Convolution example
Suppose f(t) = cos(at), g(t) = sin(at).Then
f∗g(t) = Zt
0
cos(at −aτ) sin(aτ)dτ
=Zt
0
cos(at) cos(aτ) sin(aτ) + sin(at) sin(aτ ) sin(aτ )dτ
= cos(at)Zt
0
cos(aτ) sin(aτ)dτ + sin(at)Zt
0
sin2(aτ)dτ
= cos(at)Zt
0
cos(aτ)dcos(aτ)
a+ sin(at)Zt
0
1−cos(2aτ)
2dτ
= cos(at)·cos2(aτ)
2a¸t
0
+ sin(at)·1
2−sin(2aτ)
4a¸t
0
= cos(at)·cos2(at)
2a
−1
2a¸+ sin(at)·1
2t−sin(2at)
4a¸
= cos(at)·cos2(at)
2a
−1
2a¸+ sin(at)·1
2t+2 sin(at) cos(at)
4a¸
=cos2(at)
2acos(at) + sin2(at)
2acos(at)−1
2acos(at) + 1
2tsin(at)
=cos2(at) + sin2(at)
2a
−1
2acos(at) + 1
2tsin(at)
=1
2tsin(at).
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