Question

A signal x(t) is given by
$x\left(t\right)=\{\begin{array}{c}1,-T/4<t\le 3T/4\\ -1,3T/4<t\le 7T/4\\ -x(t+T)\end{array}$
which among the following gives the fundamental fourier term of x(t)?

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

Answer 1

The correct option is B $\frac{4}{\pi }\cos (\frac{\pi t}{T}-\frac{\pi }{4})$


According to definition of signal given in question the x(t) will be as

So it is periodic with period,
$T_0=\frac{7T}{4}-(-\frac{T}{4})=2T$
$\therefore$ Fundamental angular frequency
${\omega }_0=\frac{2\pi }{T_0}=\frac{2\pi }{2T}=\frac{\pi }{T}$
Now,
$C_1=$ (exponential series coefficient for k =1)

$C_1=\frac{1}{T_0}{\int }_{T_0}x\left(t\right)e^{-j{\omega }_{0}t}dt$

$C_1=\frac{1}{T_0}[{\int }_{-T/4}^{3T/4}{e}^{-j{\omega }_{0}t}dt+{\int }_{3T/4}^{7T/4}-1e^{-j{\omega }_{0}t}dt]$

$C_1=\frac{1}{j2\pi }[2e^{j\pi /4}+2e^{j\pi /4}]$

$C_1=\frac{2}{j\pi }{e}^{j\pi /4}=\frac{2}{\pi }{e}^{-j\pi /4}=\frac{2}{\pi }[\frac{1}{\sqrt{2}}-\frac{j}{\sqrt{2}}]$

Comparing it with
$C_1=\frac{a_1}{2}-\frac{j{b}_1}{2}$

$a_1=\frac{4}{\sqrt{2}\pi },b_1=\frac{4}{\sqrt{2}\pi }$

$\therefore$ fundamental Fourier term will be

$=a_1\cos {\omega }_{0}t+b_1\sin {\omega }_{0}t$

$\frac{4}{\sqrt{2}\pi }\cos \left(\frac{\pi }{T}t\right)+\frac{4}{\sqrt{2}\pi }\sin \left(\frac{\pi }{T}t\right)$

$=\frac{4}{\pi }\cos [\frac{\pi }{T}t-\frac{\pi }{4}]$