Topic: Signals and Systems
The system with the transfer function $\frac{Y(s)}{X(s)}=\frac{s}{s+p}$ has an output $y(t)=\cos(2t-\frac{\pi}{3})$ for the input signal $x(t)=p\cos(2t-\frac{\pi}{2})$. Then, the system parameter p is
- A. $\sqrt{3}$
- B. 1
- C. $2/\sqrt{3}$
- D. $\sqrt{3}/2$
Answer: C. $2/\sqrt{3}$
Explanation: $\frac{Y(s)}{X(s)}=\frac{s}{s+p}=H(s)$
$\angle H(s)=90^\circ-\tan^{-1}(\frac{\omega}{p})$
$\angle y(t)=\angle x(t)+\angle H(s)$
$-\frac{\pi}{3}=-\frac{\pi}{2}+90^\circ-\tan^{-1}(\frac{\omega}{p})$
where $\omega=2$. So, $-\frac{\pi}{3}=-\frac{\pi}{2}+\frac{\pi}{2}-\tan^{-1}(\frac{2}{p})$
$\tan^{-1}(\frac{2}{p})=\frac{\pi}{3}$
$\frac{2}{p}=\tan\frac{\pi}{3}$
$p=\frac{2}{\sqrt{3}}$