Question

In the system shown in figure, the input $x\left(t\right)=sint$. In the steady-state, the response $y\left(t\right)$ will be

• A. $\frac{1}{\sqrt{2}}sin(t-{45}^{\circ })$
• B. $\frac{1}{\sqrt{2}}sin(t+{45}^{\circ })$
• C. $sin(t-{45}^{\circ })$
• D. $sin(t+{45}^{\circ })$

Answer 1

The correct option is B $\frac{1}{\sqrt{2}}sin(t+{45}^{\circ })$

$x\left(t\right)=sint=1<0$
$\omega =1rad/sec$
$Transferfunction=G\left(s\right)=\frac{s}{s+1}$
For sinusoidal input,
$s=j\omega =j$
$G\left(j\omega \right){|}_{\omega =1}=\frac{j\omega }{j\omega +1}=\frac{j}{j+1}=\frac{1}{\sqrt{2}}\angle {45}^{\circ }$
$y\left(t\right)=\frac{1}{\sqrt{2}}\angle {45}^{\circ }.1\angle 0^{\circ }=\frac{1}{\sqrt{2}}\angle {45}^{\circ }$
$y\left(t\right)=\frac{1}{\sqrt{2}}sin(t+{45}^{\circ })$