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I have been trying to figure out one of the homework assignments for my DSP class, and have been spending quite a lot of time figuring out a particular problem. The solution to this problem was given to us, but I just do not understand the reasoning behind it.

The main idea was that we were asked to construct a causal and stable filter. One part of the solution has stated that $g[k]$ is composed of an FIR part and an IIR part, with

$$\sigma[k] \triangleq \begin{cases} 1 \qquad & k \ge 0 \\ 0 \qquad & k \lt 0 \\ \end{cases} $$

(unit step signal).

I understand FIR and IIR, and I think that IIR should be recursive and the output is computed by using the current and previous inputs and previous outputs, whereas FIR depends on previous inputs and current inputs.

$$ g[k]=\underbrace{\tfrac{1}{8} \left(-\tfrac{5}{4}\right)^{k-2} \cdot \sigma[-(k-1)] \cdot \sigma[k]}_{g_\text{FIR}[k]} + \underbrace{\tfrac{1}{8} \left(-\tfrac{3}{4}\right)^{k-2} \sigma[k-2]}_{g_\text{IIR}[k]} $$

Yet I simply do not know how to tell FIR and IIR apart in the time domain for this particular problem?

If you could help me to clarify this, that would be great help!

After Dan's clarification, I used $k = {-1, 0, 1, 2}$ to check my understanding of the FIR section as follows:

$$k=-1: g_{\text{FIR}}[k] = \tfrac{1}{8}{\left(\tfrac{-5}{4}\right)}^{-3}\sigma[-(-2)]\sigma[-1]$$ $$k=0: g_{\text{FIR}}[k] = \tfrac{1}{8}{\left(\tfrac{-5}{4}\right)}^{-2}\sigma[1]\sigma[0]$$ $$k=1: g_{\text{FIR}}[k] = \tfrac{1}{8}{\left(\tfrac{-5}{4}\right)}^{-1}\sigma[0]\sigma[1]$$ $$k=2: g_{\text{FIR}}[k] = \tfrac{1}{8}{\left(\tfrac{-5}{4}\right)}^{0}\sigma[-1]\sigma[2]$$

Hence, we can see that we obtain non-zero values of $g_{\text{FIR}}[k]$ only when k is 0 or 1, and so it is FIR.

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  • $\begingroup$ Can you learn a little $\LaTeX$ and render your equation in that? I can't read that symbol. Is it $\sigma$? $\endgroup$ Commented Jan 6, 2023 at 19:13
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    $\begingroup$ i just shown you how to do it. for future reference. $\endgroup$ Commented Jan 6, 2023 at 19:30
  • $\begingroup$ Thank you I will pay more attention to it. $\endgroup$
    – Meow _J
    Commented Jan 8, 2023 at 11:33

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The formula is the impulse response for the filter and is broken down into a finite impulse response section and an infinite impulse response section.

Look carefully at what the function given by $\sigma[k-2]$ would represent for all $k$: basically the result of this is $1$ for $k\ge2$, so as $k$ which is implied to be a time variable increases toward infinity, it will remain as $1$ that case while the multiplier $\frac{1}{8}(-3/4)^{k-2}$ gets smaller in magnitude with increasing $k$ but never reaches 0- so has an infinite time response. This is not at all the case for $\sigma[-(k-1)]\sigma[k]$. I will leave the details of showing that it is only 1 over a finite time interval as part of the homework assignment.

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  • $\begingroup$ Thank you so much. So I tried to interpret the FIR section, and just to check my understanding, since the unit step function is 0 for k<0, we will only have a real value when k equals to 0, hence finite impulse response. For any other values of k, the FIR section gives 0, am I correct? $\endgroup$
    – Meow _J
    Commented Jan 8, 2023 at 11:32
  • $\begingroup$ I don’t think so—- carefully list out each result for the product in the FIR part; edit your question to show your work at the bottom and I will take a look at what you or I are missing $\endgroup$ Commented Jan 8, 2023 at 11:40
  • $\begingroup$ And I think you mean a “nonzero value” rather than a “real value”. (Zero is a real number) $\endgroup$ Commented Jan 8, 2023 at 11:41
  • $\begingroup$ Yes! I meant non-zero values, and I just added the computations in the edits. Thanks! $\endgroup$
    – Meow _J
    Commented Jan 8, 2023 at 12:06
  • $\begingroup$ Yes! That is what I concluded as well, when k=0 OR when k=1 it is non-zero, and zero everywhere else. Good job $\endgroup$ Commented Jan 8, 2023 at 14:31

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