Example 2.2.2: Convolution of two rectangular pulses Let A, -T0 < t < T0
f(t) =
0, |t| > T0
= g(t) .
f(u) g(t-u) du
(f*g)(t) =
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 2 (F(v))2 = 2AT0 sin(2 T0v) / 2 T0v
= A2sin2(2 T0v) / 2v2.
03/09/2014 04:37 p.m.
Example 2.2.2: Convolution of two rectangular pulses
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