} \cdot \frac{\omega_0}{2 \alpha \omega_0}) $$ - $$ T_0 = \frac{2 \alpha}{\omega_0} $$

# Section 3: Phase Detector Implementation

1. **Phase Detector Implementation 1: Phase Detector** - [DIAGRAM 2] - $$ V_{FM}(t) \times V_{FM}(t - T_0) $$ - $$ V_{P}(t) = V_{FM}(t) V_{FM}(t - T_0) = $$ - $$ V_{FM}^2 \cos(\omega_c t + m_F \sin(\omega_c t - m_F \Omega T_0 \cos(\omega_c t - \omega_c T_0)) $$ - Assuming $$ T_0 \leq 0.2 \Omega $$, after passing through a low-pass filter: - $$ V_{P}(t) = \frac{1}{2} V_{FM}^2 \cos(\omega_c t + m_F \Omega T_0 \cos(\omega_c t)) $$ - $$ = -\frac{1}{2} V_{FM}^2 \sin(\omega_c T_0 \cos(\omega_c t)) \quad (\omega_c T_0 \neq 0) $$ - $$ \approx -\frac{1}{2} V_{FM}^2 m_F \Omega T_0 \cos(\omega_c t) \quad (m_F \Omega T_0 < 0.5) $$ - $$ = -\frac{1}{2} V_{FM}^2 K_F T_0 \cos(\omega_c t) $$

2. **Phase Detector Implementation 2: Phase Detector** - [DIAGRAM 3] - $$ V_{FM}(t) + V_{FM}(t - T_0) $$ - $$ V_{FM}(t - T_0) $$

# Section 4: Zero Crossing Detector

- **Zero Crossing Detector** - Utilizes the zero crossing of the FM wave to achieve frequency demodulation. - Linear Range (f ~ V) is large, easy to integrate. - **Feedback Phase Detector:** First phase detection, then differentiation. - See PLL, PD output is a phase difference voltage.

3. Digital Modulation ASK. - Frequency Demodulator - Sampler - Decision器 FSK (fm = fc + af: 信号; fs = fc - af: 信号) MF = 2fd / fs DBPSK. - 1bit延时 QPSK. - 跳并转换 - sinout - cosout

7. Loop Filter 1. Overview. Feedback control system: AGC, AFC, APC → PLL. Among them, PLL: narrowband (noise, frequency combination), loop filter, signal detection.

2. PLL Basic Principles. 1. PD: Phase detector, low-pass filter 2. LF: Loop filter, low-pass filter in the loop, part of the loop filter memory 3. VCO: Voltage-controlled oscillator

3. PLL Linear Analysis. When θe is small, f[θe(t)] ≈ k_d θe(t) + s[θe(t)] + k_w k_d H[θe(t)] = s[θe(t)] θe(t) + k_w k_d H[θe(t)] = s[θe(t)] θe(t) + k_w k_d H[θe(t)] = s[θe(t)] θe(t) + k_w k_d H[θe(t)] = s[θe(t)] θe(t) + k_w k_d H[θe(t)] = s[θe(t)] θe(t) + k_w k_d H[θe(t)] = s[θe(t)] θe(t) + k_w k_d H[θe(t)] = s[θe(t)] θe(t) + k_w k_d H[θe(t)] = s[θe(t)] θe(t) + k_w k_d H[θe(t)] = s[θe(t)] θe(t) + k_w k_d H[θe(t)] = s[θe(t)] θe(t) + k_w k_d H[θe(t)] = s[θe(t)] θe(t) + k_w k_d H[θe(t)] = s[θe(t)] θe(t) + k_w k_d H[θe(t)] = s[θe(t)] θe(t) + k_w k_d H[θe(t)] = s[θe(t)] θe(t) + k_w k_d H[θe(t)] = s[θe(t)] θe(t) + k_w k_d H[θe(t)] = s[θe(t)] θe(t) + k_w k_d H[θe(t)] = s[θe(t)] θe(t) + k_w k_d H[θe(t)] = s[θe(t)] θe(t) + k_w k_d H[θe(t)] = s[θe(t)] θe(t) + k_w k_d H[θe(t)] = s[θe(t)] θe(t) + k_w k_d H[θe(t)] = s[θe(t)] θe(t) + k_w k_d H[θe(t)] = s[θe(t)] θe(t) + k_w k_d H[θe(t)] = s[θe(t)] θe(t) + k_w k_d H[θe(t)] = s[θe(t)] θe(t) + k_w k_d H[θe(t)] = s[θe(t)] θe(t) + k_w k_d H[θe(t)] = s[θe(t)] θe(t) + k_w k_d H[θe(t)] = s[θe(t)] θe(t) + k_w k_d H[θe(t)] = s[θe(t)] θe(t) + k_w k_d H[θe(t)] = s[θe(t)] θe(t) + k_w k_d H[θe(t)] = s[θe(t)] θe(t) + k_w k_d H[θe(t)] = s[θe(t)] θe(t) + k_w k_d H[θe(t)] = s[θe(t)] θe(t) + k_w k_d H[θe(t)] = s[θe(t)] θe(t) + k_w k_d H[θe(t)] = s[θe(t)] θe(t) + k_w k_d H[θe(t)] = s[θe(t)] θe(t) + k_w k_d H[θe(t)] = s[θe(t)] θe(t) + k_w k_d H[θe(t)] = s[θe(t)] θe(t) + k_w k_d H[θe(t)] = s[θe(t)] θe(t) + k_w k_d H[θe(t)] = s[θe(t)] θe(t) + k_w k_d H[θe(t)] = s[θe(t)] θe(t) + k_w k_d H[θe(t)] = s[θe(t)] θe(t) + k_w k_d H[θe(t)] = s[θe(t)] θe(t) + k_w k_d H[θe(t)] = s[θe(t)] θe(t) + k_w k_d H[θe(t)] = s[θe(t)] θe(t) + k_w k_d H[θe(t)] = s[θe(t)] θe(t) + k_w k_d H[θe(t)] = s[θe(t)] θe(t) + k_w k_d H[θe(t)] = s[θe(t)] θe(t) + k_w k_d H[θe(t)] = s[θe(t)] θe(t) + k_w k_d H[θe(t)] = s[θe(t)] θe(t) + k_w k_d H[θe(t)] = s[θe(t)] θe(t) + k_w k_d H[θe(t)] = s[θe(t)] θe(t) + k_w k_d H[θe(t)] = s[θe(t)] θe(t) + k_w k_d H[θe(t)] = s[θe(t)] θe(t) + k_w k

**Mathematical Formulas:**

1) RC Integrator LF: $$H_{IF}(s) = \frac{1}{1+s\tau}$$ $$\tau = RC$$ $$H(s) = \frac{\omega_n^2}{s^2 + 2\zeta\omega_n s + \omega_n^2}$$ $$\omega_n^2 = \frac{K_p}{2}$$ $$H_{IF}(s) = \frac{s^2}{s^2 + 2\zeta\omega_n s + \omega_n^2}$$ $$2\zeta\omega_n = \frac{1}{2}$$

2) RC Differentiator LF: $$H_{IF}(s) = \frac{1+s\tau}{1+s(\tau_1+\tau_2)}$$ $$\tau = RC, \tau_2 = RC$$ $$H(s) = \frac{(2\zeta-\omega_n/K_p)\omega_n s + \omega_n^2}{s^2 + 2\zeta\omega_n s + \omega_n^2}$$ $$\omega_n^2 = \frac{K_p}{\tau_1+\tau_2}$$ $$H_{IF}(s) = \frac{s^2 + s\omega_n^2/K_p}{s^2 + 2\zeta\omega_n s + \omega_n^2}$$ $$2\zeta\omega_n = \frac{1+K_p\tau_2}{\tau_1+\tau_2}$$

3) Ideal Integrator LF: $$H_{IF}(s) = \frac{1+s\tau}{s\tau}$$ $$\tau = RC, \tau_2 = RC$$ $$H(s) = \frac{\omega_n^2}{s^2 + 2\zeta\omega_n s + \omega_n^2}$$ $$\omega_n^2 = \frac{K_p}{\tau_1}$$ $$H_{IF}(s) = \frac{s^2}{s^2 + 2\zeta\omega_n s + \omega_n^2}$$ $$2\zeta\omega_n = \frac{K_p\tau_2}{\tau_1}$$

4) Ideal Differentiator LF: $$H_{IF}(s) = 1$$ $$H(s) = \frac{K_p}{s + K_p}$$ $$H_{IF}(s) = \frac{s}{s + K_p}$$

5) PLL Frequency Characteristics: $$\varphi_{IF}(s) = \frac{\Delta\Omega}{s}$$ $$\varphi_{IF}(s) = \frac{\Delta\Omega}{s} \cdot H(s)$$ $$\varphi_{IF}(s) = \varphi_{IF} = 0$$

6) PLL Phase Characteristics: $$\varphi_{IF}(s) = \frac{\Delta\Omega}{s}$$ $$\varphi_{IF}(s) = \frac{\Delta\Omega}{s} \cdot H(s)$$ $$\varphi_{IF}(s) = \varphi_{IF} = 0$$

7) PLL Frequency Characteristics: $$\varphi_{IF}(s) = \frac{\Delta\Omega}{s}$$ $$\varphi_{IF}(s) = \frac{\Delta\Omega}{s} \cdot H(s)$$ $$\varphi_{IF}(s) = \varphi_{IF} = 0$$

8) PLL Phase Characteristics: $$\varphi_{IF}(s) = \frac{\Delta\Omega}{s}$$ $$\varphi_{IF}(s) = \frac{\Delta\Omega}{s} \cdot H(s)$$ $$\varphi_{IF}(s) = \varphi_{IF} = 0$$

**Chinese Text:**

(Subject: ) Math Homework Paper

(Subject: ) Stability

(Subject: ) Closed-loop Transfer Function

(Subject: ) Ideal Integrator LF

(Subject: ) Ideal Differentiator LF

(Subject: ) PLL Frequency Characteristics

(Subject: ) PLL Phase Characteristics

5. PLL's stability. - First-order loop stable. 90° - Second-order loop due to parasitic capacitance may not be stable. 180°

6. PLL's noise characteristics. - Input noise < closed-loop transfer function. H(s) - VCO phase noise < open-loop transfer function. H(s)

4. PLL's nonlinearity analysis. - Tracking: lock → relock = linear (ωio = ωoo) → capture - Capture: unlock → lock = nonlinear. (ωio ≠ ωoo) → capture - Fast capture range ≤ capture range ≤ lock range. - Capture range: maximum initial frequency difference that can enter the lock range. - Fast capture range: within 2π can enter the lock range.

7. First-order loop nonlinearity analysis (eg. phase detection). $$\varphi_e(t) + K_p \sin(\varphi_e(t)) = \Delta \omega_i = \omega_{io} - \omega_{oo}$$ $$\varphi_e(t) - \varphi_e(t) \text{ graph: }$$

$$\varphi_{eq} = \arcsin \frac{\Delta \omega_i}{K_p} + 2n\pi. (\varphi_e = 0)$$

5. Integrated phase-locked loop. 6. Phase-locked loop applications. 1. Narrowband tracking filter - narrowband tracking loop (carrier tracking). 2. Solve the problem of the center frequency change in general receivers and the widening of the intermediate frequency bandwidth, which affects the signal-to-noise ratio.

**2. Carrier Recovery - Wideband Loop (Carrier Tracking)**

1. **Direct Frequency Synthesis**: Mixer + Divider / Divider + Filter

1) **Phase Loop**: Phase Locked Loop (PLL) - Advantage: Fast - Disadvantage: High frequency noise, phase shift large, complex structure

2) **Costas Loop**: Various digital modulation signals - Advantage: Simple structure, high purity, high frequency - Disadvantage: Slow tracking

3. **Phase Locked Loop (PLL)**

- Advantage: Simple structure, high purity, high frequency - Disadvantage: Slow tracking

4. **Direct Digital Frequency Synthesis**

**2. Main Technical Indicators**

1. Frequency Range, Frequency Resolution ~ Reference Frequency (Reference), Conversion Time ~ Reference Frequency (PLL), Frequency Stability, Spectral Purity

$$ H_n(\cos\alpha + \alpha) = \frac{1}{1 + \alpha^2} \left( \frac{1}{2} \cos\alpha \right)^2 $$

2. AFC

- Advantage: High output frequency, low phase noise - Disadvantage: High phase noise

3. **Phase Locked Loop (PLL)**

1. **Phase Locked Loop (PLL)**

- Advantage: High output frequency, low phase noise - Disadvantage: High phase noise

2. **Phase Locked Loop (PLL)**

- Advantage: High output frequency, low phase noise - Disadvantage: High phase noise

3. **Phase Locked Loop (PLL)**

- Advantage: High output frequency, low phase noise - Disadvantage: High phase noise

4. **Phase Locked Loop (PLL)**

- Advantage: High output frequency, low phase noise - Disadvantage: High phase noise

**8. Frequency Synthesis Technology**

1. **Phase Locked Loop (PLL)**

- Advantage: High output frequency, low phase noise - Disadvantage: High phase noise

2. **Phase Locked Loop (PLL)**

- Advantage: High output frequency, low phase noise - Disadvantage: High phase noise

3. **Phase Locked Loop (PLL)**

- Advantage: High output frequency, low phase noise - Disadvantage: High phase noise

4. **Phase Locked Loop (PLL)**

- Advantage: High output frequency, low phase noise - Disadvantage: High phase noise

**8. Frequency Synthesis Technology**

1. **Phase Locked Loop (PLL)**

- Advantage: High output frequency, low phase noise - Disadvantage: High phase noise

2. **Phase Locked Loop (PLL)**

- Advantage: High output frequency, low phase noise - Disadvantage: High phase noise

3. **Phase Locked Loop (PLL)**

- Advantage: High output frequency, low phase noise - Disadvantage: High phase noise

4. **Phase Locked Loop (PLL)**

- Advantage: High output frequency, low phase noise - Disadvantage: High phase noise

**8. Frequency Synthesis Technology**

1. **Phase Locked Loop (PLL)**

- Advantage: High output frequency, low phase noise - Disadvantage: High phase noise

2. **Phase Locked Loop (PLL)**

- Advantage: High output frequency, low phase noise - Disadvantage: High phase noise

3. **Phase Locked Loop (PLL)**

- Advantage: High output frequency, low phase noise - Disadvantage: High phase noise

4. **Phase Locked Loop (PLL)**

- Advantage: High output frequency, low phase noise - Disadvantage: High phase noise

**8. Frequency Synthesis Technology**

1. **Phase Locked Loop (PLL)**

- Advantage: High output frequency, low phase noise - Disadvantage: High phase noise

2. **Phase Locked Loop (PLL)**

- Advantage: High output frequency, low phase noise - Disadvantage: High phase noise

3. **Phase Locked Loop (PLL)**

- Advantage: High output frequency, low phase noise - Disadvantage: High phase noise

4. **Phase Locked Loop (PLL)**

- Advantage: High output frequency, low phase noise - Disadvantage: High phase noise

Title Page: Angli Liu (Angli Liu)

**Review of Electromagnetism**

**I. Introduction**

Radiation: Source → Space Absorption: Space → Source

**II. Radiation Field**

1. Vector Quantities of Electric Current Source and Magnetic Current Source

Let $$\vec{H}_e = \frac{1}{\mu} \nabla \times \vec{A}_e$$

Substitute $$\nabla \times \vec{E}_e = -j\omega \mu \vec{H}_e$$

Result in $$\nabla \times (\vec{E}_e + j\omega \vec{A}_e) = 0$$

Let $$\vec{E}_e + j\omega \vec{A}_e = -\nabla \phi_e$$

Substitute $$\nabla \times \vec{H}_e = j\omega \varepsilon \vec{E}_e + \vec{j}_e$$

Result in $$\frac{1}{\mu} \nabla \times \nabla \times \vec{A}_e = j\omega \varepsilon (-\nabla \phi_e - j\omega \vec{A}_e) + \vec{j}_e$$

Thus $$\nabla^2 \vec{A}_e + k^2 \vec{A}_e = -\mu \vec{j}_e + \nabla (\nabla \cdot \vec{A}_e + j\omega \varepsilon \mu \phi_e)$$

Lorentz gauge $$\nabla \cdot \vec{A}_e + j\omega \varepsilon \mu \phi_e = 0$$

Result in $$\nabla^2 \vec{A}_e + k^2 \vec{A}_e = -\mu \vec{j}_e$$ $$\phi_e = -\frac{1}{j\omega \varepsilon \mu} \nabla \cdot \vec{A}_e$$

Substitute $$\vec{E}_e + j\omega \vec{A}_e = -\nabla \phi_e$$

Result in $$\vec{E}_e = -j\omega \vec{A}_e - \frac{1}{c \omega \mu} \nabla (\nabla \cdot \vec{A}_e)$$

Solve (2) to get $$\vec{A}_e(\vec{r}) = \frac{\mu}{4\pi} \int \frac{\vec{e}_k (\vec{r} - \vec{r}')}{|\vec{r} - \vec{r}'|} \vec{j}(\vec{r}') d\vec{r}'$$

Substitute (5) into (1) and (4) to obtain the electric field $$\vec{E}_e$$ and magnetic field $$\vec{H}_e$$ generated by the electric current source $$\vec{j}_e$$.

