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Zoe Liu authored and milesdai committed Mar 24, 2022
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Expand Up @@ -720,111 +720,207 @@ \chapter{Solutions for Chapter 1}
When $f>f_0$, $\omega > \omega_0$ and $Z_\text{L} > Z_\text{C}$, which means the inductor has more impact on the general response of the whole circuit.
Thus, the circuit responses more like inductors.
\end{enumerate}
Thus, we see the response plot captured in the previous figure ($Figure 1.109$)
Thus, we see the response plot captured in Figure 1.109 from the textbook.

\ex{1.36}
\begin{circuit}{fig:1.36.1}{Generalization of "three-way" Switch Wiring with Two SPDT Switches And $N-2$ DPDT Switches}
(-1,0) to [short](0,0)
% [dashed](2.6,0.7)to[short](2.6,-0.3)
(0.6,0)node[spdt](S0){}
(S0.out 1) node[right] {}
to [short] (2,0.3)
to [short] (2,0.6)
(S0.out 2) node[right]{}
to [short](2,-0.3)
to [short] (2,-0.6)
(2.6,-0.6)node[spdt](S2){}
(2.6,0.6)node[spdt](S1){}
(S1.out 1) to [short] (4.5,0.9)
to [short](5.5,0.9)
to[short](5.5,0.6)
to [short,-o](6.5,0.6)
to [short,-o](7.2,0.3)
to [short](8,0.3)
(4.7,0.9)to [short,*-*](4.7,-0.9)
(S1.out 2) to [short] (4.5,0.3)
to [short,*-*](4.5,-0.3)
(S2.out 1) to [short] (4.5,-0.3)
to [short](5.5,-0.3)
to [short](5.5,0)
to [short,-o](6.5,0)
(S2.out 2) to [short] (4.7,-0.9)
\end{circuit}

\ex{1.37}
Recall that $I_\text{Norton} = I_\text{short circuit the load}$, so we can transform the circuit as shown below.
\begin{circuit}{fig:1.37.1}{Circuit to find $I_\text{Norton}$}
(0,0) to [V,l=10V]
(0,-2.5) node[ground](gnd){}
to [short](3,-2.5)
(0,0) to [R,l=10k](3,0)
to [R,l=10k](3,-2.5)
(3,0) to [short,*-](4,0)
to [short](4,-2.5)
to [short,-*](3,-2.5)
\end{circuit}
Then, $I_\text{Norton} = \frac{\SI{10}{V}}{\SI{10}{\kohm}}=\SI{1}{mA}$
To find $R_\text{Norton}$, it's very similar to how we find $R$ in superposition where we remove all power sources. Then, the circuit is transformed as shown below.
\begin{circuit}{fig:1.37.2}{Circuit to find $R_\text{Norton}$}
(0,0) to [R,l=10k](3,0)
to [R,l=10k](3,-3)
to [short, -o](4,-3)
(0,0) to[short](0,-3)
to [short](3,-3)
(3,0) to[short,-o](4,0)
(4,-1.5) node[right]{$R_\text{Norton}$}
\end{circuit}
Based on the circuit, $R_\text{Norton} = \frac{1}{\frac{1}{10}+\frac{1}{10}} = \SI{5}{\kohm}$
Thus, the Norton equivalent circuit is shown below.
\begin{circuit}{fig:1.37.3}{Circuit to find $R_\text{Norton}$}
(0,-2.5) to [I,l=\SI{1}{mA}](0,0)
to [short](3,0)
(0,-2.5) node[ground](gnd){}
to [short](3,-2.5)
to [R,l=\SI{5}{\kohm}](3,0)
to [short,-o](4,0)
(3,-2.5) to [short,-o](4,-2.5)
(4,-1.25)node[right]{$load$}
\end{circuit}

When the $R_\text{load} = \SI{5}{\kohm}$, $V_\text{out} = \frac{\SI{5}{\kohm}}{2}*\SI{1}{mA}=\SI{2.5}{V}$
And, the original circuit is as shown below.
\begin{circuit}{fig:1.37.4}{Original Circuit with a $\SI{5}{\kohm}$ load}
(0,0) to [V,l=10V]
(0,-2.5) node[ground](gnd){}
to [short](3,-2.5)
(0,0) to [R,l=10k](3,0)
to [R,l=10k](3,-2.5)
(3,0) to [short,*-](5,0)
to [R,l=$5k$](5,-2.5)
to [short,-*](3,-2.5)
\end{circuit}
Thus, $V_\text{out} = \frac{V_\text{in}*\frac{1}{\frac{1}{10}+\frac{1}{5}}}{10+\frac{1}{\frac{1}{10}+\frac{1}{5}}}
=\SI{2.5}{V}$


