# Electronic Circuits Homework 4 > 劉子雍．108502523 > 資訊工程學系三年級 A 班 ## 1. ### (a) $L_T = L_1 \mid\mid L_2 + L_3 \mid\mid L_4 = (1/100 + 1/50)^{-1} + (1/60 + 1/40)^{-1}\ (mH) \approx 57.33\ (mH)$ ==**ANS**==: $57.33\ (mH)$ ### (b) $L_T = (L_1 + L_2) \mid\mid (L_3 + L_4) = 12 \mid\mid 6\ (mH) = (1/12 + 1/6)^{-1}\ (mH) = 4\ (mH)$ ==**ANS**==: $4\ (mH)$ ### \(c\) $L_T = L_1 + (L_3 + L_4) \mid\mid (L_2 + L_5) = 4 + 2 \mid\mid 4\ (mH) = 4 + (1/2 + 1/4)^{-1}\ (mH) \approx 5.33\ (mH)$ ==**ANS**==: $5.33\ (mH)$ ## 2. ### (a) $\tau = \frac{L}{R} = \frac{100}{100}\ (\mu s) = 1\ (\mu s)$ ==**ANS**==: $1\ (\mu s)$ ### (b) $\tau = \frac{L}{R} = \frac{10}{4.7}\ (\mu s) \approx 2.13\ (\mu s)$ ==**ANS**==: $2.13\ (\mu s)$ ### \(c\) $\tau = \frac{L}{R} = \frac{3}{1.5}\ (\mu s) = 2\ (\mu s)$ ==**ANS**==: $2\ (\mu s)$ ## 3. <!-- TODO --> ### (a) $\tau = \frac{L}{R} = \frac{10}{1}\ (\mu s) = 10\ (\mu s)$ $i(t) = I_F + (I_i - I_F)e^{-\frac{t}{\tau}}$ $v_R(t) = R \cdot i(t) = R(I_F + (I_i - I_F)e^{-\frac{t}{\tau}}) = V_{R_F} + (V_{R_i} - V_{R_F})e^{-\frac{t}{\tau}} \\ \qquad = V_S + (0 - V_S)e^{-\frac{t}{\tau}} = V_S(1 - e^{-\frac{t}{\tau}})$ $v_L(t) = V_S - v_R(t) = V_S - R \cdot i(t) = V_S - V_S(1 - e^{-\frac{t}{\tau}}) = V_S e^{-\frac{t}{\tau}} \\ \qquad = 15 e^{-\frac{t}{10\ (\mu s)}}\ (V)$ $v_L(10\ (\mu s)) = 15 e^{-1}\ (V) \approx 5.52\ (V)$ ==**ANS**==: $5.52\ (V)$ ### (b) $v_L(20\ (\mu s)) = 15 e^{-2}\ (V) \approx 2.03\ (V)$ ==**ANS**==: $2.03\ (V)$ ### \(c\) $v_L(30\ (\mu s)) = 15 e^{-3}\ (V) \approx 0.75\ (V)$ ==**ANS**==: $0.75\ (V)$ ### (d) $v_L(40\ (\mu s)) = 15 e^{-4}\ (V) \approx 0.27\ (V)$ ==**ANS**==: $0.27\ (V)$ ### (e) $v_L(50\ (\mu s)) = 15 e^{-5}\ (V) \approx 0.10\ (V)$ ==**ANS**==: $0.10\ (V)$ ## 4. <!-- TODO --> ### (a) $R_T = 47 + 10\ (\Omega) = 57\ (\Omega)$ $L_T = 50 + 100\ (mH) = 150\ (mH)$ $X_L = 2\pi fL_T = 2\pi \cdot 100 \cdot 150\ (m\Omega) \approx 94.25\ (\Omega)$ $Z = \sqrt{R_T^2 + X_L^2} \approx \sqrt{57^2 + 94.25^2}\ (\Omega) \approx 110.14\ (\Omega)$ $\theta = \tan^{-1}{\left(\frac{X_L}{R_T}\right)} \approx \tan^{-1}{\left(\frac{94.25}{57}\right)} \approx 58.84^\circ$ ==**ANS**==: $Z \approx 110.14\ (\Omega),\ \theta \approx 58.84^\circ$ ### (b) $L_T = 5 \mid\mid 8\ (mH) = (1/5 + 1/8)^{-1} \approx 3.08\ (mH)$ $X_L = 2\pi fL_T = 2\pi \cdot 20 \cdot 3.08\ (\Omega) \approx 387.04\ (\Omega)$ $Z = \sqrt{R^2 + X_L^2} \approx \sqrt{470^2 + 387.04^2}\ (\Omega) \approx 608.85\ (\Omega)$ $\theta = \tan^{-1}{\left(\frac{X_L}{R}\right)} \approx \tan^{-1}{\left(\frac{387.04}{470}\right)} \approx 39.47^\circ$ ==**ANS**==: $Z \approx 608.85\ (\Omega),\ \theta \approx 39.47^\circ$ ## 5. <!-- TODO --> $X_L = 2\pi fL = 2\pi \cdot 60 \cdot 0.1\ (\Omega) \approx 37.70\ (\Omega)$ $\theta = \tan^{-1}{\left(\frac{X_L}{R}\right)} \approx \tan^{-1}{\left(\frac{37.70}{47}\right)} \approx 38.73^\circ \approx 0.676$ $v_S(t\ (s)) = V_S \cdot \sin{(2\pi ft)}\ (V) = \sin{(2\pi \cdot 60 \cdot t)}\ (V)$ $v_R(t\ (s)) = V_S \cos{\theta} \cdot \sin{(2\pi ft - \theta)}\ (V) \approx 0.780 \cdot \sin{(2\pi \cdot 60 \cdot t - 0.676)}\ (V)$ $v_L(t\ (s)) = V_S \sin{\theta} \cdot \sin{(2\pi ft + (\pi/2 - \theta))}\ (V) \approx 0.626 \cdot \sin{(2\pi \cdot 60 \cdot t + 0.895)}\ (V)$ ==**ANS**==: - $x$: time in $s$, $y(x)$: voltage in $V$ - 🔵 $V_S: y(x) = \sin{(2\pi \cdot 60 \cdot x)}$, Blue Curve - 🟢 $V_R: y(x) = 0.780 \cdot \sin{(2\pi \cdot 60 \cdot x - 0.676)}$, Green Curve - 🔴 $V_L: y(x) = 0.626 \cdot \sin{(2\pi \cdot 60 \cdot x + 0.895)}$, Red Curve - 🟢$V_R$ lags 🔵$V_S$ by about $38.73^\circ$. 🔵$V_S$ lags 🔴$V_L$ by about $90^\circ - 38.73^\circ = 51.27^\circ$. 🟢$V_R$ lags 🔴$V_L$ by about $90^\circ$. ![](https://i.imgur.com/lornCss.png) ## 6. <!-- TODO --> $V_{out} = \frac{R}{R + Z}V_{in} \Rightarrow V_{out} = \frac{R}{R + j\omega L} V_{in} \Rightarrow |V_{out}| = \frac{1}{\sqrt{(2\pi fL/R)^2 + 1}} |V_{in}| \\ \Rightarrow |V_{out}|(f\ (kHz)) = \frac{1}{\sqrt{(2\pi f \cdot 10 / 39)^2 + 1}} |V_{in}|\ (V)$ ==**ANS**==: - $x$: frequency in $kHz$, $y(x)$: voltage in $V$ - 🔵 $|V_{out}|: y(x) = \frac{1}{\sqrt{(2\pi x \cdot 10 / 39)^2 + 1}}$, Blue Curve | $x$ | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ | $(kHz)$ | |:------:|:------:|:------:|:------:|:------:|:------:|:------:|:-------:| | $y(x)$ | $1.00$ | $0.53$ | $0.30$ | $0.20$ | $0.15$ | $0.12$ | $(V)$ | ![](https://i.imgur.com/9PfX8j0.png) ## 7. <!