Chapter 15, Problem 7E

For the circuit in Fig. 15.57, (a) determine the transfer function H(s) = V_out/V_in in terms of circuit parameters R₁, R₂, R₃, L₁, and L₂; (b) determine the magnitude and phase of the transfer function at ω = 0, 3 × 10³ rad/s, and as ω → ∞ for the case where circuit values are R₁ = 2 kΩ, R₂ = 2 kΩ, R₃ = 20 kΩ, L₁ = 2 H, and L₂ = 2 H.

FIGURE 15.57

To determine

The transfer function $H\left(j\omega \right)$.

Answer to Problem 7E

The transfer function $H\left(j\omega \right)$ is $\frac{\frac{-j\omega L_2}{R_1}}{1-{\omega }^2\left[\left(\frac{L_1{L}_2}{R_1{R}_3}\right)+\left(\frac{L_1{L}_2}{R_2{R}_3}\right)\right]+j\omega \left(\frac{L_1+L_2}{R_3}\right)}$.

Explanation of Solution

Given data:

The required diagram is shown in Figure 1.

 

Calculation:

Assign the node intersecting the two resistors $R_1$ and $R_2$ and the two inductors $L_1$ and $L_2$ as ${\text{v}}_{\text{x}}$.

The KCL equation at node ${\text{v}}_{\text{x}}$ is written as,

$\frac{{\text{v}}_{\text{x}}-{\text{v}}_{\text{in}}}{R_1}+\frac{{\text{v}}_{\text{x}}}{R_2}+\frac{{\text{v}}_{\text{x}}{\text{-v}}_{\text{out}}}{j\omega L_2}+\frac{{\text{v}}_{\text{x}}}{j\omega L_1}=0$        (1)

The KCL equation at node ${\text{v}}_{\text{out}}$ is written as

$\frac{{\text{v}}_{\text{x}}}{j\omega L_1}+\frac{{\text{v}}_{\text{out}}}{R_3}=0$        (2)

Here,

$R_1$ is the resistor.

$R_2$ is the resistor.

$L_1$ is the inductor.

$L_2$ is the inductor.

$\omega$ is the angular frequency.

${\text{v}}_{\text{out}}$ is the output voltage.

${\text{v}}_{\text{in}}$ is the input current.

Simplifying the equation (2) as,

${\text{v}}_{\text{x}}=\frac{-j\omega L_1}{R_3}{\text{v}}_{\text{out}}$

The transfer function $H\left(j\omega \right)$ is written as,

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

Substitute $\frac{-j\omega L_1}{R_3}{\text{v}}_{\text{out}}$ for ${\text{v}}_{\text{x}}$ in equation (1).

$\begin{array}{l}\frac{\left(\frac{-j\omega L_1}{R_3}{\text{v}}_{\text{out}}\right)-{\text{v}}_{\text{in}}}{R_1}+\frac{\left(\frac{-j\omega L_1}{R_3}{\text{v}}_{\text{out}}\right)}{R_2}+\frac{\left(\frac{-j\omega L_1}{R_3}{\text{v}}_{\text{out}}\right){\text{-v}}_{\text{out}}}{j\omega L_2}+\frac{\left(\frac{-j\omega L_1}{R_3}{\text{v}}_{\text{out}}\right)}{j\omega L_1}=0\\ \left(\frac{-j\omega L_2}{R_1}\right){\text{v}}_{\text{in}}=\left(1-{\omega }^2\left[\left(\frac{L_1{L}_2}{R_1{R}_3}\right)+\left(\frac{L_1{L}_2}{R_2{R}_3}\right)\right]+j\omega \left(\frac{L_1+L_2}{R_3}\right)\right){\text{v}}_{\text{out}}\\ {\text{v}}_{\text{in}}=\frac{\left(1-{\omega }^2\left[\left(\frac{L_1{L}_2}{R_1{R}_3}\right)+\left(\frac{L_1{L}_2}{R_2{R}_3}\right)\right]+j\omega \left(\frac{L_1+L_2}{R_3}\right)\right)}{\left(\frac{-j\omega L_2}{R_1}\right)}{\text{v}}_{\text{out}}\end{array}$

Substitute $\frac{1-{\omega }^2\left[\left(\frac{L_1{L}_2}{R_1{R}_3}\right)+\left(\frac{L_1{L}_2}{R_2{R}_3}\right)\right]+j\omega \left(\frac{L_1+L_2}{R_3}\right)}{\frac{-j\omega L_2}{R_1}}{\text{v}}_{\text{out}}$ for ${\text{v}}_{\text{in}}$ in equation (3).

