Chapter 15, Problem 1E

For the RL circuit in Fig. 15.52, (a) determine the transer function defined as H(jω) = v_out/v_in; (b) for the case of R = 200 Ω and L = 5 mH, construct a plot of the magnitude and phase as a function of frequency; and (c) evaluate the magnitude and phase at a frequency of 10 kHz.

FIGURE 15.52

To determine

Find the transfer function $H\left(j\omega \right)$ of the RL circuit shown in Figure 15.52.

Answer to Problem 1E

The transfer function $H\left(j\omega \right)$ of the RL circuit shown in Figure 15.52 is $\left(\frac{j\omega \left(\frac{L}{R}\right)}{1+j\omega \left(\frac{L}{R}\right)}\right)$.

Explanation of Solution

Given data:

Refer to Figure 15.52 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 RL circuit is drawn as Figure 1.

The Figure 1 is redrawn as impedance circuit in Figure 2 using the equations (1) and (2).

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

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

Here,

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

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

Apply Kirchhoff’s voltage law on Figure 2 to find $V_{\text{out}}$.

$\begin{array}{c}{V}_{\text{out}}=\frac{j\omega L}{R+j\omega L}{V}_{\text{in}}\\ =\frac{j\omega L}{R\left(1+\frac{j\omega L}{R}\right)}{V}_{\text{in}}\\ =\frac{j\omega \left(\frac{L}{R}\right)}{1+j\omega \left(\frac{L}{R}\right)}{V}_{\text{in}}\end{array}$

Rearrange the above equation to find $\frac{V_{\text{out}}}{V_{\text{in}}}$.

$\frac{V_{\text{out}}}{V_{\text{in}}}=\frac{j\omega \left(\frac{L}{R}\right)}{1+j\omega \left(\frac{L}{R}\right)}$

Substitute $\frac{j\omega \left(\frac{L}{R}\right)}{1+j\omega \left(\frac{L}{R}\right)}$ for $\frac{V_{\text{out}}}{V_{\text{in}}}$ in equation (3) to find $H\left(j\omega \right)$.

$H\left(j\omega \right)=\frac{j\omega \left(\frac{L}{R}\right)}{1+j\omega \left(\frac{L}{R}\right)}$

Conclusion:

Thus, the transfer function $H\left(j\omega \right)$ of the RL circuit shown in Figure 15.52 is $\left(\frac{j\omega \left(\frac{L}{R}\right)}{1+j\omega \left(\frac{L}{R}\right)}\right)$.

To determine

Plot the magnitude and phase as a function of frequency.

Explanation of Solution

Given data:

The value of the resistor $\left(R\right)$ is $200\Omega$.

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

Calculation:

From part (a), the transfer function is,

$H\left(j\omega \right)=\frac{j\omega \left(\frac{L}{R}\right)}{1+j\omega \left(\frac{L}{R}\right)}$        (3)

Substitute $200\Omega$ for $R$ and $5\text{mH}$ for $L$ in equation (3) to find $H\left(j\omega \right)$.

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

Simplify the above equation to find $H\left(j\omega \right)$.

$H\left(j\omega \right)=\frac{j\omega \left(2.5\times {10}^{-5}\right)}{1+j\omega \left(2.5\times {10}^{-5}\right)}$

Re-write the transfer function $H\left(j\omega \right)$ using its magnitude and phase functions as follows,

$\begin{array}{c}H\left(j\omega \right)=\frac{\left|\left(2.5\times {10}^{-5}\right)\right|\left|j\omega \right|}{\left|\left(1+j\omega \left(2.5\times {10}^{-5}\right)\right)\right|}\angle \left({\tan }^{-1}\left(\frac{0}{\left(2.5\times {10}^{-5}\right)}\right)+{\tan }^{-1}\left(\frac{\omega }{0}\right)-{\tan }^{-1}\left(\frac{\omega \left(2.5\times {10}^{-5}\right)}{1}\right)\right)\\ =\frac{\left|\left(2.5\times {10}^{-5}\right)\right|\left|j\omega \right|}{\left|\left(1+j\omega \left(2.5\times {10}^{-5}\right)\right)\right|}\angle \left({\tan }^{-1}\left(0\right)+{\tan }^{-1}\left(\infty \right)-{\tan }^{-1}\left(\omega \left(2.5\times {10}^{-5}\right)\right)\right)\\ =\frac{\left|\left(2.5\times {10}^{-5}\right)\right|\left|j\omega \right|}{\left|\left(1+j\omega \left(2.5\times {10}^{-5}\right)\right)\right|}\angle \left(0+90°-{\tan }^{-1}\left(\omega \left(2.5\times {10}^{-5}\right)\right)\right)\left\{\begin{array}{l}\because \tan 0°=0\\ \tan 90°=\infty \end{array}\right\}\end{array}$