-- TODO --> $X_C = \frac{1}{\omega C} = \frac{1}{2\pi fC} = \frac{1}{2\pi \cdot 5 \cdot 0.047}\ (k\Omega) \approx 677.25\ (\Omega)$ $X_L = \omega L = 2\pi fL = 2\pi \cdot 5 \cdot 5\ (\Omega) \approx 157.08\ (\Omega)$ $X_{tot} = |X_L - X_C| = 520.17\ (\Omega)\ (\text{capacitive})$ $Z = \sqrt{R^2 + X_{tot}^2} \approx \sqrt{10^2 + 520.17^2} \approx 520.27\ (\Omega)$ $\theta = \tan^{-1}{\left(\frac{X_{tot}}{R}\right)} \approx \tan^{-1}{\left(\frac{520.17}{10}\right)} \approx 88.9^\circ$ ==**ANS**==: $X_{tot} \approx 520.17\ (\Omega)\ (\text{capacitive}),\ Z \approx 520.27\ (\Omega),\ \theta \approx 88.9^\circ$ ## 8. <!-- TODO --> $f_r = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{82 \cdot 1.5 \cdot 10^{-15}}}\ (Hz) \approx 453.80\ (kHz)$ $f_{cut} = \frac{\pm RC + \sqrt{R^2C^2 + 4LC}}{4\pi LC} = \frac{\pm 39 \cdot 1.5 \cdot 10^{-9} + \sqrt{(39 \cdot 1.5 \cdot 10^{-9})^2 + 4 \cdot 82 \cdot 1.5 \cdot 10^{-15}}}{4\pi \cdot 82 \cdot 1.5 \cdot 10^{-15}}\ (Hz) \approx \left\{\begin{array}{l} 493.23 \\ 417.53 \end{array}\right. (kHz)$ ==**ANS**==: $f_r \approx 453.80\ (kHz),\ f_{cut} \approx \left\{\begin{array}{l} 493.23 \\ 417.53 \end{array}\right. (kHz)$ ## 9. ### (a) $f_r = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{100 \cdot 2.2 \cdot 10^{-15}}}\ (Hz) \approx 339.32\ (kHz)$ $BW = \frac{R}{2\pi L} = \frac{150}{2\pi \cdot 100}\ (MHz) \approx 238.73\ (kHz)$ ==**ANS**==: $f_r \approx 339.32\ (kHz),\ BW \approx 238.73\ (kHz)$ ### (b) $f_r = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{5 \cdot 0.047 \cdot 10^{-9}}}\ (Hz) \approx 10.38\ (kHz)$ $BW = \frac{R}{2\pi L} = \frac{82}{2\pi \cdot 5}\ (kHz) \approx 2.61\ (kHz)$ ==**ANS**==: $f_r \approx 10.38\ (kHz),\ BW \approx 2.61\ (kHz)$ ## 10. <!-- TODO --> ### (a) $n = \frac{1}{2} = 0.5$ $V_{sec} = nV_{pri} = 0.5 \cdot 30\ (V) = 15\ (V)$ ==**ANS**==: $V_{sec} = 15\ (V)$ ### (b) $I_{sec} = \frac{V_{sec}}{R} = \frac{15}{300}\ (A) = 50\ (mA)$ ==**ANS**==: $I_{sec} = 50\ (mA)$ ### \(c\) $I_{pri} = nI_{sec} = 0.5 \cdot 50\ (mA) = 25\ (mA)$ ==**ANS**==: $I_{pri} = 25\ (mA)$ ### (d) $P_{pri} = P_{sec} = V_{sec}I_{sec} = 15 \cdot 50\ (mW) = 750\ (mW)$ ==**ANS**==: $P = 750\ (mW)$ ## 11. <!