Chapter 14, Problem 57P

Determine the center frequency and bandwidth of the band-pass filters in Fig. 14.88.

To determine

Find the center frequency and bandwidth of the band-pass filter shown in Figure 14.88(a).

Answer to Problem 57P

The value of the center frequency $\left({\omega }_0\right)$ and the bandwidth $\left(B\right)$ of the band-pass filter shown in Figure 14.88(a) is $1\frac{\text{rad}}{\text{s}}$ and $3\frac{\text{rad}}{\text{s}}$ respectively.

Explanation of Solution

Given data:

Refer to Figure 14.88(a) 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 (3) to find $Z_{C_1}\left(s\right)$.

$Z_{C_1}\left(s\right)=\frac{1}{s{C}_1}$

Use equation (3) to find $Z_{C_2}\left(s\right)$.

$Z_{C_2}\left(s\right)=\frac{1}{s{C}_2}$

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.

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

Here,

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

$V_s$ is the input response of the system.

Refer to Figure 2, the series connected impedances $Z_{C_2}\left(s\right)$ and $Z_{R_2}\left(s\right)$ are connected in parallel with the impedance $Z_{C_1}\left(s\right)$ and this parallel combination is connected in series with the impedance $Z_{R_1}\left(s\right)$.

Therefore, the equivalent impedance is calculated as follows.

$Z_{\text{eq}}\left(s\right)=Z_{R_1}\left(s\right)+\left(Z_{C_1}\left(s\right)\parallel \left(Z_{C_2}\left(s\right)+Z_{R_2}\left(s\right)\right)\right)$

Substitute $R_1$ for $Z_{R_1}\left(s\right)$, $R_2$ for $Z_{R_2}\left(s\right)$, $\frac{1}{s{C}_1}$ for $Z_{C_1}\left(s\right)$ and $\frac{1}{s{C}_2}$ for $Z_{C_2}\left(s\right)$ in above equation to find $Z_{\text{eq}}\left(s\right)$.

$Z_{\text{eq}}\left(s\right)=R_1+\left(\frac{1}{s{C}_1}\parallel \left(\frac{1}{s{C}_2}+R_2\right)\right)$        (5)

Refer to the given data. The value of the resistors $R_1$ and $R_2$ is same and the value of the capacitors $C_1$ and $C_2$ is same. Therefore, $R_1=R_2=R$ and $C_1=C_2=C$.

Substitute $R$ for $R_1$ and $R_2$, $C$ for $C_1$ and $C_2$ in equation (5) to find $Z_{\text{eq}}\left(s\right)$.

$\begin{array}{c}{Z}_{\text{eq}}\left(s\right)=R+\left(\frac{1}{sC}\parallel \left(\frac{1}{sC}+R\right)\right)\\ =R+\frac{\left(\frac{1}{sC}\right)\left(\frac{1}{sC}+R\right)}{\left(\frac{1}{sC}+\frac{1}{sC}+R\right)}\\ =R+\frac{\left(\frac{1}{sC}\right)\left(\frac{1+sRC}{sC}\right)}{\left(\frac{2}{sC}+R\right)}\\ =R+\frac{\left(\frac{1+sRC}{sC}\right)}{sC\left(\frac{2+sRC}{sC}\right)}\end{array}$

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

$\begin{array}{c}{Z}_{\text{eq}}\left(s\right)=R+\frac{\left(1+sRC\right)}{sC\left(2+sRC\right)}\\ =R+\frac{\left(1+sRC\right)}{sC\left(2+sRC\right)}\\ =\frac{2sRC+s^2{R}^2{C}^2+1+sRC}{sC\left(2+sRC\right)}\\ =\frac{1+3sRC+s^2{R}^2{C}^2}{sC\left(2+sRC\right)}\end{array}$

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

Refer to the Figure 3, the current through the circuit is expressed as,

$I=\frac{V_s}{Z_{\text{eq}}\left(s\right)}$

Apply current division rule on Figure 2 to find $I_1$.

$I_1=\frac{Z_{C_1}\left(s\right)}{Z_{C_1}\left(s\right)+Z_{C_2}\left(s\right)+Z_{R_2}\left(s\right)}I$

Substitute $\frac{V_s}{Z_{\text{eq}}\left(s\right)}$ for $I$, $R_2$ for $Z_{R_2}\left(s\right)$, $\frac{1}{s{C}_1}$ for $Z_{C_1}\left(s\right)$ and $\frac{1}{s{C}_2}$ for $Z_{C_2}\left(s\right)$ in above equation to find $I_1$.

