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I've found a paper with a filter described in terms of transfer function, amplitude response and difference equation:

transfer function of the second-order low-pass filter:

$$ H(z) = \frac{(1-z^{-6})^{2}}{(1-z^{-1})^{2}} $$

amplitude response (T - sampling period):

$$ |H(\omega T)| = \frac{\sin^{2}(3\omega T)}{\sin^{2}(\omega T/2)} $$

difference equation of the filter (cut-off 11 Hz, gain 36):

$$ y[nT] = 2y[nT-T] - y[nT-2T] + x[nT] - 2x[nT- 6T] + x[nT- 12T] $$

What was the process to create these equations and how to run the filter on the signal?

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start from the transfer function: $$ H(z) = \frac{(1-z^{-6})^{2}}{(1-z^{-1})^{2}} $$

expand it: $$ H(z) = \frac{ 1-2z^{-6} + z^{-12} }{ 1 -2z^-1 + z^{-2} } $$

Frequency response is: $$ H(e^{j\omega}) = \frac{ 1-2e^{-j6\omega} + e^{-j12\omega} }{ 1 -2e^{-j\omega} + e^{-j2\omega} }$$

Note: this may be simplified by proper algebraic grouping and cancellations to what you have provided.

Transform this rational H(z) directly back into its corresponding LCCDE: $$y[n]-2y[n-1]+y[n-2] = x[n]-2x[n-6]+x[n-12]$$

I can't figure out a meaning for the wT representation, nor if it is necessary...

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  • $\begingroup$ The $\omega T$ is simply what you call $\omega$. Your $\omega$ is a normalized frequency in radians, whereas the OP's $\omega$ is not normalized; it becomes normalized by multiplying it with $T$, i.e. dividing it by the sampling frequency. $\endgroup$ – Matt L. Jun 2 '15 at 11:26
  • $\begingroup$ @MattL The paper he refers possibly wanted to show the continuous time frequency response equivalent of the corresponding discrete time filter H(z), assuming all necessary conditions are met. So that one could compute the mentioned analog cutoff frequency (11 hz in his case) in physical hertz by a suitable choice of T. $\endgroup$ – Fat32 Jun 2 '15 at 12:08
  • $\begingroup$ @Fat32 thank you! can you guess what was the process to get H(z) for the 2nd order low pass filter? if we plug T into difference equation we get sampling locations of the signal right? $\endgroup$ – Chesnokov Yuriy Jun 2 '15 at 15:43
  • $\begingroup$ sorry but I'm not sure what created these equations, it is possibly described in the paper you referred to. $\endgroup$ – Fat32 Jun 2 '15 at 21:08
  • $\begingroup$ @Fat32 I meant what is the typical process to create a filter in H(z) form? I remember some filter design toolboxes that output just filter coefficients. There is no description of how H(z) have been created in the paper. $\endgroup$ – Chesnokov Yuriy Jun 3 '15 at 7:59

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