Chapter 15, Problem 15E

(a)

To determine

Find the value of $\xi$ and ${\omega }_0$ in terms of $R$, $L$ and $C$ for the series RLC circuit shown in Figure 15.53.

Answer to Problem 15E

The value of $\xi$ and ${\omega }_0$ in terms of $R$, $L$ and $C$ for the series RLC circuit shown in Figure 15.53 is $\frac{R}{2}\sqrt{\frac{C}{L}}$ and $\frac{1}{\sqrt{LC}}$ respectively.

Explanation of Solution

Given data:

Refer to Figure 15.53 in the textbook.

The transfer function of the circuit in Figure 15.53 is,

$H\left(s\right)=\frac{v_C}{v_{\text{in}}}=\frac{1}{1+2\xi \left(\frac{s}{{\omega }_0}\right)+{\left(\frac{s}{{\omega }_0}\right)}^2}$        (1)

Formula used:

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

$Z_R=R$        (2)

$Z_L=sL$        (3)

$Z_C=\frac{1}{sC}$        (4)

Here,

$\omega$ is the angular frequency,

$R$ is the value of the resistor,

$L$ is the value of the inductor, and

$C$ is the value of the capacitor.

Calculation:

Given series RLC circuit is drawn as Figure 1.

The Figure 1 is redrawn as impedance circuit in s-domain in Figure 2 using the equations (2), (3) and (4).

Write the general expression to calculate the transfer function of the circuit in Figure 2.

$H\left(s\right)=\frac{v_C}{v_{\text{in}}}$        (5)

Here,

$v_C$ is the output response of the system, and

$v_{\text{in}}$ is the input response of the system.

Apply Kirchhoff’s voltage law on Figure 2 to find $v_C$.

$\begin{array}{c}{v}_C=\frac{\left(\frac{1}{sC}\right)}{\left(R+sL+\frac{1}{sC}\right)}{v}_{\text{in}}\\ =\frac{\left(\frac{1}{sC}\right)}{\left(\frac{sRC+s^{2}LC+1}{sC}\right)}{v}_{\text{in}}\\ =\frac{1}{1+RCs+LC{s}^2}{v}_{\text{in}}\end{array}$

Rearrange the above equation to find $\frac{v_C}{v_{\text{in}}}$.

$\frac{v_C}{v_{\text{in}}}=\frac{1}{1+RCs+LC{s}^2}$

Substitute $\frac{1}{1+RCs+LC{s}^2}$ for $\frac{v_C}{v_{\text{in}}}$ in equation (5) to find $H\left(s\right)$.

$H\left(s\right)=\frac{1}{1+RCs+LC{s}^2}$

Compare the above equation with the equation (1) to obtain the following values.

$LC={\left(\frac{1}{{\omega }_0}\right)}^2$        (6)

$RC=\frac{2\xi }{{\omega }_0}$        (7)

Rearrange the equation (6).

$\sqrt{LC}=\frac{1}{{\omega }_0}$

Rearrange the above equation to find ${\omega }_0$.

${\omega }_0=\frac{1}{\sqrt{LC}}$

Rearrange the equation (7) to find $\xi$.

$\xi =\frac{RC{\omega }_0}{2}$

Substitute $\frac{1}{\sqrt{LC}}$ for ${\omega }_0$ in above equation to find $\xi$.

$\begin{array}{c}\xi =\frac{RC}{2\sqrt{LC}}\\ =\frac{R\sqrt{C}\sqrt{C}}{2\sqrt{L}\sqrt{C}}\\ =\frac{R\sqrt{C}}{2\sqrt{L}}\\ =\frac{R}{2}\sqrt{\frac{C}{L}}\end{array}$

Conclusion:

Thus, the value of $\xi$ and ${\omega }_0$ in terms of $R$, $L$ and $C$ for the series RLC circuit shown in Figure 15.53 is $\frac{R}{2}\sqrt{\frac{C}{L}}$ and $\frac{1}{\sqrt{LC}}$ respectively.

