Chapter 14, Problem 59P

Find the bandwidth and center frequency of the band-stop filter of Fig. 14.89.

Figure 14.89

To determine

Find the bandwidth and center frequency of the band-stop filter shown in Figure 14.89.

Answer to Problem 59P

The value of the bandwidth $\left(B\right)$ and the center frequency $\left({\omega }_0\right)$ of the band-stop filter shown in Figure 14.89 is $2.408\frac{\text{krad}}{\text{s}}$ and $15.811\frac{\text{krad}}{\text{s}}$ respectively.

Explanation of Solution

Given data:

Refer to Figure 14.89 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 drawn as Figure 1.

Use equation (1) to find $Z_{R_1}\left(s\right)$.

$Z_{R_1}\left(s\right)=R_1$

Use equation (1) to find $Z_{R_2}\left(s\right)$.

$Z_{R_2}\left(s\right)=R_2$

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

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

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

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

The s-domain circuit of the Figure 1 is drawn as Figure 2.

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

$H\left(s\right)=\frac{V_o}{V_i}$        (4)

Here,

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

$V_i$ is the input response of the system.

Refer to Figure 2, the series connected impedances $\frac{1}{sC}$ and $sL$ are connected in parallel with the impedance $R_2$.

Therefore, the equivalent impedance is calculated as follows.

$\begin{array}{c}Z=R_2\parallel \left(\frac{1}{sC}+sL\right)\\ =R_2\parallel \left(\frac{1+s^{2}LC}{sC}\right)\\ =\frac{R_2\left(\frac{1+s^{2}LC}{sC}\right)}{R_2+\left(\frac{1+s^{2}LC}{sC}\right)}\\ =\frac{\left(\frac{R_2\left(1+s^{2}LC\right)}{sC}\right)}{\left(\frac{s{R}_{2}C+1+s^{2}LC}{sC}\right)}\end{array}$

Simplify the above equation to find $Z$.

$Z=\frac{R_2\left(1+s^{2}LC\right)}{1+s{R}_{2}C+s^{2}LC}$

The reduced circuit of Figure 2 is drawn as Figure 3.

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

$V_o=\frac{Z}{Z+R_1}{V}_i$

Rearrange the above equation to find $\frac{V_o}{V_i}$.

$\frac{V_o}{V_i}=\frac{Z}{Z+R_1}$

Substitute $\frac{R_2\left(1+s^{2}LC\right)}{1+s{R}_{2}C+s^{2}LC}$ for $Z$ in above equation to find $\frac{V_o}{V_i}$.

$\begin{array}{c}\frac{V_o}{V_i}=\frac{\left(\frac{R_2\left(1+s^{2}LC\right)}{1+s{R}_{2}C+s^{2}LC}\right)}{\left(\frac{R_2\left(1+s^{2}LC\right)}{1+s{R}_{2}C+s^{2}LC}\right)+R_1}\\ =\frac{\left(\frac{R_2\left(1+s^{2}LC\right)}{1+s{R}_{2}C+s^{2}LC}\right)}{\left(\frac{R_2+s^2{R}_{2}LC+R_1+s{R}_1{R}_{2}C+s^2{R}_{1}LC}{1+s{R}_{2}C+s^{2}LC}\right)}\\ =\frac{R_2\left(1+s^{2}LC\right)}{R_1+s{R}_1{R}_{2}C+s^2{R}_{1}LC+R_2+s^2{R}_{2}LC}\end{array}$

Substitute $\frac{R_2\left(1+s^{2}LC\right)}{R_1+s{R}_1{R}_{2}C+s^2{R}_{1}LC+R_2+s^2{R}_{2}LC}$ for $\frac{V_o}{V_i}$ in equation (4) to find $H\left(s\right)$.

$H\left(s\right)=\frac{R_2\left(1+s^{2}LC\right)}{R_1+s{R}_1{R}_{2}C+s^2{R}_{1}LC+R_2+s^2{R}_{2}LC}$        (5)

Refer to Figure 3, the input impedance is expressed as,

$Z_{\text{in}}=R_1+Z$

Substitute $\frac{R_2\left(1+s^{2}LC\right)}{1+s{R}_{2}C+s^{2}LC}$ for $Z$ in above equation to find $Z_{\text{in}}$.

