Chapter 14, Problem 81E

(a)

To determine

Design a circuit which produces a transfer function of $H\left(s\right)=5\left(s+1\right)$.

Explanation of Solution

Given data:

The given transfer function is,

$H\left(s\right)=\frac{V_{\text{out}}}{V_{\text{in}}}=5\left(s+1\right)$

Calculation:

The transfer function of the circuit is,

$H\left(s\right)=\frac{V_{\text{out}}}{V_{\text{in}}}=5\left(s+1\right)$

The above transfer function has a zero at $s=-1$.

The Figure 14.39 (b) in the textbook, that shows a cascade two stages of the circuit with a zero at $s=\frac{-1}{R_1{C}_1}$ is redrawn as shown in Figure 1.

For a single zero,

$s=\frac{-1}{R_1{C}_1}$

Substitute $-1$ for $s$ in the above equation.

$-1=\frac{-1}{R_1{C}_1}$

$1=\frac{1}{R_1{C}_1}$        (1)

Consider the value of $R_1=1\text{k}\Omega$.

Substitute $1\text{k}$ for R in equation (1) as follows,

$\begin{array}{l}1=\frac{1}{\left(1\times {10}^3\right)C_1}\left\{\because 1\text{k}={10}^3\right\}\\ C_1=1\text{mF}\left\{\because 1\text{m}={10}^{-3}\right\}\end{array}$

Transfer function:

The input impedance of the cascaded circuit in Figure 1 is,

$\begin{array}{l}{Z}_1=R_1\parallel \left(\frac{1}{C_{1}s}\right)\\ Z_1=\frac{R_1\left(\frac{1}{C_{1}s}\right)}{R_1+\frac{1}{C_{1}s}}\\ Z_1=\frac{R_1}{R_1{C}_1\left(s+\frac{1}{R_1{C}_1}\right)}\end{array}$

$Z_1=\left(\frac{\frac{1}{C_1}}{s+\frac{1}{R_1{C}_1}}\right)$

Then, write the Formula for the transfer function for the cascaded two stage amplifier.

$H\left(s\right)=-\frac{Z_f}{Z_1}$

Substitute $\left(\frac{\frac{1}{C_1}}{s+\frac{1}{R_1{C}_1}}\right)$ for $Z_1$ and $R_f$ for $Z_f$ in above equation to find $H\left(s\right)$.

$H\left(s\right)=-\frac{R_f}{\left(\frac{\frac{1}{C_1}}{s+\frac{1}{R_1{C}_1}}\right)}$

Thus, the transfer function for $H\left(s\right)$ is,

$H\left(s\right)=-R_f{C}_1\left(s+\frac{1}{R_1{C}_1}\right)$

Substitute 1 for $\frac{1}{R_1{C}_1}$ and 1m for C in the above equation to find $H\left(s\right)$.

$H\left(s\right)=-R_f\left(1\times {10}^{-3}\right)\left(s+1\right)\left\{\because 1\text{m}={10}^{-3}\right\}$        (2)

Completing the design by letting $R_f=5\text{k}\Omega$ in the above equation as follows,

$\begin{array}{c}H\left(s\right)=-\left(5\times {10}^3\right)\left(1\times {10}^{-3}\right)\left(s+1\right)\left\{\because 1\text{k}={10}^3\right\}\\ =-5\left(s+1\right)\end{array}$

If the input will be inverted, add an inverting amplifier with gain of 1 to provide the transfer function as follows.

$H\left(s\right)=5\left(s+1\right)$

Thus, the final design of the circuit is,

$R_f=5\text{k}\Omega$, $R_1=1\text{k}\Omega$, and $C_1=1\text{mF}$.

Conclusion:

Thus, a circuit is designed which produces a transfer function of $H\left(s\right)=5\left(s+1\right)$.

(b)

To determine

Design a circuit which produces a transfer function of $H\left(s\right)=\frac{5}{\left(s+1\right)}$.

Explanation of Solution

Given data:

The given transfer function is,

$H\left(s\right)=\frac{V_{\text{out}}}{V_{\text{in}}}=\frac{5}{\left(s+1\right)}$

Calculation:

The transfer function of the circuit is,

$H\left(s\right)=\frac{V_{\text{out}}}{V_{\text{in}}}=\frac{5}{\left(s+1\right)}$

The above transfer function has pole at $s=-1$.

The Figure 14.39 (a) in the textbook, that shows a cascade two stages of the circuit with pole at $s=\frac{-1}{R_f{C}_f}$ is redrawn as shown in Figure 2.

For pole $s=-1$,

$s=\frac{-1}{R_{fA}{C}_{fA}}$

Substitute $-1$ for $s$ in the above equation.