Let $$\vec{E}_m = -\frac{1}{\varepsilon} \nabla \times \vec{A}_m$$

Substitute $$\nabla \times \vec{H}_m = j\omega \varepsilon \vec{E}_m$$

Result in $$\nabla \times (j\vec{H}_m + j\omega \vec{A}_m) = 0$$

Let $$\vec{H}_m + j\omega \vec{A}_m = -\nabla \phi_m$$

Substitute $$\nabla \times \vec{E}_m = -j\omega \mu \vec{H}_m - \vec{j}_m$$

Here's the extracted text and formulas from the handwritten note, with placeholders for the diagram:

---

\begin{equation} -\frac{1}{2} \nabla \times \nabla \times \vec{A}_m = -j \omega \mu (\nabla \phi_m - j \omega \vec{A}_m) - \vec{J}_m \end{equation}

\begin{equation} \nabla^2 \vec{A}_m + k^2 \vec{A}_m = -\epsilon \vec{J}_m + \nabla (\nabla \cdot \vec{A}_m + j \omega \epsilon \mu \phi_m) \end{equation}

\begin{equation} \nabla \cdot \vec{A}_m + j \omega \epsilon \mu \phi_m = 0 \end{equation}

\begin{equation} \nabla^2 \vec{A}_m + k^2 \vec{A}_m = -\epsilon \vec{J}_m = \nabla \cdot \vec{E}_m \end{equation}

\begin{equation} \phi_m = -\frac{1}{j \omega \epsilon \mu} \nabla \cdot \vec{A}_m \end{equation}

\begin{equation} \vec{H}_m + j \omega \vec{A}_m = -\nabla \phi_m \end{equation}

\begin{equation} \vec{H}_m = -j \omega \vec{A}_m - \frac{1}{\epsilon \mu} \nabla (\nabla \cdot \vec{A}_m) \end{equation}

\begin{equation} \vec{A}_m(\vec{r}) = \frac{1}{4 \pi} \int \frac{\vec{J}_m(\vec{r}')}{|\vec{r} - \vec{r}'|} dV' \end{equation}

\begin{equation} \text{Substitute (12) into (11) and (10) to obtain the electric field } \vec{E}_m \text{ and magnetic field } \vec{H}_m \text{ due to the magnetic current source } \vec{J}_m. \end{equation}

---

\section{Electric Dipole Radiation Field}

\subsection{Electric Dipole: A long, thin wire with a uniform current density I.}

\begin{equation} \vec{A}(\vec{r}) = \frac{\mu I e^{-jkr}}{4 \pi r} \hat{z} = A_z \hat{z} \end{equation}

\begin{equation} A_z = A_2 \cos \theta = \frac{\mu I e^{-jkr}}{4 \pi r} \cos \theta \end{equation}

\begin{equation} A_\theta = -A_2 \sin \theta = -\frac{\mu I e^{-jkr}}{4 \pi r} \sin \theta \end{equation}

\begin{equation} A_\phi = 0 \end{equation}

\begin{equation} \vec{E}_\phi = \vec{H}_\theta = \vec{H}_\phi = 0 \end{equation}

\begin{equation} \text{Near-field } kr < 1: \end{equation}

\begin{equation} \begin{cases} E_r \approx -j \eta \frac{2 I e^{-jkr}}{4 \pi k r^3} \cos \theta \\ E_\theta \approx -j \eta \frac{2 I e^{-jkr}}{4 \pi r^2} \frac{2}{k r} \cos \theta \\ E_\phi \approx \frac{2 I e^{-jkr}}{4 \pi r^2} \left( \frac{j \omega \mu}{r} + \frac{1}{r^2} + \frac{j \omega \epsilon r^3}{\eta} \right) \sin \theta \end{cases} \end{equation}

\begin{equation} \text{Far-field } kr >> 1: \end{equation}

\begin{equation} \begin{cases} E_r \approx 0 \\ E_\theta \approx j \eta \frac{k I e^{-jkr}}{4 \pi r^2} \sin \theta \\ E_\phi \approx \frac{E_\theta}{\eta} \end{cases} \end{equation}

The note appears to be a detailed mathematical derivation related to electromagnetic fields and antenna theory. Here's the extracted text and formulas:

---

By

$$\vec{S} = \frac{1}{2} (\vec{E} \times \vec{H}^*) = \hat{r} S_r + \hat{\theta} S_\theta + \hat{\phi} S_\phi$$

We get

$$\begin{cases} S_r = \frac{1}{2} E_0 H_\phi^* = \frac{\eta}{8} \left(\frac{2\pi}{\lambda}\right)^2 \frac{\sin^2\theta}{r^2} \left[1 - j\left(\frac{1}{kr}\right)^3\right] \\ S_\theta = -\frac{1}{2} E_r H_\phi^* = j\eta \left[\frac{kI r^2 \cos\theta \sin\theta}{16\pi^2 r^3}\right] \left[1 + \left(\frac{1}{kr}\right)^2\right] \end{cases}$$

The total radiated power

$$P = \oint \vec{S} \cdot d\vec{S} = \int_0^{2\pi} d\phi \int_0^\pi \vec{S} \cdot \hat{r} r^2 \sin\theta d\theta$$

$$= \eta \frac{\pi}{3} \left(\frac{2\pi}{\lambda}\right)^2 \left[1 - j\left(\frac{1}{kr}\right)^3\right]$$

The real part is the electric dipole radiated power

$$P_{rad} = \eta \frac{\pi}{3} \left(\frac{2\pi}{\lambda}\right)^2$$

$$= \frac{1}{2} \left[\eta \frac{2\pi}{3} \left(\frac{2\pi}{\lambda}\right)^2\right] I^2$$

$$= \frac{1}{2} \left[80\pi^2 \left(\frac{2\pi}{\lambda}\right)^2\right] I^2$$

---

Three. Antenna Basic Parameters (1)

1. **Radiation Pattern**

1. **Far Field Region**

In the far field, E, H only have θ, φ components

$$\vec{E} = \hat{\theta} E_\theta + \hat{\phi} E_\phi$$

$$\vec{H} = \frac{1}{\eta} (\hat{\phi} E_\theta - \hat{\theta} E_\phi)$$

$$\vec{S} = \frac{1}{2} (\vec{E} \times \vec{H}^*) = \frac{1}{2} \left( |E_\theta|^2 + |E_\phi|^2 \right)$$

2. **Near Field Region**

$$E_\theta = \frac{f_0 (\cos\theta) e^{-jkr}}{r}$$

$$E_\phi = \frac{f_0 (\cos\phi) e^{-jkr}}{r}$$

$$\therefore S_r = \frac{1}{2\eta r^2} \left( |H_\theta (\cos\theta) e^{-jkr}|^2 + |H_\phi (\cos\phi) e^{-jkr}|^2 \right)$$

**Radiated Power Pattern**

$$F(\theta, \phi) = r^2 S_r = \frac{1}{2\eta} \left( |f_\theta (\cos\theta) e^{-jkr}|^2 + |f_\phi (\cos\phi) e^{-jkr}|^2 \right)$$

**Normalized Radiation Pattern**

$$F_n(\theta, \phi) = \frac{F(\theta, \phi)}{F_{max}(\theta, \phi)} = S_n / S_{max}$$

(3D Diagram, Polar Diagram)

**For Dipole Antenna**

$$F_n(\theta, \phi) = \sin^2\theta$$

$$\vec{E} \text{ field} \quad \phi = const$$

$$\vec{H} \text{ field} \quad \theta = 90^\circ$$

---

2. **Wire Antenna Radiation Pattern**

The note discusses various aspects of antenna theory, including:

1. **Solid Angles:** - **Total Solid Angle:** \( \Omega_T = \int_{4\pi} F_n(\theta, \phi) d\Omega \) - **Main Solid Angle:** \( \Omega_M = \int_{\Omega} F_n(\theta, \phi) d\Omega \) - **Normal Solid Angle:** \( \Omega_N = \int_{\Omega} F_n(\theta, \phi) d\Omega \)

2. **Directionality:** - **Radiation Power Density:** \( D(\theta, \phi) = \frac{S_r(\theta, \phi)}{P_r / 4\pi r^2} \) - **Antenna Efficiency:** \( \eta_e = \frac{P_e}{P_m} \) - **Maximum Directionality:** \( D_0 = \frac{4\pi}{\Omega_T} \)

3. **Gain:** - **Efficiency:** \( \eta_e = \frac{P_e}{P_m} \)

4. **Antenna Polarization:** - **Maximum Gain Direction:** \( \frac{S_r(\theta, \phi)}{F_n(\theta, \phi)} \)

5. **Antenna Input Impedance:** - **Input Impedance:** \( R_{in} \) depends on the radiation resistance and the resistance of the antenna.

The page contains a detailed mathematical analysis of antenna radiation patterns and impedance calculations. It discusses the far-field radiation pattern of a dipole antenna and the radiation resistance of a half-wave dipole antenna. The formulas include expressions for the electric field, magnetic field, and radiation resistance. The text also mentions the concept of a two-element array and its radiation pattern. The page is filled with complex mathematical derivations and equations, including integrals and trigonometric functions. The bottom part of the page discusses the radiation resistance of a dipole antenna and the input impedance of a two-element array. The formulas are written in a structured format, with each step clearly delineated. The page is a comprehensive study of antenna theory, focusing on the radiation pattern and impedance of various antenna configurations.

$$\approx j\frac{\eta kZ\varphi}{4\pi r}e^{jkr}\sin\theta\left[e^{j\left(kd\cos\theta+\beta\right)/2}+e^{-j\left(kd\cos\theta+\beta\right)/2}\right]$$ $$=j\frac{\eta kZ\varphi}{4\pi r}e^{jkr}\sin\theta\cdot2\cdot\cos\left(kd\cos\theta+\beta\right)/2$$ $$=j\frac{\eta kZ\varphi}{4\pi r}e^{jkr}f(\theta,\varphi)$$

2. N-element array $$f_A(\psi)=a_0+a_1e^{j\psi}+\ldots+a_Ne^{j(N-1)\psi},\quad\psi=kd\cos\theta+\beta$$ When equal amplitude: $$f_A(\psi)=\frac{\sin\left(N\psi/2\right)}{\sin\left(\psi/2\right)}$$ Normalization: $$f_A(\psi)=\frac{1}{N}\frac{\sin\left(N\psi/2\right)}{\sin\left(\psi/2\right)}=\text{Sa}\left(N\psi/2\right)$$ Main lobe maximum: $$\psi=0$$ Null points: $$\frac{N\psi}{2}=p\pi,\quad p=\pm1,\pm2,\ldots,\pm\frac{N}{2}$$ Side lobe maximum: $$\frac{N\psi}{2}=2p+1\pi,\quad p=1,\pm2,\ldots,-\frac{N}{2}$$

7. Horn antenna $$\begin{cases} \text{Azimuth: } r\approx r_0-r'\cos\beta=r_0-\vec{r'}\cdot\hat{F} \\ \text{Elevation: } r\approx r_0 \end{cases}$$ $$\begin{cases} \vec{A_e}=\frac{\mu}{4\pi}\int_S\vec{J_{es}}\frac{e^{jkr}}{r}ds' \approx\frac{\mu e^{jkr_0}}{4\pi r_0}\int_S\vec{J_{es}}\cdot e^{jkr_0}ds' \\ \vec{A_m}=\frac{\varepsilon}{4\pi}\int_S\vec{J_{ms}}\frac{e^{jkr}}{r}ds' \approx\frac{\varepsilon e^{-jkr_0}}{4\pi r_0}\int_S\vec{J_{ms}}\cdot e^{jkr_0}ds' \end{cases}$$ $$r'\cos\beta=\vec{r'}\cdot\hat{F}=x'\sin\theta\cos\varphi+y'\sin\theta\sin\varphi$$ By the principle of equivalent current density: $$\vec{J_{ms}}=-2\hat{x}\times\vec{E_a},\quad\vec{J_{es}}=0$$

Rectangular horn, uniform field distribution: $$\vec{J_{ms}}=-2\hat{x}\times\vec{E_a}=\hat{x}\cdot2\vec{E_0}$$ $$\vec{A_m}=\hat{x}\frac{\varepsilon e^{-jkr_0}}{4\pi r_0}\int_{-\frac{a}{2}}^{\frac{a}{2}}\int_{-\frac{b}{2}}^{\frac{b}{2}}2\vec{E_0}e^{-jkr_0}(\cos\varphi\cos\varphi+\sin\varphi\sin\varphi)dx'dy'$$ $$\vec{A_e}=\hat{x}\frac{\mu e^{jkr_0}}{4\pi r_0}\int_{-\frac{a}{2}}^{\frac{a}{2}}\int_{-\frac{b}{2}}^{\frac{b}{2}}2\vec{E_0}e^{jkr_0}(\cos\varphi\cos\varphi+\sin\varphi\sin\varphi)dx'dy'$$ $$\vec{E_\varphi}\approx-j\omega\eta A_m\varphi=j\frac{\omega\mu kE_0e^{-jkr_0}}{2\pi r_0}\cos\varphi\cos\varphi\sin\varphi\sin\varphi$$ $$H_\varphi\approx-\vec{E_\varphi}/\eta$$

Chapter 13. Microwave Antennas

# 1. Rectangular Microstrip Structure

- **Resonant Frequency**: $$ f_{res} = \frac{c}{2\pi\sqrt{\epsilon_r} \cdot w} $$ Where \( c \) is the speed of light, \( \epsilon_r \) is the relative permittivity of the substrate, and \( w \) is the width of the microstrip line.

