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This module relates circular convolution of periodic signals in the time domain to multiplication in the frequency domain.
Given a signal
$f(t)$ with Fourier coefficients
${c}_{n}$ and a signal
$g(t)$ with Fourier coefficients
${d}_{n}$ ,
we can define a new signal,
$v(t)$ ,
where
$v(t)=\u229b(f(t), g(t))$ We find that the
Fourier
Series representation of
$v(t)$ ,
${a}_{n}$ ,
is such that
${a}_{n}={c}_{n}{d}_{n}$ .
$\u229b(f(t), g(t))$ is the
circular convolution of two periodic signals and is equivalent to the convolution
over one interval,
Take a look at a square pulse with a period of T.
For this signal $${c}_{n}=\begin{cases}\frac{1}{T} & \text{if $n=0$}\\ \frac{1}{2}\frac{\sin (\frac{\pi}{2}n)}{\frac{\pi}{2}n} & \text{otherwise}\end{cases}$$
Take a look at a triangle pulse train with a period of T.
This signal is created by circularly convolving the square pulse with itself. The Fourier coefficients for this signal are ${a}_{n}={c}_{n}^{2}=\frac{1}{4}\frac{\sin (\frac{\pi}{2}n)^{2}}{(\frac{\pi}{2}n)^{2}}$
Find the Fourier coefficients of the signal that is created when the square pulse and the triangle pulse are convolved.
$${a}_{n}=\left\{\begin{array}{cc}\text{undefined}\hfill & n=0\hfill \\ \frac{1}{8}\frac{si{n}^{3}\left(\frac{\pi}{2}n\right)}{{\left(\frac{\pi}{2}n\right)}^{3}}\hfill & \text{otherwise}\hfill \end{array}\right)$$
Circular convolution in the time domain is equivalent to multiplication of the Fourier coefficients in the frequency domain.
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