Chapter 14, Problem 79P

Refer to the network in Fig. 14.96.

1. (a) Find Z_in(s).
2. (b) Scale the elements by K_m = 10 and K_f = 100. Find Z_in(s) and ω₀.

Figure 14.96

To determine

Find the value of the input impedance $Z_{\text{in}}\left(s\right)$ for the network shown in Figure 14.96.

Answer to Problem 79P

The value of the input impedance $Z_{\text{in}}\left(s\right)$ for the network shown in Figure 14.96 is $\left(5+8s+\frac{10}{s}\right)$.

Explanation of Solution

Given data:

Refer to Figure 14.96 in the textbook.

Formula used:

Write the expression to calculate the impedance of the passive elements resistor, inductor and capacitor in s-domain.

$Z_R\left(s\right)=R$        (1)

$Z_L\left(s\right)=sL$        (2)

$Z_C\left(s\right)=\frac{1}{sC}$        (3)

Here,

$R$ is the value of the resistor,

$L$ is the value of the inductor, and

$C$ is the value of the capacitor.

Calculation:

The given circuit is redrawn as Figure 1.

 

Use equation (1) to find $Z_R\left(s\right)$.

$Z_R\left(s\right)=R$

Use equation (1) to find $Z_{R_o}\left(s\right)$.

$Z_{R_o}\left(s\right)=R_o$

Use equation (2) to find $Z_L\left(s\right)$.

$Z_L\left(s\right)=sL$

Use equation (3) to find $Z_C\left(s\right)$.

$Z_C\left(s\right)=\frac{1}{sC}$

Insert a $1\text{A}$ voltage source at the terminals and redraw the circuit in Figure 1 using s-domain as Figure 2.

 

Modify the Figure 2 with the representation of supernode and current direction as shown in Figure 3.

 

Apply Kirchhoff’s current law to the supernode in Figure 3.

$\begin{array}{l}\frac{V_1-1}{R}+\frac{V_2}{\left(sL+\frac{1}{sC}\right)}=0\\ \frac{V_1-1}{R}=-\frac{V_2}{\left(sL+\frac{1}{sC}\right)}\\ -\left(\frac{1-V_1}{R}\right)=-\frac{V_2}{\left(sL+\frac{1}{sC}\right)}\end{array}$

$\left(\frac{1-V_1}{R}\right)=\frac{V_2}{\left(sL+\frac{1}{sC}\right)}$        (4)

The Figure 3 is reduced as following Figure 4 to show the voltage relation.

 

Apply Kirchhoff’s voltage law to the circuit in Figure 4 to obtain the relationship between voltages.

$\begin{array}{l}-V_1+3V_o+V_2=0\\ 3V_o+V_2=V_1\end{array}$

$V_2=V_1-3V_o$        (5)

Apply voltage division rule on Figure 3 to find $V_o$.

$V_o=\frac{sL}{\left(sL+\frac{1}{sC}\right)}{V}_2$

Rearrange the above equation.

$\frac{V_o}{sL}=\frac{V_2}{\left(sL+\frac{1}{sC}\right)}$        (6)

Rearrange the equation (6) to find $V_2$.

$\begin{array}{c}{V}_2=\frac{\left(sL+\frac{1}{sC}\right)}{sL}{V}_o\\ =\frac{\left(\frac{s^{2}LC+1}{sC}\right)}{sL}{V}_o\\ =\left(\frac{1+s^{2}LC}{s^{2}LC}\right)V_o\end{array}$

Substitute $\left(\frac{1+s^{2}LC}{s^{2}LC}\right)V_o$ for $V_2$ in equation (5).

$\begin{array}{l}\left(\frac{1+s^{2}LC}{s^{2}LC}\right)V_o=V_1-3V_o\\ V_o+s^{2}LC{V}_o=s^{2}LC{V}_1-3s^{2}LC{V}_o\\ V_o+s^{2}LC{V}_o+3s^{2}LC{V}_o=s^{2}LC{V}_1\\ V_o\left(1+4s^{2}LC\right)=s^{2}LC{V}_1\end{array}$

Rearrange the above equation to find $V_o$.

$V_o=\frac{s^{2}LC}{1+4s^{2}LC}{V}_1$

Compare the equations (4) and (6) to obtain the following equation.

$\begin{array}{l}\frac{1-V_1}{R}=\frac{V_o}{sL}\\ 1-V_1=\frac{R{V}_o}{sL}\end{array}$

Substitute $\frac{s^{2}LC}{1+4s^{2}LC}{V}_1$ for $V_o$ in above equation.