$I_1=\frac{\left(\frac{1}{s{C}_1}\right)}{\left(\frac{1}{s{C}_1}\right)+\left(\frac{1}{s{C}_2}\right)+R_2}\left(\frac{V_s}{Z_{\text{eq}}\left(s\right)}\right)$

Substitute $\frac{1+3sRC+s^2{R}^2{C}^2}{sC\left(2+sRC\right)}$ for $Z_{\text{eq}}\left(s\right)$, $R$ for $R_2$, $C$ for $C_1$ and $C_2$ in above equation to find $I_1$.

$\begin{array}{c}{I}_1=\left(\frac{\left(\frac{1}{sC}\right)}{\left(\frac{1}{sC}\right)+\left(\frac{1}{sC}\right)+R}\right)\left(\frac{V_s}{\left(\frac{1+3sRC+s^2{R}^2{C}^2}{sC\left(2+sRC\right)}\right)}\right)\\ =\left(\frac{\left(\frac{1}{sC}\right)}{\frac{2}{sC}+R}\right)\left(V_s\left(\frac{sC\left(2+sRC\right)}{1+3sRC+s^2{R}^2{C}^2}\right)\right)\\ =\left(\frac{\left(\frac{1}{sC}\right)}{\frac{2+sRC}{sC}}\right)\left(V_s\left(\frac{sC\left(2+sRC\right)}{1+3sRC+s^2{R}^2{C}^2}\right)\right)\\ =\frac{V_s}{2+sRC}\left(\frac{sC\left(2+sRC\right)}{1+3sRC+s^2{R}^2{C}^2}\right)\end{array}$

Refer to Figure 2, the output voltage $V_o$ is calculated as,

$V_o=I_1{R}_2$

Substitute $\frac{V_s}{2+sRC}\left(\frac{sC\left(2+sRC\right)}{1+3sRC+s^2{R}^2{C}^2}\right)$ for $I_1$ and $R$ for $R_2$ in above equation to find $V_o$.

$\begin{array}{c}{V}_o=\left(\frac{V_s}{2+sRC}\left(\frac{sC\left(2+sRC\right)}{1+3sRC+s^2{R}^2{C}^2}\right)\right)\left(R\right)\\ =\left(\frac{R}{2+sRC}\left(\frac{sC\left(2+sRC\right)}{1+3sRC+s^2{R}^2{C}^2}\right)\right)V_s\end{array}$

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

$\begin{array}{c}\frac{V_o}{V_s}=\frac{R}{\left(2+sRC\right)}\left(\frac{sC\left(2+sRC\right)}{1+3sRC+s^2{R}^2{C}^2}\right)\\ =\left(\frac{sRC}{1+3sRC+s^2{R}^2{C}^2}\right)\end{array}$

Substitute $\left(\frac{sRC}{1+3sRC+s^2{R}^2{C}^2}\right)$ for $\frac{V_o}{V_s}$ in equation (4) to find $H\left(s\right)$.

$\begin{array}{c}H\left(s\right)=\left(\frac{sRC}{1+3sRC+s^2{R}^2{C}^2}\right)\\ =\frac{sRC}{R^2{C}^2\left(s^2+\frac{3sRC}{R^2{C}^2}+\frac{1}{R^2{C}^2}\right)}\\ =\frac{\left(\frac{3}{3}\right)s}{RC\left(s^2+\frac{3}{RC}s+\frac{1}{R^2{C}^2}\right)}\\ =\frac{1}{3}\left[\frac{\left(\frac{3}{RC}\right)s}{s^2+\left(\frac{3}{RC}\right)s+\left(\frac{1}{R^2{C}^2}\right)}\right]\end{array}$

Compare the denominator factor of above equation with the standard quadratic equation $\left(s^2+\frac{{\omega }_0}{Q}s+{\omega }_0^2\right)$.

${\omega }_0{}^2=\frac{1}{R^2{C}^2}$        (6)

$\frac{{\omega }_0}{Q}=\frac{3}{RC}$        (7)

Substitute $1\Omega$ for $R$ and $1\text{F}$ for $C$ in equation (6) to find ${\omega }_0{}^2$.

$\begin{array}{c}{\omega }_0{}^2=\frac{1}{{\left(1\Omega \right)}^2{\left(1\text{F}\right)}^2}\\ =\frac{1}{1{\Omega }^2\cdot {\text{F}}^2}\\ =\frac{1}{1{\Omega }^2\cdot \left(\frac{{\text{s}}^2}{{\Omega }^2}\right)}\left\{\because 1\text{F}=\frac{1\text{s}}{1\Omega }\right\}\\ =1\frac{1}{{\text{s}}^2}\end{array}$

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

$\begin{array}{c}\sqrt{{\omega }_0{}^2}=\sqrt{1\frac{1}{{\text{s}}^2}}\\ {\omega }_0=1\frac{\text{rad}}{\text{s}}\end{array}$

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

$B=\frac{{\omega }_0}{Q}$

Substitute $B$ for $\frac{{\omega }_0}{Q}$, $1\Omega$ for $R$ and $1\text{F}$ for $C$ in equation (7) to find $B$.

