Question 4.1: A square pulse is shown in Figure 4.3. Develop the Fourier s...

The square pulse is having a period of $T_{p}$. It can be assumed to vary from –$T_{p} /2$ to $T_{p} /2$ for convenience. The Fourier series constants can be evaluated as follows:

$a_{0}=\frac{1}{T_{p}} \int_{-T_{p}/2}^{T_{p}/p} p(t) dt$                 (i)

$a_{n}=\frac{2}{T_{p}} \int_{-T_{p}/2}^{T_{p}/2} \cos (n\bar{\omega }_{1}t ) dt$                      (ii)

and

$b_{n}=\frac{2}{T_{p}} \int_{-T_{p}/2}^{T_{p}/2} \sin (n\bar{\omega }t ) dt$                     (iii)

where $p(t)=-p_{0}$   for   $-T_{p}/2\lt t\lt 0$

$=p_{0}$   for   $0\lt t\lt T_{p}/2$            (iv)

Substituting the value of p(t) from Equation (iv) in Equations (i) and (ii), gives $a_{0}=0$ and $a_{n}=0$ because p(t) is an odd function of t, that is, p(t) = − p(−t) while $a_{0}$ and $a_{n}$ are coefficients of even terms in the Fourier series.

Equations (iii) and (iv) give

$b_{n}=\frac{4 p_{0}}{T_{p}} \int_{0}^{Tp/2} \sin (n\bar{\omega }_{1}t )dt$

But              $\bar{\omega } _{1}=\frac{2\pi }{T_{p}}$

∴               $b_{n}=- \frac{2p_{0}}{n\pi } (\cos (n\pi )-1)$

or,                   $b_{n}=\frac{4p_{0}}{n\pi }$ for n = 1, 3, 5, etc.

The Fourier series representation of the square pulse is given by

  $p(t) =\frac{4p_{0}}{\pi } \sum\limits_{n=1,3,5,}^{\infty } \frac{1}{n} (n\bar{\omega }_{1}t )$

It can be plotted for different values of n = 1, 3, 5, etc. in MS-EXCEL. The convergence of different harmonic components of sine to the square pulse can be seen in Figure 4.4. The convergence will improve further by taking more number of harmonic terms.