(b)

To determine

Find the values of inductor $\left(L\right)$ and capacitor $\left(C\right)$ to achieve ${\omega }_0$ of $2\times {10}^3\frac{\text{rad}}{\text{s}}$ and three cases of $\xi =0.1,0.5\text{and}1$.

Answer to Problem 15E

The value of inductor $\left(L\right)$ and capacitor $\left(C\right)$ to achieve ${\omega }_0$ of $2\times {10}^3\frac{\text{rad}}{\text{s}}$ at $\xi =0.1,0.5\text{and}1$ is $125\text{mH}$ and $2\mu \text{F}$, $25\text{mH}$ and $10\mu \text{F}$, $12.5\text{mH}$ and $20\mu \text{F}$ respectively.

Explanation of Solution

Given data:

The value of the resistor $\left(R\right)$ is $50\Omega$.

The value of the resonant frequency $\left({\omega }_0\right)$ is $2\times {10}^3\frac{\text{rad}}{\text{s}}$.

Calculation:

Case (i): $\xi =0.1$

From part (a),

$\xi =\frac{R}{2}\sqrt{\frac{C}{L}}$        (8)

${\omega }_0=\frac{1}{\sqrt{LC}}$        (9)

Substitute $0.1$ for $\xi$ and $50\Omega$ for $R$ in equation (8).

$\begin{array}{l}0.1=\frac{50}{2}\sqrt{\frac{C}{L}}\\ 0.1=25\sqrt{\frac{C}{L}}\end{array}$

Rearrange the above equation to find $\frac{C}{L}$.

$\begin{array}{c}\frac{C}{L}={\left(\frac{0.1}{25}\right)}^2\\ =1.6\times {10}^{-5}\end{array}$

Rearrange the above equation to find $C$.

$C=\left(1.6\times {10}^{-5}\right)L$        (10)

Rearrange the equation (9).

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

Rearrange the above equation to find $LC$.

$LC=\frac{1}{{\omega }_0{}^2}$        (11)

Substitute $\left(1.6\times {10}^{-5}\right)L$ for $C$ and $2\times {10}^3\frac{\text{rad}}{\text{s}}$ for ${\omega }_0$ in equation (11).

$\begin{array}{l}L\left(\left(1.6\times {10}^{-5}\right)L\right)=\frac{1}{{\left(2\times {10}^3\frac{\text{rad}}{\text{s}}\right)}^2}\\ \left(1.6\times {10}^{-5}\right)L^2=\frac{1}{\left(4\times {10}^6\right)}\end{array}$

Rearrange the above equation to find $L^2$.

$\begin{array}{c}{L}^2=\frac{1}{\left(4\times {10}^6\right)\left(1.6\times {10}^{-5}\right)}\\ =\frac{1}{64}\end{array}$

Take square root on both sides of the above equation to find $L$.

$\begin{array}{c}\sqrt{L^2}=\sqrt{\frac{1}{64}}\\ L=125\times {10}^{-3}\text{H}\\ =125\text{mH}\left\{\because 1\text{m}={10}^{-3}\right\}\end{array}$

Substitute $125\times {10}^{-3}\text{H}$ for $L$ in equation (10) to find $C$.

$\begin{array}{c}C=\left(1.6\times {10}^{-5}\right)\left(125\times {10}^{-3}\right)\\ =\left(2\times {10}^{-6}\right)\text{F}\\ =2\mu \text{F}\left\{\because 1\mu ={10}^{-6}\right\}\end{array}$

Case (ii): $\xi =0.5$

Substitute $0.5$ for $\xi$ and $50\Omega$ for $R$ in equation (8).

$\begin{array}{l}0.5=\frac{50}{2}\sqrt{\frac{C}{L}}\\ 0.5=25\sqrt{\frac{C}{L}}\end{array}$

Rearrange the above equation to find $\frac{C}{L}$.

