Chapter 15, Problem 5E

For the circuit in Fig. 15.55, (a) derive an algebraic expression for the transfer function H(jω) = i_out/v_in in terms of circuit components R₁, R₂, L₁, and L₂; (b) evaluate the the magnitude of H(jω) at frequencies of 10 kHz, 1 MHz, and 100 MHz for the case where R₁ = 3 kΩ, R₂ = 12 kΩ, L₁ = 5 mH, and L₂ = 8 mH; (c) qualitatively, explain the behavior of the transfer function magnitude frequency response.

■ FIGURE 15.55

To determine

Find an algebraic expression for the transfer function $H\left(j\omega \right)$ of the circuit shown in Figure 15.55.

Answer to Problem 5E

The transfer function $H\left(j\omega \right)$ of the circuit shown in Figure 15.55 is $\frac{j\omega L_1}{R_1{R}_2-{\omega }^2{L}_1{L}_2+j\omega \left(R_1{L}_1+R_1{L}_2+R_2{L}_1\right)}$.

Explanation of Solution

Given data:

Refer to Figure 15.55 in the textbook.

Formula used:

Write the expression to calculate the impedance of the passive elements resistor and inductor.

$Z_R=R$        (1)

$Z_L=\frac{1}{j\omega L}$        (2)

Here,

$\omega$ is the angular frequency,

$R$ is the value of the resistor, and

$L$ is the value of the inductor.

Calculation:

The given circuit is redrawn as Figure 1.

The impedance circuit of the Figure 1 is drawn as Figure 2 using the equations (1) and (2).

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

$H\left(j\omega \right)=\frac{i_{\text{out}}}{v_{\text{in}}}$        (3)

Here,

$i_{\text{out}}$ is the output response of the system, and

$v_{\text{in}}$ is the input response of the system.

Use nodal analysis on node $v$ in the Figure 2.

$\begin{array}{l}\frac{v-v_{\text{in}}}{R_1}+\frac{v}{j\omega L_1}+\frac{v}{R_2+j\omega L_2}=0\\ \frac{v}{R_1}-\frac{v_{\text{in}}}{R_1}+\frac{v}{j\omega L_1}+\frac{v}{R_2+j\omega L_2}=0\\ \frac{v}{R_1}+\frac{v}{j\omega L_1}+\frac{v}{R_2+j\omega L_2}=\frac{v_{\text{in}}}{R_1}\\ v\left(\frac{1}{R_1}+\frac{1}{j\omega L_1}+\frac{1}{R_2+j\omega L_2}\right)=\frac{v_{\text{in}}}{R_1}\end{array}$

Rearrange the above equation to find $v$.

$\begin{array}{c}v=\frac{v_{\text{in}}}{R_1\left(\frac{1}{R_1}+\frac{1}{j\omega L_1}+\frac{1}{R_2+j\omega L_2}\right)}\\ =\frac{v_{\text{in}}}{\left(1+\frac{R_1}{j\omega L_1}+\frac{R_1}{R_2+j\omega L_2}\right)}\end{array}$

The output current $i_{\text{out}}$ in the Figure 2 is expressed as,

$i_{\text{out}}=\frac{v}{R_2+j\omega L_2}$

Substitute $\frac{v_{\text{in}}}{\left(1+\frac{R_1}{j\omega L_1}+\frac{R_1}{R_2+j\omega L_2}\right)}$ for $v$ in above equation to find $i_{\text{out}}$.

$\begin{array}{c}{i}_{\text{out}}=\frac{\left(\frac{v_{\text{in}}}{\left(1+\frac{R_1}{j\omega L_1}+\frac{R_1}{R_2+j\omega L_2}\right)}\right)}{R_2+j\omega L_2}\\ =\frac{v_{\text{in}}}{R_2+j\omega L_2\left(1+\frac{R_1}{j\omega L_1}+\frac{R_1}{R_2+j\omega L_2}\right)}\\ =\frac{v_{\text{in}}}{\left(R_2+j\omega L_2+\frac{R_1{R}_2}{j\omega L_1}+\frac{R_1{L}_2}{L_1}+\frac{R_1\left(R_2+j\omega L_2\right)}{R_2+j\omega L_2}\right)}\\ =\frac{v_{\text{in}}}{\left(R_2+j\omega L_2+\frac{R_1{R}_2}{j\omega L_1}+\frac{R_1{L}_2}{L_1}+R_1\right)}\end{array}$

Simplify the above equation to find $\frac{i_{\text{out}}}{v_{\text{in}}}$.