$\begin{array}{c}{Z}_{\text{in}}=R_1+\frac{R_2\left(1+s^{2}LC\right)}{1+s{R}_{2}C+s^{2}LC}\\ =R_1+\frac{R_2+s^2{R}_{2}LC}{1+s{R}_{2}C+s^{2}LC}\\ =\frac{R_1+s{R}_1{R}_{2}C+s^2{R}_{1}LC+R_2+s^2{R}_{2}LC}{1+s{R}_{2}C+s^{2}LC}\end{array}$

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

$\begin{array}{c}{Z}_{\text{in}}\left(\omega \right)=\frac{R_1+\left(j\omega \right)R_1{R}_{2}C+{\left(j\omega \right)}^2{R}_{1}LC+R_2+{\left(j\omega \right)}^2{R}_{2}LC}{1+\left(j\omega \right)R_{2}C+{\left(j\omega \right)}^{2}LC}\\ =\frac{R_1+j\omega R_1{R}_{2}C+j^2{\omega }^2{R}_{1}LC+R_2+j^2{\omega }^2{R}_{2}LC}{1+j\omega R_{2}C+j^2{\omega }^{2}LC}\\ =\frac{R_1+j\omega R_1{R}_{2}C-{\omega }^2{R}_{1}LC+R_2-{\omega }^2{R}_{2}LC}{1+j\omega R_{2}C-{\omega }^{2}LC}\left\{\because j^2=-1\right\}\\ =\frac{R_1+j\omega R_1{R}_{2}C-{\omega }^2{R}_{1}LC+R_2-{\omega }^2{R}_{2}LC}{\left(1-{\omega }^{2}LC\right)+j\omega R_{2}C}\end{array}$

Simplify the above equation to find $Z_{\text{in}}\left(\omega \right)$.

$\begin{array}{c}{Z}_{\text{in}}\left(\omega \right)=\frac{R_1+j\omega R_1{R}_{2}C-{\omega }^2{R}_{1}LC+R_2-{\omega }^2{R}_{2}LC}{\left(1-{\omega }^{2}LC\right)+j\omega R_{2}C}\times \frac{\left(1-{\omega }^{2}LC\right)-j\omega R_{2}C}{\left(1-{\omega }^{2}LC\right)-j\omega R_{2}C}\\ =\frac{\left(\begin{array}{l}\left(1-{\omega }^{2}LC-j\omega R_{2}C\right)\\ \left(R_1+j\omega R_1{R}_{2}C-{\omega }^2{R}_{1}LC+R_2-{\omega }^2{R}_{2}LC\right)\end{array}\right)}{{\left(1-{\omega }^{2}LC\right)}^2-{\left(j\omega R_{2}C\right)}^2}\left\{\because a^2-b^2=\left(a+b\right)\left(a-b\right)\right\}\\ =\frac{\left(\begin{array}{l}{R}_1+j\omega R_1{R}_{2}C-{\omega }^2{R}_{1}LC+R_2-{\omega }^2{R}_{2}LC\\ -{\omega }^2{R}_{1}LC-j{\omega }^3{R}_1{R}_{2}L{C}^2+{\omega }^4{R}_1{L}^2{C}^2-{\omega }^2{R}_{2}LC+{\omega }^4{R}_2{L}^2{C}^2\\ -j\omega R_1{R}_{2}C-j^2{\omega }^2{R}_1{R}_2{}^2{C}^2+j{\omega }^3{R}_1{R}_{2}L{C}^2-j\omega R_2{}^{2}C+j{\omega }^3{R}_2{}^{2}L{C}^2\end{array}\right)}{{\left(1-{\omega }^{2}LC\right)}^2-j^2{\left(\omega R_{2}C\right)}^2}\\ =\frac{\left(\begin{array}{l}{R}_1+R_2-2{\omega }^2{R}_{1}LC-2{\omega }^2{R}_{2}LC+{\omega }^2{R}_1{R}_2{}^2{C}^2\\ +{\omega }^4{R}_1{L}^2{C}^2+{\omega }^4{R}_2{L}^2{C}^2+j\left(-\omega R_2{}^{2}C+{\omega }^3{R}_2{}^{2}L{C}^2\right)\end{array}\right)}{{\left(1-{\omega }^{2}LC\right)}^2+{\left(\omega R_{2}C\right)}^2}\left\{\because j^2=-1\right\}\end{array}$

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.

