E84 Home Work 8

1. Use Matlab to plot the frequency response functions (FRF) of the RC first-order low-pass (voltage across C is the output) and high-pass (voltage across R is the output) filters. Assume $C=1 \mu F$, $R=1 k\Omega$. Choose the range of frequency properly so that the corner frequency is around the middle region of the plots.
   • Plot the linear gain and phase as a function of $\omega$ of the two filters.
   • Make the Bode plots (without using the built-in function Bode) of both the gain (in log magnitude $20\log_{10}\vert H(\omega)\vert$ and phase (between $-\pi$ and $\pi$) of the two filters (log scale in frequency). The Matlab function semilogx can plot a function $f(x)$ with log scale in $x$.

   Solution: The first-order plots should be the same as those shown in the lecture notes shown here with the only difference that the cornor (cut-off) frequency is at $\omega_C=1/\tau=1/RC=1/10^{-3}=10^3\;rad/sec$.

2. Use Matlab to plot the frequency response functions (FRF) of the RLC second-order low-pass (voltage across C as the output), band-pass (voltage across R as the output), and high-pass (voltage across L as the output) filters. Assume $L=1 mH$, $C=1 \mu F$, $R=1 k\Omega$. Choose the range of frequency properly so that the natural frequency is around the middle region of the plots.
   • Plot the linear gain and phase as a function of $\omega$ of the three filters.
   • Make the Bode plots (without using the built-in function Bode) of both the gain and phase of the three filters (log scale in frequency).

   Solution:

3. In the circuit below, $R_1=20\Omega$, $R_2=10\Omega$, $C=318\mu F$, and the sinusoidal voltage source is $v_0(t)=14.1 \cos(314 t-45^{\circ})$. Find the complete voltage response $v(t)=v_C(t)$ across $C$ and $R_2$ after the switch closes at $t=0$.

   

   Solution:

   • First find phasor representation of the voltage source: $\dot{V}_0=10\angle{-45^{\circ}}$ and the impedance of the capacitor: $Z_C=1/j\omega C=-j/(314\times 318\times 10^{-6})=-j10$.
   • Find $v_\infty(t)$:
     
     $$\dot{V}=\dot{V}_0 \frac{R_2\vert\vert Z_C}{R_1+R_2\vert\vert ... ...ert\vert 10)}{20+(-j10 \vert\vert 10)} =2.77\angle{-79^\circ}$$
     
     
     
     
     $$v(t)=2.77\sqrt{2} \cos(314 t-79^\circ)=3.92 \cos(314 t-79^\circ)$$
     
     

   • Find $v(0)=v_C(0)=v_C(0+)=0$

   • Find $\tau=R_{eq}C=(R_1\vert\vert R_2) C=20 \vert\vert 10 \times 318\times 10^{-6}=2.12\times 10^{-3}$
   Now we get:
   

   $$v(t) = v_\infty(t)+[v(0)-v_\infty(0)]e^{-t/\tau} =2.77\sqrt{2} \cos(314 t-79^\circ)+[0-2.77\sqrt{2}cos(-79^\circ)] e^{-10^3 t/2.12}$$

   $$= 3.92 \cos(314 t-79^\circ)-0.75 e^{-10^3 t/2.12}$$

   
   

4. The RCL series circuit can be used as a filter when the voltage across all three components is the input and the voltage across any of the three components is treated as output. Find the resonant frequency $\omega_r$ (at which the output is maximized) in terms of the natural frequency $\omega_n=1/\sqrt{LC}$, when (1) voltage $v_C(t)$ across C is treated as output, and (2) voltage $v_L(t)$ across L is treated as output.

   The frequency response functions $H_C(\omega)$ and $H_L(\omega)$ may have a peak, i.e., $\vert H(\omega_r)\vert>\vert H(\omega)\vert$ for any $\omega\ne \omega_r$, when $\zeta$ is small enough, but it may not have such a peak if $\zeta$ is too large. Find the critical value $\zeta_c$ so that for any $\zeta<\zeta_c$ $\vert H_C(\omega)\vert$ and $H_L(\omega)\vert$ will have a peak at $\omega=\omega_r$, but such a peak no longer exist when $\zeta>\zeta_c$.

   Solution: See here

5. An RCL series circuit composed of $R=10\Omega$, $L=10 mH$ and $C=1 \mu F$ is connected to an input AC voltage $v_{in}(t)=cos\omega t$.
   • Find the quality factor $Q$ and resonant frequency $\omega_0$.
   • Assume the voltage $v_R(t)$ across $R$ is taken as the output voltage. Find the bandwidth of the circuit. Also, use any software (e.g., Matlab) to plot the ratio of the magnitudes between the output and input voltages $v_R/v_{in}$ as a function of frequency $\omega$.
   • Assume the voltage $v_L(t)$ across $L$ is taken as the output voltage. plot the ratio $v_L/v_{in}$ as a function of frequency $\omega$.
   • Assume the voltage $v_C(t)$ across $C$ is taken as the output voltage. plot the ratio $v_C/v_{in}$ as a function of frequency $\omega$.