- **Substrate**: $$ \epsilon_r > 1 $$

- **Ground Plane**: $$ \Delta \phi = 0^\circ $$

# 2. Advantages

- a. Has a planar structure, suitable for missiles, satellites, and other flat objects. - b. Suitable for printed circuit board production. - c. Suitable for forming antenna arrays. - d. Wide bandwidth, low power density, low radiation efficiency.

# 3. Analysis Methods and Radiation Patterns

- **Transmission Line Model**: - Treats the microstrip as a single antenna (half-wave dipole). - Radiation efficiency is low.

- **Electric Field Distribution**: - For \( \theta = 90^\circ \), the electric field distribution is: $$ E = \frac{1}{2} \cos\left(\frac{\pi b}{2} \sin\theta\right) - 2 \cos\left(\frac{\pi d}{2} \sin\theta\right) $$

- **Radiation Pattern**: - For \( \theta = 90^\circ \), the radiation pattern is: $$ f_{rad} = f_{rad}(0) \cdot 2 = S \cos\left(\frac{\pi b}{2} \sin\theta\right) - 2 \cos\left(\frac{\pi d}{2} \sin\theta\right) $$

4. Impedance Matching and Input Impedance

A. Impedance Matching Network and Its Input Impedance

$$ \frac{R_{c}}{2} \quad \text{and} \quad \left(\frac{R_{c}}{2}\right) $$

$$ \text{Input Impedance} \quad Z_{in} = \frac{R_{c}}{2} \quad (\because l = \frac{\lambda}{2}, \text{antenna resonant when } X_{in} = 0, \text{two } R_{c} \text{ in parallel}) $$

$$ R_{c} \approx 120 \frac{\lambda}{W} \Omega \quad (W \geq 2\lambda) $$

$$ R_{c} \approx 90 \left(\frac{\lambda}{W}\right)^2 \Omega \quad (W < 0.35\lambda) $$

$$ R_{c} \approx \frac{W}{120\lambda - \frac{1}{60\pi^2}} \Omega \quad (0.35\lambda \leq W < 2\lambda) $$

Changing \( R_{c} \) can adjust \( W \), thus adjusting the frequency.

5. Transmission Line Types

a) Sideband

$$ Z_{c} = 50 \Omega \quad \text{and} \quad \text{antenna bandwidth} $$

b) Bottomband

$$ Z_{c} = 50 \Omega \quad \text{and} \quad \text{antenna bandwidth} $$

6. Polarization Techniques

a) T-shaped branch line

b) Branch line bridge

c) 4-element array circular polarization antenna

7. Circular polarization measurement method. Rotate the circular polarization antenna to measure the radiation intensity at different angles.

8. Broadband, dual-band antenna. Stacking method for broadband: - When the two boards are close, it's broadband. - When the two boards are far apart, it's dual-band.

9. Antenna array. 16-element array antenna, T-shaped branch and parallel connection

$$E_1$$ $$E_2$$ $$E_3$$ Circular polarization.

1. Introduction

# 1.1. Parabolic Reflector

- **Parabolic Reflector**: - **Focus Point**: The point where all reflected rays converge. - **Focal Length**: The distance from the focus to the vertex of the parabola. - **Aperture**: The opening of the parabolic reflector. - **Wavefront**: The surface of constant phase of a wave.

# 1.2. Optical Properties of Parabolic Reflectors (Applicable to Centimeter and Millimeter Waves)

1. **Reflection of Parallel Rays**: Parallel rays entering the parabolic reflector are reflected and converge at the focus. 2. **Focal Length**: The distance from the focus to the vertex of the parabola is 2f.

2. Various Types of Parabolic Reflectors

# 2.1. (a) Pinhole Beam - **Applications**: Radar, satellite communication. - **Beam Pattern**: The beam pattern is a circular pattern.

# 2.2. (b) Vertical Fan Beam - **Applications**: Radar, phased array. - **Beam Pattern**: The beam pattern is a vertical fan pattern.

# 2.3. (c) Slotted Parabolic Reflectors - **Applications**: Through the slotted beam. - **Beam Pattern**: The beam pattern is a slotted beam pattern.

# 2.4. (d) Parabolic Cylinder - **Applications**: Radar, phased array. - **Beam Pattern**: The beam pattern is a cylindrical beam pattern.

3. Methods to reduce wind resistance and weight.

Metal plate structure. E: vertical polarization or reflection. Wavelength shorter, hole smaller.

$$\int |E| \propto e^{-\alpha l}$$ $$P \propto e^{-2\alpha l}$$

Reduce wind resistance. $$\alpha < \frac{1}{2} \Rightarrow \alpha < \frac{1}{4} (\frac{1}{8} \sim \frac{1}{10})$$ Drill holes. $$\alpha < \frac{1}{2} \Rightarrow \alpha < \frac{1}{4} (\frac{1}{8} \sim \frac{1}{10})$$

4. Different distances of parabolic reflector antenna. (a) Medium distance. $$\psi_0 = \frac{\pi}{2}, f/D = 0.25$$ (b) Long distance. $$\psi_0 < \frac{\pi}{2}, f/D > 0.25$$ (c) Short distance. $$\psi_0 > \frac{\pi}{2}, f/D < 0.25$$

5. Different distances of parabolic reflector antenna (polarization distribution). (a) Medium distance. Mainly linear polarization. Polarization: at the edge. (b) Long distance. Linear polarization less. (c) Short distance. Linear polarization more. Reverse polarization less.

6. Load power and leakage power. $$\eta_A = P_{rs} (\text{receiver power}) / P_r (\text{transmitter power})$$ $$\eta_A = P_{rs} (\text{receiver power}) / P_r (\text{transmitter power})$$

7. Interference and elimination of sources on the same side of the wavefront. Problem: shadowing, reflection, and interference of sources.

1. Install a hole in the wavefront. 2. Install a compensating disc before installation. 3. Eliminate the source method. S = π²/4C (source area) d = √(4π²/λ), l = (2n+1)π/4 - π/4. Φ = a1(π/4)² + a2(π/4) - a3(π)

8. Find the far-field radiation pattern of a parabolic antenna. 1. Near-field method. E.H = 0. E + H ∈ plane of the hole. Jes = H × Ha = Jms = H × Ea Am - A E - H

a. Find the surface of E and H in the near-field and far-field regions. b. In the plane of the hole, E and H are zero. c. Use the principle of electric field source and image theorem to find Jes and Jms. d. Use Jms to find Am and Jms. e. Use Am to find A and calculate E and H.

II. Mirror surface current flow: a. Js = (n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) = 2(n × H) + (n × H) =

**815. Cassegrainian Telescope**

**1. Structural Design Principles**

(1) To (8) Wavefronts (Real Source) (2) Virtual Source (3) Secondary Reflective Surface (Double Surface)

**2. Advantages:**

1) The surface distribution is adjusted to the best, with two reflective surfaces adjustable. 2) Compact structure, light weight - tail 3) Good resistance - double surface for broad reflection. 4) "Long distance parabolic surface optical performance" and "mechanical performance" are good.

**3. Two Analysis Methods:**

a. Virtual Source Method (Equivalent Source Method). b. Parabolic Surface Method.

**4. The distance of the secondary reflective surface can be solved by "rotating 90°" method.**

The page contains a series of diagrams and notes. Here's the extracted text and formulas:

1. Diagram 1: - The diagram shows a circular object with a vertical line through its center. There are arrows pointing to different parts of the circle, indicating various measurements and angles. One of the arrows is labeled "α<π/2". There's also a smaller diagram within the main circle, showing a vertical line and a horizontal line intersecting at a right angle. The text next to this smaller diagram says "Plane: Polarization rotated 90°".

2. Diagram 2: - This diagram appears to be a 3D geometric figure, possibly a cube or a rectangular prism, with various lines and arrows indicating different planes and axes. The text next to it is partially obscured but seems to describe the relationships between the different planes and axes.

3. Diagram 3: - This diagram shows a rectangular object with a curved line drawn on it. There are arrows pointing to different parts of the rectangle and the curved line. The text next to it is not fully legible but seems to describe the relationship between the rectangle and the curved line.

4. Diagram 4: - This diagram shows a circular object with a vertical line through its center. There are arrows pointing to different parts of the circle, indicating various measurements and angles. One of the arrows is labeled "α<π/2". There's also a smaller diagram within the main circle, showing a vertical line and a horizontal line intersecting at a right angle. The text next to this smaller diagram says "Plane: Polarization rotated 90°".

5. Diagram 5: - This diagram appears to be a 3D geometric figure, possibly a cube or a rectangular prism, with various lines and arrows indicating different planes and axes. The text next to it is partially obscured but seems to describe the relationships between the different planes and axes.

15

**Initial Review**

**1. Transfer Function:**

$$ \frac{Z_{1}dz}{C_{1}dz} \quad R_{1}dz \quad Z $$

When open: \( R_{1} = C_{1} = 0 \)

$$ C_{1}dz = \frac{F}{F} \quad G_{1}dz \quad V $$

**Domain:**

Capacitor: \( V(j\omega) = -\frac{1}{j\omega C} I(j\omega) \)

Inductor: \( V(j\omega) = j\omega L I(j\omega) \)

Resistor: \( V(j\omega) = R I(j\omega) \)

**Domain Transfer Function:**

$$ dV(j\omega, z) = I(j\omega) j\omega L dz $$

$$ dI(j\omega, z) = V(j\omega, z) j\omega C_{1} dz $$

Solve

$$ V(j\omega, z) = V_{0+}(j\omega) e^{j\omega\sqrt{L C_{1}} z} + V_{0-}(j\omega) e^{-j\omega\sqrt{L C_{1}} z} $$

$$ I(j\omega, z) = I_{0+}(j\omega) e^{j\omega\sqrt{L C_{1}} z} + I_{0-}(j\omega) e^{-j\omega\sqrt{L C_{1}} z} $$

$$ Z_{c} = \frac{1}{j\omega C_{1}}, \quad \omega = \frac{1}{\sqrt{L C_{1}}} $$

$$ V_{+}(j\omega, z) \triangleq V_{0+}(j\omega) e^{j\omega\sqrt{L C_{1}} z}, \quad V_{-}(j\omega, z) \triangleq V_{0-}(j\omega) e^{-j\omega\sqrt{L C_{1}} z} $$

$$ I_{+}(j\omega, z) \triangleq \frac{V_{+}(j\omega, z)}{Z_{c}}, \quad I_{-}(j\omega, z) \triangleq \frac{V_{-}(j\omega, z)}{Z_{c}} $$

Then

$$ V(j\omega, z) = V_{+}(j\omega, z) + V_{-}(j\omega, z) $$

$$ I(j\omega, z) = I_{+}(j\omega, z) + I_{-}(j\omega, z) $$

Boundary condition

$$ \begin{cases} V(j\omega, 0) = V_{0+}(j\omega) + V_{0-}(j\omega) = I(j\omega, 0) Z_{c}(j\omega) \\ Z(j\omega, 0) = \frac{1}{Z_{c}} \left[ V_{0+}(j\omega) - V_{0-}(j\omega) \right] \end{cases} $$

$$ \begin{cases} V(j\omega, t) = V_{0+}(j\omega) e^{j\omega\sqrt{L C_{1}} t} + V_{0-}(j\omega) e^{-j\omega\sqrt{L C_{1}} t} = V_{g}(j\omega) - I(j\omega) Z_{c}(j\omega) \\ Z(j\omega, t) = \frac{1}{Z_{c}} \left[ V_{0+}(j\omega) - V_{0-}(j\omega) \right] \end{cases} $$

$$ \frac{V_{0+}(j\omega) + V_{0-}(j\omega)}{Z_{c}(j\omega)} = \frac{V_{g}(j\omega)}{Z_{c}(j\omega)} - \frac{V_{0+}(j\omega) - V_{0-}(j\omega)}{Z_{c}(j\omega)} $$

Time Domain Operational Rules: (Time as variable, ω as constant).

# Voltage: $$ V = V_0 \cos(\omega t) = Re(V_0 e^{j\omega t}) \quad , \quad \tilde{V} = V_0 e^{j\omega t} $$ $$ I = I_0 \cos(\omega t) = Re(I_0 e^{j\omega t}) \quad , \quad \tilde{I} = I_0 e^{j\omega t} $$

# Capacitance: $$ \frac{dV(t)}{dt} = C \frac{dI(t)}{dt} $$ $$ I(t) = C \frac{dV(t)}{dt} = j\omega C \cdot Re(V_0 e^{j\omega t}) $$ $$ = j\omega C \cdot V_0 e^{j\omega t} $$ $$ \tilde{I}(t) = j\omega C \cdot V_0 e^{j\omega t} = j\omega C \cdot \tilde{V} $$ $$ V(t) = L \frac{dI(t)}{dt} $$ $$ \tilde{V}(t) = j\omega L \cdot \tilde{I} $$

# Inductance: $$ V(t) = R I(t) $$ $$ \tilde{V}(t) = R \tilde{I}(t) $$

$$ e^{j\omega t} \text{ factor is linearly invariant, so } \tilde{V}(x, t), \tilde{I}(x, t) \text{ can be directly substituted into } V(x, t), I(x, t). $$

The above rules are the rules of "fields", similar to the rules of "circuits". Substitute directly into V(x, t), I(x, t).

Microwave Network Research Method.

To define the voltage and current of a single-mode waveguide mode on the reference surface of the waveguide and the mode current. Under the following rules: only consider propagation in one direction + or -.

(1) $$ V(x), \text{ and } E_T, \text{ and } I_T \text{ and } H_T $$

(2) $$ P = V(x) \cdot I(x) $$

(3) $$ Z_C = \frac{V(x)}{I(x)} $$

Entering the network: inward wave $$ a = V_+ / \sqrt{Z_C} $$, outward wave $$ a' = a e^{j\alpha}, b = b e^{j\beta} $$

Leaving the network: outward wave $$ b = V_- / \sqrt{Z_C} $$, inward wave $$ b' = b e^{j\beta}, a' = a e^{j\alpha} $$

For N-terminal networks, define:

$$ \begin{pmatrix} b_1 \\ \vdots \\ b_n \end{pmatrix} = \begin{pmatrix} s_{11} & s_{1n} \\ \vdots & \vdots \\ s_{n1} & s_{nn} \end{pmatrix} \begin{pmatrix} a_1 \\ \vdots \\ a_n \end{pmatrix} \quad \text{and} \quad \vec{b} = [s] \vec{a} $$

For i in the network: s_{ii} = 0 For i in the network: r_i = 0

(1) Symmetry: s_{ij} = s_{ji}, [s] = [s]^{T} (2) Unitarity: [s]^{H} [s] = I

For two-terminal networks:

$$ \begin{array}{c} a_1 \rightarrow \frac{u_1}{v_1} \\ b_1 \leftarrow \frac{u_2}{v_2} \rightarrow b_2 \end{array} $$

(1) Symmetry: s_{12} = s_{21} (2) Unitarity: |s_{12}| = |s_{21}| > |s_{11}| = |s_{22}|, |s_{11}|^2 + |s_{22}|^2 = 1; |s_{12}|^2 + |s_{21}|^2 = 1

1) From port 1 to port 2, the voltage transmission coefficient is 1: |s_{11}| = |s_{22}| = 0, |s_{12}| = |s_{21}| = 1. 2) Cannot achieve unidirectional isolation, isolation is bidirectional: |s_{12}| = |s_{21}| = 0. 3) Bidirectional isolation, complete reflection: |s_{12}| = |s_{21}| = 0, |s_{11}| = |s_{22}| = 1.