$\begin{array}{l}\frac{-{\omega }_0{R}_2{}^{2}C+{\omega }_0{}^3{R}_2{}^{2}L{C}^2}{{\left(1-{\omega }_0{}^{2}LC\right)}^2+{\left({\omega }_0{R}_{2}C\right)}^2}=0\\ -{\omega }_0{R}_2{}^{2}C+{\omega }_0{}^3{R}_2{}^{2}L{C}^2=0\\ {\omega }_0{}^3{R}_2{}^{2}L{C}^2={\omega }_0{R}_2{}^{2}C\\ {\omega }_0{}^{2}LC=1\end{array}$

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

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

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

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

Substitute $1\text{mH}$ for $L$ and $\text{4}\mu \text{F}$ for $C$ in above equation to find ${\omega }_0$.

$\begin{array}{c}{\omega }_0=\frac{1}{\sqrt{\left(1\text{mH}\right)\left(\text{4}\mu \text{F}\right)}}\\ =\frac{1}{\sqrt{\left(1\times {10}^{-3}\right)\left(\text{4}\times {\text{10}}^{-6}\right)\text{H}\cdot \text{F}}}\left\{\because 1\text{m}={10}^{-3},1\mu ={10}^{-6}\right\}\\ =\frac{1}{\sqrt{\left(4\times {10}^{-9}\right)\left(\frac{{\text{s}}^2}{\text{F}}\cdot \text{F}\right)}}\left\{\because 1\text{H}=\frac{1{\text{s}}^2}{1\text{F}}\right\}\\ =15.811\times {10}^3\frac{\text{rad}}{\text{s}}\end{array}$

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

${\omega }_0=15.811\frac{\text{krad}}{\text{s}}\left\{\because 1\text{k}={10}^3\right\}$

Substitute $j\omega$ for $s$ in equation (5) to find $H\left(\omega \right)$.

$\begin{array}{c}H\left(\omega \right)=\frac{R_2\left(1+{\left(j\omega \right)}^{2}LC\right)}{R_1+j\omega R_1{R}_{2}C+{\left(j\omega \right)}^2{R}_{1}LC+R_2+{\left(j\omega \right)}^2{R}_{2}LC}\\ =\frac{R_2\left(1+j^2{\omega }^{2}LC\right)}{R_1+j\omega R_1{R}_{2}C+j^2{\omega }^2{R}_{1}LC+R_2+j^2{\omega }^2{R}_{2}LC}\end{array}$

$H\left(\omega \right)=\frac{R_2\left(1-{\omega }^{2}LC\right)}{R_1+j\omega R_1{R}_{2}C-{\omega }^2{R}_{1}LC+R_2-{\omega }^2{R}_{2}LC}\left\{\because j^2=-1\right\}$        (6)

Substitute $0$ for $\omega$ in equation (6) to find the minimum value $H\left(0\right)$.

$\begin{array}{c}H\left(0\right)=\frac{R_2\left(1-{\left(0\right)}^{2}LC\right)}{R_1+j\left(0\right)R_1{R}_{2}C-{\left(0\right)}^2{R}_{1}LC+R_2-{\left(0\right)}^2{R}_{2}LC}\\ =\frac{R_2\left(1-0\right)}{R_1+0-0+R_2-0}\\ =\frac{R_2}{R_1+R_2}\end{array}$

Substitute $\infty$ for $\omega$ in equation (6) to find the maximum value $H\left(\infty \right)$.

$\begin{array}{c}H\left(\infty \right)=\underset{\omega \to \infty }{\lim }\left(\frac{R_2\left(1-{\omega }^{2}LC\right)}{R_1+j\omega R_1{R}_{2}C-{\omega }^2{R}_{1}LC+R_2-{\omega }^2{R}_{2}LC}\right)\\ =\underset{\omega \to \infty }{\lim }\left(\frac{{\omega }^2{R}_2\left(\frac{1}{{\omega }^2}-LC\right)}{{\omega }^2\left(\frac{R_1+R_2}{{\omega }^2}+j\frac{\omega R_1{R}_{2}C}{{\omega }^2}-LC\left(R_1+R_2\right)\right)}\right)\\ =\underset{\omega \to \infty }{\lim }\left(\frac{R_2\left(\frac{1}{{\omega }^2}-LC\right)}{\left(\frac{R_1+R_2}{{\omega }^2}+j\frac{R_1{R}_{2}C}{\omega }-LC\left(R_1+R_2\right)\right)}\right)\\ =\frac{R_2\left(\frac{1}{{\left(\infty \right)}^2}-LC\right)}{\left(\frac{R_1+R_2}{{\left(\infty \right)}^2}+j\left(\frac{R_1{R}_{2}C}{\infty }\right)-LC\left(R_1+R_2\right)\right)}\end{array}$

Simplify the above equation to find $H\left(\infty \right)$.

