Question

1. a) the block diagram of a control system is shown in fig below; if the maximum overshoot is no more than 25%, and the settling time of the unit - step response of the closed - loop system is 0.1 sec, find the gain k and pole location p of the compensator. (10 pt)

Explanation:

Step1: Find the closed - loop transfer function

The open - loop transfer function $G(s)=\frac{s + K}{s + p}\cdot\frac{10}{s + 25}=\frac{10(s + K)}{(s + p)(s + 25)}$. The closed - loop transfer function $T(s)=\frac{G(s)}{1+G(s)}=\frac{10(s + K)}{(s + p)(s + 25)+10(s + K)}=\frac{10(s + K)}{s^{2}+(p + 25 + 10)s+(25p+10K)}$.

Step2: Relate overshoot to damping ratio

The maximum overshoot $M_p = e^{-\frac{\pi\zeta}{\sqrt{1-\zeta^{2}}}}$. Given $M_p\leq0.25$, then $e^{-\frac{\pi\zeta}{\sqrt{1-\zeta^{2}}}}\leq0.25$. Taking the natural logarithm of both sides: $-\frac{\pi\zeta}{\sqrt{1-\zeta^{2}}}\leq\ln(0.25)$. Solving for $\zeta$, we get $\zeta\geq0.403$.

Step3: Relate settling time to natural frequency

The settling time $t_s=\frac{4}{\zeta\omega_n}$ (for a $2\%$ criterion). Given $t_s = 0.1$ s, then $\zeta\omega_n=\frac{4}{t_s}=40$.

Step4: For the closed - loop transfer function $s^{2}+(p + 35)s+(25p + 10K)$

We know that for a second - order system $s^{2}+2\zeta\omega_ns+\omega_n^{2}$, $2\zeta\omega_n=p + 35$ and $\omega_n^{2}=25p + 10K$. Since $\zeta\omega_n = 40$, then $p+35 = 80$ (from $2\zeta\omega_n=p + 35$), so $p = 45$.
Substitute $p = 45$ into $\omega_n^{2}=25p + 10K$. Since $\zeta\omega_n = 40$ and $\zeta\geq0.403$, assume $\zeta = 0.403$, then $\omega_n=\frac{40}{\zeta}\approx99.25$. $\omega_n^{2}=25\times45+10K$. So $99.25^{2}=1125 + 10K$. Solving for $K$:

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Answer:

$K\approx872.56$, $p = 45$