Question 4.3: Express the square pulse of Example (4.1) in exponential for...
Express the square pulse of Example (4.1) in exponential form.
The Fourier pair is given by Equation (4.9):
p(t) =\sum\limits_{n=-\infty }^{\infty } C_{n}e^{in \bar{\omega }_{1}t } (4.9a)
where, C_{n}=\frac{1}{T_{p}} \int_{0}^{T_{p}}p(t) e^{-in\bar{\omega }_{1}t } dt (4.9b)
The constant term C_{n} can be written as follows:
C_{n}=\frac{1}{T_{p}} \int_{0}^{T_{p/2}}p_{0}e^{-i\bar{\omega }t } dt +\frac{1}{T_{p}} \int_{T_{p/2}}^{T_{p}}(-p_{0}) e^{i\bar{\omega }t }dt
= \frac{-p_{0}}{i\bar{\omega }T_{p} } \left [\left|e^{-i\bar{\omega }t }\right|^{T_{p/2}}_{0}+\left|-e^{-i\bar{\omega }t }\right| ^{T_{p}}_{T_{p/2}} \right]
But \bar{\omega } =n \bar{\omega } _{1}=n\frac{2\pi }{T_{p}}
∴ e^{-in\bar{\omega }_{1}T_{p}/2 }=e^{-in\pi }=1 for n even
and e^{-in\bar{\omega }_{1}T_{p} }=e^{-i2\pi n }=-1 for n odd
Hence, C_{n}=\frac{ip_{0}}{2\pi n} [2e^{-in\pi }-2]
or, C_{n}=\frac{-2p_{0}i}{\pi n} for n odd
= 0 for n even
This can also be written as follows as amplitude of imaginary and real terms:
I(C_{n})=\frac{-2p_{0}}{n\pi } for n odd (i)
R(C_{n})=0 for n even (ii)
The expression (i) and the amplitude of the Fourier force function can be depicted as shown in Figures 4.6(a) and 4.6(b).