$\begin{array}{c}\frac{i_{\text{out}}}{v_{\text{in}}}=\frac{1}{\left(R_2+j\omega L_2+\frac{R_1{R}_2}{j\omega L_1}+\frac{R_1{L}_2}{L_1}+R_1\right)}\\ =\frac{1}{\left(\frac{j\omega R_2{L}_1+{\left(j\omega \right)}^2{L}_1{L}_2+R_1{R}_2+j\omega R_1{L}_2+j\omega R_1{L}_1}{j\omega L_1}\right)}\\ =\frac{j\omega L_1}{j\omega R_2{L}_1-{\omega }^2{L}_1{L}_2+R_1{R}_2+j\omega R_1{L}_2+j\omega R_1{L}_1}\left\{\because j^2=-1\right\}\\ =\frac{j\omega L_1}{R_1{R}_2-{\omega }^2{L}_1{L}_2+j\omega \left(R_1{L}_1+R_1{L}_2+R_2{L}_1\right)}\end{array}$

Substitute $\frac{j\omega L_1}{R_1{R}_2-{\omega }^2{L}_1{L}_2+j\omega \left(R_1{L}_1+R_1{L}_2+R_2{L}_1\right)}$ for $\frac{i_{\text{out}}}{v_{\text{in}}}$ in equation (3) to find $H\left(j\omega \right)$.

$H\left(j\omega \right)=\frac{j\omega L_1}{R_1{R}_2-{\omega }^2{L}_1{L}_2+j\omega \left(R_1{L}_1+R_1{L}_2+R_2{L}_1\right)}$

Conclusion:

Thus, the transfer function $H\left(j\omega \right)$ of the circuit shown in Figure 15.55 is $\frac{j\omega L_1}{R_1{R}_2-{\omega }^2{L}_1{L}_2+j\omega \left(R_1{L}_1+R_1{L}_2+R_2{L}_1\right)}$.

To determine

Find the value of magnitude of $H\left(j\omega \right)$ at frequencies $10\text{kHz}$, $1\text{MHz}$ and $100\text{MHz}$.

Answer to Problem 5E

The value of magnitude of the transfer function $\left|H\left(j\omega \right)\right|$ at frequencies $10\text{kHz}$, $1\text{MHz}$ and $100\text{MHz}$ is $8.636\mu \text{S}$, $18.88\mu \text{S}$ and $0.198\mu \text{S}$ respectively.

Explanation of Solution

Given data:

The value of the resistor 1 $\left(R_1\right)$ is $3\text{k}\Omega$.

The value of the resistor 2 $\left(R_2\right)$ is $12\text{k}\Omega$.

The value of the inductor 1 $\left(L_1\right)$ is $5\text{mH}$.

The value of the inductor 2 $\left(L_2\right)$ is $8\text{mH}$.

Formula used:

Write the expression to calculate the angular frequency.

$\omega =2\pi f$

Here,

$f$ is the value of the frequency.

Calculation:

From part (a), the transfer function is,

$H\left(j\omega \right)=\frac{j\omega L_1}{R_1{R}_2-{\omega }^2{L}_1{L}_2+j\omega \left(R_1{L}_1+R_1{L}_2+R_2{L}_1\right)}$

Substitute $2\pi f$ for $\omega$ in above equation to find $H\left(j\omega \right)$.

