[PDF] EE2 Mathematics Solutions to Example Sheet 4: Fourier Transforms PDF

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[PDF] EE2 Mathematics Solutions to Example Sheet 4: Fourier Transforms

EE2 Mathematics Solutions to Example Sheet 4: Fourier Transforms 1) Because f(t) = e?t = { e?t t > 0 et t < 0 } the Fourier transform of f(t) is

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EE2 Mathematics

Solutions to Example Sheet 4: Fourier Transforms

1)Becausef(t) =e-|t|=?e-t, t >0

e t, t <0? the Fourier transform off(t) is f(ω) =? e-iωt-|t|dt=? 0 et(1-iω)dt+? 0 e-t(1+iω)dt=21 +ω2

2)(i) DesignateF{f(t)}=

f(ω) withaa real constant of either sign. ThenF{f(at)}=?∞

-∞e-iωtf(at)dt. Defineτ=atsodτ=adt. Whena >0 the limits (-∞,∞) forτcorrespond

to those fort, but whena <0 the direction reverses. Thus

F{f(at)}=|a|-1?∞

e-i(ω a)τf(τ)dτ=|a|-1f?ωa? (ii) The 'shift property" (in the formula sheets)F{f(t-a)}=e-iωa f(ω) can simply be proved by definingτ=t-athen

F{f(t-a)}=?

e-iωtf(t-a)dt=e-iωa?∞ e-iωτf(τ)dτ=e-iωa f(ω)

3)To find the Fourier transform of the non-normalized Gaussianf(t) =e-t2we first complete

the square in the exponential f(ω) =? e-iωt-t2dt=e-14ω2?∞ e-(t+12iω)2dt=⎷πe-14ω2

The normalized auto-correlation function ofe-t2is

γ(t) =?

-∞e-u2e-(t-u)2du -∞e-2u2du=e-1

2t2?∞

-∞e-2(u-12t)2du?∞ -∞e-2u2du=e-12t2 The integrals in the numerator & denominator cancel becausethey are equal; the origin of the former is shifted w.r.t. to the latter on the infiniteu-axis but its value is not affected.

4)Withf(t) =e-at2andg(t) =e-bt2, a minor re-scaling of the results of Q3 shows that

f(ω) =?π ae-ω2

4ag(ω) =?π

be-ω2 4b

The convolution theorem says thatF??∞

-∞f(t?)g(t-t?)dt?? f(ω)g(ω). Therefore, with h(t) =? exp?-at?2?exp?-b(t-t?)2?dt?=f?g we have h(ω) =π⎷abe-ω2(a+b) 4ab. 1

5)Withf(t) =e-t2for whichf(ω) =⎷πe-14ω2andg(t) = cosatfor which

g(ω) =? e-iωtcosatdt=12? e-iωt(eiat+e-iat)dt 1 2? e-it(ω-a)+e-it(ω+a)? dt=π{δ(ω-a) +δ(ω+a)} Thus f(t)g(t)dt=π⎷

2π?

e-1

4ω2{δ(ω-a) +δ(ω+a)}dω=⎷πe-14a2.

6)Writef(t) = (1 +t2)-1so that

-∞dt (1 +t2)2=? |f(t)|2dt=12π? |f(ω)|2dω

Hence we want to evaluate

f(ω)|2dω. To do this we must first findf(ω) =?∞ -∞e-iωtdt1+t2. To apply Jordan"s Lemma it is necessary to consider the two separate casesω <0 andω >0. (i)ω <0: Consider the complex integral? C Ue -iωzdz

1+z2withCUa semi-circle in theupper12-plane

in which there is a simple pole atz=i. The residue at this pole iseω/2iand the integral over H

R→0 asR→ ∞by Jordan"s Lemma.

R -R

ω <0

?H+ R -→i•For the contourCUin the upper1

2-plane (ω <0):

πe C Ue -iωzdz

1 +z2=?

R -Re -iωxdx1 +x2+? H Re -iωzdz1 +z2 (ii)ω >0: Consider the complex integral? C Le -iωzdz

1+z2withCLa semi-circle in thelower12-plane

in which there is a simple pole atz=-i. The residue at this pole is-e-ω/2i& the integral overH-

R→0 asR→ ∞by Jordan"s Lemma.

R-R←-

?H- R-i

ω >0•For the contourCLin the lower1

2-plane (ω >0):

-πe-ω=? C Le -iωzdz

1 +z2=?

-R Re -iωxdx1 +x2+? H Re -iωzdz1 +z2 Note the reverse order of the limits in the real integral.

Thus, in the limitR→ ∞, we have

f(ω) =?πe-ω, ω >0 πe

ω, ω <0?

. Finally we can now calculate 1

2π?

|f(ω)|2dω=π22π? ?0 e2ωdω+? 0 e-2ωdω?

12π.

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[PDF] [PDF] EE2 Mathematics Solutions to Example Sheet 4: Fourier Transforms

EE2 Mathematics

Solutions to Example Sheet 4: Fourier Transforms

1)Becausef(t) =e-|t|=?e-t, t >0

2)(i) DesignateF{f(t)}=

F{f(at)}=|a|-1?∞

F{f(t-a)}=?

3)To find the Fourier transform of the non-normalized Gaussianf(t) =e-t2we first complete

The normalized auto-correlation function ofe-t2is

γ(t) =?

2t2?∞

4)Withf(t) =e-at2andg(t) =e-bt2, a minor re-scaling of the results of Q3 shows that

4ag(ω) =?π

The convolution theorem says thatF??∞

5)Withf(t) =e-t2for whichf(ω) =⎷πe-14ω2andg(t) = cosatfor which

2π?

4ω2{δ(ω-a) +δ(ω+a)}dω=⎷πe-14a2.

6)Writef(t) = (1 +t2)-1so that

Hence we want to evaluate

1+z2withCUa semi-circle in theupper12-plane

R→0 asR→ ∞by Jordan"s Lemma.

ω <0

2-plane (ω <0):

1 +z2=?

1+z2withCLa semi-circle in thelower12-plane

R→0 asR→ ∞by Jordan"s Lemma.

R-R←-

ω >0•For the contourCLin the lower1

2-plane (ω >0):

1 +z2=?

Thus, in the limitR→ ∞, we have

ω, ω <0?

2π?

12π.