Signals and Systems - Electrical Engineering

10.4 Discrete Fourier Transform 619

n On the other hand, the harmonic frequencies of a periodic signal of periodNare fixed to

2 πk/Nfor 0≤k<N. In such a case we cannot pad the given period of the signal with an

arbitrary number of zeros, because such zeros are not part of the periodic signal. As an alter-

native, to increase the frequency resolution of a periodic signal we consider more periods

which give the same harmonic frequencies as for one period, but add zeros in between the

harmonic frequencies when considering more than one period.

n Frequency scales:When computing the DFT we obtain a sequence of complex valuesX[k] for
k=0, 1,...,N−1 corresponding to an input signalx[n] of lengthN. Since each of thekval-
ues corresponds to a discrete frequency 2πk/Nthek=0, 1,...,N−1 scale is converted into a
discrete frequency scale [0 2π(N− 1 )/N] (rad) (the last value is always smaller than 2πto keep
the periodicity in frequency ofX[k]) by multiplying the integer scale{ 0 ≤k≤N− 1 }by 2π/N.
Subtractingπto this frequency scale we obtain discrete frequencies [−π π− 2 π/N] (rad) where
again the last frequency does not coincide withπin order to keep the periodicity of 2πof the
X[k]. We obtain a normalized discrete-frequency scale by dividing the above scale byπso as to
obtain a nonunits normalized scale of [0 1− 2 (N− 1 )/N] or [−1 1− 2 /N]. Finally, if the signal
is the result of sampling and we wish to display the analog frequency, we then use the relation
whereTsis the sampling period andfsis the sampling frequency:

=

ω

Ts

=ωfs(rad/sec) or f=

ω

2 πTs

=

ωfs

2 π

(Hz) (10.54)

giving analog scales [0πfs] (rad/sec) and [0 fs/2] (Hz).

nExample 10.20

Consider the DFT computation via the FFT of a causal signalx[n]=(sin(πn/ 32 ))(u[n]−u[n−34])

and of its advanced versionx[n+16]. To improve its frequency resolution compute FFTs of length

N=512. Explain the difference between computing the FFTs of the causal and the noncausal

signals.

Solution

As indicated above, when computing the FFT of a causal signal the signal is simply inputted into

the function. However, to improve the frequency resolution of the FFT we attach zeros to the signal.

These zeros make it possible to have additional values of the frequency response of the signal, with

no effect on the frequency content of the signal.

For the noncausal signalx[n+16], we need to recall that the DFTs of an aperiodic signal were

computed by extending the signal into a periodic signal with an arbitrary periodN, which exceeds

the length of the signal. Thus, the periodic extension ofx[n+16] can be obtained by creating an

input vector consisting ofx[n],n=0,..., 16;N−33 zeros (Nbeing the length of the FFT and 33

the length of the signal) to improve the frequency resolution, andx[n],n=−16,...,−1.

In either case, the output of the FFT is available as an array of lengthN=512 values. This array

X[k],k=0,...,N−1 can be understood as values of the spectrum at frequencies 2πk/N—that is,