$-1=\frac{-1}{R_{fA}{C}_{fA}}$

$1=\frac{1}{R_{fA}{C}_{fA}}$        (3)

Let arbitrarily consider $R_{fA}=5\text{k}\Omega$

Substitute $5\text{k}$ for $R_{fA}$ in equation (3) as follows,

$\begin{array}{l}1=\frac{1}{\left(5\times {10}^3\right)C_{fA}}\left\{\because 1\text{k}={10}^3\right\}\\ C_{fA}=2\times {10}^{-4}\times {10}^{-2}\times {10}^2\\ C_{fA}=200\times {10}^{-6}\\ C_{fA}=200\mu \text{F}\left\{\because 1\mu ={10}^{-6}\right\}\end{array}$

Transfer function:

Find the feedback impedance of the cascaded circuit in Figure 2.

$\begin{array}{l}{Z}_f=R_f\parallel \frac{1}{C_{f}s}\\ Z_f=\frac{R_f\left(\frac{1}{C_{f}s}\right)}{R_f+\frac{1}{C_{f}s}}\\ Z_f=\frac{\frac{1}{C_f}}{s+\frac{1}{C_f{R}_f}}\end{array}$

Write the formula for the transfer function of the cascaded circuit in Figure 2 as follows

$H\left(s\right)=-\frac{Z_f}{Z_1}$

Substitute $\frac{\frac{1}{C_f}}{s+\frac{1}{C_f{R}_f}}$ for $Z_f$ and $R_1$ for $Z_1$ in above equation to find the transfer function of the cascaded circuit in Figure 2.

$\begin{array}{l}H\left(s\right)=-\frac{\left(\frac{\frac{1}{C_f}}{s+\frac{1}{C_f{R}_f}}\right)}{R_1}\\ H\left(s\right)=-\frac{\left(\frac{\frac{1}{C_f}}{s+\frac{1}{C_f{R}_f}}\right)}{R_1}\end{array}$

Therefore, consider the transfer function $H\left(s\right)$ is,

$H\left(s\right)=-\frac{\frac{1}{R_{1A}{C}_{fA}}}{\left(s+\frac{1}{R_{fA}{C}_{fA}}\right)}$

Substitute 1 for $\frac{1}{R_{fA}{C}_{fA}}$ and $200\mu$ for $C_{fA}$ in the above equation to find $H\left(s\right)$.

$H\left(s\right)=-\frac{\frac{1}{R_{1A}\left(200\times {10}^{-6}\right)}}{\left(s+1\right)}\left\{\because 1\mu ={10}^{-6}\right\}$        (4)

Completing the design by letting $R_{1A}=1\text{k}\Omega$ in equation (4) as follows,

$\begin{array}{c}H\left(s\right)=-\frac{\frac{1}{\left(1\times {10}^3\right)\left(200\times {10}^{-6}\right)}}{\left(s+1\right)}\left\{\because 1\text{k}={10}^3\right\}\\ =-\frac{5}{\left(s+1\right)}\end{array}$

If the input will be inverted, add an inverting amplifier with gain of 1 to provide the transfer function as follows.

$H\left(s\right)=\frac{5}{\left(s+1\right)}$

Thus, the final design of the circuit is,

$R_{fA}=5\text{k}\Omega$, $C_{fA}=200\mu \text{F}$, and $R_{1A}=1\text{k}\Omega$.

Conclusion:

Thus, a circuit is designed which produces a transfer function of $\frac{5}{\left(s+1\right)}$.

(c)

To determine

Design a circuit which produces a transfer function of $H\left(s\right)=5\frac{\left(s+1\right)}{\left(s+2\right)}$.

Explanation of Solution

Given data:

The given transfer function is,

$H\left(s\right)=\frac{V_{\text{out}}}{V_{\text{in}}}=5\frac{\left(s+1\right)}{\left(s+2\right)}$

Calculation:

The transfer function of the circuit is,

$H\left(s\right)=\frac{V_{\text{out}}}{V_{\text{in}}}=5\frac{\left(s+1\right)}{\left(s+2\right)}$

For the above transfer function, it has a zero at $s=-1$

Refer to Figure 1 in Part (a), that shows a cascade two stages of the circuit with a zero at $s=\frac{-1}{R_1{C}_1}$.

For a single zero,

$s=\frac{-1}{R_1{C}_1}$

Substitute $-1$ for $s$ in the above equation.

