Signals and Systems - Electrical Engineering

434 CHAPTER 7: Sampling Theory

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Interpolated sinc

Analog signal

Sampled signal

FIGURE 7.7

No aliasing: sampling simulation ofx(t)=cos( 2000 πt)usingfs= 6000 samples/sec. (a) Plots are of the signal

x(t)and the sampled signaly(t), and their spectra (|Y()|is periodic and so a period is shown). (b) The top plot

illustrates the sinc interpolation of three samples, and the bottom plot is the sinc-interpolated signalxr(t)and the

sampled signal. In this casexr(t)is very close to the original signal.

Sampling with aliasing—In Figure 7.8 we show the case when the sampling frequency isfs= 800 <

2 fs=2000, so that in this case we have aliasing. This can be seen in the sampled signaly(t)in the

top plot of Figure 7.8(a), which appears as if we were sampling a sinusoid of lower frequency. It can

also be seen in the spectra ofx(t)andy(t):|X()|is the same as in the previous case, but now|Y()|,

which is a period of the spectrum of the sampled signaly(t), displays a frequency of 200 Hz, lower

than that ofx(t), within the frequency range [−400, 400] Hz or [−fs/2,fs/2]. Aliasing has occurred.

Finally, the sinc interpolation gives a sinusoid of frequency 0.2 KHz, different fromx(t).

Similar situations occur when a more complex signal is sampled. If the signal to be sampled is

x(t)= 2 −cos(πf 0 t)−sin( 2 πf 0 t)wheref 0 =500 Hz, if we use a sampling frequency offs= 6000 > 2

fmax= 2 f 0 =1000 Hz, there will be no aliasing. On the other hand, if the sampling frequency is

fs= 800 < 2 fmax= 2 f 0 =1000 Hz, frequency aliasing will occur. In the no aliasing sampling, the

spectrum|Y()|(in a frequency range [−3000, 3000]=[−fs/2,fs/2]) corresponding to a period of

the Fourier transform of the sampled signaly(t)shows the same frequencies as|X()|. The recon-

structed signal equals the original signal. See Figure 7.9(a). When we usefs=800 Hz, the given

signalx(t)is undersampled and aliasing occurs. The spectrum|Y()|corresponding to a period of

the Fourier transform of the undersampled signaly(t)does not show the same frequencies as|X()|.

The reconstructed signal shown in the bottom right plot of Figure 7.9(b) does not resemble the

original signal.