(3) Symmetry: s_{11} = s_{22}, s_{12} = s_{21}. 1) Characteristic equation: \begin{pmatrix} s_{11} - s_j & s_{12} \\ s_{12} & s_{11} - s_j \end{pmatrix} \begin{pmatrix} u_i \\ u_j \end{pmatrix} = 0 2) Eigenvalues: s_1 = s_{11} + s_{22} \Rightarrow s_1 = \frac{s_1 + s_2}{2} s_2 = s_{11} - s_{12} \Rightarrow s_2 = \frac{s_1 - s_2}{2} 0 = s_1 - s_2 s_1 = s_2

Eigenvalues: s_1 \Rightarrow u_1 = \frac{1}{\sqrt{2}} (1, 1)^T s_2 \Rightarrow u_2 = \frac{1}{\sqrt{2}} (1, -1)^T

Meaning: Using u_1 as the two-terminal network's reflection coefficient b = [s] u_1 = s_1 u_1, the reflection coefficient is s_1. Using u_2 as the two-terminal network's reflection coefficient b = [s] u_2 = s_2 u_2, the reflection coefficient is s_2. For u_1, the two components of the two-terminal network are in the same direction, corresponding to the middle reflection situation.

(4) Indicators: 1) Voltage transmission coefficient \( T_v = S_{21} \) 2) Power transmission coefficient \( T_p = |S_{21}|^2 \) 3) Insertion loss \( L = \frac{1}{15.41} \) 4) Insertion phase \( \phi_{21} = \arg S_{21} \) 5) Group delay \( t_d = \frac{d}{d\omega} \phi_{21} \) 6) Insertion loss ratio \( \rho = \frac{1}{1 - |T_{21}|} = \frac{1 + |S_{21}|}{1 - |S_{21}|} \)

(5) Impedance Matrix: Series: \( [Z] = \sum [z] \) \( V_p = V_q = 0 \): Effective series Parallel: \( [Y] = \sum [y] \) \( V_p = V_q = 0 \): Effective parallel Cascade: \( [A] = \Pi [A_i] \) (where \( V_1, I_1, V_2, I_2 \) are represented) Transmission Matrix: Series: \( [T] = \Pi [t] \) (where \( a_1, b_1, a_2, b_2 \) are represented) Series-parallel: \( [H] = \sum [h] \) Parallel-series: \( [G] = \sum [g] \)

IV. Waveguide to Transmission Line (Time Delay, Phase, Lossless): 1. Cross relationship. (Example: \( \nabla = \nabla_1 + \nabla_2 \); \( \nabla = \nabla_1 + \nabla_2 \)). Unknown, time delay: Forward transmission (wave number \( k_1 \)): \( \vec{H}_T = -\frac{jk_2}{k_1^2 - k_2^2} \nabla \cdot \vec{E}_2 - jk_2 \nabla \cdot \vec{H}_2 \) \( \vec{E}_T = j\omega \mu \frac{1}{k_1^2 - k_2^2} \nabla \times \vec{H}_2 + jk_2 \nabla \times \vec{E}_2 \) Backward transmission (wave number \( k_2 \)): \( \vec{H}_T = j\omega \mu \frac{1}{k_1^2 - k_2^2} \nabla \times \vec{E}_2 + jk_2 \nabla \times \vec{H}_2 \) \( \vec{E}_T = -j\omega \mu \frac{1}{k_1^2 - k_2^2} \nabla \times \vec{H}_2 + jk_2 \nabla \times \vec{E}_2 \)

2. TE mode: \( E_z = 0 \). Forward transmission: \( \vec{E}_T = -\frac{jk_2}{k_1^2 - k_2^2} \nabla \times \vec{H}_2 - \eta_{TE} \nabla \times \vec{H}_T \) \( \vec{H}_T = \eta_{TE} \nabla \times \vec{E}_T \)

Boundary condition: \( \frac{\partial H_2}{\partial n} \mid_{z=0} = 0 \).

The note discusses wave equations and their solutions in a medium, specifically focusing on TE and TM waves. It includes the following key points:

1. TE wave: E2 = 0. The wave equation is ∇²H2 + k²H2 = 0. The solution is H2 = H2(τ) * e^(-jkz). The boundary condition is ∂H2/∂n = 0 at the boundary. The reflection and transmission coefficients are defined as ηTE(∞) = w/k.

2. TM wave: H2 = 0. The wave equation is ∇²E2 + k²E2 = 0. The solution is E2 = E2(τ) * e^(-jkz). The boundary condition is E2 = 0 at the boundary. The reflection and transmission coefficients are defined as ηTM(∞) = k/2w.

3. TEM wave: E2 = 0, H2 = 0. The wave equation is ∇²W + k²W = 0. The solution is W = W(τ) * e^(-jkz). The boundary condition is ∂W/∂n = 0 at the boundary. The reflection and transmission coefficients are defined as ηTEM = √(w/k).

4. The wave equations for TE and TM waves are given as: - TE wave: (k² - k²)H = -jwε/2 * ∇E - jk²∇H. - TM wave: (k² - k²)E = jwε/2 * ∇H + jk²∇E.

5. The wave number k is defined as k = √(k² - k²). The cutoff wave number is defined as k = √(k² - k²). The cutoff frequency is defined as f = k²/c.

The note discusses the transmission line model and wave equations. It starts by defining the forward and backward waves as \( e^{-jkz} \) and \( e^{jkz} \) respectively. The phase condition is given as \( \Delta \phi = -2kz = 2k\pi \), leading to a wave number \( k = \frac{\pi}{z} \).

The note then delves into the wave equation for the electric field \( E \) and magnetic field \( H \): $$ \frac{\partial^2 E}{\partial z^2} + k^2 E = 0 $$ $$ \frac{\partial^2 H}{\partial z^2} + k^2 H = 0 $$

The solution is expressed as: $$ E = E_T e^{jkz} + E_{T+} e^{-jkz} $$ $$ H = H_T e^{jkz} + H_{T+} e^{-jkz} $$

The note introduces the concept of voltage \( V \) and current \( I \) as: $$ V = V_T e^{jkz} + V_{T+} e^{-jkz} $$ $$ I = I_T e^{jkz} + I_{T+} e^{-jkz} $$

The impedance \( Z \) is defined as: $$ Z = \frac{V}{I} = \frac{V_T}{I_T} = \frac{V_{T+}}{I_{T+}} $$

The reflection coefficient \( \Gamma \) is given by: $$ \Gamma = \frac{V_{T+} - V_T}{V_{T+} + V_T} = \frac{I_T - I_{T+}}{I_T + I_{T+}} $$

The note also discusses the power ratio \( p \): $$ p = \frac{V_{max}}{V_{min}} = \frac{1 + \Gamma}{1 - \Gamma} $$

The minimum phase angle \( \phi \) is defined as: $$ \phi = \frac{\pi}{2kz} $$

The reflection coefficient \( \Gamma \) is related to the impedance \( Z \) and admittance \( Y \) as: $$ \Gamma = \frac{Z - Z_0}{Z + Z_0} = \frac{Y - Y_0}{Y + Y_0} $$

The note concludes with the reflection and transmission coefficients for the wave: $$ \Gamma = \frac{V_{T+} - V_T}{V_{T+} + V_T} = \frac{I_T - I_{T+}}{I_T + I_{T+}} $$ $$ \Gamma = \frac{Z - Z_0}{Z + Z_0} = \frac{Y - Y_0}{Y + Y_0} $$

Here's the extracted text and formulas from the handwritten note, with placeholders for the diagrams:

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**2. Pure Reflection State**

1) End Short Circuit: $$\overline{Z_F} \triangleq \infty / \overline{Y_F} \triangleq 0 \Rightarrow \Gamma_F = +1$$ $$V(z) = 2V_0 + \cos(k_2z) \quad |V(z)| = 2|V_0| |\cos(k_2z)|$$ $$I(z) = 2j\frac{V_0}{Z_0} \sin(k_2z) \quad |I(z)| = 2\left|\frac{V_0}{Z_0}\right| |\sin(k_2z)|$$ $$\Gamma(z) = +e^{-j2k_2z}$$ $$\overline{Z_1}(z) = -j\cot(k_2z) \quad \overline{y_1}(z) = j\tan(k_2z)$$

2) End Open Circuit: $$\overline{Z_F} \triangleq 0 / \overline{Y_F} \triangleq \infty \Rightarrow \Gamma_F = -1$$ $$V(z) = j2V_0 + \sin(k_2z) \quad |V(z)| = 2|V_0| |\sin(k_2z)|$$ $$I(z) = 2j\frac{V_0}{Z_0} \cos(k_2z) \quad |I(z)| = 2\left|\frac{V_0}{Z_0}\right| |\cos(k_2z)|$$ $$\Gamma(z) = -e^{-j2k_2z}$$ $$\overline{Z_1}(z) = j\tan(k_2z) \quad \overline{y_1}(z) = -j\cot(k_2z)$$

3) End Impedance: $$\overline{Z_F} \triangleq jx \quad \overline{y_F} \triangleq jb \quad |\Gamma_F| = 1 \quad (\Gamma_F = e^{j\phi_F})$$ $$V(z) = 2V_0 + e^{j\phi_F} \cos(k_2z - \frac{\phi_F}{2}) \quad |V(z)| = 2|V_0| |\cos(k_2z - \frac{\phi_F}{2})|$$ $$I(z) = 2j\frac{V_0}{Z_0} e^{j\frac{\phi_F}{2}} \sin(k_2z - \frac{\phi_F}{2}) \quad |I(z)| = 2\left|\frac{V_0}{Z_0}\right| |\sin(k_2z - \frac{\phi_F}{2})|$$ $$\Gamma(z) = e^{j(\phi_F - 2k_2z)}$$ $$\overline{Z_1}(z) = -j\cot(k_2z - \frac{\phi_F}{2}) \quad \overline{y_1}(z) = j\tan(k_2z - \frac{\phi_F}{2})$$

**3. Mixed Reflection State**

$$V(z) = V_0(1 + \Gamma_F e^{-j2k_2z}) \quad |V(z)| = |V_0| (1 + |\Gamma_F| e^{j(\phi_F - 2k_2z)})$$ $$I(z) = \frac{V_0}{Z_0} (1 - \Gamma_F e^{-j2k_2z}) \quad |I(z)| = \left|\frac{V_0}{Z_0}\right| (1 - |\Gamma_F| e^{j(\phi_F - 2k_2z)})$$ $$\Gamma(z) = |\Gamma_F| e^{j(\phi_F - 2k_2z)}$$ $$\overline{Z_1}(z) = \frac{1 - \Gamma_F e^{-j2k_2z}}{1 + \Gamma_F e^{-j2k_2z}} \quad \overline{y_1}(z) = \frac{1 - \Gamma_F e^{-j2k_2z}}{1 + \Gamma_F e^{-j2k_2z}}$$

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**1. Basic Parameters:**

- **Radiation Pattern:** Far field region has E and H components. The power density: $$ S = \frac{1}{2} (\vec{E} \times \vec{H}^*) = \frac{1}{r} \frac{1}{2\eta} (|E_{\theta}|^2 + |E_{\phi}|^2) $$ $$ E_{\theta} = \frac{f_{\theta}(\theta, \phi)}{r} e^{-jkr}, E_{\phi} = \frac{f_{\phi}(\theta, \phi)}{r} e^{-jkr}, \vec{H} = \frac{1}{r} \hat{\phi} \times \vec{E} $$ $$ \therefore S = \frac{1}{2\eta r^2} (|f_{\theta}(\theta, \phi)|^2 + |f_{\phi}(\theta, \phi)|^2) $$ $$ F(\theta, \phi) \triangleq r^2 S = \frac{1}{2\eta} (|f_{\theta}(\theta, \phi)|^2 + |f_{\phi}(\theta, \phi)|^2) $$ $$ F_{\mathrm{m}}(\theta, \phi) \triangleq F(\theta, \phi) / F_{\mathrm{max}}(\theta, \phi) $$

- **E Plane Pattern:** E field in plane F, e.g. φ = const for a disk. - **H Plane Pattern:** H field in plane F, e.g. θ = 90° for a disk.

**2. Directional Coefficient (Radiation Power Density):**

$$ D(\theta, \phi) \triangleq \frac{S_r(\theta, \phi)}{P_r / 4\pi r^2} $$

**3. Antenna Gain (Input Power is the same):**

$$ G(\theta, \phi) \triangleq \frac{S_r(\theta, \phi)}{P_{\mathrm{in}} / 4\pi r^2} $$

**4. Polarization:**

- Linear Polarization: Parallel polarization (parallel to the plane), Vertical polarization (perpendicular to the plane). - Circular Polarization: Right-hand circular polarization (right-hand rule), Left-hand circular polarization (left-hand rule). - Elliptical Polarization: The ratio of the major axis to the minor axis of the ellipse.

**5. Input Impedance:**

$$ Z_{\mathrm{in}} = \text{standing wave ratio} / \text{incident current} \text{ (on the surface)} $$

**6. Radiation Resistance:**

$$ R_{\mathrm{rad}} = \frac{2P_r}{I^2} $$

**7. Bandwidth:**

The frequency range that meets the radiation resistance and gain requirements.

3. Point Source (Z = 2.0 e^(jπx))

Near Field: S is complex, no radiation, no field. Far Field: S is real, radiated field, spherical wave.

Hφ ≈ $$\frac{2\phi}{4\pi} e^{-jk\phi} \frac{jk}{r} \sin\theta \approx E_0 / \eta$$ E0 ≈ $$\frac{2\phi}{4\pi} e^{-jk\phi} j\frac{4\pi}{k} \sin\theta$$ S_r = $$\frac{1}{2} (E_0 \times H_\phi^*) \cdot \hat{r} = \frac{\eta}{8} \left(\frac{2\phi}{\lambda}\right)^2 \frac{\sin^2\theta}{r^2}$$ P_r = $$\phi_s S_r ds = 40\pi^2 \left(\frac{2\phi}{\lambda}\right)^2$$ R_r = $$2P_r / Z^2 = 80\pi^2 \left(\frac{\lambda}{\phi}\right)^2$$

f_n(θ) = sinθ, F_n(θ) = sin^2θ f_n(θ) = 1/2 => |θ2 - θ1| = 90°: half-power angle. D(θ,φ) = $$\frac{S_r}{P_r / 4\pi r^2} = \frac{3}{2} \sin^2\theta$$ θ = π/2 1.5 = 1.76 dB.

