Question 10.6: A transfer function is given by H(s)= 72s/(s+3)(s+6).Sketch ......
A transfer function is given by H(s)={\frac{72s}{(s+3)(s+6)}}. Sketch the Bode plots.
Substituting s = jω in the transfer function yields,
H(\omega)={\frac{72(j\omega)}{(j\omega+3)(j\omega+6)}} (10.190)
H(\omega)=\frac{72(j\omega)}{3(\frac{j\omega}{3}+1)\times6\left(\frac{j\omega}{6}+1\right)} (10.191)
H(\omega)=\frac{4(j\omega)}{(\frac{j\omega}{3}+1)\left(\frac{j\omega}{6}+1\right)} (10.192)
The expressions of the magnitude and phase are,
H_{\mathrm{dB}}=20\log_{10}(4)+20\log_{10}(j\omega)-20\log_{10}\biggl|\frac{j\omega}{3}+1\biggr|-20\log_{10}\biggl|\frac{j\omega}{6}+1\biggr| (10.193)
\phi=90^{\circ}-\tan^{-1}\left({\frac{\omega}{3}}\right)-\tan^{-1}\left({\frac{\omega}{6}}\right) (10.194)
The magnitude and phase plots are shown in Figs. 10.34 and 10.35, respectively.