$\begin{array}{c}1-V_1=\frac{R\left(\frac{s^{2}LC}{1+4s^{2}LC}{V}_1\right)}{sL}\\ =\frac{s^{2}RLC}{sL\left(1+4s^{2}LC\right)}{V}_1\\ =\frac{sRC}{\left(1+4s^{2}LC\right)}{V}_1\end{array}$

Rearrange the above equation.

$\begin{array}{l}{V}_1+\frac{sRC}{\left(1+4s^{2}LC\right)}{V}_1=1\\ V_1\left(1+\frac{sRC}{\left(1+4s^{2}LC\right)}\right)=1\\ V_1\left(\frac{1+4s^{2}LC+sRC}{1+4s^{2}LC}\right)=1\\ V_1\left(1+4s^{2}LC+sRC\right)=1+4s^{2}LC\end{array}$

Simplify the above equation to find $V_1$.

$V_1=\frac{1+4s^{2}LC}{1+4s^{2}LC+sRC}$

Refer to Figure 3, the current $I_o$ is calculated as follows.

$I_o=\frac{1-V_1}{R}$

Substitute $\frac{1+4s^{2}LC}{1+4s^{2}LC+sRC}$ for $V_1$ in above equation to find $I_o$.

$\begin{array}{c}{I}_o=\left(\frac{1-\left(\frac{1+4s^{2}LC}{1+4s^{2}LC+sRC}\right)}{R}\right)\\ =\left(\frac{\left(\frac{1+4s^{2}LC+sRC-1-4s^{2}LC}{1+4s^{2}LC+sRC}\right)}{R}\right)\\ =\frac{sRC}{R\left(1+4s^{2}LC+sRC\right)}\\ =\frac{sC}{\left(1+4s^{2}LC+sRC\right)}\end{array}$

The input impedance of the circuit in Figure 2 is calculated as follows.

$Z_{\text{in}}\left(s\right)=\frac{1\text{V}}{I_o}$

Substitute $\frac{sC}{\left(1+4s^{2}LC+sRC\right)}$ for $I_o$ in above equation to find $Z_{\text{in}}\left(s\right)$.

$\begin{array}{c}{Z}_{\text{in}}\left(s\right)=\frac{1}{\left(\frac{sC}{\left(1+4s^{2}LC+sRC\right)}\right)}\\ =\frac{1+4s^{2}LC+sRC}{sC}\\ =\frac{1}{sC}+\frac{4s^{2}LC}{sC}+\frac{sRC}{sC}\end{array}$

$Z_{\text{in}}\left(s\right)=R+4sL+\frac{1}{sC}$        (7)

Substitute $5\Omega$ for $R$, $2\text{H}$ for $L$ and $0.1\text{F}$ for $C$ in equation (7) to find $Z_{\text{in}}\left(s\right)$.

$\begin{array}{c}{Z}_{\text{in}}\left(s\right)=5+4s\left(2\right)+\frac{1}{s\left(0.1\right)}\\ =\left(5+8s+\frac{10}{s}\right)\end{array}$

Substitute $j\omega$ for $s$ in equation (7) to find $Z_{\text{in}}\left(\omega \right)$.

$\begin{array}{c}{Z}_{\text{in}}\left(\omega \right)=R+4\left(j\omega \right)L+\frac{1}{\left(j\omega \right)C}\\ =R+j4\omega L+\frac{1}{j\omega C}\\ =R+j4\omega L-\frac{j}{\omega C}\\ =R+j\left(4\omega L-\frac{1}{\omega C}\right)\end{array}$

At resonance condition, the imaginary part of the impedance should be equal to zero. Therefore, equate the imaginary part of the above equation to zero.

$\begin{array}{l}4{\omega }_{0}L-\frac{1}{{\omega }_{0}C}=0\\ \frac{4{\omega }_0{}^{2}LC-1}{{\omega }_{0}C}=0\\ 4{\omega }_0{}^{2}LC-1=0\\ 4{\omega }_0{}^{2}LC=1\end{array}$

Simplify the above equation to find ${\omega }_0{}^2$.

${\omega }_0{}^2=\frac{1}{4LC}$

Take square root on both sides of the above equation to find ${\omega }_0$.

$\begin{array}{c}\sqrt{{\omega }_0{}^2}=\sqrt{\frac{1}{4LC}}\\ {\omega }_0=\frac{1}{\sqrt{4LC}}\end{array}$

${\omega }_0=\frac{1}{2\sqrt{LC}}$        (8)

Substitute $2\text{H}$ for $L$ and $0.1\text{F}$ for $C$ in equation (8) to find ${\omega }_0$.

