ECE 320

Sampling Example #2

This example is a continuation of the first example on sampling. A continuous-time signal which is the sum of sinusoids, with frequencies of 7 r/s and 23 r/s, is sampled at three different rates T_s1= 0.05 seconds, T_s2= 0.1 seconds, and T_s3= 0.2 seconds. This example will look at what happens when these three sampled signals are passed though a low-pass filter in at attempt to reconstruct the original x(t). For reference, the signal x(t) is shown in the following figure.
Original Signal x(t)

Butterworth filters constitute a family of filters which share certain characteristics. We will consider only low-pass Butterworth filters in this example. The characteristic shared by these filters is that the locations of the poles of the transfer functions fall into a certain pattern, called the "Butterworth pattern". The poles of different order filters are all the same distance away from the origin of the s-plane, with that distance being the cutoff frequency w_n r/s. The result is that at the cutoff frequency, the filter's magnitude is -3 db, regardless of the order of the filter. The rolloff of the magnitude curve for frequencies above the cutoff frequency is -20n db/decade, where n is the number of poles in the filter (there are no zeros). Bode plots for the magnitudes and phases of Butterworth filters for orders 1-4 are shown below.
Butterworth Filter Magnitude Curves
Butterworth Filter Phase Curves

Note that the higher the order of the filter, the steeper the roll-off in the magnitude is. This means that there can be a narrower band of frequencies between those which are passed by the filter and those which are rejected. This is normally a benefit. There is a cost associated with the steeper roll-off, however. The cost is more phase shift in the higher order filters. This phase shift corresponds to a time delay between input and output. Ideally, we would like to have a filter which has a magnitude = 1 for frequencies you want to pass and magnitude = 0 for frequencies you want to reject and have no phase shift from input to output. The transfer functions for the first 4 Butterworth filters (assuming the cutoff frequency = 1 r/s) and the general equation for the transfer function for an n^th-order filter are

We will start by passing the signal which was sampled with T_s1= 0.05 seconds through the 4th-order Butterworth filter. Remember that both of the sinusoidal frequencies which make up x(t), w₁= 7 r/s and w₂= 23 r/s, are less than 1/2 of the sampling frequency (w_s1/2 = 62.8319 r/s) and that the first few frequencies which appear in X*(jw) when T_s= 0.05 s are:

[-132.6637   -118.6637   -102.6637  -23.0000   -7.0000   7.0000   23.0000  
         102.6637  118.6637  132.6637] r/s

The cutoff frequency for the filter is w_s1/2. In the figure below, you can see that the filtered signal approximates the original x(t) quite well. There is a slight time delay between the original and reconstructed signal, but not much distortion. One reason for this is that with T_s= 0.05 s, there is a big gap between the frequencies in x(t) and the next frequency in the sampled signal.
Signal Reconstruction of x(t) with T_s1= 0.05 s

Next, consider T_s2= 0.1 seconds (w_s2 = 62.8319 r/s). Both of the frequencies in x(t) are less than w_s2/2 = 31.4159 r/s, but now the higher frequency (23 r/s) is closer to the top of the primary strip than with the first sampling period. The first few frequencies which appear in X*(jw) when T_s= 0.1 s are:

[-69.8319  -55.8319  -39.8319  -23.0000   -7.0000    7.0000   23.0000   
        39.8319   55.8319   69.8319   85.8319] r/s

We will pass this sampled signal through the 4th-order Butter worth filter. The cutoff frequency is w_s2/2 r/s. The filtered signal approximates the original x(t) fairly well, in that most of the peaks and valleys of x(t) are also in the filtered signal. Now there is more time delay and more distortion in the signal. There is less gap between the frequencies in x(t) and the next frequency in the sampled signal.
Signal Reconstruction of x(t) with T_s2= 0.1 s

Now, consider T_s3= 0.2 seconds (w_s3 = 31.4159 r/s). Aliasing has occurred in sampling this signal because the sampling frequency is less than twice the highest frequency in x(t). The first few frequencies which appear in X*(jw) when T_s= 0.2 s are:

[-38.4159  -24.4159  -23.0000   -8.4159   -7.0000    7.0000    8.4159   
       23.0000   24.4159   38.4159] r/s

A 4th-order Butterworth filter with a cut-off frequency of w_s3/2 = 15.7080 r/s is now used to process the discrete-time samples. The next figure shows that there is significant distortion in the filtered signal relative to x(t). The higher frequency oscillations in x(t) are absent in the filtered version of the sampled signal. This filtered signal does not represent the characteristics of the original x(t) very well at all.
Signal Reconstruction of x(t) with T_s3= 0.2 s

If we compare the filtered signal just produced from the signal sampled at T_s3= 0.2 seconds with the signal x₁(t) described in the Example #1 on sampling (x₁(t) has frequencies of 7 r/s and 8.4159 r/s), we see that they are very similar except for the time delay. The aliasing has produced a sampled signal which accurately describes a different signal, not the original signal. The original x(t) cannot be recovered from its samples when T_s3= 0.2 seconds is used as the sampling period. The false signal x₁(t) is the result of low-pass filtering the sampled signal.
Comparison of Filtered Signal with x1(t) with T_s3= 0.2 s

MATLAB Code

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Latest revision on 05/07/01 11:44 AM