Test Problems (don't click until you start taking the test)

1. (33 points) A crystal oscillator is an electronic oscillator circuit that uses the mechanical resonance of a vibrating crystal of piezoelectric material to create an electrical signal with a precise frequency. The crystal can be modeled by the RLC circuit shown in the figure. We assume $R$ can be approximated to be zero.
   • Find the parallel resonant frequency $\omega_p$ at which the impedance of the model is maximized.
   • Find the series resonant frequency $\omega_s$ at which the impedance of the model is minimized.

   

   Solution The total admittance is
   

   $$Y(\omega) = \frac{1}{j\omega L+1/j\omega C}+j\omega C_0 =\frac{j\omega C}{1-\omega^2LC}+j\omega C_0 =\frac{j\omega C+j\omega C_0(1-\omega^2LC)}{1-\omega^2LC}$$

   
   
   • The magnitude of $Y(\omega)$ is minimized to zero if
     
     $$j\omega C+j\omega C_0(1-\omega^2LC)=0$$
     
     
     i.e.,
     
     $$\omega^2=\frac{C+C_0}{L(CC_0)},\;\;\;\;\;\mbox{or} \;\;\;\;\omega=\frac{1}{\sqrt{LC_p}}$$
     
     
     where
     
     $$C_p=C\vert\vert C_0=\frac{CC_0}{C+C_0}$$
     
     
     This is the parallel resonant frequency.
   • The magnitude of $Y(\omega)$ is maximized to infinity if $\omega=1/\sqrt{LC}$ and the denominator becomes zero. This is the series resonant frequency.

2. (33 points) In the circuit below, the filter composed of $L$, $C_1$ and $C_2$ between the source $v(t)$ and the load $R_L=100\Omega$ is to pass the fundamental frequency $\omega_0=1000$ without attenuation but completely block the 2nd harmonic $2\omega_0=2000$. Given $L=25\,mH$, find $C_1$ and $C_2$.

   

   Solution Find the total impedance of the filter branch:
   

   $$Z(\omega) = \frac{j\omega L/j\omega C_1}{j\omega L+1/j\omega C_1}+\frac{1}{j\omega C_2} =\frac{j\omega L}{1-\omega^2LC_1}-\frac{j}{\omega C_2}$$

   $$= j\frac{\omega^2LC_2+\omega^2LC_1-1}{(1-\omega^2LC_1)\omega C_2} =j\frac{\omega^2L(C_1+C_2)-1}{(1-\omega^2LC_1)\omega C_2}$$

   
   
   When the denominator is zero, i.e., $\omega^2=1/LC_1$, we get $\vert Z(\omega)\vert=\infty$, the filter is an open circuit. Therefore, to completely block $2\omega_0=2000$, $C_1$ needs to satisfy:
   
   $$\frac{1}{\sqrt{LC_1}}=2\times 10^3,\;\;\;\;\mbox{i.e.,}\;\;... ...c{1}{4\times 10^6 \times 25\times 10^-3}=10^{-5}\,F=10\,\mu F$$
   
   
   When the numerator is zero, i.e., $\omega^2L(C_1+C_2)=1$, $\vert Z(\omega)\vert=0$, the filter is a short circuit. Therefore, to pass $\omega_0=1000$ without attenuation, $C_1+C_2$ needs to satisfy
   
   $$C_1+C_2=\frac{1}{\omega_0^2L}=\frac{1}{10^6\times 25\times 10^{-3}}=40\times 10^{-6}=40\,\mu F$$
   
   
   i.e.,
   
   $$C_2=40\,\mu F-C_1=40\,\mu F-10\,\mj F=30\,\mu F$$
   
   

3. (34 points) In the circuit below, $R_1=3\Omega$, $R_2=6\Omega$, $R_3=2\Omega$, $L=0.5\,H$, $V_1=6V$, $V_2=3V$. The circuit is in steady state when $t<0$. Find current $i_3(t)$ through $R_3$ when switch is closed at $t=0$.

   

   Solution: Current through $L$ at $t=0$ is
   

   $$i_L(0^-)=i_L(0^+)=3/6=0.5\,A$$
   
   
   We have $\tau=L/(R_1\vert\vert R_2\vert\vert R_3)=0.5/(3\vert\vert 6\vert\vert 2)=0.5\,s$ and $i_3(\infty)=0$. To find $i_3(0^+)$ after the switch is closed at $t=0$, we use superposition
   • $V_1=6V$ alone: Total current through $V_1$:
     
     $$\frac{V_1}{R_1+R_2\vert\vert R_3}=\frac{6}{3+1.5}=\frac{4}{3}$$
     
     
     By current divider:
     
     $$i'_3(0^+)=\frac{4}{3}\;\frac{2}{2+6}=\frac{4\times 6}{3\times 8}=1\;\;\;\;\;\mbox{(down)}$$
     
     

   • $V_2$ alone:
     
     $$i''_3(0^+)=\frac{V_2}{R_1\vert\vert R_2+R_3}=\frac{3}{2+2}=\frac{3}{4}=0.75\;\;\;\;\;\mbox{(down)}$$
     
     

   • $i_L(0^+)$ alone:
     
     $$i'''_3(0^+)=i_L(0^+)\frac{R_1\vert\vert R_2}{R_3+R_1\vert\vert R_2}=0.5\frac{2}{2+2}=0.25\;\;\;\;\;\mbox{(up)}$$
     
     

   
   
   $$i_3(0^+)=i'_3(0^+)+i''_3(0^+)+i'''_3(0^+)=1+0.75-0.25=1.5\,A$$
   
   
   
   
   $$i_3(t)=0-(1.5-0)e^{-t/\tau}=1.5e^{-2t}\,A$$
   
   

   Alternative method: $i_L(0)=0.5\,A$, applying KCL to the middle point a:
   

   $$\frac{V_a}{R_3}+\frac{V_a-V_1-V_2}{R_1}+\frac{V_a-V_2}{R_2}+i_L =\frac{V_a}{2}+\frac{V_a-9}{3}+\frac{V_a-3}{6}+0.5=0$$
   
   
   Solving this we get $V_a=3\,V$, and $i_3(0^+)=V_a/R_3=3/2=1.5\,A$, same as above.