$\begin{array}{c}H\left(\infty \right)=\frac{R_2\left(0-LC\right)}{\left(0+j\left(0\right)-LC\left(R_1+R_2\right)\right)}\\ =\frac{-R_{2}LC}{-LC\left(R_1+R_2\right)}\\ =\frac{R_2}{R_1+R_2}\end{array}$

At corner frequency ${\omega }_1$ and ${\omega }_2$, the transfer function is expressed as,

$\left|H\left(\omega \right)\right|=\frac{1}{\sqrt{2}}H\left(\infty \right)$

Substitute $\frac{R_2\left(1-{\omega }^{2}LC\right)}{R_1+j\omega R_1{R}_{2}C-{\omega }^2{R}_{1}LC+R_2-{\omega }^2{R}_{2}LC}$ for $H\left(\omega \right)$ and $\frac{R_2}{R_1+R_2}$ for $H\left(\infty \right)$ in above equation.

$\begin{array}{l}\left|\frac{R_2\left(1-{\omega }^{2}LC\right)}{R_1+j\omega R_1{R}_{2}C-{\omega }^2{R}_{1}LC+R_2-{\omega }^2{R}_{2}LC}\right|=\frac{1}{\sqrt{2}}\left(\frac{R_2}{R_1+R_2}\right)\\ \left|\frac{R_2\left(1-{\omega }^{2}LC\right)}{\left(R_1+R_2-{\omega }^{2}LC\left(R_1+R_2\right)\right)+j\omega R_1{R}_{2}C}\right|=\frac{R_2}{\sqrt{2}\left(R_1+R_2\right)}\\ \frac{\sqrt{{\left(R_2\left(1-{\omega }^{2}LC\right)\right)}^2+0^2}}{\sqrt{{\left(R_1+R_2-{\omega }^{2}LC\left(R_1+R_2\right)\right)}^2+{\left(\omega R_1{R}_{2}C\right)}^2}}=\frac{R_2}{\sqrt{2}\left(R_1+R_2\right)}\\ \frac{R_2\left(1-{\omega }^{2}LC\right)}{\sqrt{{\left(R_1+R_2-{\omega }^{2}LC\left(R_1+R_2\right)\right)}^2+{\left(\omega R_1{R}_{2}C\right)}^2}}=\frac{R_2}{\sqrt{2}\left(R_1+R_2\right)}\end{array}$

Simplify the above equation.

$\begin{array}{l}\frac{R_2\left(1-{\omega }^{2}LC\right)\left(R_1+R_2\right)}{\left(R_2\right)\sqrt{{\left(R_1+R_2-{\omega }^{2}LC\left(R_1+R_2\right)\right)}^2+{\left(\omega R_1{R}_{2}C\right)}^2}}=\frac{1}{\sqrt{2}}\\ \frac{\left(R_1+R_2\right)\left(1-{\omega }^{2}LC\right)}{\sqrt{{\left(R_1+R_2-{\omega }^{2}LC\left(R_1+R_2\right)\right)}^2+{\left(\omega R_1{R}_{2}C\right)}^2}}=\frac{1}{\sqrt{2}}\end{array}$

Substitute $6\Omega$ for $R_1$, $4\Omega$ for $R_2$, $1\text{mH}$ for $L$ and $4\mu \text{F}$ for $C$ in above equation.

$\begin{array}{l}\frac{\left(6+4\right)\left(1-{\omega }^2\left(1\text{m}\right)\left(4\mu \right)\right)}{\sqrt{{\left(6+4-{\omega }^2\left(1\text{m}\right)\left(4\mu \right)\left(6+4\right)\right)}^2+{\left(\omega \left(6\right)\left(4\right)\left(4\mu \right)\right)}^2}}=\frac{1}{\sqrt{2}}\\ \frac{\left(6+4\right)\left(1-{\omega }^2\left(1\times {\text{10}}^{-3}\right)\left(4\times {\text{10}}^{-6}\right)\right)}{\sqrt{{\left(6+4-{\omega }^2\left(1\times {\text{10}}^{-3}\right)\left(4\times {\text{10}}^{-6}\right)\left(6+4\right)\right)}^2+{\left(\omega \left(6\right)\left(4\right)\left(4\times {\text{10}}^{-6}\right)\right)}^2}}=\frac{1}{\sqrt{2}}\left\{\begin{array}{l}\because 1\text{m}={10}^{-3},\\ 1\mu ={10}^{-6}\end{array}\right\}\\ \frac{\left(10\right)\left(1-{\omega }^2\left(4\times {\text{10}}^{-9}\right)\right)}{\sqrt{{\left(10-{\omega }^2\left(4\times {\text{10}}^{-8}\right)\right)}^2+{\left(\omega \left(96\times {\text{10}}^{-6}\right)\right)}^2}}-\frac{1}{\sqrt{2}}=0\\ \frac{\sqrt{2}\left(10-\left(4\times {\text{10}}^{-8}\right){\omega }^2\right)-\sqrt{{\left(10-\left(4\times {\text{10}}^{-8}\right){\omega }^2\right)}^2+{\left(\left(96\times {\text{10}}^{-6}\right)\omega \right)}^2}}{\left(\sqrt{2}\right)\sqrt{{\left(10-\left(4\times {\text{10}}^{-8}\right){\omega }^2\right)}^2+{\left(\left(96\times {\text{10}}^{-6}\right)\omega \right)}^2}}=0\end{array}$

