---
# System prepended metadata

title: EC Fin Past (6/13)

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# EC Fin Past (6/13)

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[TOC]

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## 2021 Mid



## 2020 Mid



## 2019 Mid



## 2018 Mid



## 2020
### 1.
#### (1) ==TODO==
$L = 200 \mid\mid 500\ (mH) = (1/200 + 1/800)^{-1}\ (mH) = 160\ (mH)$
$\tau = \frac{L}{R} = \frac{160}{1} = 160\ (\mu s)$
==**ANS**==: $\tau = 160\ (\mu s)$

#### (2)
$X_L = 2\pi fL = 2\pi \cdot 1 \cdot 160\ (\Omega) \approx 1005.31\ (\Omega)$
$\theta = \tan^{-1}{\left(\frac{X_L}{R}\right)} \approx \tan^{-1}{\left(\frac{1005.31}{1000}\right)} \approx 45.15^\circ \approx 0.788$
==**ANS**==: $X_L \approx1005.31\ (\Omega),\ \theta \approx 45.15^\circ$

#### (3)
$v_{in}(t\ (s)) = V_{p(in)} \cdot \sin{(2\pi ft)}\ (V) = 10 \cdot \sin{(2\pi \cdot 1000 \cdot t)}\ (V)$
$v_{out}(t\ (s)) = V_{p(in)}\sin{\theta} \cdot \sin{(2\pi ft + (\pi/2 - \theta))}\ (V) \\
\qquad\qquad \approx 10 \sin{(0.788)} \cdot \sin{(2\pi \cdot 1000 \cdot t + (\pi/2 - 0.788))}\ (V) \\
\qquad\qquad \approx 7.089 \cdot \sin{(2\pi \cdot 1000 \cdot t + 0.783)}\ (V)$
==**ANS**==:
- $x$: time in $s$, $y(x)$: voltage in $V$
- 🔵 $V_{in}: y(x) = 10 \cdot \sin{(2\pi \cdot 1000 \cdot x)}$
- 🔴 $V_{out}: y(x) = 7.089 \cdot \sin{(2\pi \cdot 1000 \cdot x + 0.783)}$
- 🔵$V_{in}$ lags 🔴$V_{out}$ by about $90^\circ - 45.15^\circ = 44.85^\circ$.

![](https://i.imgur.com/7Fz1DSr.png)

#### (4)
$f_C = \frac{1}{2\pi L/R} = \frac{1}{2\pi (0.16/1000)}\ (Hz) \approx 994.72\ (Hz)$
==**ANS**==: $f_c \approx 994.72\ (Hz)$

#### (5)
$V_{out} = \frac{Z}{R + Z}V_{in} \Rightarrow V_{out} = \frac{j\omega L}{R + j\omega L} V_{in} \Rightarrow |V_{out}| = \frac{2\pi fL/R}{\sqrt{(2\pi fL/R)^2 + 1}} |V_{in}| \\
\Rightarrow \left|\frac{V_{out}}{V_{in}}\right|(f) = \frac{2\pi fL/R}{\sqrt{(2\pi fL/R)^2 + 1}} = \frac{2\pi f (0.16/1000)}{\sqrt{(2\pi f (0.16/1000))^2 + 1}}$
==**ANS**==:
- $x$: frequency in $kHz$, $y(x)$: response
- 🔵 $\left|\frac{V_{out}}{V_{in}}\right|: y(x) = \frac{2\pi x (0.16/1000)}{\sqrt{(2\pi x (0.16/1000))^2 + 1}}$

![](https://i.imgur.com/Qvtvgfm.png)

#### (6)
==**ANS**==: High-pass filter.


### 2. ==FUCK==
<!-- #### (1)
==TOOD==

#### (2)
==TODO==
$X_L = 2\pi fL = 2\pi \cdot 2 \cdot 50\ (\Omega) \approx 628.32\ (\Omega)$
$X_C = \frac{1}{2\pi fC} = \frac{1}{2\pi \cdot 2 \cdot 20}\ (k\Omega) \approx 3.97\ (\Omega)$
$X_{tot} = |X_L - X_C| \approx |628.32 - 3.97| = 624.35\ (\Omega),\ (\text{inductive})$
$Z = \sqrt{R^2 + X_{tot}^2} \approx \sqrt{1000^2 + 624.35^2}\ (\Omega) \approx 1178.90\ (\Omega)$
$\theta = \tan^{-1}{\left(\frac{X_{tot}}{R}\right)} \approx \tan^{-1}{\left(\frac{624.35}{1000}\right)} \approx 31.98^\circ$
==**ANS**==: $Z \approx 1178.90\ (\Omega),\ \theta \approx 31.98^\circ$

