Question

Consider a continuous time LTI system whose frequency response is $H\left(j\omega \right)=\{\begin{array}{cc}1;& -\pi \le \omega \le \pi \\ 0;& otherwise\end{array}$

If the input to this system is a periodic signal
$x\left(t\right)=\{\begin{array}{cc}1;& 0\le t<4\\ -1;& 4\le t<8\end{array}$

with period T=8, the corresponding output of the system is y(t). Then the value of y(2) = ______.(Answer upto two decimal places)

• A. 0.85

Answer 1

The correct option is A 0.85








Fourier series for the square wave shown above is

$x\left(t\right)={\sum }_{n=1,2,3}^{\infty }\frac{4}{n\pi }sin.n{\omega }_{0}t$

where, ${\omega }_0=\frac{2\pi }{T}=\frac{2\pi }{8}=\frac{\pi }{4}$

$x\left(t\right)={\sum }_{n=1,3,5}^{\infty }\frac{4}{n\pi }sin\frac{n\pi }{4}t$

$x\left(t\right)=\frac{4}{\pi }sin\frac{\pi }{4}t+\frac{4}{3\pi }sin\frac{3\pi }{4}t+\frac{4}{5\pi }sin\frac{5\pi }{4}t+...$

The given system H($j\omega$) is a low pass filter,
It allows frequencies less than $\pi$ rad/sec

$\therefore y\left(t\right)=\frac{4}{\pi }sin\frac{\pi }{4}t+\frac{4}{3\pi }sin\frac{3\pi }{4}t$

$y\left(2\right)=\frac{4}{\pi }sin\frac{\pi }{2}+\frac{4}{3\pi }sin\frac{3\pi }{2}=\frac{4}{\pi }-\frac{4}{3\pi }$

$y\left(2\right)=0.848$