$\begin{array}{c}\frac{C}{L}={\left(\frac{0.5}{25}\right)}^2\\ =4\times {10}^{-4}\end{array}$

Rearrange the above equation to find $C$.

$C=\left(4\times {10}^{-4}\right)L$        (12)

Substitute $\left(4\times {10}^{-4}\right)L$ for $C$ and $2\times {10}^3\frac{\text{rad}}{\text{s}}$ for ${\omega }_0$ in equation (11).

$\begin{array}{l}L\left(\left(4\times {10}^{-4}\right)L\right)=\frac{1}{{\left(2\times {10}^3\frac{\text{rad}}{\text{s}}\right)}^2}\\ \left(4\times {10}^{-4}\right)L^2=\frac{1}{\left(4\times {10}^6\right)}\end{array}$

Rearrange the above equation to find $L^2$.

$\begin{array}{c}{L}^2=\frac{1}{\left(4\times {10}^6\right)\left(4\times {10}^{-4}\right)}\\ =\frac{1}{1600}\end{array}$

Take square root on both sides of the above equation to find $L$.

$\begin{array}{c}\sqrt{L^2}=\sqrt{\frac{1}{1600}}\\ L=25\times {10}^{-3}\text{H}\\ =25\text{mH}\left\{\because 1\text{m}={10}^{-3}\right\}\end{array}$

Substitute $25\times {10}^{-3}\text{H}$ for $L$ in equation (12) to find $C$.

$\begin{array}{c}C=\left(4\times {10}^{-4}\right)\left(25\times {10}^{-3}\right)\\ =\left(10\times {10}^{-6}\right)\text{F}\\ =10\mu \text{F}\left\{\because 1\mu ={10}^{-6}\right\}\end{array}$

Case (iii): $\xi =1$

Substitute $1$ for $\xi$ and $50\Omega$ for $R$ in equation (8).

$\begin{array}{l}1=\frac{50}{2}\sqrt{\frac{C}{L}}\\ 1=25\sqrt{\frac{C}{L}}\end{array}$

Rearrange the above equation to find $\frac{C}{L}$.

$\begin{array}{c}\frac{C}{L}={\left(\frac{1}{25}\right)}^2\\ =1.6\times {10}^{-3}\end{array}$

Rearrange the above equation to find $C$.

$C=\left(1.6\times {10}^{-3}\right)L$        (13)

Substitute $\left(1.6\times {10}^{-3}\right)L$ for $C$ and $2\times {10}^3\frac{\text{rad}}{\text{s}}$ for ${\omega }_0$ in equation (11).

$\begin{array}{l}L\left(\left(1.6\times {10}^{-3}\right)L\right)=\frac{1}{{\left(2\times {10}^3\frac{\text{rad}}{\text{s}}\right)}^2}\\ \left(1.6\times {10}^{-3}\right)L^2=\frac{1}{\left(4\times {10}^6\right)}\end{array}$

Rearrange the above equation to find $L^2$.

$\begin{array}{c}{L}^2=\frac{1}{\left(4\times {10}^6\right)\left(1.6\times {10}^{-3}\right)}\\ =\frac{1}{6400}\end{array}$

Take square root on both sides of the above equation to find $L$.

$\begin{array}{c}\sqrt{L^2}=\sqrt{\frac{1}{6400}}\\ L=12.5\times {10}^{-3}\text{H}\\ =12.5\text{mH}\left\{\because 1\text{m}={10}^{-3}\right\}\end{array}$

Substitute $12.5\times {10}^{-3}\text{H}$ for $L$ in equation (13) to find $C$.

$\begin{array}{c}C=\left(1.6\times {10}^{-3}\right)\left(12.5\times {10}^{-3}\right)\\ =\left(20\times {10}^{-6}\right)\text{F}\\ =20\mu \text{F}\left\{\because 1\mu ={10}^{-6}\right\}\end{array}$

Conclusion:

Thus, the value of inductor $\left(L\right)$ and capacitor $\left(C\right)$ to achieve ${\omega }_0$ of $2\times {10}^3\frac{\text{rad}}{\text{s}}$ at $\xi =0.1,0.5\text{and}1$ is $125\text{mH}$ and $2\mu \text{F}$, $25\text{mH}$ and $10\mu \text{F}$, $12.5\text{mH}$ and $20\mu \text{F}$ respectively.

