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

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24.11 Using the d.f.t.to estimate a Fourier transform 791

Table24.2

n,k f[n] = e −0.2n F[k] F(ω k ) 0.1F[k]

0 1.0000 5.2918 0.5000 0.5292

1 0.8187 1.4835 −1.9082j 0.1030 −0.2022j 0.1484 −0.1908j

2 0.6703 0.7882 −1.0837j 0.0304 −0.1196j 0.0788 −0.1084j

3 0.5488 0.6311 −0.6952j 0.0140 −0.0825j 0.0631 −0.0695j

4 0.4493 0.5743 −0.4702j 0.0080 −0.0626j 0.0574 −0.0470j

5 0.3679 0.5485 −0.3159j 0.0051 −0.0504j 0.0548 −0.0316j

6 0.3012 0.5355 −0.1964j 0.0036 −0.0421j 0.0536 −0.0196j

7 0.2466 0.5293 −0.0944j 0.0026 −0.0362j 0.0529 −0.0094j

8 0.2019 0.5274 0.0020 −0.0317j 0.0527

9 0.1653 0.5293 +0.0944j 0.0016 −0.0282j 0.0529 +0.0094j

10 0.1353 0.5355 +0.1964j 0.0013 −0.0254j 0.0536 +0.0196j

11 0.1108 0.5485 +0.3159j 0.0011 −0.0231j 0.0548 +0.0316j

12 0.0907 0.5743 +0.4702j 0.0009 −0.0212j 0.0574 +0.0470j

13 0.0743 0.6311 +0.6952j 0.0008 −0.0196j 0.0631 +0.0695j

14 0.0608 0.7882 +1.0837j 0.0007 −0.0182j 0.0788 +0.1084j

15 0.0498 1.4835 +1.9082j 0.0006 −0.0170j 0.1484 +0.1908j

also sampled at intervals of 2π . Recall that to obtain an estimate of the Fourier

NT

transform of f (t) we must multiply the d.f.t. values byT = 0.1. This is shown in

the fifth column of the table.

Figure24.19showsgraphsof |0.1F[k]|and |F(ω)|.Valuesof |F[k]|obtainedbeyond

k = 8aremirrorimagesoftheearliervalues.ThisisbecauseF[N−k] =F[k]asproved

inQuestion5inExercises24.9.1.Thisisaphenomenon ofthed.f.t.andarisesbecause

ofitsperiodicityandsymmetryproperties,furtherdetailsofwhicharebeyondthescope

of this book. The interested reader is referred to a text on signal processing for further

details.

0.6

0.5

0.4

F ( k) = F

0.3

(

2 pk

u v u

N T )

0.1uF[k]u

0.2

0.1

0 2 4 6 8 10 12 14

k

Figure24.19

Comparisonoftruevalues ofaFourier

transformand approximate values obtained

usingad.f.t.

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