$\begin{array}{c}{\omega }_0=\frac{1}{2\sqrt{\left(2\text{H}\right)\left(0.1\text{F}\right)}}\\ =\frac{1}{2\sqrt{0.2\left(\frac{{\text{s}}^2}{\text{F}}\cdot \text{F}\right)}}\left\{\because 1\text{H}=\frac{1{\text{s}}^2}{1\text{F}}\right\}\\ =1.118\frac{\text{rad}}{\text{s}}\end{array}$

Conclusion:

Thus, the value of the input impedance $Z_{\text{in}}\left(s\right)$ for the network shown in Figure 14.96 is $\left(5+8s+\frac{10}{s}\right)$.

To determine

Find the value of the input impedance $Z_{\text{in}}\left(s\right)$ and resonant frequency after scaling the elements by $K_m=10$ and $K_f=100$.

Answer to Problem 79P

The value of the input impedance $Z_{\text{in}}\left(s\right)$ and resonant frequency $\left({\omega }_0\right)$ after scaling of the elements is $50+0.8s+\frac{{10}^4}{s}$ and $111.8\frac{\text{rad}}{\text{s}}$ respectively.

Explanation of Solution

Given data:

The value of the magnitude scaling factor $\left(K_m\right)$ is $10$.

The value of the frequency scaling factor $\left(K_f\right)$ is $100$.

Formula used:

Consider the equations used in magnitude and frequency scaling.

Write the expression to calculate the scaled resistor.

$R^{\prime }=K_{m}R$        (9)

Here,

$K_m$ is the magnitude scaling factor.

Write the expression to calculate the scaled inductor.

$L^{\prime }=\frac{K_m}{K_f}L$        (10)

Here,

$K_f$ is the frequency scaling factor.

Write the expression to calculate the scaled capacitor.

$C^{\prime }=\frac{1}{K_m{K}_f}C$        (11)

Calculation:

Substitute $5\Omega$ for $R$ and $10$ for $K_m$ in equation (9) to find $R^{\prime }$.

$\begin{array}{c}{R}^{\prime }=\left(10\right)\left(5\Omega \right)\\ =50\Omega \end{array}$

Substitute $\text{2}\text{H}$ for $L$, $10$ for $K_m$ and $100$ for $K_f$ in equation (10) to find $L^{\prime }$.

$\begin{array}{c}{L}^{\prime }=\frac{10}{100}\left(\text{2}\text{H}\right)\\ =0.2\text{H}\end{array}$

Substitute $\text{0}\text{.1}\text{F}$ for $C$, $10$ for $K_m$ and $100$ for $K_f$ in equation (11) to find $C^{\prime }$.

$\begin{array}{c}{C}^{\prime }=\frac{\text{0}\text{.1}\text{F}}{\left(10\right)\left(100\right)}\\ =1\times {10}^{-4}\text{F}\end{array}$

Substitute $50\Omega$ for $R^{\prime }$, $0.2\text{H}$ for $L^{\prime }$ and $1\times {10}^{-4}\text{F}$ for $C^{\prime }$ in equation (7) to find $Z_{\text{in}}\left(s\right)$.

$\begin{array}{c}{Z}_{\text{in}}\left(s\right)=\left(50\right)+4s\left(0.2\right)+\frac{1}{s\left({10}^{-4}\right)}\\ =\left(50+0.8s+\frac{{10}^4}{s}\right)\end{array}$

Substitute $0.2\text{H}$ for $L^{\prime }$ and $1\times {10}^{-4}\text{F}$ for $C^{\prime }$ in equation (8) to find ${\omega }_0$.

$\begin{array}{c}{\omega }_0=\frac{1}{2\sqrt{\left(0.2\text{H}\right)\left(1\times {10}^{-4}\text{F}\right)}}\\ =\frac{1}{2\sqrt{\left(2\times {10}^{-5}\right)\left(\frac{{\text{s}}^2}{\text{F}}\cdot \text{F}\right)}}\left\{\because 1\text{H}=\frac{1{\text{s}}^2}{1\text{F}}\right\}\\ =111.8\frac{\text{rad}}{\text{s}}\end{array}$

Conclusion:

Thus, the value of the input impedance $Z_{\text{in}}\left(s\right)$ and resonant frequency $\left({\omega }_0\right)$ after scaling of the elements is $50+0.8s+\frac{{10}^4}{s}$ and $111.8\frac{\text{rad}}{\text{s}}$ respectively.