Chapter 14, Problem 83E

Design a circuit which produces the transfer function

To determine

Design a circuit which produces a transfer function of $H\left(s\right)=3\frac{\left(s+50\right)}{{\left(s+75\right)}^2}$.

Explanation of Solution

Given data:

The given transfer function is,

$H\left(s\right)=\frac{V_{\text{out}}}{V_{\text{in}}}=3\frac{\left(s+50\right)}{{\left(s+75\right)}^2}$

Calculation:

The transfer function of the circuit is,

$H\left(s\right)=\frac{V_{\text{out}}}{V_{\text{in}}}=3\frac{\left(s+50\right)}{{\left(s+75\right)}^2}$        (1)

Equation (1) is written as,

$H\left(s\right)=3\frac{\left(s+50\right)}{\left(s+75\right)\left(s+75\right)}$        (2)

For numerator:

From equation (2), for the numerator $\left(s+50\right)$, it has a zero at $s=-50$.

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.

From the Figure 1, a single zero can be written as,

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

Substitute $-50$ for $s$ in the above equation.

$-50=\frac{-1}{R_1{C}_1}$

$50=\frac{1}{R_1{C}_1}$        (3)

Consider the value of $R_1=20\Omega$.

Substitute $20$ for $R_1$ in equation (3) as follows,

$\begin{array}{c}50=\frac{1}{20\times C_1}\\ C_1=\frac{1}{20\times 50}\\ =1\times {10}^{-3}\text{F}\\ =1\text{mF}\left\{\because 1\text{m}={10}^{-3}\right\}\end{array}$

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 $H\left(s\right)$.

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

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

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

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

Substitute 50 for $\frac{1}{R_1{C}_1}$ and $1\text{m}$ for C in the above equation to find $H_1\left(s\right)$.

$H_1\left(s\right)=-R_f\left(1\times {10}^{-3}\right)\left(s+50\right)\left\{\because 1\text{m}={10}^{-3}\right\}$        (4)

Completing the design by letting $R_f=3\text{k}\Omega$ in the above equation as follows,

$\begin{array}{c}{H}_1\left(s\right)=-\left(3\times {10}^3\right)\left(1\times {10}^{-3}\right)\left(s+50\right)\left\{\because 1\mu ={10}^{-6}\right\}\\ =-3\left(s+50\right)\end{array}$

Thus, the final design of the circuit is,

$R_f=3\text{k}\Omega$, $R_1=20\Omega$, and $C_1=1\text{mF}$.

For denominator:

From the transfer function shown in equation (2), it has two repeated poles at $s=-75$.

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.

Consider that the cascaded circuit for the two poles representation as shown in Figure 2.

From Figure 2, and the denominator of given transfer function a first pole at $s=-75$,

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

Where, the circuit parameters are considered as $R_{fA},C_{fA}$ to represent the first pole in the first stage of the cascaded circuit.

Substitute $-75$ for $s$ in the above equation.

$-75=\frac{-1}{R_{fA}{C}_{fA}}$

$75=\frac{1}{R_{fA}{C}_{fA}}$        (5)

Let arbitrarily consider $R_{fA}=13.3\Omega$

Substitute 13.3 for $R_{fA}$ in equation (5) as follows,

$\begin{array}{l}75=\frac{1}{13.3\times C_{fA}}\\ C_{fA}=\frac{1}{75\times 13.3}\\ C_{fA}=1\times {10}^{-3}\\ C_{fA}=1\text{mF}\left\{\because 1\text{m}={10}^{-3}\right\}\end{array}$

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 $H_A\left(s\right)$ as,

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

Substitute 75 for $\frac{1}{R_{fA}{C}_{fA}}$ and 1m for $C_{fA}$ in the above equation to find $H_A\left(s\right)$.

$H_A\left(s\right)=-\frac{\frac{1}{R_{1A}\left(1\times {10}^{-3}\right)}}{\left(s+75\right)}\left\{\because 1\text{m}={10}^{-3}\right\}$        (6)

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

$\begin{array}{c}{H}_A\left(s\right)=-\frac{\frac{1}{\left(1\times {10}^3\right)\left(1\times {10}^{-3}\right)}}{\left(s+75\right)}\left\{\because 1\text{k}={10}^3\right\}\\ =-\frac{1}{\left(s+75\right)}\end{array}$

Since, two repeated poles are at same $s=-75$, therefore, consider the second stage circuit in the two stages cascaded circuit for the denominator similar for two poles. Therefore,

$H_A\left(s\right)=H_B\left(s\right)=-\frac{1}{\left(s+75\right)}$

Therefore, the complete transfer function for the denominator part is,

$H_2\left(s\right)=H_A\left(s\right)H_B\left(s\right)$

Substitute $-\frac{1}{\left(s+75\right)}$ for $H_A\left(s\right)$ and $-\frac{1}{\left(s+75\right)}$ for $H_B\left(s\right)$ in the above equation.

$\begin{array}{c}{H}_2\left(s\right)=\left(-\frac{1}{s+75}\right)\left(-\frac{1}{s+75}\right)\\ =\frac{1}{{\left(s+75\right)}^2}\end{array}$

Thus, the final design of the two stages cascaded circuit for the denominator part is,

$R_{fA}=13.3\Omega$, $C_{fA}=1\text{mF}$, $R_{1A}=1\text{k}\Omega$, $R_{fB}=13.3\Omega$, $C_{fB}=1\text{mF}$, and $R_{2B}=1\text{k}\Omega$.

Thus, the overall transfer function of the complete circuit using the transfer function of numerator $H_1\left(s\right)$ and denominator $H_2\left(s\right)$ is,

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

Substitute $-3\left(s+50\right)$ for $H_1\left(s\right)$ and $\frac{1}{{\left(s+75\right)}^2}$ for $H_2\left(s\right)$ in the above equation.

$\begin{array}{c}H\left(s\right)=-3\left(s+50\right)\frac{1}{{\left(s+75\right)}^2}\\ =\frac{-3\left(s+50\right)}{{\left(s+75\right)}^2}\end{array}$

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

$H\left(s\right)=\frac{3\left(s+50\right)}{{\left(s+75\right)}^2}$

Conclusion:

Thus, a circuit is designed which produces a transfer function of $\frac{3\left(s+50\right)}{{\left(s+75\right)}^2}$.