Example 2.2.2: Convolution of two rectangular pulses
Let

f(t) = A, -T0 < t < T0    = g(t) .
0, |t| > T0

(f*g)(t) = f(u) g(t-u) du
= A · A du, -2T0 < t < 0
A · A du, 0 < t < 2T0
= A2 (t+2T0 ), -2T0 < t < 0
A2 (2T0 -t), 0 < t < 2T0      .
0, | t | > 2T0

The convolution of two rectangular pulses = triangular pulse

The Fourier transform of f * g i.e. of f * f is [F(v)]2, where F(v) is the Fourier transform of f, that is

(F(v))2 = 2AT0 sin(2T0v) / 2T0v 2
= A2sin2(2T0v) / 2v2.

Thus


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