How do we add the affect of the sampling rate to the difference equation?

  • #1

Main Question or Discussion Point

I designed four different transfer function to differentiate and accumulate a given function. In my example, the given function is sin() between [0, 6*PI].

  1. Dashed blue line: The original input function (i.e.; sin(t))
  2. Red line: Integration of sin(t) with initial value of -1.0. (H(s) = 1/s)
  3. Green line: Differentiation of sin(t). (H(s) = s)
  4. Magenta line: Integration of sin(t) with initial value of -1.0. (H(z) = 1/(1-z^(1)))
  5. Black line: Differentiation of sin(t). (H(z) = 1 - z^(-1))

My algorithm is working in computer, so it uses z-transform. I transformed continuous time transfer functions to discrete time using bilinear transformation.
[tex]s = \frac{2}{T_d}\times\frac{1-z^{-1}}{1+z^{-1}}[/tex]

As you can see from the graph, the continuous time operations have the same amplitude with the input signal, because we add the affect of sampling rate during bilinear transformation. But the discrete time systems have different amplitudes with the input signal. For instance, the amplitude of the output of the integration operation increases with increasing sample rate as expected, because the number of samples accumulated increases. Similarly, the amplitude of the output of the differentiating system decreases with decreasing sampling rate.

My question arises at this point.
How do I normalize the output of a discrete time system? I mean, I want to make the amplitude of the output signal independent of sampling rate.

Sampling rate = 1 samples/second
[PLAIN]http://img267.imageshack.us/img267/3425/t10r.png [Broken]
Unitary sampling rate, all the signals have the same amplitude.

Sampling rate = 2 samples/second
[PLAIN]http://img403.imageshack.us/img403/2950/t05.png [Broken]

Sampling rate = 10 samples/second
[PLAIN]http://img155.imageshack.us/img155/2510/t01.png [Broken]
Notice that, output of discrete time differentiator is almost a horizontal line.
And also notice that, originally continuous systems are well behaving.

Sampling rate = 100 samples/second
[PLAIN]http://img842.imageshack.us/img842/370/t001.png [Broken]
Output of integrator is so large that it cover all the graph!

And a little side question; why does discrete time differentiated signal (the green one) has noise (another sinusoidal signal with higher frequency) on it?
 
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Answers and Replies

  • #2
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As you approach the Nyquist sampling rate, results become sensitive to sampling frequency. If you keep the sampling frequency fast enough, the sensitivity should disappear.


nd a little side question; why does discrete time differentiated signal (the green one) has noise (another sinusoidal signal with higher frequency) on it?
That is inherent in all differentiion, analog or digital. Differentiation amplifies high frequency noise. Integration filters out high frequency noise. For that reason, differentiation is more difficult.

The usual remedy is an analog low pass filter before sampling that attenuates high frequencies.
 

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