# Final Exam: CE2009 Electronic Circuit<br>Answer Sheet
> 劉子雍．108502523
> 資訊工程學系三年級 A 班

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## 1.
### (1)
$V_R = \frac{R}{R + Z} V_S \Rightarrow V_R = \frac{R}{R + 1/j\omega C} V_S \Rightarrow |V_R| = \frac{2\pi fRC}{\sqrt{(2\pi fRC)^2 + 1}} |V_S| \\
\Rightarrow \left|\frac{V_R}{V_S}\right|(f) = \frac{2\pi fRC}{\sqrt{(2\pi fRC)^2 + 1}} = \frac{2\pi f \cdot 1 \cdot 0.16 \cdot 10^{-3}}{\sqrt{(2\pi f \cdot 1 \cdot 0.16 \cdot 10^{-3})^2 + 1}}$
==**ANS**==:
- $x$: frequency in $Hz$, $y(x)$: response
- 🔵 $\left|\frac{V_R}{V_S}\right|: y(x) = \frac{2\pi x \cdot 1 \cdot 0.16 \cdot 10^{-3}}{\sqrt{(2\pi x \cdot 1 \cdot 0.16 \cdot 10^{-3})^2 + 1}}$

| $x$ | $10$ | $100$ | $1000$ | $10000$ | $(Hz)$ |
|:------:|:------:|:------:|:------:|:-------:|:-----------------:|
| $y(x)$ | $0.01$ | $0.10$ | $0.71$ | $1.00$ | $\text{Response}$ |


### (2)
$f_{cut} = \frac{1}{2\pi RC} = \frac{1}{2\pi \cdot 1 \cdot 0.16 \cdot 10^{-3}} \approx 994.72\ (Hz)$
This is a high-pass filter.
==**ANS**==: $f_{cut} \approx 994.72\ (Hz).$ This is a high-pass filter.

### (3)
$\tau = RC = 1 \cdot 0.16 \cdot 10^{-3}\ (s) = 160\ (\mu s)$
$V_C: v_C(t) = V_F + (V_i - V_F)e^{-\frac{t}{\tau}}$
$V_R: v_R(t) = v_S(t) - v_C(t) = v_S(t) - (V_F + (V_i - V_F)e^{-\frac{t}{\tau}}) = \left\{\begin{array}{l}
5 - (5 + (0 - 5)e^{-\frac{t}{\tau}}) \\
0 - (0 + (5 - 0)e^{-\frac{t}{\tau}})
\end{array}\right. (V)$
==**ANS**==: $\tau = 160\ (\mu s)$
- $x$: time in $s$, $y(x)$: voltage in $V$
- 🔵 $V_R: y(x) = \left\{\begin{array}{l}
5 - (5 + (0 - 5)e^{-\frac{t + 0.001 \cdot floor(\frac{t}{0.001}\ )}{0.00016}}\quad ),\ v_S(t) = 5 \\
0 - (0 + (5 - 0)e^{-\frac{t + 0.001 \cdot floor(\frac{t}{0.001}\ )}{0.00016}}\quad ),\ v_S(t) = 0
\end{array}\right.$


## 2.
### (1)
$X_L = 2\pi fL = 2\pi \cdot 1 \cdot 100\ (\Omega) \approx 628.32\ (\Omega)$
$Z = \sqrt{R^2 + X_L^2} \approx \sqrt{100^2 + 628.32^2}\ (\Omega) \approx 636.23\ (\Omega)$
$\theta = \tan^{-1}{\left(\frac{X_L}{R}\right)} \approx \tan^{-1}{\left(\frac{628.32}{100}\right)} \approx 80.96^\circ \approx 1.413$
==**ANS**==: $Z \approx 636.23\ (\Omega),\ \theta \approx 80.96^\circ$

### (2)
$V_{in}: v_{in}(t\ (s)) = V_{p(in)} \cdot \sin{(2\pi ft)}\ (V) = 10 \cdot \sin{(2\pi \cdot 1000 \cdot t)}\ (V)$
$V_{out}: v_{out}(t\ (s)) = V_{p(in)} \cos{\theta} \cdot \sin{(2\pi ft - \theta)}\ (V) \approx 10\cos{(1.413)} \cdot \sin{(2\pi \cdot 1000 \cdot t - 1.413)}\ (V)$
==**ANS**==:
- $x$: time in $s$, $y(x)$: voltage in $V$
- 🔵 $V_{in}: y(x) = 10 \cdot \sin{(2\pi \cdot 1000 \cdot x)}$
- 🔴 $V_{out}: y(x) = 10\cos{(1.413)} \cdot \sin{(2\pi \cdot 1000 \cdot x - 1.413)}$
- 🔴$V_{out}$ lags 🔵$V_{in}$ by about $80.96^\circ$.