4. Wire Antenna

Far Field: α = j $$\frac{Id^3}{4\pi r} e^{-jk\phi} \frac{1}{r} \sin\theta = dE_0 / \eta$$ dE0 = $$\frac{Id^3}{4\pi r} e^{-jk\phi} j\frac{4\pi}{k} \sin\theta$$

① Symmetric Wire: $$\begin{cases} I(x) = 2.0 \sin k(\frac{l}{2} - z) & 0 \leq z \leq \frac{l}{2} \\ I(x') = I_0 \sin k(\frac{l}{2} + z) & -\frac{l}{2} \leq z' \leq 0 \end{cases}$$ R ≈ r Hφ = j $$\frac{2\phi}{2\pi r} \cdot \frac{\cos(\frac{k\phi}{2} \cos\theta) - \cos\frac{k\phi}{2}}{\sin\theta} = E_0 / \eta$$ E0 = j $$\frac{2\phi}{2\pi r} \cdot \frac{\cos(\frac{k\phi}{2} \cos\theta) - \cos\frac{k\phi}{2}}{\sin\theta}$$ S_r ≈ $$\eta \frac{2\phi^2}{8\pi^2 r^2} \left[\frac{\cos(\frac{k\phi}{2} \cos\theta) - \cos\frac{k\phi}{2}}{\sin\theta}\right]^2$$

l = λ/2: P_r = 36.54 Z^2 R_r = 73 Ω = Zin. f_n(θ) = $$\frac{\cos(\frac{\pi}{2} \cos\theta)}{\sin\theta}$$, F_n(θ) = $$\frac{f_n(0)}{(1 - \cos\frac{\pi}{2})^2}$$ f_n(θ) = 1/2 => |θ2 - θ1| = 78°: half-power angle. D(θ,φ) = $$\frac{80}{\pi^2}$$ 1.64 = 2.15 dB.

l = λ: R_r = 200 Ω, f = λ/2: P_r = 106 Ω.

2. Folded dipole $$\rho_{r} \approx 2 \times 73 \Omega$$ $$Z_{in} \approx 4 \times 73 \Omega$$

3. Monopole antenna $$D = 2D_{folded}$$ $$\lambda = \frac{\lambda}{4}$$

4. Inverted L antenna $$\Delta R_{end}$$

5. Dipole antenna - Yagi-Uda

6. Planar antenna Far field: Conductors/ferromagnetic materials + image principle. $$\vec{J}_{ms} = -\hat{n} \times \vec{E}$$ $$\vec{J}_{es} = \hat{n} \times \vec{H}$$

1. Guided wave. ($\vec{E}_{y} = \hat{y} E_{0}$) $$\vec{J}_{ms} = \hat{x} \vec{J}_{m} = \hat{x} \cdot 2 \vec{E}_{0}$$ $$\vec{H}_{mo} = -j \omega \epsilon_{0} \mu_{0} \vec{E}_{0} \cdot ab \sin \left(\frac{\pi a}{2} \cos \phi \right) \sin \left(\frac{\pi b}{2} \cos \phi \right)$$ $$\vec{H}_{mp} = -j \omega \epsilon_{0} \mu_{0} \vec{E}_{0} \cdot ab \sin \left(\frac{\pi a}{2} \cos \phi \right) \sin \left(\frac{\pi b}{2} \cos \phi \right)$$ $$\vec{H}_{mp} = -j \omega \epsilon_{0} \mu_{0} \vec{E}_{0} \cdot ab \sin \left(\frac{\pi a}{2} \cos \phi \right) \sin \left(\frac{\pi b}{2} \cos \phi \right)$$

Here's the extracted text and formulas from the handwritten note, with placeholders for the diagrams:

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**φ = 90°:**

$$ f_{n,E} = S a\left(\frac{k b}{2} \sin \phi\right) \triangleq \frac{1}{\sqrt{2}} \quad 2 \theta_{0.5,E} \approx 51^{\circ} \cdot \frac{\lambda}{5} $$

**φ = 0°:**

$$ f_{n,H} = S a\left(\frac{k a}{2} \cos \phi\right) \triangleq \frac{1}{\sqrt{2}} \quad 2 \theta_{0.5,H} \approx 51^{\circ} \cdot \frac{\lambda}{a} $$

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**② Rectangular Waveguide (Ez = Eo sin(π/αx))**

$$ 2 \theta_{0.5,E} = 68^{\circ} \cdot \frac{\lambda}{6} $$

$$ 2 \theta_{0.5,H} = 68^{\circ} \cdot \frac{\lambda}{a} $$

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**③ Parabolic Reflector**

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**④ Cardioid Array: Using "polarization" to achieve beamwidth reduction**

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**6. Array**

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**① Array**

1) Binary:

$$ E_{0} = j \frac{k n 2 d}{4 \pi} \frac{e^{j k r}}{r} \sin \phi \cdot 2 \cos \frac{k d \cos \phi + \beta}{2} $$

$$ = j \frac{k n 2 d}{4 \pi} \frac{e^{j k r}}{r} f_{e}(\theta, \phi) - f_{a}(\theta, \phi) $$

2) N-element:

$$ f_{n} = \sin \frac{N \beta}{2} / N \sin \frac{\beta}{2} \quad N \text{ is an integer, } \beta = k d \cos \phi + \beta $$

$$ \theta_{\max} = \cos^{-1} \left( \frac{\beta}{k d} \right), \phi = 0 $$

$$ \theta_{0} = \cos^{-1} \left( \frac{2 p \pi}{N k d} - \frac{\beta}{k d} \right), \phi = \frac{2 p \pi}{N}, p = \pm 1, \pm 2, \ldots $$

$$ \theta_{\max, \phi} = \cos^{-1} \left( \frac{2 p + \frac{1}{2} \pi}{N k d} - \frac{\beta}{k d} \right), \phi = \frac{2 p + \frac{1}{2} \pi}{2 N}, p = (1, -2), (2, -3), (3, -1), \ldots $$

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**3) Microwave Array: Two-element array**

$$ f_{E, a} = S a\left(\frac{k b}{2} \sin \phi\right) \cdot 2 \cos \left( \frac{\beta}{2} \cos \phi \right) $$

$$ f_{H, a} = S a\left(\frac{k a}{2} \cos \phi\right) \cdot 2 \cos \left( \frac{\beta}{2} \cos \phi \right) $$

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$$ Z_{in} = \frac{R_{r}}{2} $$

$$ R_{r} \approx \left\{ \begin{array}{ll} 120 \frac{\lambda}{W} & W \geq 2\lambda \\ R_{0} \left(\frac{\lambda}{W}\right)^2 & W < 0.35\lambda \\ \left[\frac{W}{120\lambda} - \frac{1}{60\pi}\right]^{-1} & 0.25\lambda \leq W < 2\lambda \end{array} \right. $$

4) Waveguide slot antenna: cutting the same direction current. Top figure:

2) Face array: \( f_{a} = f_{ax} \cdot f_{ay} \)

3) Body array: \( f_{a} = f_{ax} \cdot f_{ay} \cdot f_{az} \)

Review Outline

1. Transmission Line

1. Transmission line theory formulas: $$ \Gamma = \frac{Z_{s} - Z_{c}}{Z_{s} + Z_{c}} $$ $$ Z_{s} = \frac{1 + \Gamma}{1 - \Gamma} \quad (\Gamma = \frac{1}{2}) $$ $$ Z_{c} = \frac{1}{\Gamma} $$ $$ P = \frac{4\pi I_{0}^{2}}{1 - |\Gamma|^{2}} \quad (\Gamma = \frac{V_{max}}{V_{min}}) $$ $$ Z = Z_{in} / Z_{c} $$ $$ \Gamma_{F} = \frac{2\Gamma - 1}{2\Gamma + 1} $$ $$ \Gamma_{i} = \frac{1 - \Gamma}{1 + \Gamma} \quad (\Gamma = \frac{1}{2}) $$ $$ Y_{c} $$ $$ k = \frac{1}{\rho} $$ $$ \bar{y} = \frac{1}{\bar{z}} $$

2. Smith chart construction, chart rules. Mark the meaning of each line on the chart, and solve the problem. Resistance circle: $$ \Gamma = |\Gamma_{F}| e^{j(\phi_{F} - 2k\pi)} $$ $$ \Gamma = -1 $$ $$ r < 1 $$ $$ r > 1 $$ $$ \Gamma = 1 $$ $$ \Gamma = \infty $$ $$ Z = Z_{0} + Z_{-} = 0 $$ $$ \Gamma = 1 $$ $$ \Gamma = -1 $$ $$ r = 1 $$ $$ \Gamma = 1 $$ $$ \Gamma = -1 $$ $$ r = 1 $$ $$ \Gamma = 1 $$ $$ \Gamma = -1 $$ $$ r = 1 $$ $$ \Gamma = 1 $$ $$ \Gamma = -1 $$ $$ r = 1 $$ $$ \Gamma = 1 $$ $$ \Gamma = -1 $$ $$ r = 1 $$ $$ \Gamma = 1 $$ $$ \Gamma = -1 $$ $$ r = 1 $$ $$ \Gamma = 1 $$ $$ \Gamma = -1 $$ $$ r = 1 $$ $$ \Gamma = 1 $$ $$ \Gamma = -1 $$ $$ r = 1 $$ $$ \Gamma = 1 $$ $$ \Gamma = -1 $$ $$ r = 1 $$ $$ \Gamma = 1 $$ $$ \Gamma = -1 $$ $$ r = 1 $$ $$ \Gamma = 1 $$ $$ \Gamma = -1 $$ $$ r = 1 $$ $$ \Gamma = 1 $$ $$ \Gamma = -1 $$ $$ r = 1 $$ $$ \Gamma = 1 $$ $$ \Gamma = -1 $$ $$ r = 1 $$ $$ \Gamma = 1 $$ $$ \Gamma = -1 $$ $$ r = 1 $$ $$ \Gamma = 1 $$ $$ \Gamma = -1 $$ $$ r = 1 $$ $$ \Gamma = 1 $$ $$ \Gamma = -1 $$ $$ r = 1 $$ $$ \Gamma = 1 $$ $$ \Gamma = -1 $$ $$ r = 1 $$ $$ \Gamma = 1 $$ $$ \Gamma = -1 $$ $$ r = 1 $$ $$ \Gamma = 1 $$ $$ \Gamma = -1 $$ $$ r = 1 $$ $$ \Gamma = 1 $$ $$ \Gamma = -1 $$ $$ r = 1 $$ $$ \Gamma = 1 $$ $$ \Gamma = -1 $$ $$ r = 1 $$ $$ \Gamma = 1 $$ $$ \Gamma = -1 $$ $$ r = 1 $$ $$ \Gamma = 1 $$ $$ \Gamma = -1 $$ $$ r = 1 $$ $$ \Gamma = 1 $$ $$ \Gamma = -1 $$ $$ r = 1 $$ $$ \Gamma = 1 $$ $$ \Gamma = -1 $$ $$ r = 1 $$ $$ \Gamma = 1 $$ $$ \Gamma = -1 $$ $$ r = 1 $$ $$ \Gamma = 1 $$ $$ \Gamma = -1 $$ $$ r = 1 $$ $$ \Gamma = 1 $$ $$ \Gamma = -1 $$ $$ r = 1 $$ $$ \Gamma = 1 $$ $$ \Gamma = -1 $$ $$ r = 1 $$ $$ \Gamma = 1 $$ $$ \Gamma = -1 $$ $$ r = 1 $$ $$ \Gamma = 1 $$ $$ \Gamma = -1 $$ $$ r = 1 $$ $$ \Gamma = 1 $$ $$ \Gamma = -1 $$ $$ r = 1 $$ $$ \Gamma = 1 $$ $$ \Gamma = -1 $$ $$ r = 1 $$ $$ \Gamma = 1 $$ $$ \Gamma = -1 $$ $$ r = 1 $$ $$ \Gamma = 1 $$ $$ \Gamma = -1 $$ $$ r = 1 $$ $$ \Gamma = 1 $$ $$ \Gamma = -1 $$ $$ r = 1 $$ $$ \Gamma = 1 $$ $$ \Gamma = -1 $$ $$ r = 1 $$ $$ \Gamma = 1 $$ $$ \Gamma = -1 $$ $$ r = 1 $$ $$ \Gamma = 1 $$ $$ \Gamma = -1 $$ $$ r = 1 $$ $$ \Gamma = 1 $$ $$ \Gamma = -1 $$ $$ r = 1 $$ $$ \Gamma = 1 $$ $$

1. Find the area of the double line transformer (Example 2.9 on page 49).

2. Diagram 1: Capacitive and inductive components.

3. Diagram 2: Capacitive and inductive components with a short circuit.

4. Diagram 3: Capacitive and inductive components with a short circuit and a voltage distribution line.

5. Diagram 4: Capacitive and inductive components with a voltage distribution line and a short circuit.

6. Diagram 5: Capacitive and inductive components with a voltage distribution line and a short circuit, and a ZF transformer.

7. Diagram 6: Capacitive and inductive components with a voltage distribution line and a short circuit, and a ZF transformer, with a ZF transformer at 0.3-j0.5.

3. Impedance Example

P48. Example 2.8 (Delta Impedance).

$$ Z_{c} = \frac{1 + \Gamma_{F}}{1 - \Gamma_{F}} = \frac{1 + \Gamma_{F} e^{-j2k_{a}z}}{1 - \Gamma_{F} e^{j2k_{a}z}} $$

$$ = \frac{1 + \Gamma_{F} e^{-j\pi}}{1 - \Gamma_{F} e^{-j\pi}} = \frac{1 - \Gamma_{F}}{1 + \Gamma_{F}} = \frac{1}{2\Gamma_{F}} $$

P76. 2.10.

$$ Z_{F} = \frac{20 - j30}{30} = 0.67 - j1 $$

4. Voltage Distribution

P46. Example 2.6. Before.

$$ \Gamma_{F} = \frac{4 - 1}{4 + 1} = \frac{2 - 1}{2 + 1} = \frac{1}{3} = \frac{V_{0-}}{V_{0+}} $$

$$ Z = 0 \text{ when } |V| = |V_{0+}|(1 + \frac{1}{3}) $$

(b) $$\Gamma_F = \frac{2-1}{2+1} = \frac{1}{3}$$ $$|V| = |V_0| + \frac{1}{3} e^{-j\frac{2\pi}{3}}$$

$$Z_{\text{ref}}' = \frac{1-\Gamma_F}{1+\Gamma_F} = \frac{1-\frac{1}{3} e^{-j\frac{2\pi}{3}}}{1+\frac{1}{3} e^{-j\frac{2\pi}{3}}} = \frac{1-\frac{1}{3}}{1+\frac{1}{3}} = 2Z_c$$

$$\therefore Z_{\text{ref}} = 2Z_c // 2Z_c = Z_c \therefore \Gamma_{\text{ref}} = 0$$

(c) $$\Gamma_{F_1} = \frac{1}{3}, \Gamma_{F_2} = \frac{1}{3}$$

$$Z_{\text{ref}} = 2Z_c // \frac{1}{2}Z_c = \frac{2}{3}Z_c$$

$$\therefore \Gamma_{\text{ref}} = \frac{1}{1+\frac{2}{3}} = -\frac{3}{7}$$

$$\therefore |V| = |V_0| + \frac{3}{7} e^{-j\frac{2\pi}{3}}$$

5. Incident voltage, reflected voltage, total voltage, reflection coefficient concept.

$$V = V_+ + V_-$$

$$\Gamma = \frac{V_-}{V_+}$$

$$\overline{Z}_{PA} = 0.5, \overline{Z}_{PB} = 2, \therefore \overline{Z}_{\text{ref}} = 2//2 = 1$$

$$\therefore \overline{Z}_{FA} = 1, \overline{Z}_{FB} = 50\Omega$$

$$\therefore \Gamma_C = 0, V_B' = V_B e^{-j\frac{2\pi}{3}} = -jV_B, P_{FB} = \frac{V_B^2}{2R_{FB}}$$

$$\& \overline{Z}_{FA} = 2, \therefore \Gamma_B = \frac{2-1}{2+1} = \frac{1}{3}$$

$$\therefore V_- = \frac{3}{4}V_B, V_+ = \frac{1}{4}V_B$$

$$\therefore V_A = V_- e^{-j\frac{2\pi}{3}} + V_+ e^{j\frac{2\pi}{3}} = (-j\frac{3}{4} + j\frac{1}{4})V_B$$

$$= -\frac{1}{2}V_B = -\frac{1}{2}V_B^*, P_{FA} = \frac{V_B^2}{8R_{FA}}$$

Directivity and transience of the reflection coefficient!

