563489578934

Sec. 2–8 Discrete Fourier Transform 101
samples is obtained. Second, the DFT and the IDFT are periodic with periods f s = 1/¢t
and T, respectively. The parameters ¢ t, T, and N are selected with the following considerations
in mind:
• ∆t is selected to satisfy the Nyquist sampling condition, f s = 1/¢t 7 2B, where B is
the highest frequency in the waveform. ¢ t is the time between samples and is called the
time resolution as well. Also, t = k¢t.
• T is selected to give the desired frequency resolution, where the frequency resolution is
¢f = 1/T. Also, f = n/T.
• N is the number of data points and is determined by N = T/¢t.
N depends on the values used for ¢ t and T. The computation time increases as N is increased. †
The N-point DFT gives the spectra of N frequencies over the frequency interval (0, f s ) where
f s = 1/¢t = N/T. Half of this frequency interval represents positive frequencies and half represents
negative frequencies, as described by Eq. (2–184). This is illustrated in the following
example.
Example 2–20 USING THE FFT TO CALCULATE SPECTRA
Using the FFT, Eq. (2–178), evaluate the magnitude spectrum and the phase spectrum for a rectangular
pulse. See Example2_20.m for the solution. Table 2–3 lists this MATLAB file. The computed
results are shown in Fig. 2–21. Note that Eqs. (2–178) and (2–183) are used to relate the FFT
results to the spectrum (i.e., the Continuous Fourier Transform). The parameters M, tend, and T
are selected so that the computed magnitude and phase-spectral results match the true spectrum of
the rectangular pulse as given by Eqs. (2–59) and (2–60) of Example 2–6. (T in Example 2–6 is
equivalent to tend in Table 2–3.)
The rectangular pulse is not absolutely bandlimited. However, for a pulse width of
tend = 1, the magnitude spectrum becomes relatively small at 5/tend = 5Hz = B. Thus, we
need to sample the waveform at a rate of 2B = 10 Hz or greater. For T = 10 and N = 128,
¢t = 0.08 or f s = 1/¢t = 12.8 Hz. Therefore, the values of T and N have been selected to satisfy
the Nyquist rate of f s 7 2B. The frequency resolution is ¢f = 1/T = 0.1 Hz.
Consequently, a good spectral representation is obtained by using the FFT.
In the bottom plot of Fig. 2–21, the magnitude spectrum is shown over the whole range
of the FFT vector—that is, for 0 6 f 6 f s , where f s = 12.8 Hz. Because Eq. (2–184b) was
not used in the MATLAB program, the plot for 0 6 f 6 6.8 (f s /2 = 6.8 Hz) corresponds to
the magnitude spectrum of the CFT over positive frequencies, and the plot for
6.4 6 f 6 12.8 corresponds to the negative frequencies of the CFT. The reader should try
other values of M, tend, and T to see how leakage, aliasing, and picket-fence errors become
large if the parameters are not carefully selected. Also, note that significant errors occur if
tend = T. Why?
† The fast Fourier transform (FFT) algorithms are fast ways of computing the DFT. The number of complex
multiplications required for the DFT is N 2 , whereas the FFT (with N selected to be a power of 2) requires only (N/2)
log 2 N complex multiplications. Thus, the FFT provides an improvement factor of 2N/(log 2 N) compared with the
DFT, which gives an improvement of 113.8 for an N = 512 point FFT.