### (3)
$V_{out} = \frac{R}{R + Z}V_{in} \Rightarrow V_{out} = \frac{R}{R + j\omega L} V_{in} \Rightarrow |V_{out}| = \frac{1}{\sqrt{(2\pi fL/R)^2 + 1}} |V_{in}| \\
\Rightarrow \left|\frac{V_{out}}{V_{in}}\right|(f\ (kHz)) = \frac{1}{\sqrt{(2\pi fL/R)^2 + 1}} = \frac{1}{\sqrt{(2\pi f(0.1/100))^2 + 1}}$
==**ANS**==:
- $x$: frequency in $Hz$, $y(x)$: response
- 🔵 $\left|\frac{V_{out}}{V_{in}}\right|: y(x) = \frac{1}{\sqrt{(2\pi x(0.1/100))^{2} + 1}}$


### (4)
$f_{cut} = \frac{1}{2\pi L/R} = \frac{1}{2\pi (0.1/100)}\ (Hz) \approx 159.15\ (Hz)$
This is a low-pass filter.
==**ANS**==: $f_{cut} \approx 159.15\ (Hz).$ This is a low-pass filter.

## 3.
### (1)
==**ANS**==: $V_{in} = v_S,\ V_{out} = v_R,\ v_S = v_R + v_C + v_L,\ v_S = RC\frac{dv_C}{dt} + v_C + LC\frac{d^2v_C}{dt^2}$

### (2)
$X_L = 2\pi fL = 2\pi \cdot 1 \cdot 10\ (\Omega) \approx 62.83\ (\Omega)$
$X_C = \frac{1}{2\pi fC} = \frac{1}{2\pi \cdot 1 \cdot 1}\ (k\Omega) \approx 159.15\ (\Omega)$
$X_{tot} = |X_L - X_C| = |62.83 - 159.15|\ (\Omega) = 96.32\ (\Omega), (\text{conductive})$
$Z = \sqrt{R^2 + X_{tot}^2} \approx \sqrt{100^2 + 96.32^2}\ (\Omega) \approx 138.84\ (\Omega)$
$\theta = \tan^{-1}{\left(\frac{X_{tot}}{R}\right)} = \tan^{-1}{\left(\frac{96.32}{100}\right)} \approx 43.93^\circ \approx 0.767$
==**ANS**==: $Z \approx 138.84\ (\Omega),\ \theta \approx 43.93^\circ$

### (3)
$f_r = \frac{1}{2\pi \sqrt{LC}} = \frac{1}{2\pi \sqrt{10 \cdot 1 \cdot 10^{-9}}}\ (Hz) \approx 1591.55\ (Hz)$
$f_{cut} = \frac{\pm RC + \sqrt{R^2C^2 + 4LC}}{4\pi LC} = \frac{\pm 100 \cdot 1 \cdot 10^{-6} + \sqrt{(100 \cdot 1 \cdot 10^{-6})^2 + 4 \cdot 10 \cdot 1 \cdot 10^{-9}}}{4\pi \cdot 10 \cdot 1 \cdot 10^{-9}} \approx \left\{\begin{array}{l} 2575.18 \\ 983.63 \end{array}\right. (Hz)$
==**ANS**==: $f_r \approx 1591.55\ (Hz),\ f_{cut} \approx \left\{\begin{array}{l} 2575.18 \\ 983.63 \end{array}\right. (Hz)$

### (4)
$BW = \frac{R}{2\pi L} = \frac{100}{2\pi \cdot 0.01}\ (Hz) \approx 1591.55\ (Hz)$
This is a band-pass filter.
==**ANS**==: $BW \approx 1591.55\ (Hz).$ This is a band-pass filter.

### (5)
$\left|\frac{V_{out}}{V_{in}}\right|(f) = \left|\frac{R}{R + j(\omega L - 1/\omega C)}\right| = \frac{R}{\sqrt{R^2 + (\omega L - 1/\omega C)^2}} = \frac{100}{\sqrt{100^2 + (2\pi f \cdot 0.01 - (1/(2\pi f \cdot 0.000001)))^2}}$
==**ANS**==:
- $x$: frequency in $Hz$, $y(x)$: response
- 🔵 $\left|\frac{V_{out}}{V_{in}}\right|: y(x) = \frac{100}{\sqrt{100^2 + (2\pi x \cdot 0.01 - (1/(2\pi x \cdot 0.000001)))^2}}$


