ECE 320
Sampling Example
This is an example of ideal impulse sampling of a continuous-time signal to
produce a discrete-time signal. Several different sampling periods Ts will
be used to study the effect of the sampling period on the ability of the
continuous-time signal to be reconstructed from the discrete-time samples.
With each sample period Ts, the values of the samples exactly equal the
values of the continuous-time signal at the corresponding sample points in
time. The continuous-time signal is the sum of two pure cosine terms as
shown by the equation below and illustrated in the figure.
Original Continuous-Time Signal x(t)
The three sample periods which will be used are Ts1 = 0.05 s, Ts2
= 0.1 s, and Ts3 = 0.2 s. The corresponding sampling frequencies are
ws1 = 125.6637 r/s, ws2 = 62.8319 r/s, ws3 = 31.4159 r/s. The effect of sampling x(t) at each
of these rates will be studied in turn.
First, consider Ts1 = 0.05 seconds (ws1 = 125.6637 r/s). We will call the
sampled signal x*(t). The frequencies that appear in the sampled signal
will be the frequencies in the original signal (w1 = 7 r/s and w2
= 23 r/s), the
negatives of those values (-7 r/s and -23 r/s), and integer multiples of
the sampling frequency added to those frequencies in accordance with the
following equation, where X(jw) is the Fourier Transform of x(t) and
X*(jw) is the Fourier Transform of x*(t)
Note 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 (ws1/2 = 62.8319
r/s). The first few frequencies which appear in X*(jw) when Ts =
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
Note that the original frequencies are there, and also note that there are
no frequencies between -ws1/2 and +ws1/2 other than the original frequencies
(and their negative values). Therefore, the original continuous-time signal
x(t) could be recovered from the samples using an ideal low-pass filter with
a bandwidth of ws1/2. Even with practical filters, x(t) could be recovered
fairly well since the next higher frequency is 102.7 r/s. In the figure
below, note that the sample values are close enough together in time to
give an accurate picture of the continuous-time x(t).
Samples of x(t) with Ts1 = 0.05 s
Next, consider Ts2 = 0.1 seconds (ws2 = 62.8319 r/s). The frequency content
of the sampled signal x*(t) is computed in the same way as before. Both
of the frequencies in x(t) are less than ws2/2 = 31.4159 r/s, but now the
higher frequency (23 r/s) is closer to the top of the primary strip (defined
by +/- ws/2) than with the first sampling period. The first few frequencies
which appear in X*(jw) when Ts = 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
Note that there are no frequencies between -ws2/2 and +ws2/2 except the
original frequencies, but now the filter used to recover x(t) from the
samples must be much closer to ideal than before since the next frequencies
outside the band of +/- ws2/2 are at +/- 39.8 r/s. In the figure below,
note that the samples exactly equal the x(t) values at the sample points,
but if you looked only at the sample values, you would not have as clear
an idea of what x(t) actually looks like as you would be the Ts1=0.05
sampling period.
Samples of x(t) with Ts2 = 0.1 s
Now, consider Ts3 = 0.2 seconds (ws3 = 31.4159 r/s). Now the higher frequency
in x(t) (23 r/s) is larger than ws3/2 = 15.7080 r/s. Aliasing occurs
in this situation. This means that there will be frequencies in X*(jw)
which have values between -ws/2 and +ws/2 which are not in the original
signal. "High" frequencies in the original signal (23 r/s in this example)
are appearing under the "alias" of low frequencies. For our example, the
frequency content in X*(jw) will have a term (23 r/s - 31.4159 r/s) in it,
so the frequency -8.4159 r/s will appear in X*(jw) (as well as +8.4159 r/s).
The first few frequencis which appear in X*(jw) when Ts = 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
If an ideal low-pass filter with a cut-off frequency of ws3/2 = 15.7080 r/s
is used to process the discrete-time samples, the original frequency of 23
r/s would be eliminated, and the frequency of 8.4159 r/s would be included.
This produces a completely different signal than the original x(t), even
though the sample values correspond to x(t) exactly. If we define the
new signal by x1(t)
we can see that the samples with Ts3 = 0.2 seconds exactly equal the values
of x(t) and x1(t) at the sample points even though they are two completely
different signals. This illustrates the fact that in order to recover the
original continuous-time signal from discrete-time samples, the sampling
frequency must be at least twice as high as the highest frequency in the
continuous-time signal. The higher the sampling frequency, the more
reliable and easier will be the recovery -- compare the sampled signals using ws1 vs.
ws2. When the sampling frequency is less than twice the highest
frequency in the continuous-time signal, recovery cannot be accomplished
using only the sample values, even with ideal filtering.
Samples of x(t) with Ts3 = 0.2 s
Comparing samples of x(t) and x1(t) with
Ts3 = 0.2 s
MATLAB Code
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Latest revision on 05/07/01 11:09 AM