-- TODO --> $R_{pri} = R_{sec} = 16\ (\Omega)$ $R = R_{pri} + R_{sec} = 32\ (\Omega)$ $P = \frac{V^2}{R} = \frac{25^2}{32}\ (W) \approx 19.53\ (W)$ $P_{sec} = \frac{P}{2} \approx \frac{19.53}{2}\ (W) \approx 9.76\ (W)$ ==**ANS**==: $9.76\ (W)$ ## 12. $\tau = \frac{L}{R} = \frac{10}{10}\ (\mu s) = 1\ (\mu s)$ $i(t) = I_F + (I_i - I_F)e^{-\frac{t}{\tau}}$ $v_R(t) = R \cdot i(t) = R(I_F + (I_i - I_F)e^{-\frac{t}{\tau}}) = V_{R_F} + (V_{R_i} - V_{R_F})e^{-\frac{t}{\tau}}$ $v_1(t)= 8 + (0 - 8)e^{-\frac{t}{\tau}}\ (V) = 8(1 - e^{-\frac{t}{\tau}})\ (V)$ $v_1(\tau) = 8(1 - e^{-\frac{\tau}{\tau}})\ (V) = 8(1 - e^{-1})\ (V) \approx 5.057\ (V)$ $v_2(t) = 0 + (v_1(\tau) - 0)e^{-\frac{t - \tau}{\tau}}\ (V) = v_1(\tau)e^{-\frac{t - \tau}{\tau}}\ (V) \approx 5.057e^{-\frac{t - \tau}{\tau}}\ (V)$ ==**ANS**==: - $x$: time in $\mu s$, $y(x)$: voltage in $V$ - $V_{out} = \left\{\begin{array}{l} v_1,\ 0 \le x \le 1 \\ v_2,\ x > 1 \end{array}\right.$ - 🔵 $v_1: y(x) = 8(1 - e^{-x}),\ \forall x \in [0,1]$, Blue Curve - 🟢 $v_2: y(x) = 5.057e^{-(x - 1)},\ \forall x > 1$, Green Curve ![](https://i.imgur.com/1Hnc40F.png) ## 13. <!-- TODO --> ### (a) $\tau = \frac{L}{R} = \frac{100}{2.2}\ (ns) \approx 45.45\ (ns)$ ==**ANS**==: $45.45\ (ns)$ ### (b) $i(t) = I_F + (I_i - I_F)e^{-\frac{t}{\tau}}$ $v_R(t) = R \cdot i(t) = R(I_F + (I_i - I_F)e^{-\frac{t}{\tau}}) = V_{R_F} + (V_{R_i} - V_{R_F})e^{-\frac{t}{\tau}}$ $v_L(t) = V_S - v_R(t) = V_S - R \cdot i(t) = V_S - (V_{R_F} + (V_{R_i} - V_{R_F})e^{-\frac{t}{\tau}})$ $v_1(t) = 10 - (10 + (0 - 10)e^{-\frac{t}{\tau}})\ (V) = 10 - (10(1-e^{-\frac{t}{\tau}}))\ (V) = 10e^{-\frac{t}{\tau}}\ (V)$ $v_R(100\ (ns)) = 10(1 - e^{-\frac{100}{45.45}})\ (V) \approx 8.89\ (V)$ $v_2(t) = 0 - (0 + (v_R(100\ (ns)) - 0)e^{-\frac{t - 100\ (ns)}{\tau}})\ (V) = -v_R(100\ (ns))e^{-\frac{t - 100\ (ns)}{\tau}}\ (V)\\ \qquad \approx -8.89e^{-\frac{t - 100\ (ns)}{\tau}}\ (V)$ ==**ANS**==: - $x$: time in $ns$, $y(x)$: voltage in $V$ - $V_{out} = \left\{\begin{array}{l} v_1,\ 0 \le x \le 100 \\ v_2,\ x > 100 \end{array}\right.$ - 🔵 $v_1: y(x) = 10e^{-\frac{x}{45.45}},\ \forall x \in [0,100]$, Blue Curve - 🟢 $v_2: y(x) = -8.89e^{-\frac{x - 100}{45.45}},\ \forall x > 100$, Green Curve - 🔴 $x(y) = 100$, Red Dotted Line, used for annotating a specific time point: $100\ (ns)$ ![](https://i.imgur.com/m8n0inu.png)