(c)

To determine

Construct the magnitude Bode plots for the three cases $\xi =0.1,0.5\text{and}1$ using MATLAB.

Explanation of Solution

Calculation:

Simplify the equation (1) to find $H\left(s\right)$.

$\begin{array}{c}H\left(s\right)=\frac{1}{1+2\xi \left(\frac{s}{{\omega }_0}\right)+{\left(\frac{s}{{\omega }_0}\right)}^2}\\ =\frac{1}{1+\frac{2\xi }{{\omega }_0}s+\frac{s^2}{{\omega }_0{}^2}}\\ =\frac{1}{\left(\frac{{\omega }_0{}^2+2\xi {\omega }_{0}s+s^2}{{\omega }_0{}^2}\right)}\end{array}$

$H\left(s\right)=\frac{{\omega }_0{}^2}{s^2+2\xi {\omega }_{0}s+{\omega }_0{}^2}$        (14)

Case (i): $\xi =0.1$

Substitute $0.1$ for $\xi$ and $2\times {10}^3\frac{\text{rad}}{\text{s}}$ for ${\omega }_0$ in equation (14) to find $H\left(s\right)$.

$H\left(s\right)=\frac{{\left(2\times {10}^3\right)}^2}{s^2+2\left(0.1\right)\left(2\times {10}^3\right)s+{\left(2\times {10}^3\right)}^2}$

$H\left(s\right)=\frac{\left(4\times {10}^6\right)}{s^2+400s+\left(4\times {10}^6\right)}$        (15)

Case (ii): $\xi =0.5$

Substitute $0.5$ for $\xi$ and $2\times {10}^3\frac{\text{rad}}{\text{s}}$ for ${\omega }_0$ in equation (14) to find $H\left(s\right)$.

$H\left(s\right)=\frac{{\left(2\times {10}^3\right)}^2}{s^2+2\left(0.5\right)\left(2\times {10}^3\right)s+{\left(2\times {10}^3\right)}^2}$

$H\left(s\right)=\frac{\left(4\times {10}^6\right)}{s^2+2000s+\left(4\times {10}^6\right)}$        (16)

Case (iii): $\xi =1$

Substitute $0.5$ for $\xi$ and $2\times {10}^3\frac{\text{rad}}{\text{s}}$ for ${\omega }_0$ in equation (14) to find $H\left(s\right)$.

$H\left(s\right)=\frac{{\left(2\times {10}^3\right)}^2}{s^2+2\left(1\right)\left(2\times {10}^3\right)s+{\left(2\times {10}^3\right)}^2}$

$H\left(s\right)=\frac{\left(4\times {10}^6\right)}{s^2+4000s+\left(4\times {10}^6\right)}$        (17)

The equations (15), (16) and (17) are the transfer function of the given series RLC circuit at three different cases $\xi =0.1,0.5\text{and}1$.

The MATLAB code is given below to sketch the magnitude Bode plots for the three cases using the equations (15), (16) and (17).

MATLAB Code:

clc;

clear all;

close all;

sys1=tf([(4*10^6)],[1 400 (4*10^6)]);

sys2=tf([(4*10^6)],[1 2000 (4*10^6)]);

sys3=tf([(4*10^6)],[1 4000 (4*10^6)]);

bode(sys1,sys2,sys3)

legend({'sys1','sys2','sys3'},'Location','best')

Output:

The MATLAB output Bode plot of the three transfer functions is shown Figure 1.

Conclusion:

Thus, the magnitude Bode plot for the three cases $\xi =0.1,0.5\text{and}1$ is constructed using MATLAB.