Waveguides

# 1. Solution Methods for Waveguides, Cylindrical Waveguides, and Coaxial Cables

**Circular Waveguide:** - TE mode: $$\left\{\begin{array}{l} \nabla^2 H_z(x, y) + k_0^2 H_z(x, y) = 0 \\ \left.\frac{\partial H_z}{\partial n}\right|_{z=0} = 0 \end{array}\right.$$ $$\eta_{TE} = \frac{\omega \mu}{k_0^2}$$

- TM mode: $$\left\{\begin{array}{l} \nabla^2 E_z(x, y) + k_0^2 E_z(x, y) = 0 \\ \left.\frac{\partial E_z}{\partial n}\right|_{z=0} = 0 \end{array}\right.$$ $$\eta_{TM} = \frac{k_0^2}{\omega \mu}$$

**Rectangular TE mode:** $$\left\{\begin{array}{l} \frac{\partial^2}{\partial x^2} H_z + \frac{\partial^2}{\partial y^2} H_z + k_0^2 H_z = 0 \\ \left.\frac{\partial H_z}{\partial x}\right|_{x=0, a} = 0, \left.\frac{\partial H_z}{\partial y}\right|_{y=0, b} = 0 \end{array}\right.$$ $$\therefore H_z(x, y) = H_{mn} \cos\left(\frac{m\pi}{a} x\right) \cos\left(\frac{n\pi}{b} y\right)$$ $$H_x(x, y) = \frac{j k_0}{k_0^2} \left(\frac{m\pi}{a}\right) H_{mn} \sin\left(\frac{m\pi}{a} x\right) \cos\left(\frac{n\pi}{b} y\right)$$ $$H_y(x, y) = \frac{j k_0}{k_0^2} \left(\frac{n\pi}{b}\right) H_{mn} \cos\left(\frac{m\pi}{a} x\right) \sin\left(\frac{n\pi}{b} y\right)$$ $$E_x(x, y) = \eta_{TE} H_y(x, y)$$ $$E_y(x, y) = -\eta_{TE} H_x(x, y)$$

**Rectangular TM mode:** $$\left\{\begin{array}{l} \frac{\partial^2}{\partial x^2} E_z + \frac{\partial^2}{\partial y^2} E_z + k_0^2 E_z = 0 \\ \left.\frac{\partial E_z}{\partial x}\right|_{x=0, a} = 0, \left.\frac{\partial E_z}{\partial y}\right|_{y=0, b} = 0 \end{array}\right.$$ $$\therefore E_z(x, y) = E_{mn} \sin\left(\frac{m\pi}{a} x\right) \sin\left(\frac{n\pi}{b} y\right)$$ $$E_x(x, y) = -\frac{j k_0}{k_0^2} \left(\frac{m\pi}{a}\right) E_{mn} \cos\left(\frac{m\pi}{a} x\right) \sin\left(\frac{n\pi}{b} y\right)$$ $$E_y(x, y) = -\frac{j k_0}{k_0^2} \left(\frac{n\pi}{b}\right) E_{mn} \sin\left(\frac{m\pi}{a} x\right) \cos\left(\frac{n\pi}{b} y\right)$$ $$H_x(x, y) = -\frac{1}{\eta_{TM}} E_y(x, y)$$ $$H_y(x, y) = \frac{1}{\eta_{TM}} E_x(x, y)$$

**Circular TE mode:** $$\left\{\begin{array}{l} \frac{1}{r} \frac{\partial}{\partial r} \left(r \frac{\partial H_z}{\partial r}\right) + \frac{1}{r^2} \frac{\partial^2 H_z}{\partial \varphi^2} + k_0^2 H_z = 0 \end{array}\right.$$ $$H_z(r, \varphi) = H_{mn} J_n(k_0 r) \cos(n\varphi)$$ $$H_r(r, \varphi) = -\frac{j k_0 n}{k_0^2 r} H_{mn} J_n'(k_0 r) \cos(n\varphi)$$ $$H_\varphi(r, \varphi) = \frac{j k_0 n}{k_0^2 r} H_{mn} J_n(k_0 r) \sin(n\varphi)$$ $$E_r(r, \varphi) = \eta_{TE} H_\varphi(r, \varphi)$$ $$E_\varphi(r, \varphi) = -\eta_{TE} H_r(r, \varphi)$$

The notebook page contains a detailed set of equations and explanations related to electromagnetic wave propagation in coaxial cables and waveguides. Here's a breakdown of the content:

---

**Circular TM Mode:**

$$\left\{\begin{array}{l} \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial E_{z}}{\partial r} \right) + \frac{1}{r^{2}} \frac{\partial^{2} E_{z}}{\partial \varphi^{2}} + k_{c}^{2} E_{z} = 0 \\ E_{z} \mid_{r=R}=0. \end{array}\right.$$

$$E_{z}(r,\varphi) = E_{m} J_{n}(k_{c} r) \cos(n\varphi)$$

$$E_{r}(r,\varphi) = -\frac{j k_{c}^{2} n}{k_{c} r} E_{m} J_{n}^{\prime}(k_{c} r) \cos(n\varphi)$$

$$E_{\varphi}(r,\varphi) = \frac{j k_{c}^{2} n}{k_{c} r} E_{m} J_{n}(k_{c} r) \sin(n\varphi)$$

$$H_{r}(r,\varphi) = -\frac{1}{\eta_{TM}} E_{\varphi}(r,\varphi)$$

$$H_{\varphi}(r,\varphi) = \frac{1}{\eta_{TM}} E_{r}(r,\varphi).$$

**Coaxial TEM Mode:**

$$\nabla_{T} \cdot \vec{E}(x,y) = 0 \Rightarrow \nabla_{T}^{2} \psi = -\frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial \psi}{\partial r} \right) + \frac{1}{r^{2}} \frac{\partial^{2} \psi}{\partial \varphi^{2}} = 0.$$

$$\psi = \frac{V}{\ln(a/b)} \ln r.$$

$$\therefore \vec{E}_{T}(x,y) = -\frac{\partial \psi}{\partial r} \hat{r} = -\frac{V}{r \ln(a/b)} \hat{r}$$

$$\vec{H}_{T}(x,y) = \frac{1}{\eta_{TEM}} \hat{z} \times \vec{E}_{T}(x,y)$$

$$\eta_{TEM} = \frac{\omega \mu}{k} = \sqrt{\frac{\mu}{\epsilon}}.$$

**Coaxial TE Mode:**

$$\left\{\begin{array}{l} \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial H_{z}}{\partial r} \right) + \frac{1}{r^{2}} \frac{\partial^{2} H_{z}}{\partial \varphi^{2}} + k_{c}^{2} H_{z} = 0 \\ \frac{\partial H_{z}}{\partial r} \mid_{r=a} = \frac{\partial H_{z}}{\partial r} \mid_{r=b} = 0. \end{array}\right.$$

**Coaxial TM Mode:**

$$\left\{\begin{array}{l} \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial E_{z}}{\partial r} \right) + \frac{1}{r^{2}} \frac{\partial^{2} E_{z}}{\partial \varphi^{2}} + k_{c}^{2} E_{z} = 0 \\ E_{z} \mid_{r=a} = E_{z} \mid_{r=b} = 0. \end{array}\right.$$

---

**Question 2:**

*What mode (or modes) does each of the above transmission lines and waveguides support? How does the electromagnetic field distribution look for each mode? How can we ensure single-mode transmission?*

*Solution:*

*TE10:*

$$\lambda_{c} = 2 \sqrt{\left(\frac{m}{a}\right)^{2} + \left(\frac{n}{b}\right)^{2}}, \quad m=1, n=0 \sim m=2, n=0.$$

*The single-mode TE10 works when*

$$a < \lambda_{c} < 2a, \quad \frac{\lambda_{c}}{2} < a < \lambda_{c}.$$

*General working frequency range*

$$1.05a \leq \lambda_{c} \leq 0.8 \times 2a.$$

No. Date

1. TE01 mode and TM01 mode

Circle: TE11 mode: $$\lambda_c = \frac{2\pi a}{V_{ni}}$$ $$n=1, i=1 \rightarrow TE11$$ $$\lambda_c = \frac{2\pi a}{V_{ni}}$$ $$n=0, i=1 \rightarrow TM10$$

The single mode working area for TE11 mode is: $$\frac{\lambda_c}{3.41} < a < \frac{\lambda_c}{2.61}$$ $$a \approx \frac{2}{3}$$

Coaxial line: TEM mode: $$\lambda_c = \pi(a+b)$$ The single mode working area for TEM mode is: $$\lambda_c \geq \pi(a+b) \Rightarrow \lambda_c \geq 1.1\pi(a+b)$$

Waveguide: Quasi-TEM mode: $$\lambda_g = \frac{\lambda_0}{\sqrt{\epsilon_r}}$$

3. TE01 mode internal surface current distribution. Which slots will cut these surface currents and cause radiation?

The page contains handwritten notes and diagrams related to electromagnetic wave modes in rectangular and circular waveguides. Here's a breakdown:

**Rectangular Waveguide Diagrams:**

- **TEmn:** Represents the electric field component in the y-direction, with arrows indicating the direction of the field. The subscript 'm' and 'n' denote the number of half-periods of the field in the x and y directions, respectively. - **TMmn:** Represents the magnetic field component in the x-direction, with arrows indicating the direction of the field. The subscript 'm' and 'n' denote the number of half-periods of the field in the x and y directions, respectively.

**Circular Waveguide Diagrams:**

- **TEmi:** Represents the electric field component in the y-direction, with arrows indicating the direction of the field. The subscript 'm' and 'n' denote the number of half-periods of the field in the x and y directions, respectively. - **TMmi:** Represents the magnetic field component in the x-direction, with arrows indicating the direction of the field. The subscript 'm' and 'n' denote the number of half-periods of the field in the x and y directions, respectively.

The notes also mention that the field components are periodic and that the field values are zero at the boundaries of the waveguide. The diagrams illustrate the field patterns in both rectangular and circular waveguides, with arrows indicating the direction of the field components.

5. Impedance Matching Methods and Rules.

Excitation Methods: Electric Field Excitation, Magnetic Field Excitation, Electromagnetic Excitation Equivalent J_e Equivalent J_m Equivalent J_e J_m

Excitation Rule: ∫(E·J_e + H·J_m) dV ≠ 0 Excitation (field and source interaction) (Condition) Voltage frequency and current zero condition! ∫ = 0 Not excitable (field and source no interaction).

III. Microwave Network.

1. S-parameters Definition, Properties, Physical Meaning?

Definition: S_ij = b_i / a_j | a_k ≠ i = 0; According to this, S_ii = Γ_i [b_1] = [S_11 ... S_1n] [a_1] [b_n] = [S_n1 ... S_nn] [a_n] a = I + √2c = V_+ / √2c. b = -I - √2c = V_- / √2c

Properties: N-ports: Symmetry: S_ij = S_ji, [S] = [S]^T. Non-reciprocity: [S]^H [S] = I, ∴ Symmetric [S] [S]^H [S] = I

2. Endports: Symmetry: S_12 = S_21 (unknown) Non-reciprocity: |S_12| = |S_21|, |S_11| = |S_22|, |S_11|^2 + |S_22|^2 = 1, |S_12|^2 + |S_22|^2 = 1 (unknown). (Values: unknown) (Magnitudes: unknown)

Physical Meaning: S_ii: Port connected to the source, other port connected to the load or load, i-port reflection coefficient.

2. Reference plane transformation of [S]

$$ S_{11}' = \frac{b_1 e^{-j\theta_1}}{a_1 e^{j\theta_1}} |_{a_2=0} = S_{11} e^{-j2\theta_1} $$ $$ S_{22}' = \frac{b_2 e^{-j\theta_2}}{a_2 e^{j\theta_2}} |_{a_1=0} = S_{22} e^{-j2\theta_2} $$ $$ S_{21}' = \frac{b_2 e^{-j\theta_2}}{a_1 e^{j\theta_1}} |_{a_2=0} = S_{21} e^{-j(\theta_1+\theta_2)} $$ $$ S_{12}' = \frac{b_1 e^{-j\theta_1}}{a_2 e^{j\theta_2}} |_{a_1=0} = S_{12} e^{-j(\theta_1+\theta_2)} $$

$$ \therefore [S'] = \begin{bmatrix} S_{11} e^{-j2\theta_1} & S_{12} e^{-j(\theta_1+\theta_2)} \\ S_{21} e^{-j(\theta_1+\theta_2)} & S_{22} e^{-j2\theta_2} \end{bmatrix} = \begin{bmatrix} e^{-j\theta_1} & 0 \\ 0 & e^{-j\theta_2} \end{bmatrix} \begin{bmatrix} S_{11} & S_{12} \\ S_{21} & S_{22} \end{bmatrix} \begin{bmatrix} e^{-j\theta_1} & 0 \\ 0 & e^{-j\theta_2} \end{bmatrix} $$

Note: [S'] = [P] [S] [P] , [P] = diag{e^{-j\theta_i}}. i=1,...,n.