Finally, we showed that the Norton equivalent gives the same output voltage as the actual circuit when loaded by a $5k$ resistor.
\ex{1.38}
Based on the circuit shown in this excercise, $V_\text{Th} = V_\text{open}=\SI{0.5}{mA}*\SI{10}{\kohm}=\SI{5}{V}$

$I_\text{short} = \SI{0.5}{mA}$, so $R_\text{Th} = \frac{\SI{5}{V}}{\SI{0.5}{mA}}= \SI{10}{\kohm}$

Similarly, we can find the Thévenin equivalent for the previous excercise.

\[V_\text{Th} = \frac{\SI{10}{\kohm}}{10+10}*V_\text{in} = \SI{5}{V}\]
\[I_\text{short} =\frac{\SI{10}{V}}{\SI{10}{\kohm}} = \SI{1}{mA}\]
\[R_\text{Th} = \frac{\SI{5}{V}}{\SI{1}{mA}} = \SI{5}{\kohm}\]
Since the $R_\text{Th}$ is different from the previous result, we can see that these Thévenin circuits are not the same
We can see from textbook Figure 1.122 that two SPDT switches can control the lamp independently.
We need to explore wirings that allow DPDT switches to switch from two states. The following wire shows how to turn DPDT switches to work as described.
\begin{circuit}{fig:1.36.1}{Wiring a DPDT switch to switch from two states}
(2.6,0.7)to[short](2.6,-0.5)
(2.6,-0.6)node[spdt](S2){}
(2.6,0.6)node[spdt](S1){}
(S1.out 1) to [short] (4.5,0.9)
to [short,-o](5.5,0.9) node[right]{state 1}
(4.7,0.9)to [short,*-*](4.7,-0.9)
(S1.out 2) to [short] (4.5,0.3)
to [short,*-*](4.5,-0.3)
(S2.out 1) to [short] (4.5,-0.3)
to [short,-o](5.5,-0.3) node[right]{state 2}
(S2.out 2) to [short] (4.7,-0.9)
\end{circuit}

\begin{circuit}{fig:1.36.2}{Generalization of ``three-way" Switch Wiring with Two SPDT Switches And $N-2$ DPDT Switches}
(-1,0) to [short](0,0)
(2.6,0.7)to[short](2.6,-0.5)
(0.6,0)node[spdt](S0){}
(S0.out 1) node[right] {}
to [short] (2,0.3)
to [short] (2,0.6)
(S0.out 2) node[right]{}
to [short](2,-0.3)
to [short] (2,-0.6)
(2.6,-0.6)node[spdt](S2){}
(2.6,0.6)node[spdt](S1){}
(S1.out 1) to [short] (4.5,0.9)
to [short](5.5,0.9)
to[short](5.5,0.6)
to [short,-o](6.5,0.6)
to [short,-o](7.2,0.3)
to [short](8,0.3)
(4.7,0.9)to [short,*-*](4.7,-0.9)
(S1.out 2) to [short] (4.5,0.3)
to [short,*-*](4.5,-0.3)
(S2.out 1) to [short] (4.5,-0.3)
to [short](5.5,-0.3)
to [short](5.5,0)
to [short,-o](6.5,0)
(S2.out 2) to [short] (4.7,-0.9)
\end{circuit}

\ex{1.37}
Recall that $I_\text{Norton} = I_\text{short circuit the load}$, so we can transform the circuit as shown below.
\begin{circuit}{fig:1.37.1}{Circuit to find $I_\text{Norton}$}
(0,0) to [V,l=10V]
(0,-2.5) node[ground](gnd){}
to [short](3,-2.5)
(0,0) to [R,l=10k](3,0)
to [R,l=10k](3,-2.5)
(3,0) to [short,*-](4,0)
to [short](4,-2.5)
to [short,-*](3,-2.5)
\end{circuit}
Then, $I_\text{Norton} = \frac{\SI{10}{V}}{\SI{10}{\kohm}}=\SI{1}{mA}$
To find $R_\text{Norton}$, it's very similar to how we find $R$ in superposition where we remove all power sources. Then, the circuit is transformed as shown below.
\begin{circuit}{fig:1.37.2}{Circuit to find $R_\text{Norton}$}
(0,0) to [R,l=10k](3,0)
to [R,l=10k](3,-3)
to [short, -o](4,-3)
(0,0) to[short](0,-3)
to [short](3,-3)
(3,0) to[short,-o](4,0)
(4,-1.5) node[right]{$R_\text{Norton}$}
\end{circuit}
Based on the circuit, $R_\text{Norton} = \frac{1}{\frac{1}{10}+\frac{1}{10}} = \SI{5}{\kohm}$
Thus, the Norton equivalent circuit is shown below.
\begin{circuit}{fig:1.37.3}{Circuit to find $R_\text{Norton}$}
(0,-2.5) to [I,l=\SI{1}{mA}](0,0)
to [short](3,0)
(0,-2.5) node[ground](gnd){}
to [short](3,-2.5)
to [R,l=\SI{5}{\kohm}](3,0)
to [short,-o](4,0)
(3,-2.5) to [short,-o](4,-2.5)
(4,-1.25)node[right]{$load$}
\end{circuit}

