Sampling rate decimation Compressing the sampling rate by an integer factor in the discrete-time domain is defined in the provided block diagram. The down arrow in ↓K shows down sampling by a part of K. The filter H(z) is a digital anti-aliasing filter whose output v(n) is a low pass filtered type of x(n).
If the filter H(z) is explained as a linear phase FIR filter with (M+1) coefficients defined as {br, r = 0 to M}, (some call it "Mth order"), then
We requires the output y(n) to be a down-sampled version of x(n), that is
Example Suppose the 4 Hz signal x(t) = cos 2π4t which is obviously band-limited to Fmax = 4 Hz. It is more sufficient to sample it at 8 Hz. Consider instead that it has been over sampled, say, by a factor of 6 at Fs = 48 Hz to give x(n) = cos 2π4n(1/48) = cos (πn/6).
Can we create from x(n) another signal x3(n) that is a discrete-time version of x(t) sampled at Fs3 = Fs/8 = 6 Hz? This is down sampling by a part of 8. We simply redefine t by nT = n(1/6) to get
x3(n) = cos 2π4n(1/6) = cos (8πn/6) = x(8n) = {x(0), x(8), x(16), ...}
In other parts, x3(n) is prepared up of every 8 sample of x(n). For each sample value of x(n) we take we replaced the next 7 samples. We know, however, that a sampling frequency of 6 Hz does not consider the sampling theorem; in that case down sampling has been going too far.
We defines below three plots: (1) The sampled (at 48 Hz) version x(n) - that is replaced from above, (2) x(2n), the 2-fold down-sampled version of x(n); that is same to sampling x(t) at 24 Hz, and (3) x(8n), the 8-fold down-sampled version of x(n); that is equivalent to sampling x(t) at the unacceptably low rate of 6 Hz.
t1 = 0 : 1/48: 0.5; xn = cos (2*pi*4*t1); %Sampled at 48 Hz
subplot(3, 1, 1), stem(t1, xn); legend ('x(n) at 48 Hz');
xlabel ('time, sec.'), ylabel('x(n)'); grid; title ('x(nT) at T = 1/48')
%
t2 = 0 : 1/24: 0.5; xt2 = cos (2*pi*4*t2); %Sampled at 24 Hz
subplot(3, 1, 2), stem(t2, xt2); xlabel ('time, sec.'), ylabel('x(2n)');
grid; title ('2-fold down-sampled, x(nT) at T = 1/24 sec.')
%
t4 = 0 : 1/6: 0.5; x3n = cos (2*pi*4*t4); %Sampled at 6 Hz
subplot(3, 1, 3), stem(t4, x3n); xlabel ('time, sec.'), ylabel('x(8n)');
grid; title ('8-fold down-sampled, x(nT) at T = 1/6 sec.')
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