Multirate DSP, part 4: Bandpass undersampling

Part 3 looks at oversampling in analog-to-digital convertors. It applies these principles to a sigma-delta ADC and revisits the CD player case study.

12.5 Undersampling of Bandpass Signals
As we discussed in Chapter 2, the sampling theorem requires that the sampling rate be twice as large as the highest frequency of the analog signal to be sampled. The sampling theorem ensures the complete reconstruction of the analog signal without aliasing distortion. In some applications, such as modulated signals in communications systems, the signal exists in only a small portion of the bandwidth. Figure 12-34 shows an amplitude-modulated (AM) signal in both time domain and frequency domain. Assuming that the message signal has a bandwidth of 4 kHz and a carrier frequency of 96 kHz, the upper frequency edge is therefore 100 kHz. Then the traditional sampling process requires that the sampling rate be larger than 200 kHz at a high cost. The baseband signal of 4 kHz requires a sampling rate of only 8 kHz.

Figure 12-34. Message signal, modulated signal, and their spectra.

If a certain condition is satisfied at the undersampling stage, we are able to make use of the aliasing signal to recover the message signal, since the aliasing signal contains the folded original message information (which we used to consider as distortion). The reader is referred to the undersampling technique discussed in Ifeachor and Jervis (2002) and Porat (1997). Let the message to be recovered have a bandwidth of B, the theoretical minimum sampling rate be f_s = 2B, and the carrier frequency of the modulated signal be f_c. We discuss the following cases.

Case 1.
If f_c = even integer × B and f_s = 2B, the sampled spectrum with all the replicas will be as shown in Figure 12-35.(a).

Figure 12-35. Spectrum of the undersampled signal.

As an illustrative example in time domain for Case 1, suppose we have a bandpass signal with a carrier frequency of 20 Hz; that is,

x(t) = cos (2π × 20t)m(t), (12.39)

where m(t) is the message signal with a bandwidth of 2 Hz. Using a sampling rate of 4 Hz by substituting t = nT, where T = 1/f_s into Equation (12.39), we get the sampled signal as

x(nT) = cos (2π × 20t)m(t)|_t–nT = cos (2π × 20n/4)m(nT). (12.40)

Since 10nπ = 5n(2π) is the multiple of 2π,

cos (2π – 20n/4) = cos (10πn) = 1, (12.41)

we obtained the undersampled signal as

x(nT) = cos (2π × 20n/4)m(nT) = m(nT), (12.42)

which is a perfect digital message signal. Figure 12-36 shows the bandpass signal and its sampled signal when the message signal is 1 Hz, given as

m(t) = cos (2πt). (12.43)

Figure 12-36. Plots of the bandpass signal and sampled signal for Case 1.