$\begin{array}{c}B=\frac{3}{\left(1\Omega \right)\left(1\text{F}\right)}\\ =\frac{3}{1\Omega \cdot \left(\frac{\text{s}}{\Omega }\right)}\left\{\because 1\text{F}=\frac{1\text{s}}{1\Omega }\right\}\\ =\frac{3}{1\text{s}}\\ =3\frac{\text{rad}}{\text{s}}\end{array}$

Conclusion:

Thus, the value of the center frequency $\left({\omega }_0\right)$ and the bandwidth $\left(B\right)$ of the band-pass filter shown in Figure 14.88(a) is $1\frac{\text{rad}}{\text{s}}$ and $3\frac{\text{rad}}{\text{s}}$ respectively.

To determine

Find the center frequency and bandwidth of the band-pass filter shown in Figure 14.88(b).

Answer to Problem 57P

The value of the center frequency $\left({\omega }_0\right)$ and the bandwidth $\left(B\right)$ of the band-pass filter shown in Figure 14.88(b) is $1\frac{\text{rad}}{\text{s}}$ and $3\frac{\text{rad}}{\text{s}}$ respectively.

Explanation of Solution

Given data:

Refer to Figure 14.88(b) in the textbook.

Calculation:

The given circuit is drawn as Figure 4.

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 $Z_{L_1}\left(s\right)$.

$Z_{L_1}\left(s\right)=s{L}_1$

Use equation (2) to find $Z_{L_2}\left(s\right)$.

$Z_{L_2}\left(s\right)=s{L}_2$

The s-domain circuit of the Figure 4 is drawn as Figure 5.

Write the general expression to calculate the transfer function of the circuit in Figure 5.

$H\left(s\right)=\frac{V_o}{V_s}$        (8)

Refer to Figure 5, the series connected impedances $Z_{R_2}\left(s\right)$ and $Z_{L_2}\left(s\right)$ are connected in parallel with the impedance $Z_{R_1}\left(s\right)$ and this parallel combination is connected in series with the impedance $Z_{L_1}\left(s\right)$.

Therefore, the equivalent impedance is calculated as follows.

$Z_{\text{eq}}\left(s\right)=Z_{L_1}\left(s\right)+\left(Z_{R_1}\left(s\right)\parallel \left(Z_{R_2}\left(s\right)+Z_{L_2}\left(s\right)\right)\right)$

Substitute $R_1$ for $Z_{R_1}\left(s\right)$, $R_2$ for $Z_{R_2}\left(s\right)$, $s{L}_1$ for $Z_{L_1}\left(s\right)$ and $s{L}_2$ for $Z_{L_2}\left(s\right)$ in above equation to find $Z_{\text{eq}}\left(s\right)$.

$Z_{\text{eq}}\left(s\right)=s{L}_1+\left(R_1\parallel \left(R_2+s{L}_2\right)\right)$        (9)

Refer to the given data. The value of the resistors $R_1$ and $R_2$ is same and the value of the inductors $L_1$ and $L_2$ is same. Therefore, $R_1=R_2=R$ and $L_1=L_2=L$.

Substitute $R$ for $R_1$ and $R_2$, $L$ for $L_1$ and $L_2$ in equation (9) to find $Z_{\text{eq}}\left(s\right)$.

$\begin{array}{c}{Z}_{\text{eq}}\left(s\right)=sL+\left(R\parallel \left(R+sL\right)\right)\\ =sL+\frac{R\left(R+sL\right)}{\left(R+R+sL\right)}\\ =sL+\frac{R^2+sRL}{\left(2R+sL\right)}\\ =\frac{2sRL+s^2{L}^2+R^2+sRL}{2R+sL}\end{array}$

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

$Z_{\text{eq}}\left(s\right)=\frac{R^2+3sRL+s^2{L}^2}{2R+sL}$

The reduced circuit of Figure 5 is drawn as Figure 6.

Refer to the Figure 6, the current through the circuit is expressed as,

$I=\frac{V_s}{Z_{\text{eq}}\left(s\right)}$

Apply current division rule on Figure 5 to find $I_1$.

$I_1=\frac{Z_{R_1}\left(s\right)}{Z_{R_1}\left(s\right)+Z_{R_2}\left(s\right)+Z_{L_2}\left(s\right)}I$

Substitute $\frac{V_s}{Z_{\text{eq}}\left(s\right)}$ for $I$, $R_1$ for $Z_{R_1}\left(s\right)$, $R_2$ for $Z_{R_2}\left(s\right)$ and $s{L}_2$ for $Z_{L_2}\left(s\right)$ in above equation to find $I_1$.

