Chapter 14, Problem 21P

Sketch the magnitude Bode plot for

$H(s)=\frac{10s(s+20)}{(s+1)(s^2+60s+400)},$ s = jω

To determine

Sketch the Bode magnitude plots for given transfer function.

Explanation of Solution

Given data:

The transfer function is,

$H\left(s\right)=\frac{10s\left(s+20\right)}{\left(s+1\right)\left(s^2+60s+400\right)}$        (1)

Calculation:

Compare the denominator factor of equation (1) with the standard quadratic equation $\left(s^2+2\xi {\omega }_{n}s+{\omega }_n^2\right)$.

${\omega }_n^2=400$

${\omega }_n=20$

For the quadratic factor, the corner frequency is ${\omega }_n=20$.

Substitute $j\omega$ for $s$ in equation (1) to find $H\left(\omega \right)$.

$\begin{array}{c}H\left(\omega \right)=\frac{10j\omega \left(j\omega +20\right)}{\left(j\omega +1\right)\left({\left(j\omega \right)}^2+60j\omega +400\right)}\\ =\frac{10j\omega \left(j\omega +20\right)}{\left(j\omega +1\right)\left(j^2{\omega }^2+60j\omega +400\right)}\\ =\frac{10j\omega \left(j\omega +20\right)}{\left(j\omega +1\right)\left(\left(-1\right){\omega }^2+60j\omega +400\right)}\left\{\because j^2=-1\right\}\\ =\frac{10j\omega \left(j\omega +20\right)}{\left(j\omega +1\right)\left(400-{\omega }^2+60j\omega \right)}\end{array}$

Simplify the above equation as follows:

$\begin{array}{c}H\left(\omega \right)=\frac{10\left(20\right)j\omega \left(1+\frac{j\omega }{20}\right)}{400\left(1+j\omega \right)\left(1-\frac{{\omega }^2}{400}+\frac{60}{400}j\omega \right)}\\ =\frac{\left(0.5\right)\left(j\omega \right)\left(1+\frac{j\omega }{20}\right)}{\left(1+j\omega \right)\left(1-\frac{{\omega }^2}{400}+\frac{3}{20}j\omega \right)}\end{array}$

From the above equation, the corner frequencies $\left({\omega }_c\right)$ are $1$ and $20$.

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

$\begin{array}{c}H\left(\omega \right)=\frac{\left|0.5\right|\left|j\omega \right|\left|1+\frac{j\omega }{20}\right|}{\left(\left|1+j\omega \right|\right)\left(\left|1-\frac{{\omega }^2}{400}+\frac{3}{20}j\omega \right|\right)}\angle \left(\begin{array}{l}{\tan }^{-1}\left(\frac{0}{0.5}\right)+{\tan }^{-1}\left(\frac{\omega }{0}\right)+{\tan }^{-1}\left(\frac{\left(\frac{\omega }{20}\right)}{1}\right)\\ -{\tan }^{-1}\left(\frac{\omega }{1}\right)-{\tan }^{-1}\left(\frac{\left(\frac{3\omega }{20}\right)}{1-\frac{{\omega }^2}{400}}\right)\end{array}\right)\\ =\frac{\left|0.5\right|\left|j\omega \right|\left|1+\frac{j\omega }{20}\right|}{\left(\left|1+j\omega \right|\right)\left(\left|1-\frac{{\omega }^2}{400}+\frac{3}{20}j\omega \right|\right)}\angle \left(\begin{array}{l}{\tan }^{-1}\left(0\right)+{\tan }^{-1}\left(\infty \right)+{\tan }^{-1}\left(\frac{\omega }{20}\right)\\ -{\tan }^{-1}\left(\omega \right)-{\tan }^{-1}\left(\frac{3\omega }{20\left(1-\frac{{\omega }^2}{400}\right)}\right)\end{array}\right)\\ =\frac{\left|0.5\right|\left|j\omega \right|\left|1+\frac{j\omega }{20}\right|}{\left(\left|1+j\omega \right|\right)\left(\left|1-\frac{{\omega }^2}{400}+\frac{3}{20}j\omega \right|\right)}\angle \left(\begin{array}{l}0°+90°+{\tan }^{-1}\left(\frac{\omega }{20}\right)\\ -{\tan }^{-1}\left(\omega \right)\\ -{\tan }^{-1}\left(\frac{3\omega }{20\left(1-\frac{{\omega }^2}{400}\right)}\right)\end{array}\right)\left\{\begin{array}{l}\because \tan 0°=0\\ \tan 90°=\infty \end{array}\right\}\end{array}$

$H\left(\omega \right)=\frac{\left|0.5\right|\left|j\omega \right|\left|1+\frac{j\omega }{20}\right|}{\left(\left|1+j\omega \right|\right)\left(\left|1-\frac{{\omega }^2}{400}+\frac{3}{20}j\omega \right|\right)}\angle \left(\begin{array}{l}90°+{\tan }^{-1}\left(\frac{\omega }{20}\right)-{\tan }^{-1}\left(\omega \right)\\ -{\tan }^{-1}\left(\frac{3\omega }{20\left(1-\frac{{\omega }^2}{400}\right)}\right)\end{array}\right)$        (2)

From equation (2), the magnitude function of $H\left(\omega \right)$ is expressed as follows:

$\left|H\left(\omega \right)\right|=\frac{\left|0.5\right|\left|j\omega \right|\left|1+\frac{j\omega }{20}\right|}{\left(\left|1+j\omega \right|\right)\left(\left|1-\frac{{\omega }^2}{400}+\frac{3}{20}j\omega \right|\right)}$

Write the above equation in decibel (dB).