$H\left(j\omega \right)=\frac{\left|\left(2.5\times {10}^{-5}\right)\right|\left|j\omega \right|}{\left|\left(1+j\omega \left(2.5\times {10}^{-5}\right)\right)\right|}\angle \left(90°-{\tan }^{-1}\left(\omega \left(2.5\times {10}^{-5}\right)\right)\right)$        (4)

From equation (4), the magnitude function of $H\left(j\omega \right)$ is expressed as follows:

$\left|H\left(j\omega \right)\right|=\frac{\left|\left(2.5\times {10}^{-5}\right)\right|\left|j\omega \right|}{\left|\left(1+j\omega \left(2.5\times {10}^{-5}\right)\right)\right|}$

Write the above equation in decibel (dB).

$\begin{array}{c}{H}_{\text{dB}}=\left[20{\log }_{10}\left|\left(2.5\times {10}^{-5}\right)\right|+20{\log }_{10}\left|j\omega \right|-20{\log }_{10}\left|\left(1+j\omega \left(2.5\times {10}^{-5}\right)\right)\right|\right]\text{dB}\\ \text{=}\left[20{\log }_{10}\sqrt{{\left(2.5\times {10}^{-5}\right)}^2+0^2}+20{\log }_{10}\sqrt{0^2+{\omega }^2}-20{\log }_{10}\sqrt{1^2+{\left(\omega \left(2.5\times {10}^{-5}\right)\right)}^2}\right]\text{dB}\end{array}$$H_{\text{dB}}=\left[20{\log }_{10}\left(2.5\times {10}^{-5}\right)+20{\log }_{10}\omega -20{\log }_{10}\sqrt{1+{\omega }^2{\left(2.5\times {10}^{-5}\right)}^2}\right]\text{dB}$        (5)

From equation (4), the phase angle is expressed as follows:

$\varphi =90°-{\tan }^{-1}\left(\omega \left(2.5\times {10}^{-5}\right)\right)$        (6)

Substitute $0.1$ for $\omega$ in equation (5) to find $H_{\text{dB}}$.

$\begin{array}{c}{H}_{\text{dB}}=\left[20{\log }_{10}\left(2.5\times {10}^{-5}\right)+20{\log }_{10}\left(0.1\right)-20{\log }_{10}\sqrt{1+{\left(0.1\right)}^2{\left(2.5\times {10}^{-5}\right)}^2}\right]\text{dB}\\ =\left[20{\log }_{10}\left(2.5\times {10}^{-5}\right)-20-20{\log }_{10}\sqrt{1+\left(0.01\right){\left(2.5\times {10}^{-5}\right)}^2}\right]\text{dB}\\ =\left[20{\log }_{10}\left(2.5\times {10}^{-5}\right)-20-20{\log }_{10}\sqrt{1}\right]\text{dB}\\ =-112\text{dB}\end{array}$

Substitute $0.1$ for $\omega$ in equation (6) to find $\varphi$.

$\begin{array}{c}\varphi =90°-{\tan }^{-1}\left(\left(0.1\right)\left(2.5\times {10}^{-5}\right)\right)\\ =90°-{\tan }^{-1}\left(2.5\times {10}^{-6}\right)\\ =90°\end{array}$

Similarly, by substituting various values for $\omega$ including the corner frequencies in equation (5) and equation (6), the Table (1) and (2) are obtained.