==TODO==
$X_L = 2\pi fL = 2\pi \cdot 2 \cdot 50\ (\Omega) \approx 628.32\ (\Omega)$
$X_C = \frac{1}{2\pi fC} = \frac{1}{2\pi \cdot 2 \cdot 20}\ (k\Omega) \approx 3.97\ (\Omega)$
$B_L = \frac{1}{X_L} = \frac{1}{628.32}\ (1/\Omega)$
$B_C = \frac{1}{X_C} = \frac{1}{3.97}\ (1/\Omega)$
$B_{tot} = |B_L - B_C| = |\frac{1}{628.32} - \frac{1}{3.97}|\ (1/\Omega) \approx 0.25\ (1/\Omega),\ (\text{conductive})$
$X_{tot} = \frac{1}{B_{tot}} \approx \frac{1}{0.25}\ (\Omega) = 4.00\ (\Omega)$
$Z = \sqrt{R^2 + X_{tot}^2} \approx \sqrt{1000^2 + 4.00^2}\ (\Omega) \approx 1000.01\ (\Omega)$
$\theta = \tan^{-1}{\left(\frac{X_{tot}}{R}\right)} \approx \tan^{-1}{\left(\frac{4.00}{1000}\right)} \approx 0.23^\circ$
==**ANS**==: $Z \approx 1000.01\ (\Omega),\ \theta \approx 0.23^\circ$

#### (3)
$f_r = \frac{1}{2\pi \sqrt{LC}} = \frac{1}{2\pi \sqrt{0.05 \cdot 0.00002}}\ (Hz) \approx 159.15\ (Hz)$
==**ANS**==: $159.15\ (Hz)$

#### (4)
$f_{cut} = \frac{\pm RC + \sqrt{R^2C^2 + 4LC}}{4\pi LC} = \frac{\pm 1 \cdot 20 \cdot 10^{-3} + \sqrt{(1 \cdot 20 \cdot 10^{-3})^2 + 4 \cdot 50 \cdot 20 \cdot 10^{-9}}}{4\pi \cdot 50 \cdot 20 \cdot 10^{-9}}\ (Hz) \approx \left\{\begin{array}{l} 3191.04 \\ 7.94 \end{array}\right. (Hz)$
$BW \approx 3191.04 - 7.94\ (Hz) = 3183.1\ (Hz)$
==**ANS**==: $f_{cut} \approx \left\{\begin{array}{l} 3191.04 \\ 7.94 \end{array}\right. (Hz),\ BW \approx 3183.1\ (Hz)$

#### (5)
$V_{out} = \frac{-j(1/(\omega C - 1/\omega L))}{R - j(1/(\omega C - 1/\omega L))} V_{in} = \frac{j\omega L/(1 - \omega^2 LC)}{R + j\omega L/(1 - \omega^2 LC)} V_{in} \\
\Rightarrow \left|\frac{V_{out}}{V_{in}}\right| = \left|\frac{-j(1/(\omega L - 1/\omega C))}{R - j(1/(\omega L - 1/\omega C))}\right| = \frac{\omega L / (1 - \omega^2LC)}{\sqrt{R^2 + (\omega L / (1 - \omega^2LC))^2}} = \frac{2\pi fL / (1 - (2\pi f)^2LC)}{\sqrt{R^2 + (2\pi fL / (1 - (2\pi f)^2LC))^2}}$