3. Time-domain network [S] solution method: open circuit → S1; short circuit → S2. Page 187 Example 4.5. $$ S_{21} = S_{11} = \frac{S_1 + S_2}{2} $$ $$ S_{12} = S_{21} = \frac{S_1 - S_2}{2} $$

$$ \frac{(Z_c) Y}{(Z_c)} \Rightarrow \frac{(Z_c) Y}{\frac{1}{2} (Z_c)} $$

Short: $$ S_2 = -1 $$

Open: $$ S_1 = \frac{Y/2 \cdot Z_c + 1}{Y/2 \cdot Z_c + 1} = \frac{2Yc - Y}{2Yc + Y} $$

$$ \therefore S_{11} = S_{22} = \frac{S_1 + S_2}{2} = \frac{1}{2} \cdot \frac{-2Y}{2Yc + Y} = -\frac{Y}{2Yc + Y} $$

$$ S_{12} = S_{21} = \frac{S_1 - S_2}{2} = \frac{1}{2} \cdot \frac{4Yc}{2Yc + Y} = \frac{2Yc}{2Yc + Y} $$

$$ S_{11} = \frac{b_1}{a_1} |_{a_2=0} = \Gamma_1 = \frac{1 - Y_1}{1 + Y_1} = \frac{Y_c - (R + Y_c)}{Y_c + (R + Y_c)} = -\frac{Y}{R + 2Y_c} $$

$$ S_{21} = \frac{b_2}{a_1} |_{a_2=0} = \frac{a_1 + a_1 \Gamma_1}{a_1} = 1 + \Gamma_1 = \frac{2Yc}{Y + 2Y_c} $$

Lesson Example:

$$\Gamma_{\phi 2} = -e^{-j\kappa_2 \cdot 2 \cdot \frac{\lambda}{8}} = -(-j) = j$$ $$\Gamma_{\phi 1} = e^{-j\kappa_2 \cdot 2 \cdot \frac{\lambda}{8}} = -j$$ $$\Sigma_{\phi 2} = \frac{1 + \Gamma_{\phi 1}}{1 - \Gamma_{\phi 1}} = \frac{1 + j}{1 - j}$$ $$\Sigma_{\phi 1} = \frac{1 + \Gamma_{\phi 2}}{1 - \Gamma_{\phi 2}} = \frac{1 - j}{1 + j}$$ $$\therefore Z_{\phi 2} = \frac{1}{Y_{C2}} - \frac{1 + j}{1 - j}, Z_{\phi 1} = \frac{1}{Y_{C1}} + \frac{1 - j}{1 + j}$$ $$\therefore Y_{\phi} = Y_{C2} \frac{1 + j}{1 + j} + Y_{C1} \frac{1 + j}{1 - j}$$ $$\therefore \Gamma_{\phi} = \frac{1 - Y_{\phi}}{1 + Y_{\phi}} = \frac{Y_{C2} \frac{1 - j}{1 + j} + Y_{C1} \frac{1 + j}{1 - j}}{Y_{C2} \frac{1 - j}{1 + j} + Y_{C1} \frac{1 + j}{1 - j}} = S_1$$ $$\therefore S_{11} = S_{22} = \frac{S_1 + S_2}{2}, S_{12} = S_{21} = \frac{S_1 - S_2}{2}$$

Short Circuit: $$\Gamma_{\phi 2} = -e^{-j\kappa_2 \cdot 2 \cdot \frac{\lambda}{8}} = -(-j) = j$$ $$\Gamma_{\phi 1} = -j$$ $$\Sigma_{\phi 2} = -j$$ $$\Sigma_{\phi 1} = -j$$ $$\therefore Y_{\phi} = +j(Y_{C1} + Y_{C2})$$ $$\therefore \Gamma_{\phi} = \frac{Y_{C} - Y_{\phi}}{Y_{C} + Y_{\phi}} = \frac{Y_{C} - j(Y_{C1} + Y_{C2})}{Y_{C} + j(Y_{C1} + Y_{C2})} = S_1$$ $$\therefore S_{11} = S_{22} = \frac{S_1 + S_2}{2}, S_{12} = S_{21} = \frac{S_1 - S_2}{2}$$

Open Circuit: $$\Gamma_{\phi 2} = e^{-j\kappa_2 \cdot 2 \cdot \frac{\lambda}{8}} = -j$$ $$\Gamma_{\phi 1} = -j$$ $$\Sigma_{\phi 2} = -j$$ $$\Sigma_{\phi 1} = -j$$ $$\therefore Y_{\phi} = +j(Y_{C1} + Y_{C2})$$ $$\therefore \Gamma_{\phi} = \frac{Y_{C} - Y_{\phi}}{Y_{C} + Y_{\phi}} = \frac{Y_{C} - j(Y_{C1} + Y_{C2})}{Y_{C} + j(Y_{C1} + Y_{C2})} = S_1$$ $$\therefore S_{11} = S_{22} = \frac{S_1 + S_2}{2}, S_{12} = S_{21} = \frac{S_1 - S_2}{2}$$

$$\Gamma_{\phi 2} = -e^{-j\kappa_2 \cdot 2 \cdot \frac{\lambda}{8}} = -(-j) = j$$ $$\Gamma_{\phi 1} = -j$$ $$\Sigma_{\phi 2} = -j$$ $$\Sigma_{\phi 1} = -j$$ $$\therefore Y_{\phi} = +j(Y_{C1} + Y_{C2})$$ $$\therefore \Gamma_{\phi} = \frac{Y_{C} - Y_{\phi}}{Y_{C} + Y_{\phi}} = \frac{Y_{C} - j(Y_{C1} + Y_{C2})}{Y_{C} + j(Y_{C1} + Y_{C2})} = S_1$$ $$\therefore S_{11} = S_{22} = \frac{S_1 + S_2}{2}, S_{12} = S_{21} = \frac{S_1 - S_2}{2}$$

$$\Gamma_{\phi 2} = e^{-j\kappa_2 \cdot 2 \cdot \frac{\lambda}{8}} = -j$$ $$\Gamma_{\phi 1} = -j$$ $$\Sigma_{\phi 2} = -j$$ $$\Sigma_{\phi 1} = -j$$ $$\therefore Y_{\phi} = +j(Y_{C1} + Y_{C2})$$ $$\therefore \Gamma_{\phi} = \frac{Y_{C} - Y_{\phi}}{Y_{C} + Y_{\phi}} = \frac{Y_{C} - j(Y_{C1} + Y_{C2})}{Y_{C} + j(Y_{C1} + Y_{C2})} = S_1$$ $$\therefore S_{11} = S_{22} = \frac{S_1 + S_2}{2}, S_{12} = S_{21} = \frac{S_1 - S_2}{2}$$

$$\Gamma_{\phi 2} = -e^{-j\kappa_2 \cdot 2 \cdot \frac{\lambda}{8}} = -(-j) = j$$ $$\Gamma_{\phi 1} = -j$$ $$\Sigma_{\phi 2} = -j$$ $$\Sigma_{\phi

4. Interconnected matrices [A], [a] [A] ~ [S] transformation.

Interconnected matrix $$\frac{z_1}{v_1} \rightarrow \begin{bmatrix} [A] \\ \downarrow v_2 \end{bmatrix} \frac{z_2}{v_2}$$

$$\begin{bmatrix} v_1 \\ z_1 \end{bmatrix} = \begin{bmatrix} A & B \\ C & D \end{bmatrix} \begin{bmatrix} v_2 \\ z_2 \end{bmatrix} \quad [A] = \sum_{i=1}^{n} [A_i].$$

Normalized interconnected matrix (normalized matrix). $$\frac{z_1}{v_1} \rightarrow \begin{bmatrix} \bar{a} \\ \downarrow v_2 \end{bmatrix} \frac{z_2}{v_2} (z_2). \quad \left\{ \begin{array}{l} z = I \cdot \sqrt{2} e \\ v = V / \sqrt{2} e. \end{array} \right.$$

$$\begin{bmatrix} \bar{a} & \bar{b} \\ \bar{c} & \bar{d} \end{bmatrix} = \begin{bmatrix} A \sqrt{\frac{2}{2c_1}} & B \sqrt{\frac{2}{2c_1c_2}} \\ C \sqrt{\frac{2}{2c_1c_2}} & D \sqrt{\frac{2}{2c_1c_2}} \end{bmatrix}.$$

[A] ~ [S] transformation: $$\begin{aligned} S_{11} &= \frac{\bar{a} + \bar{b} - \bar{c} - \bar{d}}{\bar{a} + \bar{b} + \bar{c} + \bar{d}} \\ S_{12} &= \frac{2(\bar{a} - \bar{c})}{\bar{a} + \bar{b} + \bar{c} + \bar{d}} \\ S_{21} &= \frac{2(\bar{a} - \bar{c})}{\bar{a} + \bar{b} + \bar{c} + \bar{d}} \\ S_{22} &= \frac{\bar{a} + \bar{b} - \bar{c} + \bar{d}}{\bar{a} + \bar{b} + \bar{c} + \bar{d}} \\ \end{aligned}$$ $$\begin{aligned} \bar{a} &= \frac{1}{2S_{11}} (1 + S_{11} - S_{22} - |S|) \\ \bar{b} &= \frac{1}{2S_{11}} (1 + S_{11} + S_{22} + |S|) \\ \bar{c} &= \frac{1}{2S_{11}} (1 - S_{11} - S_{22} + |S|) \\ \bar{d} &= \frac{1}{2S_{11}} (1 - S_{11} + S_{22} - |S|). \end{aligned}$$

Example: Find [A] when the input impedance is matched (S11 = 1 = 0). Use a, b, c, d for convenience. $$\frac{(z_{c1})}{(z_{c1})} \frac{(z_{c2})}{(z_{c2})} \frac{(z_{c3})}{(z_{c3})}$$

First find [A2]: $$\begin{aligned} \bar{a} &= \frac{V_1}{V_2} \mid_{z_2 = 0} = \frac{V_2 \cos k_1 l}{V_2} = \cos k_1 l \\ \bar{b} &= \frac{V_1}{V_2} \mid_{z_2 = 0} = \frac{j2k_1 \sin k_1 l}{\sqrt{2c_1}} = j \sin k_1 l. \end{aligned}$$ Since symmetric: $\bar{a} = \bar{a}$, $\bar{b} = \bar{c}$.

Find [A3]: $$\begin{aligned} \bar{a} &= \frac{V_1}{V_2} \mid_{z_2 = 0} = \frac{V_2 \cos k_1 l}{V_2} = \cos k_1 l \\ \bar{b} &= \frac{V_1}{V_2} \mid_{z_2 = 0} = \frac{j2k_1 \sin k_1 l}{\sqrt{2c_1}} = j \sin k_1 l. \end{aligned}$$ Since symmetric: $\bar{a} = \bar{a}$, $\bar{b} = \bar{c}$. $$\therefore [A] = \begin{bmatrix} \sqrt{\frac{2}{2c_1}} & 0 \\ 0 & \sqrt{\frac{2}{2c_1}} \end{bmatrix} = \begin{bmatrix} \sqrt{\frac{2c_1}{2c_1}} & 0 \\ 0 & \sqrt{\frac{2c_1}{2c_1}} \end{bmatrix}.$$

$$\therefore [a] = [a_1] \cdot [a_2] \cdot [a_3]$$ $$= \begin{bmatrix} \frac{z_{c_3}}{z_{c_1}} \cos k_1 & \frac{z_{c_2}}{z_{c_1} z_{c_3}} \sin k_1 \\ \frac{z_{c_2}}{z_{c_1}} \sin k_1 & \frac{z_{c_1}}{z_{c_3}} \cos k_1 \end{bmatrix}$$

Solve for S11: $$S_{11} = \frac{\bar{a} + \bar{b} - \bar{c} - \bar{d}}{\bar{a} + \bar{b} + \bar{c} + \bar{d}} = \frac{1}{K} \left[ (\sqrt{\frac{z_{c_3}}{z_{c_1}}} - \sqrt{\frac{z_{c_1}}{z_{c_3}}}) \cos k_1 + j (\frac{z_{c_2}}{\sqrt{z_{c_1} z_{c_3}}} - \frac{\sqrt{z_{c_1} z_{c_3}}}{z_{c_2}}) \sin k_1 \right]$$ (K ≠ 0).

$$\left\{ \begin{array}{ll} z_{c_1} = z_{c_2} = z_{c_3} & S_{11} = 0 \quad \text{Uniform Line} \\ z_{c_1} = z_{c_3}, l = \frac{\lambda}{4} & S_{11} = 0 \quad \text{Quarter-Wave Transformer} \\ z_{c_2} = \sqrt{z_{c_1} z_{c_3}}, l = \frac{\lambda}{4} & S_{11} = 0 \quad \text{Quarter-Wave Transformer} \end{array} \right.$$

Antenna

1. **Antenna Pattern**: Main Beam, Side Lobe, Tail, Half Power Angle, Main Beam Half Angle.

$$f_n(\theta, \phi)$$

2. **Antenna Directivity Coefficient D and Gain G**: What are they? What is the difference between them?

Directivity Coefficient: $$D_n(\theta, \phi) = \left| \frac{S_{\mathrm{r}, \max}(\theta, \phi)}{P_r / 4 \pi r^2} \right|$$ Gain: $$G_n(\theta, \phi) = \left| \frac{S_{\mathrm{r}, \max}(\theta, \phi)}{P_r / 4 \pi r^2} \right|$$

$$P_{\mathrm{r}} < P_{\mathrm{in}}$$

$$P_{\mathrm{in}}$$

$$P_{\mathrm{r}} = P_{\mathrm{in}}$$

3. **What is the polarization of an antenna, polarization matching, and polarization correction?**

Polarization (Maximum Gain Direction): - Linear Polarization: Parallel (Parallel to the receiving surface), Vertical (Perpendicular to the receiving surface). - Circular Polarization: Right-hand (Right-hand circular), Left-hand (Left-hand circular).

Circular Polarization: Equal to the right-hand and left-hand circularly polarized wave components.

The page contains a series of notes and diagrams related to antenna theory. Here's a breakdown of the content:

1. **Polarization Matching:** - **Polarization Matching:** The receiving and transmitting antennas should have the same polarization. - **Polarization Misalignment:** Polarization mismatch.