$\begin{array}{c}H\left(j\omega \right)=\frac{{\text{v}}_{\text{out}}}{\frac{1-{\omega }^2\left[\left(\frac{L_1{L}_2}{R_1{R}_3}\right)+\left(\frac{L_1{L}_2}{R_2{R}_3}\right)\right]+j\omega \left(\frac{L_1+L_2}{R_3}\right)}{\frac{-j\omega L_2}{R_1}}{\text{v}}_{\text{out}}}\\ =\frac{\frac{-j\omega L_2}{R_1}}{1-{\omega }^2\left[\left(\frac{L_1{L}_2}{R_1{R}_3}\right)+\left(\frac{L_1{L}_2}{R_2{R}_3}\right)\right]+j\omega \left(\frac{L_1+L_2}{R_3}\right)}\end{array}$

Conclusion:

Therefore, the transfer function $H\left(j\omega \right)$ is. $\frac{\frac{-j\omega L_2}{R_1}}{1-{\omega }^2\left[\left(\frac{L_1{L}_2}{R_1{R}_3}\right)+\left(\frac{L_1{L}_2}{R_2{R}_3}\right)\right]+j\omega \left(\frac{L_1+L_2}{R_3}\right)}$.

To determine

The magnitude of transfer function $\left|H\left(j\omega \right)\right|$ and phase of transfer function $\angle H\left(j\omega \right)$ for the given condition.

Answer to Problem 7E

The magnitude of transfer function $\left|H\left(j\omega \right)\right|$ at $0\text{rad/s}$ is $0$, at $3\times {10}^4\text{rad/s}$ is $3$ and at $\infty \text{rad/s}$ is $0$. The phase of transfer function $\angle H\left(j\omega \right)$ at $0\text{rad/s}$ is $-90°$, at $3\times {10}^4\text{rad/s}$ is $-113.80°$ and at $\infty \text{rad/s}$ is $-90°$.

Explanation of Solution

Given data:

The resistance $R_1$ is $2\text{k}\Omega$.

The resistance $R_2$ is $2\text{k}\Omega$.

The resistance $R_3$ is $20\text{k}\Omega$.

The inductance $L_1$ is $2\text{H}$.

The inductance $L_2$ is $2\text{H}$.

The angular frequency ${\omega }_1$ is $0$.

The angular frequency ${\omega }_2$ is $3\times {10}^3\text{rad/s}$.

The angular frequency ${\omega }_3$ is $\infty$.

Calculation:

The conversion of $\text{k}\Omega$ to $\Omega$ is written as,

$1\text{k}\Omega =1\times {10}^3\Omega$

The conversion of $\text{2}\text{k}\Omega$ to $\Omega$ is written as,

$2\text{k}\Omega =2\times {10}^3\Omega$

The conversion of $\text{20}\text{k}\Omega$ to $\Omega$ is written as,

$20\text{k}\Omega =20\times {10}^3\Omega$

The magnitude of transfer function $\left|H\left(j{\omega }_1\right)\right|$ is written as,

$\left|H\left(j{\omega }_1\right)\right|=\frac{\frac{{\omega }_1{L}_2}{R_1}}{\sqrt{{\left[1-{\omega }_1{}^2\left(\frac{L_1{L}_2}{R_1{R}_3}+\frac{L_1{L}_2}{R_2{R}_3}\right)\right]}^2+{\left[{\omega }_1\left(\frac{L_1+L_2}{R_3}\right)\right]}^2}}$        (4)

The phase of transfer function $\angle H\left(j{\omega }_1\right)$ is written as,