$\begin{array}{c}{H}_{\text{dB}}=\left|H\left(\omega \right)\right|=\frac{\left|0.5\right|\left|j\omega \right|\left|1+\frac{j\omega }{20}\right|}{\left(\left|1+j\omega \right|\right)\left(\left|1-\frac{{\omega }^2}{400}+\frac{3}{20}j\omega \right|\right)}\\ =\left[\begin{array}{l}20{\log }_{10}\left|0.5\right|+20{\log }_{10}\left|j\omega \right|+20{\log }_{10}\left|1+\frac{j\omega }{20}\right|\\ -20{\log }_{10}\left|1+j\omega \right|-20{\log }_{10}\left|1-\frac{{\omega }^2}{400}+\frac{3}{20}j\omega \right|\end{array}\right]\text{dB}\\ \text{=}\left[\begin{array}{l}20{\log }_{10}0.5+20{\log }_{10}\sqrt{0^2+{\omega }^2}+20{\log }_{10}\sqrt{1^2+{\left(\frac{\omega }{20}\right)}^2}\\ -20{\log }_{10}\sqrt{1^2+{\omega }^2}-20{\log }_{10}\sqrt{{\left(1-\frac{{\omega }^2}{400}\right)}^2+{\left(\frac{3\omega }{20}\right)}^2}\end{array}\right]\text{dB}\\ =\left[\begin{array}{l}20{\log }_{10}0.5+20{\log }_{10}\omega \\ +20{\log }_{10}\sqrt{1+\frac{{\omega }^2}{400}}-20{\log }_{10}\sqrt{1+{\omega }^2}\\ -20{\log }_{10}\sqrt{1^2-2\left(1\right)\frac{{\omega }^2}{400}+\frac{{\omega }^4}{160000}+\frac{{\omega }^2}{400}}\end{array}\right]\text{dB}\left\{\begin{array}{l}\because {\left(a+b\right)}^2\\ =a^2+2ab+b^2\end{array}\right\}\end{array}$$H_{\text{dB}}\text{=}\left[\begin{array}{l}20{\log }_{10}0.5+20{\log }_{10}\omega +20{\log }_{10}\sqrt{1+\frac{{\omega }^2}{400}}\\ -20{\log }_{10}\sqrt{1+{\omega }^2}-20{\log }_{10}\sqrt{1-\frac{{\omega }^2}{400}+\frac{{\omega }^4}{160000}}\end{array}\right]\text{dB}$        (3)

From equation (2), the phase angle is expressed as follows:

$\varphi =90°+{\tan }^{-1}\left(\frac{\omega }{20}\right)-{\tan }^{-1}\left(\omega \right)-{\tan }^{-1}\left(\frac{3\omega }{20\left(1-\frac{{\omega }^2}{400}\right)}\right)\text{for }\omega \le {\omega }_n$        (4)

$\varphi =90°+{\tan }^{-1}\left(\frac{\omega }{20}\right)-{\tan }^{-1}\left(\omega \right)-\left({\tan }^{-1}\left(\frac{3\omega }{20\left(1-\frac{{\omega }^2}{400}\right)}\right)+180°\right)\text{for}\omega \ge {\omega }_n$        (5)

Substitute $0.1$ for $\omega$ in equation (3) to find $H_{\text{dB}}$.

$\begin{array}{c}{H}_{\text{dB}}\text{=}\left[\begin{array}{l}20{\log }_{10}0.5+20{\log }_{10}0.1+20{\log }_{10}\sqrt{1+\frac{{0.1}^2}{400}}\\ -20{\log }_{10}\sqrt{1+{0.1}^2}-20{\log }_{10}\sqrt{1-\frac{{0.1}^2}{400}+\frac{{0.1}^4}{160000}}\end{array}\right]\text{dB}\\ =\left[\begin{array}{l}20{\log }_{10}0.5+20{\log }_{10}0.1+20{\log }_{10}\sqrt{1.000025}\\ -20{\log }_{10}\sqrt{1.01}-20{\log }_{10}\sqrt{0.999974999}\end{array}\right]\text{dB}\\ =-26.063\text{dB}\end{array}$

Substitute $0.1$ for $\omega$ in equation (4) to find $\varphi$.

$\begin{array}{c}\varphi =90°+{\tan }^{-1}\left(\frac{0.1}{20}\right)-{\tan }^{-1}\left(0.1\right)-{\tan }^{-1}\left(\frac{3\left(0.1\right)}{20\left(1-\frac{{0.1}^2}{400}\right)}\right)\\ =90°+{\tan }^{-1}\left(0.005\right)-{\tan }^{-1}\left(0.1\right)-{\tan }^{-1}\left(0.015\right)\\ =83.716°\end{array}$

Substitute $50$ for $\omega$ in equation (5) to find $\varphi$.

$\begin{array}{c}\varphi =90°+{\tan }^{-1}\left(\frac{50}{20}\right)-{\tan }^{-1}\left(50\right)-\left({\tan }^{-1}\left(\frac{3\left(50\right)}{20\left(1-\frac{{50}^2}{400}\right)}\right)+180°\right)\\ =90°+{\tan }^{-1}\left(2.5\right)-{\tan }^{-1}\left(50\right)-{\tan }^{-1}\left(-1.4285\right)-180°\\ =-55.647°\end{array}$

Similarly, by substituting various values for $\omega$ including the corner frequencies in equation (3), the Table (1) is obtained.

Table 1:

$\omega \left(\frac{\text{rad}}{\text{sec}}\right)$	0.1	1	2	10	20	50
$H_{\text{dB}}\left(\text{dB}\right)$	–26.063	–9.0092	–6.903	–4.192	–3.021	–12.709

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

Conclusion:

Thus, the Bode magnitude plot for given transfer function is sketched.