Page 501 - Advanced engineering mathematics

14.3 The Fourier Transform 481

To evaluate this integral, we must consider three cases. If t > 0 then
0 t ∞
e
e
e
24( f ∗ g)(t) = e −2|t−τ| −3|τ| dτ + e −2|t−τ| −3|τ| dτ + e −2|t−τ| −3|τ| dτ
−∞ 0 t
0 t ∞
e
e
e dτ +
= e −2(t−τ) 3τ e −2(t−τ) −3τ dτ + e −2(t−τ) −3τ dτ
−∞ 0 t
6 4
= e −2t − e −3t .
5 5
If t < 0, then
t 0 ∞

e
e
e
24( f ∗ g)(t) = e −2|t−τ| −3|τ| dτ + e −2|t−τ| −3|τ| dτ + e −2|t−τ| −3|τ| dτ
−∞ t 0
t 0 ∞
e
e dτ +
= e −2(t−τ) 3τ e 2(t−τ) 3τ e 2(t−τ) −3τ dτ
e dτ +
−∞ t 0
4 6
3t
2t
=− e + e .
5 5
Finally, if t = 0,
∞ 2

e
24( f ∗ g)(0) = e −2|τ| −3|τ| dτ = .
5
−∞
Therefore
1 1 6 4

−1
F (t) = e −2|t| − e −3|t|
(4 + ω )(9 + ω ) 24 5 5
2
2
1 1
= e −2|t| − e −3|t| .
20 30
14.3.1 Filtering and the Dirac Delta Function
The Dirac delta function δ(t) was discussed in Section 3.5. We may think of this function as the
limit of a pulse (Section 3.3.2), as the height tends to infinity and the duration to zero. In terms
of the Heaviside function H(t),
1
δ(t) = lim [H(t + a) − H(t − a)].
a→0+ 2a
In this definition, the pulse is centered at 0, extending from t − a to t + a. Often we deal with the
shifted delta function H(t − t 0 ), in which the defining pulse is centered at t 0 .
The filtering property of the delta function enables us to recover a function value f (t 0 ) by
“summing” function values when they are impacted with a shifted delta function.
THEOREM 14.3 Filtering by a Delta Function
If f (t) has a Fourier transform and is continuous at t 0 , then
∞

f (t)δ(t − t 0 )dt = f (t 0 ).
−∞
Proof To prove this, first observe that

0 for t ≤ t 0 − a and for t > t 0 + a,
H(t − t 0 + a) − H(t − t 0 − a) =
1 for t 0 − a ≤ t < t 0 + a.

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October 14, 2010 16:43 THM/NEIL Page-481 27410_14_ch14_p465-504