Table 1

$\omega \left(\frac{\text{rad}}{\text{sec}}\right)$	0.1	1	2	10	20	50	200
$H_{dB}\left(\text{dB}\right)$	–112	–92	–86	–72	–66	–58	–46

Table 2

$\omega \left(\frac{\text{rad}}{\text{sec}}\right)$	0.1	10	104	105	106
$\varphi \left(\mathrm{deg}\right)$	90	90	75.96	21.8	2.3

The Figure 1 is the magnitude plot of the given transfer function obtained using Table 1.

The Figure 2 is the phase plot of the given transfer function obtained using Table 2.

Conclusion:

Thus, the magnitude and phase as a function of frequency is plotted.

To determine

Find the value of the magnitude and phase at a frequency of $10\text{kHz}$.

Answer to Problem 1E

The value of the magnitude and phase at a frequency of $10\text{kHz}$ is $0.8436\frac{\text{rad}}{\text{s}}$ and $32.48°$ respectively.

Explanation of Solution

Given data:

The value of the frequency $\left(f\right)$ is $10\text{kHz}$.

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 expressed as,

$H\left(j\omega \right)=\frac{j\omega L}{R+j\omega L}$        (7)

From equation (7), the magnitude function is expressed as,

$\begin{array}{c}\left|H\left(j\omega \right)\right|=\frac{\left|j\omega L\right|}{\left|R+j\omega L\right|}\\ =\frac{\sqrt{{\left(\omega L\right)}^2}}{\sqrt{R^2+{\left(\omega L\right)}^2}}\\ =\frac{\omega L}{\sqrt{R^2+{\omega }^2{L}^2}}\end{array}$

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

$\begin{array}{c}\left|H\left(j\omega \right)\right|=\frac{2\pi fL}{\sqrt{R^2+{\left(2\pi f\right)}^2{L}^2}}\\ =\frac{2\pi fL}{\sqrt{R^2+4{\pi }^2{f}^2{L}^2}}\end{array}$

Substitute $10\text{kHz}$ for $f$, $200\Omega$ for $R$ and $5\text{mH}$ for $L$ in above equation to find $\left|H\left(j\omega \right)\right|$.

$\begin{array}{c}\left|H\left(j\omega \right)\right|=\frac{2\pi \left(10\times {10}^3\right)\left(5\times {10}^{-3}\right)}{\sqrt{{\left(200\right)}^2+4{\pi }^2{\left(10\times {10}^3\right)}^2{\left(5\times {10}^{-3}\right)}^2}}\left\{\because 1\text{k}={10}^3,1\text{m}={10}^{-3}\right\}\\ =\frac{314.16}{372.42}\\ =0.8436\frac{\text{rad}}{\text{s}}\end{array}$

From equation (7), the phase function is expressed as,

$\begin{array}{c}\varphi ={\tan }^{-1}\left(\frac{\omega L}{0}\right)-{\tan }^{-1}\left(\frac{\omega L}{R}\right)\\ ={\tan }^{-1}\left(\infty \right)-{\tan }^{-1}\left(\frac{\omega L}{R}\right)\\ =90°-{\tan }^{-1}\left(\frac{\omega L}{R}\right)\left\{\because \tan 90°=\infty \right\}\end{array}$

Substitute $2\pi f$ for $\omega$ in above equation to find $\varphi$.

$\varphi =90°-{\tan }^{-1}\left(\frac{2\pi fL}{R}\right)$

Substitute $10\text{kHz}$ for $f$, $200\Omega$ for $R$ and $5\text{mH}$ for $L$ in above equation to find $\varphi$.

$\begin{array}{c}\varphi =90°-{\tan }^{-1}\left(\frac{2\pi \left(10\times {10}^3\right)\left(5\times {10}^{-3}\right)}{200}\right)\left\{\because 1\text{k}={10}^3,1\text{m}={10}^{-3}\right\}\\ =90°-{\tan }^{-1}\left(\frac{314.16}{200}\right)\\ =32.48°\end{array}$

Conclusion:

Thus, the value of the magnitude and phase at a frequency of $10\text{kHz}$ is $0.8436\frac{\text{rad}}{\text{s}}$ and $32.48°$ respectively.