Simplify the above equation.

$\begin{array}{l}\sqrt{2}\left(10-\left(4\times {\text{10}}^{-8}\right){\omega }^2\right)-\sqrt{{\left(10-\left(4\times {\text{10}}^{-8}\right){\omega }^2\right)}^2+{\left(\left(96\times {\text{10}}^{-6}\right)\omega \right)}^2}=0\\ \sqrt{2}\left(10-\left(4\times {\text{10}}^{-8}\right){\omega }^2\right)=\sqrt{{\left(10-\left(4\times {\text{10}}^{-8}\right){\omega }^2\right)}^2+{\left(\left(96\times {\text{10}}^{-6}\right)\omega \right)}^2}\end{array}$

Square on both sides of the above equation to simplify it.

$\begin{array}{l}{\left(\sqrt{2}\left(10-\left(4\times {\text{10}}^{-8}\right){\omega }^2\right)\right)}^2={\left(\sqrt{{\left(10-\left(4\times {\text{10}}^{-8}\right){\omega }^2\right)}^2+{\left(\left(96\times {\text{10}}^{-6}\right)\omega \right)}^2}\right)}^2\\ 2{\left(10-\left(4\times {\text{10}}^{-8}\right){\omega }^2\right)}^2={\left(10-\left(4\times {\text{10}}^{-8}\right){\omega }^2\right)}^2+{\left(\left(96\times {\text{10}}^{-6}\right)\omega \right)}^2\\ {\left(96\times {\text{10}}^{-6}\right)}^2{\omega }^2+1{\left(10-\left(4\times {\text{10}}^{-8}\right){\omega }^2\right)}^2-2{\left(10-\left(4\times {\text{10}}^{-8}\right){\omega }^2\right)}^2=0\\ {\left(96\times {\text{10}}^{-6}\right)}^2{\omega }^2-1{\left(10-\left(4\times {\text{10}}^{-8}\right){\omega }^2\right)}^2=0\end{array}$

Simplify the above equation.

$\begin{array}{l}{\left(96\times {\text{10}}^{-6}\right)}^2{\omega }^2-{\left(10-\left(4\times {\text{10}}^{-8}\right){\omega }^2\right)}^2=0\\ \left(9.216\times {\text{10}}^{-9}\right){\omega }^2-\left({10}^2+{\left(\left(4\times {\text{10}}^{-8}\right){\omega }^2\right)}^2-2\left(10\right)\left(\left(4\times {\text{10}}^{-8}\right){\omega }^2\right)\right)=0\\ \left(9.216\times {\text{10}}^{-9}\right){\omega }^2-\left(100+\left(1.6\times {\text{10}}^{-15}\right){\omega }^4-\left(8\times {\text{10}}^{-7}\right){\omega }^2\right)=0\\ \left(9.216\times {\text{10}}^{-9}\right){\omega }^2-100-\left(1.6\times {\text{10}}^{-15}\right){\omega }^4+\left(8\times {\text{10}}^{-7}\right){\omega }^2=0\end{array}$

Simplify the above equation.

$-\left(1.6\times {\text{10}}^{-15}\right){\omega }^4+\left(8.092\times {\text{10}}^{-7}\right){\omega }^2-100=0$

$\left(1.6\times {\text{10}}^{-15}\right){\omega }^4-\left(8.092\times {\text{10}}^{-7}\right){\omega }^2+100=0$        (7)

Assume $s={\omega }^2$ and substitute in the equation (7) to simplify it.