3. **Question 4:** Discuss the concept of a circular array and its total radiation pattern. The formula provided is: $$ f_{\alpha}(\theta, \phi) = f_{\alpha}(\theta, \phi) \cdot f_{\alpha}(\theta, \phi) = f_{\alpha}(\theta, \phi) $$

4. **Question 5:** Discuss the radiation patterns of dipole, half-wave dipole, and full-wave dipole antennas in the E and H planes. The formulas provided are: - **Dipole:** $$ f_{\alpha}(\theta) = \sin(\theta) $$ - **E-plane (1st order):** $$ \text{Circular} $$ - **H-plane (1st order):** $$ \text{Circular} $$ - **Half-wave Dipole:** $$ f_{\alpha}(\theta) = \frac{\cos\left(\frac{\pi}{2} \cos(\theta)\right)}{\sin(\theta)} $$ - **E-plane (1st order):** $$ \text{Elliptical} $$ - **H-plane (1st order):** $$ \text{Circular} $$ - **Full-wave Dipole:** $$ f_{\alpha}(\theta) = \frac{\cos\left(\pi \cos(\theta)\right) + 1}{\sin(\theta)} $$ - **E-plane (1st order):** $$ \text{Elliptical} $$ - **H-plane (1st order):** $$ \text{Circular} $$

5. **Question 6:** Discuss the structure, analysis method, and equivalent circuit of a microstrip patch antenna. The formulas provided are: - **Structure:** $$ \text{Patch} $$ $$ \text{Ground Plane} $$ - **Analysis Formula:** $$ f_{\alpha, E} = S_{\alpha} \left( \frac{k \beta}{2} \sin(\theta) \right) \cdot 2 \cos\left( \frac{k \beta}{2} \sin(\theta) \right) $$ $$ f_{\alpha, H} = S_{\alpha} \left( \frac{k \beta}{2} \sin(\theta) \right) \cdot 2 \cos\left( \frac{k \beta}{2} \sin(\theta) \right) $$

**Equivalent Circuit:**

$$\frac{1}{Z_{eq}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$$

$$Z_{in} = R_{1} / 2$$

7. **Three-Element Eight-Element Analysis Method:**

$$Z_{2} \rightarrow E_{B}, E_{C} \rightarrow E_{B}^{'}, E_{C}^{'}, e_{B}, e_{C} \rightarrow Z_{1}, Z_{3}$$

$$k - \frac{\lambda}{4} \rightarrow k - \frac{\lambda}{4} \rightarrow k$$

1) **Near-field (B, C) Electric Field:**

$$E_{B, C} = j \frac{2 \pi e^{jkr}}{2 \pi r} \cdot \frac{\cos(\frac{\pi}{2} \cos \theta)}{\cos \theta} = \frac{2 \pi e^{jkr}}{\lambda} \cdot \frac{\cos(\frac{\pi}{2} \cos \theta)}{\cos \theta}$$

$$l = \frac{\lambda}{2}$$

$$E_{B}, E_{C} \text{ are in phase}$$

2) **Induced Electric Fields:**

$$E_{B} = \int_{1}^{2} E_{B} \cdot dt = - \int_{1}^{2} E_{B} \cdot dt$$

$$E_{C} = \int_{1}^{2} E_{C} \cdot dt = - \int_{1}^{2} E_{C} \cdot dt$$

3) **Left Resonator Long:**

$$B \text{ input inductive reactance (open circuit)}$$

$$C \text{ input capacitive reactance (open circuit)}$$

4) $$I_B = e_B / Z_{B,in}, I_C = e_C / Z_{C,in}, I_B \neq I_C 90°, I_B \neq I_C 90°$$

Near-field impedance matching: $$I_B \approx I_C \approx I_0$$

5) Left-hand and right-hand waves in the near field add in phase, in the far field add in phase.

Left-hand and right-hand waves in the near field add in phase, in the far field add in phase.

8. Describe several forms of parabolic surface antennas. What types of waves do they produce, and what are their uses?

a) Parabolic surface antenna (circular face) b) Slotted parabolic surface antenna (rectangular face)

c) Slotted parabolic surface antenna (elliptical face) d) Parabolic surface antenna (elliptical face)

Horizontal wave emission Vertical wave emission

Navigation radar, anti-ship radar

9. How to calculate the far-field radiation pattern of a parabolic surface antenna?

Near-field method: Use the initial wave impedance and parabolic surface reflection characteristics to find the surface currents $$E_A, H_A$$ (only for the parabolic surface).

$$H_A E_A \rightarrow J_{in} J_{es}$$

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10. Cassie's two analysis methods of the line.

Virtual Source Method (Equivalent Source Method)

Equivalent Polarization Method: Define the polarized surface, only one equivalent polarized surface corresponds to it.

17. Additional

Mode Classification:

1) Transverse Electric (TE) Wave, Transverse Magnetic (TM) Wave, Electric Wave, Magnetic Wave

2) Transverse Electric (TE) Wave, Transverse Magnetic (TM) Wave, Electric Wave, Magnetic Wave

3) Transverse Electric (TE) Wave, Transverse Magnetic (TM) Wave

4) Hybrid Wave (EH Wave): TE Wave and TM Wave in hybrid combination, e.g., hybrid.

Mode Transmission Conditions:

1) Electromagnetic excitation mode: Excitation and hybrid

2) Working wavelength: λ

3) Impedance matching

No.

Date

1. Matching Network TE10: Parallel: Helical Inductor

2. Waveguide Head TE11: Waveguide wall near strong I2, good contact

3. Short Circuit S-shaped flow stopper: $$\frac{1}{Z_{a}} = \frac{1}{Z_{c1}}, \frac{1}{Z_{b3}} = \frac{1}{Z_{a}} = \frac{Z_{c1}}{Z_{a}}, Z_{b} = \frac{Z_{c1}^2}{Z_{a}}$$ $$\frac{1}{Z_{b2}} = \frac{Z_{c1}^2}{Z_{a}Z_{c2}}, \frac{1}{Z_{c1}} = \frac{1}{Z_{b3}} = \frac{Z_{a}Z_{c1}}{Z_{c1}^2}, Z_{c} = \frac{Z_{a}Z_{c1}^2}{Z_{c1}^2}$$ When Za ≈ 0, $$\left(\frac{Z_{c1}^2}{Z_{c1}}\right) \ll 1$$, Zc ≈ 0.

4. Attenuator Rotating polarized attenuator: TE10 → TE11 → TE10 → TE11

$$[S] = \begin{bmatrix} 0 & \cos\theta \\ \cos\theta & 0 \end{bmatrix}$$

5. Mode Suppressor TE01 mode suppressor

For TE10: $$[S] = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}$$ For through mode: $$[S] = \begin{bmatrix} e^{-j\omega t} & 0 \\ 0 & e^{j\omega t} \end{bmatrix}$$

6. Waveguide T-branch: (lossless end network)

$$[S] = \frac{1}{2} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}$$

Properties of Three-Port Networks: 1. It is impossible to have all three ports matched simultaneously. 2. It is impossible to have two ports matched simultaneously, otherwise it becomes a two-port network.

7. 3dB Ideal Power Divider (Three-Port Network): $$ [S] = \begin{bmatrix} 0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & 0 & 0 \\ \frac{1}{\sqrt{2}} & 0 & 0 \end{bmatrix} $$

1 → 2, 3: Power Divider 2, 3 → 1: Combiner.

8. Magic T Branch (Four-Port Network): $$ \text{Structural Symmetry: } S_{21} = S_{31}, S_{22} = S_{33} $$ $$ S_{41} = S_{14} = 0, S_{24} = -S_{34} $$ $$ S_{ij} = S_{ji} $$ 1, 4 Port Matching: S_{11}, S_{44} = 0 $$ \text{No loss: } [S]^{H} \cdot [S] = I. $$

$$ [S] = \frac{1}{\sqrt{2}} \begin{bmatrix} 0 & 1 & 1 & 0 \\ 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & -1 \\ 0 & 1 & -1 & 0 \end{bmatrix} $$

9. Waveguide Branch Coupler (Four-Port Network): 1 → 4: Direct 2 → 1: Isolator 3 → 4: Coupler 4 → 2: Isolator $$ \text{Structural Symmetry: } S_{11} = S_{22} = S_{33} = S_{44}, S_{12} = S_{21}, S_{13} = S_{31}, S_{14} = S_{41} $$ $$ \text{No loss, no coupling: } S_{12} \text{ and } S_{13} \text{ are } 90^\circ. $$

$$ [S] = \begin{bmatrix} 0 & S_{12} & S_{13} & 0 \\ S_{12} & 0 & 0 & S_{13} \\ S_{13} & 0 & 0 & S_{12} \\ 0 & S_{13} & S_{12} & 0 \end{bmatrix} $$

10. Circular Polarizer: (Unidirectional Waveguide Network)

- Geometrically: Two-Port - Electrically: Four-Port (Horizontal Polarization TE_{11}°, Vertical Polarization TE_{11}°)

- Each Port Matching: S_{ii} = 0 - Horizontal Polarization: S_{12} = S_{14} = S_{23} = S_{34} = 0 - Symmetry: S_{ij} = S_{ji} - Vertical Polarization and Horizontal Polarization at π/2: S_{42} = j e^{-jφ} - Assume S_{31} = e^{-jφ}, then S_{12} = j e^{-jφ}

$$ [S] = \begin{bmatrix} 0 & 0 & e^{-jφ} & 0 \\ 0 & 0 & 0 & j e^{-jφ} \\ e^{-jφ} & 0 & 0 & 0 \\ 0 & j e^{-jφ} & 0 & 0 \end{bmatrix} $$

11. Rotational Symmetry at Four-Port (Unidirectional Waveguide Network)

- Each Port Matching: S_{ii} = 0 - Rotational Symmetry: S_{12} = S_{23} = S_{34} = S_{45} = S_{51} = S_{10} = S_{20} = S_{25} = S_{35} - Symmetry: S_{ij} = S_{ji} - Unidirectional: [S]^{H} [S] = I_{5}

$$ \Rightarrow |S_{12}| = |S_{13}| = \frac{1}{2} $$

$$ S_{12} = \frac{1}{2}, S_{13} = \frac{1}{2} e^{jφ}, \text{ then } $$

$$ S_{13}^{*} S_{12} + S_{12}^{*} S_{13} + |S_{12}|^{2} = 0 $$

$$ \text{Result: } \frac{1}{4} e^{jφ} + \frac{1}{4} e^{-jφ} + \frac{1}{4} = 0 $$

$$ \therefore \cos φ = -\frac{1}{2}, φ = 120° $$

12. Circular Polarizer: (Unidirectional Waveguide Network)

$$ [S] = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix} $$

The page contains a series of diagrams and notes related to digital electronics and logic gates. Here's a breakdown of the content:

1. **Logic Gates and Expressions**: - **SOP (Sum of Products)**: The smallest form of a logic expression. - **POS (Product of Sums)**: The largest form of a logic expression. - **Truth Tables**: Used to represent the truth values of logic gates. - **Karnaugh Maps**: Used to simplify logic expressions.

3. **CMOS Logic Gates**: - **CMOS (Complementary Metal-Oxide-Semiconductor)**: - Includes implementations of basic CMOS gates: NOT gate (Inverter), Transmission gate, Tri-state gate, NAND gate, NOR gate, and Wired-AND gate.

IV. Combination Circuit Analysis and Design

1. **Analysis**

- **① Encoders and Priority Encoders** - 1) Binary Encoders: \( n \rightarrow 2^n \) - 2) Priority Encoders: \( 2^n \rightarrow n \). Not completely determined. - 3) Medium-Scale Integrated Priority Encoders: 74148. Capacity expansion: \( (8-3) \times 2 \rightarrow (16-4) \)

- **② Decoders** - 1) Binary Decoders: \( n \rightarrow 2^n \) - 2) Medium-Scale Integrated Decoders: 74139(54). Capacity expansion: previous output after the last bit. - 3) Digital Display Decoders: Output truth table and simplify to the simplest expression.

- **③ Multiplexers and Demultiplexers** - 1) Multiplexer: Select data source by address code. - 2) Medium-Scale Integrated Multiplexer: 74153. Capacity expansion: hierarchical selection. - 3) Demultiplexer: Select data source by address code.

- **④ Comparator** - 1) Binary Comparison: \( (A > B) = AB, (A < B) = \overline{A}B, (A = B) = \overline{A} \overline{B} \) - 2) Binary Comparison: Can be constructed using binary comparators and other combinational logic. - 3) Medium-Scale Integrated Comparator: 7485. Capacity expansion: between "L", "E", "G" connections.

- **⑤ Adders** - 1) Full Adder: \( A_n, B_n, C_{n-1} \rightarrow S_n, C_n \) - 2) Ripple-Carry Adder: Propagate carry. - 3) Look-Ahead Carry Adder: \( P_n = A_n \oplus B_n \), \( S_n = P_n \oplus C_{n-1} \), \( C_n = P_n \cdot C_{n-1} + G_n \)

- **⑥ 8421 BCD Code Adder Circuit**: For addition of 9's complement, need to perform a 6's complement correction.

**Stability: Simultaneous toggle errors.**

**Single Change: Regardless of input changes, output changes only once.**

**Flipping: Output changes only when CP=1.**

**Design**

**Truth table → Simplify → Logical expression → Eliminate race conditions → Logical diagram.**

**Race and Hazard: Transient**

**1. Algebraic elimination: Add redundant terms to eliminate possible forms of A + A̅ or A · A̅.**

**2. Karnaugh map elimination: Add redundant terms, connect all adjacent minimal terms.**

**Sequential Logic Circuit Analysis & Design**

**1. Analysis**

**1) Flip-flops and registers**

1) Flip-flops: Edge-triggered, no CP

- RS Flip-flop: Stable state, single change, flipping - D Flip-flop: Flipping

2) Registers: Edge-triggered, with CP

- Master-slave type: - Master-slave RS Flip-flop: Stable state, single change - Master-slave JK Flip-flop: Single change - Master-slave D Flip-flop: √

- Edge-triggered type: √

**2) Registers: 1 bit register; asynchronous: independent of CP; synchronous: dependent on CP.**

**3) Counters: 74161: Asynchronous reset, synchronous count. Capacity expansion: connect to the next lower level.**

**74160: Up, down, up-down counter. Capacity expansion: same.**

**4) Shift registers: Store, serial-parallel conversion, parallel-serial conversion.**

**5) Sequential Signal Generator:**

- Register (right shift) → Feedback logic: If the feedback logic is asynchronous or synchronous, it will produce a race condition.

- Q1 → Qn+1 → out

- Feedback logic: F = f(Q1, ..., Qn)

(Finite State Machine)

**Example: Synchronous Circuit Analysis**

- **Mealy Model:** Input, State → Output - **Moore Model:** State → Output

**Trigger Types → Trigger State Equation → Output Function → State Table → Waveform Diagram**

**2. (Synchronous Circuit) Design**

- **① Draw Out Initial State Diagram and Initial State Table.** - **② State Simplification**

1) **Complete Deterministic Synchronous Circuit State Simplification:**

- **Truth Table → Association Comparison → Maximum Equivalence Class → Minimum State Table**

2) **Incomplete Deterministic Synchronous Circuit State Simplification:**

- **Truth Table → Association Comparison → Maximum Equivalence Class → Minimum State Table**

- **③ State Distribution: Mutual Exclusion Rule.**

1) **Same Input Different States.**

2) **Different Inputs Same State.**

3) **Output Same State.**

- **④ Trigger Type Selection: Write Output Function. Draw Out FSM Logic Diagram**

- **⑤ Add Reset Circuit or Additional Circuit to Make FSM Self-Start.**

**3. Race and Hazard:Essential**

- **① Add Buffer Drivers to Prevent Clock Skew.**

- **② Signal Timing Synchronization.**

- **③ Reliable Clock Signal.**

1. RAM: random access memory

$$ ext{Address} \rightarrow \left[ ext{Decode} \right] \rightarrow \left[ 2^n \right] \rightarrow \left[ ext{Storage} \right] \leftarrow \left[ m \right] \leftarrow \left[ ext{Read} \right] \leftarrow \left[ I/O \right] \rightarrow 2^n ext{Word} \times m ext{Bit} $$

① SRAM: No refresh, small capacity, high power consumption ② DRAM: Requires refresh, large capacity, low power consumption

Capacity Expansion: ① Bit expansion: One address code, multiple I/O ② Word expansion: Multiple address codes - I/O ③ Word-bit expansion

2. ROM: read only memory

① PROM: One-time programmable; fuse. ② EPROM: Multiple-time programmable; UV reset. ③ E²PROM: Multiple-time programmable; tunnel effect ④ flash memory

3. PLD: programmable logic device

① SPLD: AND array (address), OR array (storage). ② CPLD: Multiple GALs. ③ FPGA: Multiple CLBs

$$ \text{AND} \leq 2^n \rightarrow \text{OR} $$