#### (6)

 -->


### 3.
#### (1)
Reversed bias. $I = 0$
$V_R = 10 - 4\ (V) = 6\ (V)$
==**ANS**==: $V_R = 6\ (V)$

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


### 4.
#### (1)
$V_{in}: v_{in}(t) = V_{p(in)}\sin{(2\pi ft)} = 20 \cdot \sin{(2\pi ft)}\ (V)$
$V_{sec}: v_{sec}(t) = V_{p(sec)}\sin{(2\pi ft)} = \frac{20}{1} \cdot \sin{(2\pi ft)}\ (V) = 20 \cdot \sin{(2\pi ft)}\ (V)$
$V_{out}: v_{out}(t) = \left\{\begin{array}{l} v_{sec}(t) - V_B,\ v_{sec}(t) > V_B \\ 0,\ \text{otherwise} \end{array}\right. (V) \\
\qquad\qquad\quad = \left\{\begin{array}{l} 20 \cdot \sin{(2\pi ft)} - 0.7,\ 20 \cdot \sin{(2\pi ft)} > 0.7 \\ 0,\ \text{otherwise} \end{array}\right. (V)$
==**ANS**==:
- $x$: time in $s$, $y(x)$: voltage in $V$
- For $f = 1\ (Hz)$
- 🔵 $V_{out}: y(x) = \left\{\begin{array}{l} 20 \cdot \sin{(2\pi fx)} - 0.7,\ 20 \cdot \sin{(2\pi fx)} > 0.7 \\ 0,\ \text{otherwise} \end{array}\right.$

![](https://i.imgur.com/C9OVlVu.png)

#### (2)
$V_{in}: v_{in}(t) = V_{p(in)}\sin{(2\pi ft)} = 20 \cdot \sin{(2\pi ft)}\ (V)$
$V_{sec}: v_{sec}(t) = V_{p(sec)}\sin{(2\pi ft)} = \frac{20}{1} \cdot \sin{(2\pi ft)}\ (V) = 20 \cdot \sin{(2\pi ft)}\ (V)$
$V_{p(sec.h)} = \frac{V_{p(sec)}}{2} = \frac{V_{p(in)}}{2} = 10\ (V)$
$V_{out}: v_{out}(t) = \left\{\begin{array}{l} V_{p(sec.h)}|\sin{(2\pi ft)}| - V_B,\ V_{p(sec.h)}|\sin{(2\pi ft)}| > V_B \\ 0,\ \text{otherwise} \end{array}\right. (V) \\
\qquad\qquad\quad = \left\{\begin{array}{l} 10 \cdot |\sin{(2\pi ft)}| - 0.7,\ 10 \cdot |\sin{(2\pi ft)}| > 0.7 \\ 0,\ \text{otherwise} \end{array}\right. (V)$
==**ANS**==:
- $x$: time in $s$, $y(x)$: voltage in $V$
- For $f = 1\ (Hz)$
- 🔵 $V_{out}: y(x) = \left\{\begin{array}{l} 10 \cdot |\sin{(2\pi fx)}| - 0.7,\ 10 \cdot |\sin{(2\pi fx)}| > 0.7 \\ 0,\ \text{otherwise} \end{array}\right.$

![](https://i.imgur.com/CbWYw5A.png)

#### (3)
$V_{in}: v_{in}(t) = V_{p(in)}\sin{(2\pi ft)} = 20 \cdot \sin{(2\pi ft)}\ (V)$
$V_{sec}: v_{sec}(t) = V_{p(sec)}\sin{(2\pi ft)} = \frac{20}{1} \cdot \sin{(2\pi ft)}\ (V) = 20 \cdot \sin{(2\pi ft)}\ (V)$
$V_{out}: v_{out}(t) = \left\{\begin{array}{l} |v_{sec}(t)| - 2 \cdot V_B,\ |v_{sec}(t)| > 2 \cdot V_B \\ 0,\ \text{otherwise} \end{array}\right. (V) \\
\qquad\qquad\quad = \left\{\begin{array}{l} 20 \cdot |\sin{(2\pi ft)}| - 2 \cdot 0.7,\ 20 \cdot |\sin{(2\pi ft)}| > 2 \cdot 0.7 \\ 0,\ \text{otherwise} \end{array}\right. (V)$
==**ANS**==:
- $x$: time in $s$, $y(x)$: voltage in $V$
- For $f = 1\ (Hz)$
- 🔵 $V_{out}: y(x) = \left\{\begin{array}{l} 20 \cdot |\sin{(2\pi fx)}| - 2 \cdot 0.7,\ 20 \cdot |\sin{(2\pi fx)}| > 2 \cdot 0.7 \\ 0,\ \text{otherwise} \end{array}\right.$

![](https://i.imgur.com/sQXBv3c.png)