$\left(1.6\times {\text{10}}^{-15}\right)s^2-\left(8.092\times {\text{10}}^{-7}\right)s+100=0$        (8)

Write a general expression to calculate the roots of quadratic equation $\left(a{s}^2+bs+c=0\right)$.

$s_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$        (9)

Compare the equation (8) with the quadratic equation $\left(a{s}^2+bs+c=0\right)$ to obtain the following values.

$\begin{array}{l}a=\left(1.6\times {\text{10}}^{-15}\right)\\ b=-\left(8.092\times {\text{10}}^{-7}\right)\\ c=100\end{array}$

Substitute $\left(1.6\times {\text{10}}^{-15}\right)$ for $a$, $-\left(8.092\times {\text{10}}^{-7}\right)$ for $b$, and $100$ for $c$ in equation (9) to find $s_{1,2}$.

$\begin{array}{c}{s}_{1,2}=\frac{-\left(-\left(8.092\times {\text{10}}^{-7}\right)\right)\pm \sqrt{{\left(-\left(8.092\times {\text{10}}^{-7}\right)\right)}^2-4\left(1.6\times {\text{10}}^{-15}\right)\left(100\right)}}{2\left(1.6\times {\text{10}}^{-15}\right)}\\ =\frac{\left(8.092\times {\text{10}}^{-7}\right)\pm \sqrt{\left(1.48\times {\text{10}}^{-14}\right)}}{\left(3.2\times {\text{10}}^{-15}\right)}\\ =\frac{\left(8.092\times {\text{10}}^{-7}\right)\pm \left(1.2167\times {\text{10}}^{-7}\right)}{\left(3.2\times {\text{10}}^{-15}\right)}\\ =2.1471\times {10}^8,2.9109\times {10}^8\end{array}$

Substitute the roots of characteristic equation in equation (8).

$\left(s_1-\left(2.1471\times {10}^8\right)\right)\left(s_2-\left(2.9109\times {10}^8\right)\right)=0$

Substitute ${\omega }^2$ for $s$ in above equation.

$\left({\omega }_1{}^2-\left(2.1471\times {10}^8\right)\right)\left({\omega }_2{}^2-\left(2.9109\times {10}^8\right)\right)=0$        (10)

Simplify the equation (10) to find ${\omega }_1{}^2$.

$\begin{array}{l}{\omega }_1{}^2-\left(2.1471\times {10}^8\right)=0\\ {\omega }_1{}^2=\left(2.1471\times {10}^8\right)\end{array}$

Take square root on both sides of the above equation to find ${\omega }_1$.

$\begin{array}{l}\sqrt{{\omega }_1{}^2}=\sqrt{\left(2.1471\times {10}^8\right)}\\ {\omega }_1=14.653\times {10}^3\frac{\text{rad}}{\text{s}}\\ {\omega }_1=14.653\frac{\text{krad}}{\text{s}}\left\{\because 1\text{k}={10}^3\right\}\end{array}$

Simplify the equation (10) to find ${\omega }_2{}^2$.

$\begin{array}{l}{\omega }_2{}^2-\left(2.9109\times {10}^8\right)=0\\ {\omega }_2{}^2=\left(2.9109\times {10}^8\right)\end{array}$

Take square root on both sides of the above equation to find ${\omega }_2$.

$\begin{array}{c}\sqrt{{\omega }_2{}^2}=\sqrt{\left(2.9109\times {10}^8\right)}\\ {\omega }_2=17.061\times {10}^3\frac{\text{rad}}{\text{s}}\\ =17.061\frac{\text{krad}}{\text{s}}\left\{\because 1\text{k}={10}^3\right\}\end{array}$

Write the expression to calculate the bandwidth of the band-stop filter.

$B={\omega }_2-{\omega }_1$        (11)

Here,

${\omega }_1$ is the lower corner frequency, and

${\omega }_2$ is the upper corner frequency.

Substitute $14.653\frac{\text{krad}}{\text{s}}$ for ${\omega }_1$ and $17.061\frac{\text{krad}}{\text{s}}$ for ${\omega }_2$ in equation (11) to find $B$.

$\begin{array}{c}B=17.061\frac{\text{krad}}{\text{s}}-14.653\frac{\text{krad}}{\text{s}}\\ =2.408\frac{\text{krad}}{\text{s}}\end{array}$

Conclusion:

Thus, the value of the bandwidth $\left(B\right)$ and the center frequency $\left({\omega }_0\right)$ of the band-stop filter shown in Figure 14.89 is $2.408\frac{\text{krad}}{\text{s}}$ and $15.811\frac{\text{krad}}{\text{s}}$ respectively.