082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

24.8 Convolution and correlation 779
Whent isgreaterthan −1butlessthan1,asinFigure24.15(e),thereisanoverlap,
andhenceanon-zeroproduct.Thisoccursforvaluesof λbetween −1andt,thatis
in the interval −1 λ t. Within this interval f (λ) = 1 andg(t − λ) = e −(t−λ) .
So:
if −1t<1
f∗g=
∫ t
−1
∫ t
e −(t−λ) dλ = e −t e λ dλ = e −t [e λ ] t −1 = e−t [e t −e −1 ] = 1 −e −1−t
−1
Whent is greater than 1 the graphs overlap, but only for values of λ between −1
and 1,thatisfor −1 λ < 1.So:
ift>1
f∗g=
∫ 1
−1
e −(t−λ) dλ = e −t [e λ ] 1 −1 = e−t (e 1 −e −1 )
Putting all these results together
⎧
⎨0 t<−1
(f ∗g)(t)= 1−e −1−t −1t<1
⎩
e −t (e 1 −e −1 ) t 1
(b) Using Table 24.1 the Fourier transformsof f andgare
F(ω) = 2sinω
ω
and theirproduct is
F(ω)G(ω) =
2sinω
ω(1+jω)
G(ω) = 1
1+jω
(24.12)
Since the convolution is defined differently on each part of the domain, then the
Fouriertransformoftheconvolutionmustbefoundbyintegratingovereachpartof
the domain separately. You will alsoneed torecall that
So,
sinω= ejω −e −jω
2j
F{(f ∗g)(t)} =
=
=
=
∫ ∞
−∞
∫ −1
−∞
∫ ∞
+
1
∫ 1
−1
∫ 1
−1
(f ∗g)(t)e −jωt dt
0 ·e −jωt dt +
∫ 1
−1
e −t (e 1 −e −1 )e −jωt dt
(1 −e −1−t )e −jωt dt
e −jωt −e −1−t−jωt dt + (e 1 −e −1 )
∫ ∞
e −jωt −e −1 e −t(1+jω) dt+(e 1 −e −1 )
1
e −t e −jωt dt
∫ ∞
1
e −t(1+jω) dt
[ ] e
−jωt 1 [ e
= −e −1 −t(1+jω) 1 [ e
+(e
−jω
−1
−(1+jω)] 1 −e −1 −t(1+jω) ∞
)
−1
−(1+jω)]
1