$-1=\frac{-1}{R_1{C}_1}$

$1=\frac{1}{R_1{C}_1}$        (5)

Let arbitrarily consider $R_1=1\text{k}\Omega$

Substitute $1\text{k}$ for R in equation (5) as follows,

$\begin{array}{l}1=\frac{1}{\left(1\times {10}^3\right)C_1}\left\{\because 1\text{k}={10}^3\right\}\\ C_1=1\text{mF}\left\{\because 1\text{m}={10}^{-3}\right\}\end{array}$

Consider the same circuit shown in Figure 1 and the transfer function as in a cascaded circuit,

$H_1\left(s\right)=-R_f{C}_1\left(s+\frac{1}{R_1{C}_1}\right)$

Substitute 1 for $\frac{1}{R_1{C}_1}$ and 1m for C in the above equation to find $H_1\left(s\right)$.

$H_1\left(s\right)=-R_f\left(1\times {10}^{-3}\right)\left(s+1\right)\left\{\because 1\text{m}={10}^{-3}\right\}$        (6)

Completing the design by letting $R_f=5\text{k}\Omega$ in equation (6) as follows,

$\begin{array}{c}{H}_1\left(s\right)=-\left(5\times {10}^3\right)\left(1\times {10}^{-3}\right)\left(s+1\right)\left\{\because 1\text{k}={10}^3\right\}\\ =-5\left(s+1\right)\end{array}$

Thus, the final design of the circuit is,

$R_f=5\text{k}\Omega$, $R_1=1\text{k}\Omega$, and $C_1=1\text{mF}$.

The given transfer function has a pole at $s=-2$.

Refer to Figure 2 in Part (b), that shows a cascade two stages of the circuit with pole at $s=\frac{-1}{R_f{C}_f}$.

For pole $s=-2$,

$s=\frac{-1}{R_{fA}{C}_{fA}}$

Substitute $-2$ for $s$ in the above equation.

$-2=\frac{-1}{R_{fA}{C}_{fA}}$

$2=\frac{1}{R_{fA}{C}_{fA}}$        (7)

Let arbitrarily consider $R_{fA}=5\text{k}\Omega$

Substitute $5\text{k}$ for $R_{fA}$ in equation (7) as follows,

$\begin{array}{l}2=\frac{1}{\left(5\times {10}^3\right)C_{fA}}\left\{\because 1\text{k}={10}^3\right\}\\ C_{fA}=\frac{1}{2\times 5\times {10}^3}\\ C_{fA}=1\times {10}^{-4}\times {10}^{-2}\times {10}^2\\ C_{fA}=100\times {10}^{-6}\end{array}$

The above equation becomes,

$C_{fA}=100\mu \text{F}\left\{\because 1\mu ={10}^{-6}\right\}$

Consider the same circuit shown in Figure 2 and the transfer function as in a cascaded circuit,

$H_2\left(s\right)=-\frac{\frac{1}{R_{1A}{C}_{fA}}}{\left(s+\frac{1}{R_{fA}{C}_{fA}}\right)}$

Substitute 2 for $\frac{1}{R_{fA}{C}_{fA}}$ and $100\mu$ for $C_{fA}$ in the above equation to find $H_2\left(s\right)$.

$H_2\left(s\right)=-\frac{\frac{1}{R_{1A}\left(100\times {10}^{-6}\right)}}{\left(s+2\right)}\left\{\because 1\mu ={10}^{-6}\right\}$        (8)

Completing the design by letting $R_{1A}=10\text{k}\Omega$ in equation (8) as follows,

$\begin{array}{c}{H}_2\left(s\right)=-\frac{\frac{1}{\left(10\times {10}^3\right)\left(100\times {10}^{-6}\right)}}{\left(s+2\right)}\left\{\because 1\text{k}={10}^3\right\}\\ =-\frac{1}{\left(s+2\right)}\end{array}$

Thus, the final design of the circuit is,

$R_{fA}=5\text{k}\Omega$, $C_{fA}=100\mu \text{F}$, and $R_{1A}=10\text{k}\Omega$.

Therefore, the overall transfer function of the cascaded circuit is,

$H\left(s\right)=H_1\left(s\right)H_2\left(s\right)$

Substitute $-5\left(s+1\right)$ for $H_1\left(s\right)$ and $-\frac{1}{\left(s+2\right)}$ for $H_2\left(s\right)$ in the above equation as follows,

$\begin{array}{c}H\left(s\right)=\left(-5\left(s+1\right)\right)\left(-\frac{1}{\left(s+2\right)}\right)\\ =\frac{5\left(s+1\right)}{\left(s+2\right)}\end{array}$

Conclusion:

Thus, a circuit is designed which produces a transfer function of $\frac{5\left(s+1\right)}{\left(s+2\right)}$.