B. P. Lathi, Zhi Ding - Modern Digital and Analog Communication Systems-Oxford University Press (2009)

3 . 10 MATLAB Exercises 125
In this example, we knew G(f) beforehand and hence could make IN TELLIGENT choices
for B (or the sampling frequency fs). In practice, we generally do not know G(f) beforehand.
In fact, that is the very thing we are trying to determine. In such a case, we must make
an intelligent guess for B or fs from circumstantial evidence. We then continue reducing
the value of T s and recomputing the transform until the result stabilizes within the desired
number of significant digits.
Next, we compute the Fourier transform of g(t) = 8 TI (t).
COMPUTER EXAMPLE C3.2
Use DFf (implemented by the FFf algorithm) to compute the Fourier transform of 8 TT (t). Plot the
resulting Fourier spectra.
This rectangular function and its Fourier transform are shown in Fig. 3.41a and b. To
determine the value of the sampling interval T s , we must first decide on the essential
bandwidth B. From Fig. 3.41b, we see that G(f) decays rather slowly with f. Hence,
the essential bandwidth B is rather large. For instance, at B = 15.5 Hz (97.39 rad/s),
G(f) = -0.1643, which is about 2% of the peak at G(O). Hence, the essential bandwidth
may be taken as 16 Hz. However, we shall deliberately take B = 4 for two reasons: (1) to
show the effect of aliasing and (2) because the use of B > 4 will give an enormous number
of samples, which cannot be conveniently displayed on a book-sized page without losing
sight of the essentials. Thus, we shall intentionally accept approximation for the sake of
clarifying the concepts of DFT graphically.
The choice of B = 4 results in the sampling interval T s = 1 /2B = 1 /8. Looking
again at the spectrum in Fig. 3.41b, we see that the choice of the frequency resolution
J o = 1/4 Hz is reasonable. This will give four samples in each lobe of G(f). In this case
To = I /J o = 4 seconds and N o = To/T = 32. The duration of g(t) is only 1 second. We
must repeat it every 4 seconds (To = 4), as shown in Fig. 3.41c, and take samples every
0.125 second. This gives us 32 samples (No = 32). Also,
8k = T s g (kT)
1
= Sg(kT)
Since g(t) = 8 TI (t), the values of 8k are 1, 0, or 0.5 (at the points of discontinuity), as
shown in Fig. 3.41c, where for convenience, 8k is shown as a function of t as well as k.
In the derivation of the DFT, we assumed that g (t) begins at t = 0 (Fig. 3.39a), and
then took No samples over the interval (0, To). In the present case, however, g(t) begins at
-½. This difficulty is easily resolved when we realize that the DFT found by this procedure
is actually the DFT of 8k repeating periodically every T o seconds. From Fig. 3.41c, it is
clear that repeating the segment of 8k over the interval from -2 to 2 seconds periodically
is identical to repeating the segment of 8k over the interval from O to 4 seconds. Hence,
the DFT of the samples taken from -2 to 2 seconds is the same as that of the samples
taken from O to 4 seconds. Therefore, regardless of where g (t) starts, we can always take
the samples of g(t) and its periodic extension over the interval from Oto To. In the present