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Unified Engineering II Spring 2004
Problem S8 Solution
1. The convolution is given by
∞
y(t) = g(t)∗ u(t) = g(t − τ)u(τ)dτ (1)
−∞
Note that u(τ) is nonzero only for −3 ≤ τ ≤ 0, and g(t − τ) is nonzero only for
0 ≤ t − τ ≤ 3, that is, for −3 + t ≤ τ ≤ t. So there are four distinct regimes:
(a) t < −3
(b) −3 ≤ t ≤ 0
(c) 0 ≤ t ≤ 3
(d) t > 3
For cases (a) and (d), there is no overlap between g(t − τ) and u(τ), so y(t) = 0.
For case (b), the overlap is for −3 ≤ τ ≤ t. So
∞
y(t) = g(t − τ)u(τ)dτ
−∞
t
= sin(−2π(t − τ)) sin(2πτ )dτ
−3
At this point, we have to do a little trig:
sin(−2π(t − τ)) sin(2π τ ) = sin(2π(τ − t)) sin(2πτ )
= [sin(2πτ ) cos(2πt)− cos(2πτ ) sin(2πt)] sin(2πτ )
= cos(2πt) sin2(2πτ )− sin(2πt) cos(2π τ ) sin(2πτ )
sin(4πτ )
= cos(2πt)1 − cos(4πτ ) − sin(2πt)
2 2
So the integral is given by
t t t
cos(2πt) cos(2πt) sin(2πt)
y(t) = dτ − cos(4πτ)dτ − sin(4πτ )dτ
2 2 2
−3 −3 −3
t t
cos(2πt) cos(2πt) sin(2πt)
(t + 3) − sin(4πτ ) cos(4πτ )
+
= 2 8π 8π
τ=−3 τ=−3
cos(2πt) cos(2πt) sin(2πt)
= (t + 3) − sin(4πt) + [cos(4πt)− 1]
2 8π 8π
(As often happens with problems involving trig functions, there are other equiv-
alent expressions.)
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