082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017
764 Chapter 24 The Fourier transformTherefore, fromthe first shift theorem witha = 2 wehaveF{e 2jt u(t)e −3t } =13+j(ω−2)Consequently the function whose Fourier transform is13+j(ω−2) isu(t)e−(3−2j)t .Example24.6 Findthe Fourier transformof{ e−3tt0f(t)=e 3t t<0Deducethe functionwhoseFourier transformisG(ω) =∫ ∞Solution F(ω) = f (t)e −jωt dt−∞Now==∫ 0−∞∫ 0−∞e 3t e −jωt dt +e (3−jω)t dt +∫ ∞0∫ ∞0e −3t e −jωt dte −(3+jω)t dt[ e(3−jω)t 0 [ e−(3+jω)t ∞= +3−jω]−∞−(3+jω)]0= 13−jω + 13+jω= 69+ω 2G(ω) =610+2ω+ω = 62 (ω+1) 2 +9 =F(ω+1)610+2ω+ω 2.Then, using the first shift theorem F(ω + 1) will be F{e −jt f (t)}; that is, the requiredfunction is{ e(−3−j)tt 0g(t) =e (3−j)t t < 024.3.3 SecondshifttheoremIfF(ω)isthe Fourier transform of f (t)thenF{f(t −α)} =e −jαω F(ω)
24.3 Some properties of the Fourier transform 765Example24.7 Giventhatwhen f (t) ={ 1 |t|10 |t|>1tofind the Fourier transform ofg(t) ={ 1 1t30 otherwiseVerifyyour resultdirectly.2sinω,F(ω) =ω,applythesecondshifttheoremg(t)11 2 3tFigure24.3The function{ 1 1t3g(t) =0 otherwise.Solution The functiong(t) is depicted in Figure 24.3. Clearlyg(t) is the function f (t)translated2unitstotheright,thatisg(t) = f (t−2).NowF(ω) = 2sinω istheFouriertransformωof f (t).Therefore, by the second shift theoremF{g(t)} = F{f(t −2)} =e −2jω F(ω) = 2e−2jω sinωωTo verifythis resultdirectly wemustevaluate∫ ∞ ∫ 3[ ] eF{g(t)} = g(t)e −jωt dt = e −jωt −jωt 3dt =−∞ 1 −jω1as required.= e−3jω −e −jω−jω= 2e−2jω sinωω= e −2jω ( e −jω −e jω−jω)EXERCISES24.31 Provethe first shift theorem.2 Find the Fourier transform of{1−t 2 |t|1f(t)=0 |t|>1Usethe first shift theoremto deduce the Fouriertransformsof{e(a)g(t) =3jt (1−t 2 ) |t|10 |t|>1{e(b)h(t) =−t (1−t 2 ) |t|10 |t|>13 Find the inverse Fourier transformsof(a)(b)1(ω+7)j+121+2(ω−1)j4 Prove the second shift theorem.
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24.3 Some properties of the Fourier transform 765
Example24.7 Giventhatwhen f (t) =
{ 1 |t|1
0 |t|>1
tofind the Fourier transform of
g(t) =
{ 1 1t3
0 otherwise
Verifyyour resultdirectly.
2sinω
,F(ω) =
ω
,applythesecondshifttheorem
g(t)
1
1 2 3
t
Figure24.3
The function
{ 1 1t3
g(t) =
0 otherwise.
Solution The functiong(t) is depicted in Figure 24.3. Clearlyg(t) is the function f (t)translated
2unitstotheright,thatisg(t) = f (t−2).NowF(ω) = 2sinω istheFouriertransform
ω
of f (t).Therefore, by the second shift theorem
F{g(t)} = F{f(t −2)} =e −2jω F(ω) = 2e−2jω sinω
ω
To verifythis resultdirectly wemustevaluate
∫ ∞ ∫ 3
[ ] e
F{g(t)} = g(t)e −jωt dt = e −jωt −jωt 3
dt =
−∞ 1 −jω
1
as required.
= e−3jω −e −jω
−jω
= 2e−2jω sinω
ω
= e −2jω ( e −jω −e jω
−jω
)
EXERCISES24.3
1 Provethe first shift theorem.
2 Find the Fourier transform of
{
1−t 2 |t|1
f(t)=
0 |t|>1
Usethe first shift theoremto deduce the Fourier
transformsof
{
e
(a)g(t) =
3jt (1−t 2 ) |t|1
0 |t|>1
{
e
(b)h(t) =
−t (1−t 2 ) |t|1
0 |t|>1
3 Find the inverse Fourier transformsof
(a)
(b)
1
(ω+7)j+1
2
1+2(ω−1)j
4 Prove the second shift theorem.