$I_1=\frac{R_1}{R_1+R_2+s{L}_2}\left(\frac{V_s}{Z_{\text{eq}}\left(s\right)}\right)$

Substitute $\frac{R^2+3sRL+s^2{L}^2}{2R+sL}$ for $Z_{\text{eq}}\left(s\right)$, $L$ for $L_2$, $R$ for $R_1$ and $R_2$ in above equation to find $I_1$.

$\begin{array}{c}{I}_1=\frac{R}{R+R+sL}\left(\frac{V_s}{\left(\frac{R^2+3sRL+s^2{L}^2}{2R+sL}\right)}\right)\\ =\frac{R}{2R+sL}\left(V_s\left(\frac{2R+sL}{R^2+3sRL+s^2{L}^2}\right)\right)\\ =\frac{R{V}_s}{R^2+3sRL+s^2{L}^2}\end{array}$

Refer to Figure 5, the output voltage $V_o$ is calculated as,

$V_o=s{L}_2{I}_1$

Substitute $\frac{R{V}_s}{R^2+3sRL+s^2{L}^2}$ for $I_1$ and $L$ for $L_2$ in above equation to find $V_o$.

$\begin{array}{c}{V}_o=s\left(L\right)\left(\frac{R{V}_s}{R^2+3sRL+s^2{L}^2}\right)\\ =\left(\frac{sRL}{R^2+3sRL+s^2{L}^2}\right)V_s\end{array}$

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

$\frac{V_o}{V_s}=\left(\frac{sRL}{R^2+3sRL+s^2{L}^2}\right)$

Substitute $\left(\frac{sRL}{R^2+3sRL+s^2{L}^2}\right)$ for $\frac{V_o}{V_s}$ in equation (8) to find $H\left(s\right)$.

$\begin{array}{c}H\left(s\right)=\frac{sRL}{R^2+3sRL+s^2{L}^2}\\ =\frac{sRL}{L^2\left(s^2+\frac{3sRL}{L^2}+\frac{R^2}{L^2}\right)}\\ =\frac{\left(\frac{3}{3}\right)sR}{L\left(s^2+\frac{3R}{L}s+\frac{R^2}{L^2}\right)}\\ =\frac{1}{3}\left(\frac{\left(\frac{3R}{L}\right)s}{\left(s^2+\frac{3R}{L}s+\frac{R^2}{L^2}\right)}\right)\end{array}$

Compare the denominator factor of above equation with the standard quadratic equation $\left(s^2+\frac{{\omega }_0}{Q}s+{\omega }_0^2\right)$.

${\omega }_0{}^2=\frac{R^2}{L^2}$        (10)

$\frac{{\omega }_0}{Q}=\frac{3R}{L}$        (11)

Substitute $1\Omega$ for $R$ and $1\text{H}$ for $L$ in equation (10) to find ${\omega }_0{}^2$.

$\begin{array}{c}{\omega }_0{}^2=\frac{{\left(1\Omega \right)}^2}{{\left(1\text{H}\right)}^2}\\ =\frac{1{\Omega }^2}{1{\text{H}}^2}\\ =\frac{1{\Omega }^2}{1{\Omega }^2\cdot {\text{s}}^2}\left\{\because 1\text{H}=1\Omega \cdot 1\text{s}\right\}\\ =1\frac{1}{{\text{s}}^2}\end{array}$

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

$\begin{array}{c}\sqrt{{\omega }_0{}^2}=\sqrt{1\frac{1}{{\text{s}}^2}}\\ {\omega }_0=1\frac{\text{rad}}{\text{s}}\end{array}$

Substitute $B$ for $\frac{{\omega }_0}{Q}$, $1\Omega$ for $R$ and $1\text{H}$ for $L$ in equation (11) to find $B$.

$\begin{array}{c}B=\frac{3\left(1\Omega \right)}{1\text{H}}\\ =\frac{3\Omega }{1\Omega \cdot \text{s}}\left\{\because 1\text{H}=1\Omega \cdot 1\text{s}\right\}\\ =\frac{3}{1\text{s}}\\ =3\frac{\text{rad}}{\text{s}}\end{array}$

Conclusion:

Thus, the value of the center frequency $\left({\omega }_0\right)$ and the bandwidth $\left(B\right)$ of the band-pass filter shown in Figure 14.88(b) is $1\frac{\text{rad}}{\text{s}}$ and $3\frac{\text{rad}}{\text{s}}$ respectively.