When the $R_\text{load} = \SI{5}{\kohm}$, $V_\text{out} = \frac{\SI{5}{\kohm}}{2}*\SI{1}{mA}=\SI{2.5}{V}$
And, the original circuit is as shown below.
\begin{circuit}{fig:1.37.4}{Original Circuit with a $\SI{5}{\kohm}$ load}
(0,0) to [V,l=10V]
(0,-2.5) node[ground](gnd){}
to [short](3,-2.5)
(0,0) to [R,l=10k](3,0)
to [R,l=10k](3,-2.5)
(3,0) to [short,*-](5,0)
to [R,l=$5k$](5,-2.5)
to [short,-*](3,-2.5)
\end{circuit}
Thus, $V_\text{out} = \frac{V_\text{in}*\frac{1}{\frac{1}{10}+\frac{1}{5}}}{10+\frac{1}{\frac{1}{10}+\frac{1}{5}}}
=\SI{2.5}{V}$


Finally, we showed that the Norton equivalent gives the same output voltage as the actual circuit when loaded by a $5k$ resistor.
\ex{1.38}
Based on the circuit shown in this excercise, $V_\text{Th} = V_\text{open}=\SI{0.5}{mA}*\SI{10}{\kohm}=\SI{5}{V}$

$I_\text{short} = \SI{0.5}{mA}$, so $R_\text{Th} = \frac{\SI{5}{V}}{\SI{0.5}{mA}}= \SI{10}{\kohm}$

Similarly, we can find the Thévenin equivalent for the previous excercise.

\todoex{1.39}
\[V_\text{Th} = \frac{\SI{10}{\kohm}}{10+10}*V_\text{in} = \SI{5}{V}\]
\[I_\text{short} =\frac{\SI{10}{V}}{\SI{10}{\kohm}} = \SI{1}{mA}\]
\[R_\text{Th} = \frac{\SI{5}{V}}{\SI{1}{mA}} = \SI{5}{\kohm}\]
Since the $R_\text{Th}$ is different from the previous result, we can see that these Thévenin circuits are not the same.
\ex{1.39}
Based on the filter behaviors, the ``rumble filter" is essentially a high-pass filter.

\todoex{1.40}
Given $f_\text{3dB} = \SI{10}{\Hz}$ from the question:
\[R*C = \frac{1}{2\pi*\SI{10}{\Hz}} \approx \SI{16}{\ms} \]
Since the load (\SI{10}{\kohm} minimum) is in parallel with the reistor in the RC high-pass filer, we should select a relatively small resistor for the filter design
so that the load doesn't affect the filter's performance significantly.
We select a resistor: $R = \frac{\SI{10}{\kohm}}{100}=\SI{100}{\ohm}$ then:
\[C = \frac{\SI{16}{\ms}}{\SI{100}{\ohm}} = \SI{160}{\micro\farad}\]
The filter design is shown below.
\begin{circuit}{fig:1.39.1}{Rumble Filter}
(-1,2) to[V, l_=$V_\in$] (-1,0)
(-1,2) to[C=$\SI{160}{\micro\farad}$] (2,2)
to[R, l_=$\SI{100}{\ohm}$, *-*] (2,0)
to[short] (-1,0)
(2,2) to[short, -o] (4,2)
(2,0) to[short, -o] (4,0)
(4,0) to[open, v_<=$V_\load$] (4,2)
\end{circuit}

\ex{1.40}
The ``scratch filter" for audio signals is essentially a low-pass filter that filter out high-frequency sounds, such as scratches.

