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

790 Chapter 24 The Fourier transform
When we use this transform to approximate the Fourier transform of the function f (t)
sampled at intervalsT, it follows from Equation (24.13) that we must multiply by the
factor T to obtain our approximation. An example of this is given in the next
section.
24.11 USINGTHED.F.T.TOESTIMATEAFOURIERTRANSFORM
One application of the d.f.t. is to estimate the continuous Fourier transform of a signal
f (t). The following example shows how this can bedone.
Example24.22 The signal f (t) = u(t)e −2t , where u(t) is the unit step function, is shown in
Figure 24.18(a). Its Fourier transform, which can be found from Table 24.1, is
F(ω) = 1 and a graph of |F(ω)|isshown inFigure 24.18(b).
2+jω
Suppose we obtain 16 sample values of f (t) at intervals ofT = 0.1 fromt = 0 to
t = 1.5.
(a) Obtain the d.f.t. of the sampled sequence.
(b) Usethe d.f.t. toestimatethe trueFourier transform values.
(c) Plot a graph tocompare values of |F[k]| and |F(ω)|.
Solution (a) If f (t) =u(t)e −2t is sampled att =nT, that ist = 0,0.1,0.2,...,1.5, we obtain
the sequence f[n] = e −2nT = e −0.2n forn = 0,1,2,...,15. This is the sequence
given inthe second column of Table 24.2.
(b) Thecalculationofthed.f.t.isfartoolaborioustobedonebyhand.Insteadwehave
used the MATLAB ® command fft() to do the calculation and the results, F[k],
have been placed inthe third column.
(c) From Section 24.10 we know that the d.f.t. samples at intervals of 2π in the frequencydomain.SoforcomparisonthefourthcolumnshowsthetruevaluesofF(ω)
NT
1 f(t) = u(t)e – 2 t t
4 + 2 30
u F( v) u =
1
v
0.5
(a)
1
(b)
Figure24.18
The signal f (t) =u(t)e −2t and itsFourier transform.
–30
v