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

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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.

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.

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