0
$\begingroup$

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 $H(e^{j\pi})$?

$\endgroup$
  • $\begingroup$ 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?) $\endgroup$ – Marcus Müller Jul 4 '18 at 9:04
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.