$\angle H\left(j{\omega }_1\right)=-180°+90°-{\tan }^{-1}\frac{{\omega }_1\left(\frac{L_1+L_2}{R_3}\right)}{1-{\omega }_1{}^2\left(\frac{L_1{L}_2}{R_1{R}_3}+\frac{L_1{L}_2}{R_2{R}_3}\right)}$        (5)

Substitute $0$ for ${\omega }_1$, $2\times {10}^3\Omega$ for $R_1$, $2\times {10}^3\Omega$ for $R_2$, $20\times {10}^3\Omega$ for $R_3$, $2\text{H}$ for $L_1$ and $2\text{H}$ for $L_2$ in equation (4).

$\begin{array}{c}\left|H\left(j{\omega }_1\right)\right|=\frac{\frac{\left(0\right)\times \left(2\text{H}\right)}{\left(2\times {10}^3\Omega \right)}}{\sqrt{\begin{array}{l}{\left[1-{\left(0\right)}^2\left(\frac{\left(2\text{H}\right)\times \left(2\text{H}\right)}{\left(2\times {10}^3\Omega \right)\times \left(20\times {10}^3\Omega \right)}+\frac{\left(2\text{H}\right)\times \left(2\text{H}\right)}{\left(2\times {10}^3\Omega \right)\times \left(20\times {10}^3\Omega \right)}\right)\right]}^2+\\ {\left[\left(0\right)\left(\frac{\left(2\text{H}\right)+\left(2\text{H}\right)}{\left(20\times {10}^3\Omega \right)}\right)\right]}^2\end{array}}}\\ =0\end{array}$

Substitute $0$ for ${\omega }_1$, $2\times {10}^3\Omega$ for $R_1$, $2\times {10}^3\Omega$ for $R_2$, $20\times {10}^3\Omega$ for $R_3$, $2\text{H}$ for $L_1$ and $2\text{H}$ for $L_2$ in equation (5).

$\begin{array}{c}\angle H\left(j{\omega }_1\right)=-180°+90°-{\tan }^{-1}\frac{\left(0\right)\left(\frac{\left(2\text{H}\right)+\left(2\text{H}\right)}{\left(20\times {10}^3\Omega \right)}\right)}{1-{\left(0\right)}^2\left(\frac{\left(2\text{H}\right)\left(2\text{H}\right)}{\left(2\times {10}^3\Omega \right)\left(20\times {10}^3\Omega \right)}+\frac{\left(2\text{H}\right)\left(2\text{H}\right)}{\left(2\times {10}^3\Omega \right)\left(20\times {10}^3\Omega \right)}\right)}\\ =-90°-{\tan }^{-1}0\\ =-90°\end{array}$

The magnitude of transfer function $\left|H\left(j{\omega }_2\right)\right|$ is written as,

$\left|H\left(j{\omega }_2\right)\right|=\frac{\frac{{\omega }_2{L}_2}{R_1}}{\sqrt{{\left[1-{\omega }_2{}^2\left(\frac{L_1{L}_2}{R_1{R}_3}+\frac{L_1{L}_2}{R_2{R}_3}\right)\right]}^2+{\left[{\omega }_2\left(\frac{L_1+L_2}{R_3}\right)\right]}^2}}$        (6)

The phase of transfer function $\angle H\left(j{\omega }_2\right)$ is written as,

$\angle H\left(j{\omega }_2\right)=-180°+90°-{\tan }^{-1}\frac{{\omega }_2\left(\frac{L_1+L_2}{R_3}\right)}{1-{\omega }_2{}^2\left(\frac{L_1{L}_2}{R_1{R}_3}+\frac{L_1{L}_2}{R_2{R}_3}\right)}$        (7)

Substitute $3\times {10}^3\text{rad/s}$ for ${\omega }_2$, $2\times {10}^3\Omega$ for $R_1$, $2\times {10}^3\Omega$ for $R_2$, $20\times {10}^3\Omega$ for $R_3$, $2\text{H}$ for $L_1$ and $2\text{H}$ for $L_2$ in equation (6).

