Question 4.1: A square pulse is shown in Figure 4.3. Develop the Fourier s...
A square pulse is shown in Figure 4.3. Develop the Fourier series representation of p(t).
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.