$\begin{array}{c}H\left(j\omega \right)=\frac{j\left(2\pi f\right)L_1}{R_1{R}_2-{\left(2\pi f\right)}^2{L}_1{L}_2+j\left(2\pi f\right)\left(R_1{L}_1+R_1{L}_2+R_2{L}_1\right)}\\ =\frac{j2\pi f{L}_1}{R_1{R}_2-4{\pi }^2{f}^2{L}_1{L}_2+j2\pi f\left(R_1{L}_1+R_1{L}_2+R_2{L}_1\right)}\end{array}$

Substitute $3\text{k}\Omega$ for $R_1$, $12\text{k}\Omega$ for $R_2$, $5\text{mH}$ for $L_1$ and $8\text{mH}$ for $L_2$ in above equation to find $H\left(j\omega \right)$.

$\begin{array}{c}H\left(j\omega \right)=\frac{j2\pi f\left(5\times {10}^{-3}\right)}{\left(\begin{array}{l}\left(3\times {10}^3\right)\left(12\times {10}^3\right)-4{\pi }^2{f}^2\left(5\times {10}^{-3}\right)\left(8\times {10}^{-3}\right)\\ +j2\pi f\left(\begin{array}{l}\left(3\times {10}^3\right)\left(5\times {10}^{-3}\right)+\left(3\times {10}^3\right)\left(8\times {10}^{-3}\right)\\ +\left(12\times {10}^3\right)\left(5\times {10}^{-3}\right)\end{array}\right)\end{array}\right)}\left\{\begin{array}{l}\because 1\text{k}={10}^3,\\ 1\text{m}={10}^{-3}\end{array}\right\}\\ =\frac{jf\left(10\pi \times {10}^{-3}\right)}{\left(\left(36\times {10}^6\right)-f^2\left(1.6{\pi }^2\times {10}^{-4}\right)+j2\pi f\left(15+24+60\right)\right)}\end{array}$

$H\left(j\omega \right)=\frac{jf\left(10\pi \times {10}^{-3}\right)}{\left(\left(36\times {10}^6\right)-f^2\left(1.6{\pi }^2\times {10}^{-4}\right)+j198\pi f\right)}$        (4)

For frequency of 10 kHz:

Substitute $10\text{kHz}$ for $f$ in equation (4) to find $H\left(j\omega \right)$.

$\begin{array}{c}H\left(j\omega \right)=\frac{j\left(10\times {10}^3\right)\left(10\pi \times {10}^{-3}\right)}{\left(\left(36\times {10}^6\right)-{\left(10\times {10}^3\right)}^2\left(1.6{\pi }^2\times {10}^{-4}\right)+j198\pi \left(10\times {10}^3\right)\right)}\left\{\because 1\text{k}={10}^3\right\}\\ =\frac{j\left(314.159\right)}{\left(\left(36\times {10}^6\right)-\left(157.913\times {10}^3\right)+j\left(622.035\times {10}^4\right)\right)}\\ =\frac{j\left(314.159\right)}{\left(\left(358.42\times {10}^5\right)+j\left(622.035\times {10}^4\right)\right)}\end{array}$

Take magnitude for above equation to find $\left|H\left(j\omega \right)\right|$.

$\begin{array}{c}\left|H\left(j\omega \right)\right|=\left|\frac{j\left(314.159\right)}{\left(\left(358.42\times {10}^5\right)+j\left(622.035\times {10}^4\right)\right)}\right|\\ =\frac{\sqrt{{\left(314.159\right)}^2}}{\sqrt{\left({\left(358.42\times {10}^5\right)}^2+{\left(622.035\times {10}^4\right)}^2\right)}}\\ =8.636\times {10}^{-6}\text{S}\\ =8.636\mu \text{S}\left\{\because 1\mu ={10}^{-6}\right\}\end{array}$

For frequency of 1 MHz:

Substitute $1\text{MHz}$ for $f$ in equation (4) to find $H\left(j\omega \right)$.