## 4.
### (1): The left circuit
Reversed bias. $I = 0$
$V_R = 10 - 5\ (V) = 5\ (V)$
==**ANS**==: $V_R = 5\ (V)$

### (2): The right circuit
Forward bias.
$V_F \approx 0.7\ (V)$
==**ANS**==: $V_F \approx 0.7\ (V)$

## 5.
### (1): The left circuit
$V_{in}: v_{in}(t) = V_{p(in)}\sin{(2\pi ft)} = 20 \cdot \sin{(2\pi ft)}\ (V)$
$V_{sec}: v_{sec}(t) = V_{p(sec)}\sin{(2\pi ft)} = \frac{20}{1} \cdot \sin{(2\pi ft)}\ (V) = 20 \cdot \sin{(2\pi ft)}\ (V)$
$V_{out}: v_{out}(t) = \left\{\begin{array}{l} v_{sec}(t) - V_B,\ v_{sec}(t) > V_B \\ 0,\ \text{otherwise} \end{array}\right. (V) \\
\qquad\qquad\quad = \left\{\begin{array}{l} 20 \cdot \sin{(2\pi ft)} - 0.7,\ 20 \cdot \sin{(2\pi ft)} > 0.7 \\ 0,\ \text{otherwise} \end{array}\right. (V)$
==**ANS**==:
- $x$: time in $s$, $y(x)$: voltage in $V$
- For $f = 1\ (Hz)$
- 🔵 $V_{out}: y(x) = \left\{\begin{array}{l} 20 \cdot \sin{(2\pi fx)} - 0.7,\ 20 \cdot \sin{(2\pi fx)} > 0.7 \\ 0,\ \text{otherwise} \end{array}\right.$


### (2): The right circuit
$V_{in}: v_{in}(t) = V_{p(in)}\sin{(2\pi ft)} = 20 \cdot \sin{(2\pi ft)}\ (V)$
$V_{sec}: v_{sec}(t) = V_{p(sec)}\sin{(2\pi ft)} = \frac{20}{1} \cdot \sin{(2\pi ft)}\ (V) = 20 \cdot \sin{(2\pi ft)}\ (V)$
$V_{out}: v_{out}(t) = \left\{\begin{array}{l} |v_{sec}(t)| - 2 \cdot V_B,\ |v_{sec}(t)| > 2 \cdot V_B \\ 0,\ \text{otherwise} \end{array}\right. (V) \\
\qquad\qquad\quad = \left\{\begin{array}{l} 20 \cdot |\sin{(2\pi ft)}| - 2 \cdot 0.7,\ 20 \cdot |\sin{(2\pi ft)}| > 2 \cdot 0.7 \\ 0,\ \text{otherwise} \end{array}\right. (V)$
==**ANS**==:
- $x$: time in $s$, $y(x)$: voltage in $V$
- For $f = 1\ (Hz)$
- 🔵 $V_{out}: y(x) = \left\{\begin{array}{l} 20 \cdot |\sin{(2\pi fx)}| - 2 \cdot 0.7,\ 20 \cdot |\sin{(2\pi fx)}| > 2 \cdot 0.7 \\ 0,\ \text{otherwise} \end{array}\right.$


## 6.
$V_B = \left(\frac{R_2}{R_1 + R_2}\right) V_{CC} = \left(\frac{2}{6 + 2}\right) \cdot 10\ (V) = 2.5\ (V)$
$V_E = V_B - V_{BE} = 2.5 - 0.7\ (V) = 1.8\ (V)$
$I_E = \frac{V_E}{R_E} = \frac{1.8}{1000}\ (A) = 1.8\ (mA)$
$I_C = \left(\frac{\beta_{DC}}{\beta_{DC} + 1}\right) I_E \approx \left(\frac{200}{201}\right) 1.8\ (mA) \approx 1.79\ (mA)$
$I_B = I_E - I_C \approx 1.8 - 1.79 = 0.01$
$V_C = V_{CC} - I_C R_C \approx 10 - 1.79 \cdot 1\ (V) = 8.21\ (V)$

==**ANS**==: $V_B = 2.5\ (V),\ V_E = 1.8\ (V),\ V_C \approx 8.21\ (V),\ I_B\approx 0.01\ (mA) ,\ I_E \approx 1.8\ (mA),\ I_C \approx 1.79\ (mA)$

## 7.
$V_{OUT1} = -\frac{R_3}{R_1}V_1 - \frac{R_3}{R_2} V_2$
$V_{OUT2} = (1 + \frac{R_5}{R_4})V_{OUT1} = (1 + \frac{R_5}{R_4})(-\frac{R_3}{R_1}V_1 - \frac{R_3}{R_2} V_2)$
==**ANS**==: $V_{OUT1} = -\frac{R_3}{R_1}V_1 - \frac{R_3}{R_2} V_2,\ V_{OUT2} = (1 + \frac{R_5}{R_4})(-\frac{R_3}{R_1}V_1 - \frac{R_3}{R_2} V_2)$