Chapter 14, Problem 82E

(a)

To determine

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

Explanation of Solution

Problem design:

Synthesize a circuit that will yield the transfer function $H\left(s\right)=\frac{V_{\text{out}}}{V_{\text{in}}}=2{\left(s+1\right)}^2$.

Calculation:

The transfer function of the circuit is,

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

The above equation is written as,

$H\left(s\right)=2\left(s+1\right)\left(s+1\right)$        (1)

For the above transfer function, it has two repeated zeros 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)

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

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 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 $H\left(s\right)$.

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

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

Thus, consider that the transfer function for $H_1\left(s\right)$ as follows.

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

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

$H_1\left(s\right)=-R_f\left(1\times {10}^{-3}\right)\left(s+1\right)\left\{\because 1\text{m}={10}^{-3}\right\}$        (2)

Completing the design by letting $R_f=1.41\text{k}\Omega$ in the above equation as follows,

$\begin{array}{c}{H}_1\left(s\right)=-\left(1.41\times {10}^3\right)\left(1\times {10}^{-3}\right)\left(s+1\right)\left\{\because 1\text{k}={10}^3\right\}\\ =-\sqrt{2}\left(s+1\right)\end{array}$

Since, two repeated zeros at $s=-1$, therefore,

$H_1\left(s\right)=H_2\left(s\right)=-\sqrt{2}\left(s+1\right)$

Therefore,

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

Substitute $-\sqrt{2}\left(s+1\right)$ for $H_1\left(s\right)$ and $-\sqrt{2}\left(s+1\right)$ for $H_2\left(s\right)$ in the above equation.

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

Thus, the final design of the circuit is,

$R_f=1.41\text{k}\Omega$, $R_1=1\text{k}\Omega$, and $C_1=1\text{mF}$.

Conclusion:

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

(b)

To determine

Design a circuit which produces a transfer function of $H\left(s\right)=\frac{3}{\left(s+500\right)\left(s+100\right)}$

Explanation of Solution

Problem design:

Synthesize a circuit that will yield the transfer function $H\left(s\right)=\frac{V_{\text{out}}}{V_{\text{in}}}=\frac{3}{\left(s+500\right)\left(s+100\right)}$

Calculation:

The transfer function of the circuit is,

$H\left(s\right)=\frac{V_{\text{out}}}{V_{\text{in}}}=\frac{3}{\left(s+500\right)\left(s+100\right)}$

Consider the transfer function for the cascaded circuit as below.

$H\left(s\right)=H_A\left(s\right)H_B\left(s\right)$

The above transfer function, it has two poles at $s=-500$ and $s=-100$.

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.

The above Figure 2, is the representation for the first stage and second stage of a cascaded circuit to be drawn.

Consider the denominator of given transfer function and the first pole as $s=-500$.

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

Where, $R_{fA},C_{fA}$ are the parameters for first stage of the circuit in the design.

Substitute $-500$ for $s$ in the above equation.

$-500=\frac{-1}{R_{fA}{C}_{fA}}$

$500=\frac{1}{R_{fA}{C}_{fA}}$        (3)

Let arbitrarily consider $R_{fA}=2\Omega$

Substitute $2$ for $R_{fA}$ in equation (3) as follows,

$\begin{array}{l}500=\frac{1}{2\times C_{fA}}\\ C_{fA}=\frac{1}{2\times 500}\\ C_{fA}=1\times {10}^{-3}\\ C_{fA}=1\text{mF}\left\{\because 1\text{m}={10}^{-3}\right\}\end{array}$

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.

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

Therefore, consider the transfer function for the first pole $H_A\left(s\right)$ as,

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

Substitute 500 for $\frac{1}{R_{fA}{C}_{fA}}$ and 1m for $C_{fA}$ in the above equation to find $H_A\left(s\right)$.

$H_A\left(s\right)=-\frac{\frac{1}{R_{1A}\left(1\times {10}^{-3}\right)}}{\left(s+500\right)}\left\{\because 1\text{m}={10}^{-3}\right\}$        (4)

Completing the design by letting $R_{1A}=333\Omega$ in equation (4) as follows,

$\begin{array}{c}{H}_A\left(s\right)=-\frac{\frac{1}{333\left(1\times {10}^{-3}\right)}}{\left(s+500\right)}\\ =-\frac{3}{s+500}\end{array}$

Consider the denominator of given transfer function and the first pole as $s=-100$.

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

Where, $R_{fB},C_{fB}$ are the parameters for second stage of the circuit in the design.

Substitute $-100$ for $s$ in the above equation.

$-100=\frac{-1}{R_{fB}{C}_{fB}}$

$100=\frac{1}{R_{fB}{C}_{fB}}$        (5)

Arbitrarily consider $R_{fB}=10\Omega$.

Substitute $10$ for $R_{fB}$ in equation (5) as follows,

$\begin{array}{l}10=\frac{1}{100\times C_{fB}}\\ C_{fB}=\frac{1}{10\times 100}\\ C_{fB}=1\times {10}^{-3}\text{F}\\ C_{fB}=1\text{mF}\left\{\because 1\text{m}={10}^{-3}\right\}\end{array}$

Therefore, consider the transfer function for the second pole $H_B\left(s\right)$ as,

$H_B\left(s\right)=-\frac{\frac{1}{R_{2B}{C}_{fB}}}{\left(s+\frac{1}{R_{fB}{C}_{fB}}\right)}$

Substitute 100 for $\frac{1}{R_{fB}{C}_{fB}}$ and $1\text{m}$ for $C_{fB}$ in the above equation to find $H_B\left(s\right)$.

$H_B\left(s\right)=-\frac{\frac{1}{R_{2B}\left(1\times {10}^{-3}\text{F}\right)}}{\left(s+100\right)}\left\{\because 1\text{m}={10}^{-3}\right\}$        (6)

Completing the design by letting $R_{2B}=1\text{k}\Omega$ in the equation (6) as follows,

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

Then, from the transfer function of the two stages write the complete transfer function as follows.

$H\left(s\right)=H_A\left(s\right)H_B\left(s\right)$

Substitute $-\frac{3}{s+500}$ for $H_A\left(s\right)$ and $\frac{-1}{s+100}$ for $H_B\left(s\right)$ in the above equation.

$\begin{array}{l}H\left(s\right)=\left(-\frac{3}{s+500}\right)\left(\frac{-1}{s+100}\right)\\ H\left(s\right)=\frac{3}{\left(s+500\right)\left(s+100\right)}\end{array}$

Thus, the final design of the circuit is,

$R_{fA}=2\Omega$, $C_{fA}=1\text{mF}$, $R_{1A}=333\Omega$, $R_{fB}=10\Omega$, $C_{fB}=1\text{mF}$, and $R_{2B}=1\text{k}\Omega$.

Conclusion:

Thus, a circuit is designed which produces a transfer function of $\frac{3}{\left(s+500\right)\left(s+100\right)}$.