Given $f_\text{3dB} = \SI{10}{\kHz}$ from the question:
\[R*C = \frac{1}{2\pi*\SI{10}{\kHz}} = \SI{0.016}{\ms} \]
Similar to the high-pass filter design in the previous question, we need to pick a low impedance capacitor so that the load doesn't affect the filter's performance significantly.
We select a capacitor: $C =\SI{0.01}{\micro\farad}$ then:
\[R = \frac{1}{2\pi*10000*0.01*0.000001} \approx \SI{1.6}{\kohm}\]
\begin{circuit}{fig:1.40.1}{Scratch Filter}
(-1,2) to[V, l_=$V_\in$] (-1,0)
(-1,2) to[R, l=$\SI{1.6}{\kohm}$] (2,2)
to[C=$\SI{0.01}{\micro\farad}$,*-*] (2,0)
to[short] (-1,0)
(2,2) to[short, -o] (4,2)
(2,0) to[short, -o] (4,0)
(4,0) to[open, v_<=$V_\load$] (4,2)
\end{circuit}

\ex{1.41}
Since we need to design a filter using resitors and capcitors to produce the results as plotted, we need to think of RC circuits that we discussed before as building blocks for this problem.

Based on the plot, when $\omega < \omega_0$, the filter acts like a voltage divider and $\frac{V_\text{out}}{V_\text{in}} = 0.5$

\todoex{1.41}
When $\omega > \omega_0$, the filer looks like a high-pass filter.

\todoex{1.42}
We can combine voltage divider circuit and high-pass filter as shown below.
\begin{circuit}{fig:1.41.1}{High-emphasis filter}
(0,0) to[short] (0,-2)
to[C, l_=$C$] (3.5,-2)
to[short,-*](3.5,0)
(-1.5,0) node[above]{$V_\in$} to[short, o-*] (0,0)
(0,0) to [R, l=$R1$, *-*] (4,0)
(4,-2.5) node[ground](gnd){}
(4,0) to [R, l=$R2$] (4,-2.5)
(4,0) to[short, -o] (5.5,0) node[above]{$V_\out$}
\end{circuit}
when $\omega < \omega_0$, the capacitor acts as if it's open, and the circuit is a voltage devider.

Given that $\frac{V_\text{out}}{V_\text{in}} = 0.5$, $R_1 = R_2 = R$

When $\omega > \omega_0$, the high frequency signals can pass the capacitor, but low frequency signals are blocked.
We can find the capacitor value by treating the circuit as a high-pass filter.
$f_\text{3dB} = \frac{1}{2\pi*RC}$ and $\omega = 2\pi*f_\text{3dB}$ so $C = \frac{R}{\omega_0}$


\ex{1.42}
\begin{circuit}{fig:1.42.1}{Bandpass filter}
(0,-2.5) node[ground](gnd){}
(0,-2.5) to[R, l_=$R1$] (0,0)
(-4,0) node[above]{$V_\in$} to[C, l=$C1$, o-*] (0,0)
(0,0) to [R, l=$R2$, *-*] (4,0)
(4,-2.5) node[ground](gnd){}
(4,0) to [C, l=$C2$] (4,-2.5)
(4,0) to[short, -o] (6.5,0) node[above]{$V_\out$}
\end{circuit}
The bandpass filter is made of a high-pass filter and a low-pass filter. Thus, given $f_1$ and $f_2$, we have
\[f_1 = \frac{1}{2\pi*R_1*C_1}\]
\[f_2 = \frac{1}{2\pi*R_2*C_2}\]
As discussed in the session ``Driving and loading RC filters" in the book, the assumptions are small input impedance compared to the load.
We can set $R_1 = \frac{1}{10} * R_2$, and then
\[C_2 = \frac{1}{2\pi * f_2 R_2 }\]
\[C_1=\frac{1}{f_1*2\pi R_1} = \frac{1}{f_1*2\pi * 0.1R_2} = \frac{5}{\pi R_2 f_1}\]

\ex{1.43}
The circuit is the diode limiter discussed before (Figure 1.78 in the textbook), and the output is the same as the plots A and B in $Figure 1.79$.
We need to find the characters of the output signal.
Given that $f=\SI{60}{\Hz}$
\[T = \frac{1}{f} = \frac{1}{60} \approx \SI{17}{ms}\]
\[\omega = 2 \pi f = 120 \pi\]

\todoex{1.43}

\ex{1.44}
The equivalent capacitance of the oscilloscope and cable is
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