   Solution:

   • 
     
     $$Q=\frac{1}{R}\sqrt{\frac{L}{C}}=\frac{1}{10}\sqrt{\frac{10^{-2}}{10^{-6}}} =10$$
     
     
     
     
     $$\omega_0=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{10^{-2}10^{-6}}} =\frac{1}{\sqrt{10^{-8}}}=10^4$$
     
     

   • 
     
     $$\triangle \omega=\frac{\omega_0}{Q}=\frac{10^4}{10}=1,000$$
     
     

   

6. A series circuit composed of a capacitor and an inductor is to be resonant at 800 kHz with voltage input. Specify the value of $C$ for the capacitor required for the given inductor with $L=40\mu H$ and an internal resistance $R_L=4.02\Omega$, and predict the bandwidth. Assume the capacitor is ideal, i.e., it introduces no resistance.

   Solution: As $\omega_0=1/\sqrt{LC}=2\pi 8\times 10^5$, and $L=4\times 10^{-5}H$, we can find $C$ to be
   

   $$C=\frac{1}{\omega_0^2 L}=\frac{1}{(2\pi 8\times 10^5)^2\times 4\times 10^{-5}} =0.99\;nF$$
   
   
   Next find the quality factor:
   
   $$Q=\frac{\omega_0L}{R}=\frac{2\pi 8\times 10^5\times 4\times 10^{-5}}{4.02}=50$$
   
   
   then the bandwidth is
   
   $$f_2-f_2=\frac{f_0}{Q}=\frac{800\times 10^3}{50}=16\;kHz$$
   
   

7. Design a series circuit to be resonant at 800 kHz with a bandwidth of 32 kHz. The inductor has $L=40\mu H$ and $R_L=4.02\Omega$. Find the capacitance $C$ needed for the desired resonant frequency. In order to satisfy the desired bandwidth, you may also need to include a resistor in the circuit.

   Solution: Based on the desired resonant frequency and bandwidth, the quality factor needs to be
   

   $$Q=\frac{f_0}{\Triangle f}=\frac{800\times 10^3}{32 \times 10^3}=25$$
   
   
   Since $Q>20$, the resonant frequency is approximately
   
   $$\omega_0=\frac{1}{\sqrt{LC}},\;\;\;\;\mbox{i.e.,}\;\;\;\; C=... ...frac{1}{(2\pi 8\times 10^5)^2\times 4\times 10^{-5}} =0.99 nF$$
   
   
   However, the quality factor of the parallel circuit is
   
   $$Q=\frac{\omega_0L}{R}=\frac{2\pi 8\times 10^5\times 4\times 10^{-5}}{4.02}=50$$
   
   
   twice the desired $Q=25$, we have to double the resistance $R=4.02$ to $R=8.04$ to reduce $Q$ by half.

8. The function of a loudspeaker crossover network is to channel frequencies higher than a given crossover frequency $f_c$ into the high-frequency speaker (``tweeter'') and frequencies below $f_c$ into the low-frequency speaker (``woofer''). One such circuit is shown below. Assume the resistances of the tweeter is $R_1=8\Omega$ and that of the woofer is $R_2=8\Omega$, the voltage amplifier can be modeled as an ideal voltage source, and the crossover frequency is $f_c=2000\; Hz$. Design the network in terms of $L$ and $C$ so that $f_c$ is the corner freqnency or half-power point of each of the two speaker circuits. Give the expression of the power $P_1(f)$ and $P_2(f)$ of the speakers as a function of frequency $f$ and crossover frequency $f_c$, and sketch them. Assume the RMS of the input voltage is 1V.

   

   Solution

   The RMS voltage across the tweeter is
   

   $$V_1=\vert\frac{R}{1/j2\pi f C+R}\vert =\frac{2\pi f CR}{\sqrt{1+(2\pi f CR)^2}}$$
   
   
   If $f=f_c=2000$ is at half-power point ( $V_1=V_{in}/\sqrt{2}$), the real and imaginary parts of the denominator should be equal and we get
   
   $$\frac{1}{2\pi f C}=R;\;\;\;\mbox{i.e.}\;\;\;\; C=\frac{1}{2\pi f_c R}=9.95 \;\mu F$$
   
   
   The RMS voltage across the woofer is
   
   $$V_2=\vert\frac{R}{j2\pi f L+R}\vert =\frac{R}{\sqrt{R^2+(2\pi f L)^2}}$$
   
   
   If $f=f_c=2000$ is at half-power point ( $V_2=V_{in}/\sqrt{2}$), the real and imaginary parts of the denominator should be equal and we get
   
   $$2\pi f L=R;\;\;\;\mbox{i.e.}\;\;\;\; L=\frac{R}{2\pi f_c}=0.637 \;mH$$
   
   
   The power plots:
   
   $$P_1(f)=\frac{V_1^2(f)}{R}=\frac{1}{8}(\frac{1}{1+(f_c/f)^2})$$
   
   
   
   
   $$P_2(f)=\frac{V_1^2(f)}{R}=\frac{1}{8}(\frac{1}{1+(f/f_c)^2})$$