3.11 Signal sampling  (Page 3/3)

To avoid the aliasing there are two approaches: One is to raise the sampling frequency to satisfy the sampling theorem, the other is to filter off the unecessary high-frequency component from the continuous-time signal. We limit the signal frequency by an effective lowpass filter, called antialiasing prefilter , so that the remained highest frequency is less than half of the intended sampling rate. If the filter is not perfect we must give some allowance. For example in voice processing, if the lowpass filter still allows frequencies above 3,4kHz go through even at small amplitude, the sampling frequency should be 10 kHz or more instead of 8 kHz.

The aliasing phenomenon can be shown mathematically. Let’s consider a complex exponential signal at frequency F which is sampled at interwal $T_s$ to yield the samples x(nT ${}_s$ ):

$$x(t)=e^{\mathrm{j2\pi }\text{Ft}}$$ $\Rightarrow$ $x({\text{nT}}_s)=e^{\mathrm{j2\pi }{\text{FnT}}_s}$

Now consider other signals at frequency $F\pm m{f}_s$ , m = 0, 1, 2 … sampled to give $x_m(nT)$ :

$$x_m(t)=e^{j2\pi (F\pm m{f}_s)}\Rightarrow x_m(nT)=e^{j2\pi (F\pm m{f}_s)nT}$$

Because

$$f_s{T}_s=1and{e}^{j2\pi m{f}_{s}n{T}_s}=e^{j2\pi mn}=1$$

then

This result means that two signals $x_m(t)$ and x(t) at different frequencies have the same samples. When we recover the signals from these samples then those signals lie within the Nyquist interval [-fs/2, fs/2] ( [link] b) are recovered correctly, whereas the signals having frequencies outside the Nyquist interval may be aliased into this interval. In general, for an analog signal of frequency F sampled at the sampling rate fs , first we add and subltract frequencies as follows:

and then look for the frequencies lying within the Nyquist interval, they are the reconstructed frequencies.

Solution

With F = 50Hz, $f_s=80$ Hz, the signal is undersampled (not satisfied the sampling theorem). The Nyquist interval is [-40Hz, 40Hz]. The samples do not only represent the frequency F = 50Hz but all frequencies $F\pm m{f}_s=100\pm m80,m=0,1,\mathrm{2...},$ i.e. the frequencies

$$\begin{array}{l}{f}_0=50,50\pm 80,50\pm 160,50\pm \mathrm{240...}\\ =50,130,-30,210,-110,290,-\mathrm{190...}\end{array}$$

Only the frequency -30Hz lies within the Nyquist interval, then the recovered signal will be -30Hz (30Hz and phase reversal). This signal is the alias of the original signal at 50Hz. Notice that 30Hz is just the difference 80Hz – 50Hz

Now, the sampling frequency is 120Hz, the sampling theorem is statisfied, then the original frequency of 50Hz will be recovered. None of other frequencies $f_0=50\pm m120=50,170,-70,290,-\mathrm{190...}$ … lie in the Nyquist interval [-60hZ, 60Hz], except the original frequency of 50Hz as already known.

Solution

For sampling rate $f_s=20kHz$ , the Nyquist interval is [-10kHz, 10kHz]. Thus the audio frequency 0 – 10kHz will be recovered as is. The audio frequency from 10 – 20kHz will be aliased into the frequencyrange 10 – 0kHz, and the audio the audio frequency from 20 – 30kHz will be aliased into the frequency range 0 – 10kHz. The resulting audio will be distorted due to the superposition of the 3 frequency bands.

We end up this section with the block diagram of the general complete DSP system (Fig.1.20). The digital signal output y(n) from the DSP unit is converted by the digital-to-analog converter (DAC or D/A) back to a coarse analog signal which is then lowpass filtered in the postfilter . The finally reconstructed analog signal $\widehat{x}(t)$ is, ideally, the same as the original input x(t).