32
The Discrete Fourier Transform
3.1 The Fourier Analysis and Synthesis of Waveforms
Consider the waveform shown in Fig. 3.1(d) with sample values 2+^, |,
2—
^, and
—
|. The waveform is periodic with a period of 4 samples and one
cycle of which is shown. The problem of Fourier analysis is to find the
sinusoids, the superposition summation of which yields this waveform. The
solution is shown in Fig. 3.1: (a) a dc component xa(n) = 1, (the spectral
value X(0) = 1), (b) a sinusoid xb(n) = cos(f n - f), (X(l) = ^ - j\
and X(3) = ^ +
j\),
and (c) a sinusoid xc(n) = cos(7rn), (X(2) = 1).
The sums of the corresponding sample values of these three waveforms are
equal to the sample values of the waveform shown in Fig. 3.1(d), x(n) —
xa(n) + xb(n) + xc(n).
The spectrum of x(n) is displayed in Fig. 3.1(e) with a scale factor of 4.
The real part of the spectrum X(k) is marked with the symbol '•' and the
imaginary part is indicated by the symbol V. The two problems in Fourier
analysis are: (i) given the time-domain waveform of a signal, how do we get
the coefficients of the individual sinusoids (analysis) and (ii) given the coeffi-
cients of the individual sinusoids, how do we get the combined time-domain
waveform (synthesis). We have started with the problem and the solution
in order to easily understand the problem formulation and the method of
obtaining the solution. The periodic sequence shown in Fig. 3.1(d) can be
considered as the output of a single complex signal generator or, equiva-
lent^, the output of three simple signal generators connected in series as
shown, respectively, in Figs. 3.2(a) and (b).
The Fourier analysis of waveforms
Each frequency component contributes to the value of the time-domain
waveform at each sample point. By multiplying each of the complex fre-
quency coefficient by the sample value of the corresponding complex sinu-
soid and summing the products must be equal to the time-domain sample
value. Therefore, we set up four simultaneous equations with four time-
domain samples.
X(0)e^(«)(o)
+
x(l)e^(OW + X(2)eJT(°)(
2
) +X(3)e^(o)(3)
= x
(
0
)
l(0)e^«(o) +X(l)e^(i)W +I(2)e*TP)P) +X(3)e^W(3) =
x
(l)
X(0)e^(
2
)W +X(l)e^(
2
)(
1
) +X(2)e^(
2
)W +X(3)e^(2)(3)
= X
(
2
)
X(0)e^(3)(o) + X(i)e»*(3)(D + X(2)<W><
2
> +X(3)e^(3)(3)
= x(3
)