$\begin{array}{c}\left|H\left(j{\omega }_2\right)\right|=\frac{\frac{\left(3\times {10}^3\text{rad/s}\right)\times \left(2\text{H}\right)}{\left(2\times {10}^3\Omega \right)}}{\sqrt{\begin{array}{l}{\left[1-{\left(3\times {10}^3\text{rad/s}\right)}^2\left(\frac{\left(2\text{H}\right)\times \left(2\text{H}\right)}{\left(2\times {10}^3\Omega \right)\times \left(20\times {10}^3\Omega \right)}+\frac{\left(2\text{H}\right)\times \left(2\text{H}\right)}{\left(2\times {10}^3\Omega \right)\times \left(20\times {10}^3\Omega \right)}\right)\right]}^2+\\ {\left[\left(3\times {10}^3\text{rad/s}\right)\left(\frac{\left(2\text{H}\right)+\left(2\text{H}\right)}{\left(20\times {10}^3\Omega \right)}\right)\right]}^2\end{array}}}\\ =\frac{3}{\sqrt{{\left[1-\left(9\times {10}^6\right)\left({10}^{-7}+{10}^{-7}\right)\right]}^2+{\left[6\times {10}^{-1}\right]}^2}}\\ =\frac{3}{\sqrt{{\left[1-\left(9\times {10}^{-8}\right)\right]}^2+{\left[0.6\right]}^2}}\\ \approx 3\end{array}$

Substitute $3\times {10}^3\text{rad/s}$ for ${\omega }_2$, $2\times {10}^3\Omega$ for $R_1$, $2\times {10}^3\Omega$ for $R_2$, $20\times {10}^3\Omega$ for $R_3$, $2\text{H}$ for $L_1$ and $2\text{H}$ for $L_2$ in equation (7).

$\begin{array}{c}\angle H\left(j{\omega }_2\right)=-180°+90°-{\tan }^{-1}\frac{\left(3\times {10}^3\text{rad/s}\right)\left(\frac{\left(2\text{H}\right)+\left(2\text{H}\right)}{\left(20\times {10}^3\Omega \right)}\right)}{1-{\left(3\times {10}^3\text{rad/s}\right)}^2\left(\begin{array}{l}\frac{\left(2\text{H}\right)\left(2\text{H}\right)}{\left(2\times {10}^3\Omega \right)\left(20\times {10}^3\Omega \right)}+\\ \frac{\left(2\text{H}\right)\left(2\text{H}\right)}{\left(2\times {10}^3\Omega \right)\left(20\times {10}^3\Omega \right)}\end{array}\right)}\\ =-90°-{\tan }^{-1}\frac{0.6}{{\left[1-\left(9\times {10}^6\right)\left({10}^{-7}+{10}^{-7}\right)\right]}^2+{\left[6\times {10}^{-1}\right]}^2}\\ =-90°-{\tan }^{-1}\frac{0.6}{{\left[1-\left(9\times {10}^{-8}\right)\right]}^2+{\left[0.6\right]}^2}\\ =-113.80°\end{array}$

The magnitude of transfer function $\left|H\left(j{\omega }_3\right)\right|$ is written as,

$\begin{array}{c}\left|H\left(j{\omega }_3\right)\right|=\frac{\frac{{\omega }_3{L}_2}{R_1}}{\sqrt{{\left[1-{\omega }_3{}^2\left(\frac{L_1{L}_2}{R_1{R}_3}+\frac{L_1{L}_2}{R_2{R}_3}\right)\right]}^2+{\left[{\omega }_3\left(\frac{L_1+L_2}{R_3}\right)\right]}^2}}\\ =\frac{\frac{{\omega }_3{L}_2}{R_1}}{{\omega }_3{}^2\sqrt{{\left[\frac{1}{{\omega }_3{}^4}-\frac{1}{{\omega }_3{}^2}\left(\frac{L_1{L}_2}{R_1{R}_3}+\frac{L_1{L}_2}{R_2{R}_3}\right)\right]}^2+{\left[\frac{1}{{\omega }_3{}^2}\left(\frac{L_1+L_2}{R_3}\right)\right]}^2}}\\ =\frac{\frac{L_2}{R_1}}{{\omega }_3\sqrt{{\left[\frac{1}{{\omega }_3{}^4}-\frac{1}{{\omega }_3{}^2}\left(\frac{L_1{L}_2}{R_1{R}_3}+\frac{L_1{L}_2}{R_2{R}_3}\right)\right]}^2+{\left[\frac{1}{{\omega }_3{}^2}\left(\frac{L_1+L_2}{R_3}\right)\right]}^2}}\end{array}$        (8)

