Topic: Control Systems
The approximate Bode magnitude plot of a minimum-phase system is shown in the figure below. The transfer function of the system is
(Figure indicates an initial magnitude of 20 dB up to $\omega=0.1$, then a slope up to $\omega=10$ reaching 140 dB, then a slope up to $\omega=100$ reaching 160 dB, and constant thereafter).
- A. $10^8\frac{(s+0.1)^2}{(s+10)^2(s+100)}$
- B. $10^7\frac{(s+0.1)^3}{(s+10)^2(s+100)}$
- C. $10^8\frac{(s+0.1)^3}{(s+10)^2(s+100)}$
- D. $10^7\frac{(s+0.1)^2}{(s+10)^2(s+100)}$
Answer: C. $10^8\frac{(s+0.1)^3}{(s+10)^2(s+100)}$
Explanation: Slope after $\omega=0.1$ is +60 dB/dec. Thus, there are 3 zeros at 0.1.
Slope after $\omega=10$ changes from +60 dB/dec to +20 dB/dec (since it goes from 140 dB to 160 dB over one decade). This implies two poles at $\omega=10$.
Slope after $\omega=100$ is 0 dB/dec, implying 1 pole at $\omega=100$.
Gain is 20 dB for $\omega<0.1$, so $20\log_{10}K=20 \Rightarrow K=10$.
$T.F.=\frac{10(\frac{s}{0.1}+1)^3}{(\frac{s}{10}+1)^2(\frac{s}{100}+1)}=\frac{10 \times 10^3 \times (s+0.1)^3}{\frac{(s+10)^2}{100} \times \frac{(s+100)}{100}}=\frac{10^8(s+0.1)^3}{(s+10)^2(s+100)}$.