Question

The Closed loop transfer function of a control system is given by $\frac{C\left(s\right)}{R\left(s\right)}=\frac{1}{s+1}$. For the input r(t) = sin t, the steady response c(t)______.

• A. $\frac{1}{\sqrt{2}}cost$
• B. $\frac{1}{\sqrt{2}}sin(t+\frac{\pi }{4})$
• C. 1
• D. $\frac{1}{\sqrt{2}}sin(t-\frac{\pi }{4})$

Answer 1

The correct option is D $\frac{1}{\sqrt{2}}sin(t-\frac{\pi }{4})$

$Transferfunction=\frac{1}{s+1}=G\left(s\right)$

$G\left(j\omega \right)=\frac{1}{j\omega +1}=\frac{1}{\sqrt{{\omega }^2+1}}\angle -ta{n}^{-1}\omega$

$Input=\underset{A}{1}sin\left(\right(\underset{\omega }{1})t+\underset{{\varphi }_1}{0^o})$

$Output=Bsin(\omega t+{\varphi }_2)$

where
$B=A|G\left(j\omega \right)|=1\times \frac{1}{\sqrt{1+1}}=\frac{1}{\sqrt{2}}$

$and{\varphi }_2={\varphi }_1+\angle G\left(j\omega \right)=-ta{n}^{-1}1$
$={45}^o$

$\therefore$ Steady state o/p
$=\frac{1}{\sqrt{2}}sin(t-\frac{\pi }{4})$