Consider the following Constant Coefficient Difference Equation (CCDE) given below

$$y[n] - \frac{1}{2}y[n-1] = 2x[n] - 5x[n-1] - x[n-2]$$

Let $H(e^{j\omega})$ denote the transfer function of this system. What is $H(e^{j\pi})$?

  • $\begingroup$ Please pick a better question title. This whole site is about digital signal processing, so you literally picked the most redundant question title possible – a good question title describes the question (something to do with CCDE in your case?) $\endgroup$ Jul 4 '18 at 9:04

You need to follow the following steps (reponses at each steps are in a spoiler block, try to find them by yourself before checking). I will use the usual notation $z=e^{j w}$ for simplicity.

  • Find the relation between the transfer function $H(z)$ and the discrete-time Fourier Transform of the input $X(z)$ and the output $Y(z)$. This should be a basic results of your course

    $H(z) = \dfrac{Y(z)}{X(z)}$

  • Take the (discrete-time) Fourier Transform of your CCDE. For that, you will need to use two basic properties of the TF: linearity and time-shifting

    $Y(z) - \frac{1}{2} z^{-1} Y(z) = 2 X(z) - 5 z^{-1} X(z) - z^{-2} X(z)$

  • Compute $H(z)$ from the previous equation

    $H(z) = \dfrac{2 - 5z^{-1} - z^{-2}}{1-\frac{1}{2} z^{-1}}$

  • Finally take the value for $z = e^{j \pi} = -1$

    $H(e^{j \pi}) = \dfrac{2 + 5 - 1}{1+\frac{1}{2}} = \dfrac{6}{\frac{3}{2}} = 4$

This last result mean that your filter, at Nyquist frequency, has

no phase delay and a gain of 4.

  • $\begingroup$ Which definition of the Z-transform do you use? For the one that is most common in DSP, a delay corresponds to multiplication by $z^{-1}$, not by $z$ as in your answer. $\endgroup$
    – Matt L.
    Jul 4 '18 at 12:14
  • $\begingroup$ I was thinking of the standard definition, but didn't take time to ckeck the translation property. I should've.... I edited my answer $\endgroup$
    – Klaz
    Jul 4 '18 at 15:44

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