### 5.
$V_B = \left(\frac{R_2}{R_1 + R_2}\right) V_{CC} = \left(\frac{2}{6 + 2}\right) \cdot 12\ (V) = 3\ (V)$
$V_E = V_B - V_{BE} = 3 - 0.7\ (V) = 2.3\ (V)$
$I_E = \frac{V_E}{R_E} = \frac{2.3}{1000}\ (A) = 2.3\ (mA)$
$I_C = \left(\frac{\beta_{DC}}{\beta_{DC} + 1}\right) I_E \approx \left(\frac{200}{201}\right) 2.3\ (mA) \approx 2.29\ (mA)$
$V_C = V_{CC} - I_C R_C \approx 12 - 2.29 \cdot 1\ (V) \approx 9.71\ (V)$
==**ANS**==: $V_B = 3\ (V),\ V_E = 2.3\ (V),\ I_E \approx 2.3\ (mA),\ I_C \approx 2.29\ (mA),\ V_C \approx 9.71\ (V)$


### 6.
$V_{OUT1} = (1 + \frac{R_2}{R_1})V_{1}$
$V_{OUT2} = -\frac{R_5}{R_3} V_{OUT1} - \frac{R_5}{R_4} V_2 = -\frac{R_5}{R_3} ((1 + \frac{R_2}{R_1})V_{1}) - \frac{R_5}{R_4} V_2$
==**ANS**==: $V_{OUT1} = (1 + \frac{R_2}{R_1})V_{1},\ V_{OUT2} = -\frac{R_5}{R_3} ((1 + \frac{R_2}{R_1})V_{1}) - \frac{R_5}{R_4} V_2$

## 2019
### 1.
#### (1)
==TODO==
$L = 200 + 100\ (mH) = 300\ (mH)$
$\tau = \frac{L}{R} = \frac{300}{1000}\ (ms) = 300\ (\mu s)$
==**ANS**==: $\tau = 200\ (\mu s)$

#### (2)
$X_L = 2\pi fL = 2\pi \cdot 2 \cdot 300\ (\Omega) \approx 3.77\ (k\Omega)$
$\theta = \tan^{-1}{\left(\frac{X_L}{R}\right)} \approx \tan^{-1}{\left(\frac{3.77}{1}\right)} \approx 75.14^\circ \approx 1.312$
==**ANS**==: $X_L \approx 3.77\ (k\Omega),\ \theta \approx 75.14^\circ$

### (3)
==TODO==


### 2.
#### (1)
==TODO==

#### (2)
$X_L = 2\pi fL = 2\pi \cdot 2 \cdot 5\ (\Omega) \approx 62.83\ (\Omega)$
$X_C = \frac{1}{2\pi fC} = \frac{1}{2\pi \cdot 2 \cdot 2}\ (k\Omega) \approx 39.79\ (\Omega)$
$X_{tot} = |X_L - X_C| \approx |62.83 - 39.79|\ (\Omega) = 23.04\ (\Omega),\ (\text{inductive})$
$Z = \sqrt{R^2 + X_{tot}^2} \approx \sqrt{100^2 + 23.04^2}\ (\Omega) \approx 102.62\ (\Omega)$
$\theta = \tan^{-1}{\left(\frac{X_{tot}}{R}\right)} \approx \tan^{-1}{\left(\frac{23.04}{100}\right)} \approx 12.79^\circ \approx 0.226$
==**ANS**==: $Z \approx 102.62\ (\Omega),\ \theta \approx 12.79^\circ$

#### (3)
$f_r = \frac{1}{2\pi \sqrt{LC}} = \frac{1}{2\pi \sqrt{5 \cdot 2 \cdot 10^{-9}}}\ (Hz) \approx 1591.55\ (Hz)$
==**ANS**==: $f_r \approx 1591.55\ (Hz)$

#### (4)
$f_{cut} = \frac{\pm RC + \sqrt{R^2C^2 + 4LC}}{4\pi LC} = \frac{\pm 100 \cdot 2 \cdot 10^{-6} + \sqrt{(100 \cdot 2 \cdot 10^{-6})^2 + 4 \cdot 5 \cdot 2 \cdot 10^{-9}}}{4\pi \cdot 5 \cdot 2 \cdot 10^{-9}} \approx \left\{\begin{array}{l} 3842.34 \\ 659.24 \end{array}\right. (Hz)$
$BW = \frac{R}{2\pi L} = \frac{100}{2\pi \cdot 0.005}\ (Hz) \approx 3183.10\ (Hz)$
==**ANS**==: $f_{cut} \approx \left\{\begin{array}{l} 3842.34 \\ 659.24 \end{array}\right. (Hz),\ BW \approx 3183.10\ (Hz)$

#### (5)
==TODO==

#### (6)
==**ANS**==: Band-stop filter.


## 2018

