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SOURCE-SIGNAL ANALYSIS 559<br />

Since the spectrum is averaged over the duration <strong>of</strong> a record, the time<br />

resolution can be no better than the record duration. And since the FFT gives<br />

harmonic frequencies that are integral multiples <strong>of</strong> the reciprocal <strong>of</strong> the<br />

record duration, the frequency resolution can be no better than this either.<br />

Thus, time resolution multiplied by frequency resolution can be no less than<br />

unity. Another consideration is that for maximum efficiency and ease <strong>of</strong><br />

programming, the record size should be a power <strong>of</strong> two.<br />

Continuing the example used earlier (20-ks/s signal sample rate, 50-Hz<br />

frequency resolution, 100-sls spectrum sample rate), we see that the record<br />

size must be at least 400 samples to obtain the required frequency resolution.<br />

The corresponding time resolution is 20 msec, which means that the 100-s/s<br />

spectrum sample rate oversamples the changing spectrum by a factor <strong>of</strong> two.<br />

We will see later that overlap between successive records is <strong>of</strong>ten desirable and<br />

that it results in spectral oversampling as well. After rounding up to the next<br />

power <strong>of</strong> two, the FFT example used in the following discussion will assume<br />

a record size <strong>of</strong> 512, which gives a frequency resolution <strong>of</strong> 39 Hz, and a<br />

spectral sample rate <strong>of</strong> 78 sis.<br />

Equivalent Bandpass Filter<br />

Since the filterbank and the FFT methods <strong>of</strong> spectral analysis yield<br />

similar results, it makes sense to talk about an equivalent bandpass filter<br />

corresponding to each <strong>of</strong> the "harmonics" computed by the FFT. As an<br />

example, let us study the 10th FFT harmonic. Ideally, it should act as a<br />

bandpass filter with a lower cut<strong>of</strong>f <strong>of</strong> 9. 5 X 39Hz = 370 Hz and an upper<br />

cut<strong>of</strong>f <strong>of</strong> 10.5 X 39 Hz = 409 Hz. Conceptually, it is simple to plot the<br />

o<br />

-5<br />

-10<br />

-15<br />

(\<br />

~ -20<br />

..... -25<br />

~ -30 (\ nf\ f\<br />

~ -35<br />

-40<br />

-45<br />

f\f\<br />

f\{\<br />

-50<br />

-55<br />

- 60 '---'------'--:'----'_'---'------'--:'----'---,_-'----':---'::---'---,"------'------'--:'----'---,"-----..<br />

o 2 3 4 5 6 7 8 9 10 /I 12 13 14 15 16 17 18 19 20<br />

FREOUENCY (HARMONIC NUMBER)<br />

Fig. 16-10. Equivalent bandpass filter response <strong>of</strong> 10th FFT harmonic

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