$\begin{array}{c}H\left(j\omega \right)=\frac{j\left(1\times {10}^6\right)\left(10\pi \times {10}^{-3}\right)}{\left(\left(36\times {10}^6\right)-{\left(1\times {10}^6\right)}^2\left(1.6{\pi }^2\times {10}^{-4}\right)+j198\pi \left(1\times {10}^6\right)\right)}\left\{\because 1\text{M}={10}^6\right\}\\ =\frac{j\left(31.416\times {10}^3\right)}{\left(\left(36\times {10}^6\right)-\left(157.913\times {10}^7\right)+j\left(622.035\times {10}^6\right)\right)}\\ =\frac{j\left(31.416\times {10}^3\right)}{\left(-\left(1543.13\times {10}^6\right)+j\left(622.035\times {10}^6\right)\right)}\end{array}$

Take magnitude for above equation to find $\left|H\left(j\omega \right)\right|$.

$\begin{array}{c}\left|H\left(j\omega \right)\right|=\left|\frac{j\left(31.416\times {10}^3\right)}{\left(-\left(1543.13\times {10}^6\right)+j\left(622.035\times {10}^6\right)\right)}\right|\\ =\frac{\sqrt{{\left(31.416\times {10}^3\right)}^2}}{\sqrt{\left({\left(-1543.13\times {10}^6\right)}^2+{\left(622.035\times {10}^6\right)}^2\right)}}\\ =18.88\times {10}^{-6}\text{S}\\ =18.88\mu \text{S}\left\{\because 1\mu ={10}^{-6}\right\}\end{array}$

For frequency of 100 MHz:

Substitute $100\text{MHz}$ for $f$ in equation (4) to find $H\left(j\omega \right)$.

$\begin{array}{c}H\left(j\omega \right)=\frac{j\left(100\times {10}^6\right)\left(10\pi \times {10}^{-3}\right)}{\left(\left(36\times {10}^6\right)-{\left(100\times {10}^6\right)}^2\left(1.6{\pi }^2\times {10}^{-4}\right)+j198\pi \left(100\times {10}^6\right)\right)}\left\{\because 1\text{M}={10}^6\right\}\\ =\frac{j\left(31.416\times {10}^5\right)}{\left(\left(36\times {10}^6\right)-\left(1.579\times {10}^{13}\right)+j\left(6.22\times {10}^{10}\right)\right)}\\ =\frac{j\left(31.416\times {10}^5\right)}{\left(-\left(1.579\times {10}^{13}\right)+j\left(6.22\times {10}^{10}\right)\right)}\end{array}$

Take magnitude for above equation to find $\left|H\left(j\omega \right)\right|$.

$\begin{array}{c}\left|H\left(j\omega \right)\right|=\left|\frac{j\left(31.416\times {10}^5\right)}{\left(-\left(1.579\times {10}^{13}\right)+j\left(6.22\times {10}^{10}\right)\right)}\right|\\ =\frac{\sqrt{{\left(31.416\times {10}^5\right)}^2}}{\sqrt{\left({\left(-1.579\times {10}^{13}\right)}^2+{\left(6.22\times {10}^{10}\right)}^2\right)}}\\ =0.198\times {10}^{-6}\text{S}\\ =0.198\mu \text{S}\left\{\because 1\mu ={10}^{-6}\right\}\end{array}$

Conclusion:

Thus, the value of magnitude of the transfer function $\left|H\left(j\omega \right)\right|$ at frequencies $10\text{kHz}$, $1\text{MHz}$ and $100\text{MHz}$ is $8.636\mu \text{S}$, $18.88\mu \text{S}$ and $0.198\mu \text{S}$ respectively.

To determine

Explain the behavior of the transfer function magnitude frequency response.

Explanation of Solution

Discussion:

At lower frequencies $\left(f=0\text{Hz}\right)$, the inductor is short. Therefore, all the current flows through the inductor $L_1$. At higher frequencies $\left(f=\infty \right)$, the inductor is made open, therefore the current $i_{\text{out}}$ is zero.

It is clear that the current $i_{\text{out}}$ is minimum at the lowest and the highest frequencies and the current $i_{\text{out}}$ at medium frequencies provides the current flow that too less than the current through the inductor $L_1$.

The magnitude approaches zero at low and high frequencies and at medium frequencies, the magnitude approaches maximum value. It implies the characteristics of the band-pass filter.

Conclusion:

Thus, the behavior of the transfer function magnitude frequency response is explained.