Question

Consider a continuous time LTI system whose frequency response is

$H\left(j\omega \right)={\int }_{-\infty }^{\infty }h\left(t\right)e^{-j\omega t}.dt=\frac{sin4\omega }{\omega }$

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, and the cprresponding system output y(t) has Fourier series coefficient $b_k$. Which one of the following option is correct ?

• A. $b_k=0;$ for even values of k
  $b_k=0;$ for odd values of k
• B. $b_k\ne 0;$ for even values of k
  $b_k\ne 0;$ for odd values of k
• C. $b_k=0;$ for even values of k
  $b_k\ne 0;$ for odd values of k
• D. $b_k\ne 0;$ for even values of k
  $b_k=0;$ for odd values of k

Answer 1

The correct option is A $b_k=0;$ for even values of k

$b_k=0;$ for odd values of k
Let us first evaluate the Fourier series coefficients of x(t) clearly, since x(t) is real and odd , $n_k$ is purely imaginary and odd.

Therefore, $a_0=0$

Now, $a_k=\frac{1}{8}{\int }_0^{8}x\left(t\right)e^{-f\left(\frac{2\pi }{8}\right)kt}.dt$

$=\frac{1}{8}{\int }_0^4{e}^{-j\left(\frac{2\pi }{8}\right)kt}dt-\frac{1}{8}{\int }_0^8{e}^{-j\left(\frac{2\pi }{8}\right)kt}dt$

$=\frac{1}{8}{\left(\frac{e^{-j\frac{\pi kt}{4}}}{-j\frac{\pi }{4}k}\right)}_0^4-\frac{1}{8}{\left(\frac{e^{-j\frac{\pi kt}{4}}}{-j\frac{\pi }{4}k}\right)}_4^8$

$=\frac{1}{2}\frac{1(1-e^{-j\pi k})}{j\pi k}-\frac{1}{2}\frac{(e^{-j\pi k}-e^{-j2\pi k})}{j\pi k}$

$=\frac{1}{j2\pi k}(1-e^{-j\pi k}-e^{-j\pi k}+1)$

$a_k=\{\begin{array}{cc}0:& K=0,\pm 2,\pm 4,\pm \mathrm{6...},\\ \frac{2}{j\pi k};& K=\pm 1,\pm 3,\pm 5,...\end{array}$

Input of LTI system does not contain even harmonics
$\therefore$ Output also does not contain even harmonics.

$b_k=0,$ for even value of k

When x(t) is passed through an LTI system with frequency response $H\left(j\omega \right)$, the output y(t) is given by ,

$y\left(t\right)={\sum }_{K=-\infty }^{\infty }{a}_k.H\left(jk{\omega }_0\right)-e^{jk{\omega }_{0}t}$

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

Since $n_k$ is non-zero only for odd values of k, we need to evaluate the above summation only for odd k.

$H\left(kK{\omega }_0\right)=H\left(jk\frac{\pi }{4}\right)=\frac{sin\left(k\pi \right)}{\frac{k\pi }{4}}$

$H\left(jk{\omega }_0\right)$ is always zero for odd values of k

$\therefore$ $y\left(t\right)=0$

$\therefore$ $b_k=0$ for odd values of k