The phase of transfer function $\angle H\left(j{\omega }_3\right)$ is written as,

$\angle H\left(j{\omega }_3\right)=-180°+90°-{\tan }^{-1}\frac{\left(\frac{L_1+L_2}{R_3}\right)}{{\omega }_3\left[\frac{1}{{\omega }_3{}^2}-\left(\frac{L_1{L}_2}{R_1{R}_3}+\frac{L_1{L}_2}{R_2{R}_3}\right)\right]}$        (9)

Substitute $\infty$ for ${\omega }_3$, $2\times {10}^3\Omega$ for $R_1$, $2\times {10}^3\Omega$ for $R_2$, $20\times {10}^3\Omega$ for $R_3$, $2\text{H}$ for $L_1$ and $2\text{H}$ for $L_2$ in equation (8).

$\begin{array}{c}\left|H\left(j{\omega }_3\right)\right|=\frac{\frac{\left(2\text{H}\right)}{\left(2\times {10}^3\Omega \right)}}{\left(\infty \right)\sqrt{\begin{array}{l}{\left[\frac{1}{{\left(\infty \right)}^4}-\frac{1}{{\left(\infty \right)}^2}\left(\frac{\left(2\text{H}\right)\times \left(2\text{H}\right)}{\left(2\times {10}^3\Omega \right)\times \left(20\times {10}^3\Omega \right)}+\frac{\left(2\text{H}\right)\times \left(2\text{H}\right)}{\left(2\times {10}^3\Omega \right)\times \left(20\times {10}^3\Omega \right)}\right)\right]}^2+\\ {\left[\frac{1}{{\left(\infty \right)}^2}\left(\frac{\left(2\text{H}\right)+\left(2\text{H}\right)}{\left(20\times {10}^3\Omega \right)}\right)\right]}^2\end{array}}}\\ =\frac{{10}^{-3}}{\infty }\\ =0\end{array}$

Substitute $\infty$ for ${\omega }_3$, $2\times {10}^3\Omega$ for $R_1$, $2\times {10}^3\Omega$ for $R_2$, $20\times {10}^3\Omega$ for $R_3$, $2\text{H}$ for $L_1$ and $2\text{H}$ for $L_2$ in equation (9).

$\begin{array}{c}\angle H\left(j{\omega }_3\right)=-180°+90°-{\tan }^{-1}\frac{\left(\frac{\left(2\text{H}\right)+\left(2\text{H}\right)}{\left(20\times {10}^3\Omega \right)}\right)}{\left(\infty \right)\left[\frac{1}{{\left(\infty \right)}^2}-\left(\begin{array}{l}\frac{\left(2\text{H}\right)\left(2\text{H}\right)}{\left(2\times {10}^3\Omega \right)\left(20\times {10}^3\Omega \right)}+\\ \frac{\left(2\text{H}\right)\left(2\text{H}\right)}{\left(2\times {10}^3\Omega \right)\left(20\times {10}^3\Omega \right)}\end{array}\right)\right]}\\ =-90°-{\tan }^{-1}\left(0\right)\\ =-90°\end{array}$

Conclusion:

Therefore, the magnitude of transfer function $\left|H\left(j\omega \right)\right|$ at $0\text{rad/s}$ is $0$, at $3\times {10}^4\text{rad/s}$ is $3$ and at $\infty \text{rad/s}$ is $0$. The phase of transfer function $\angle H\left(j\omega \right)$ at $0\text{rad/s}$ is $-90°$, at $3\times {10}^4\text{rad/s}$ is $-113.80°$ and at $\infty \text{rad/s}$ is $-90°$.