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I am new in DSP and I need to have a clear udrastanding about below example. Consider the below $h[n]$ as the impulse response of a system:

\begin{equation} h[n] = a^n(u[n] - u[n-N]) \end{equation}

so the opened z transform can be expressed as : \begin{equation} H(z) = 1 + a^1 z^{-1} + a^2 z^{-2} + \dots + a^{N-1} z^{-(N-1)} \end{equation} and its closed form is expressed like : \begin{equation} H(z) = \frac{1}{z^{N-1}} \frac{z^N-a^N}{z-a} \end{equation} defenetly, closed form is better than the opened form for hardware implementation because as below, the closed form can be written in the time domain like this: \begin{equation} y[n] = x[n] - a^N x[n-N] + a y[n-1] \end{equation}

Now the only difference between the closed and opened form is a zero and a pole at $z=a$. Actually, it seems to me, by adding zero and pole, we obtained a better representation. But my question is,

What is the drawback of adding zero and pole to the system at $z=a$.

How can we analyze the benefits and drawbacks between closed and opened form?

I would appreciate any help you can provide.

robert bristow-johnson
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1 Answers1

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One possible difference occurs for a pole outside of unit circle. Note that the original system is a truncated IIR filter. As such, it's an FIR filter and have no stability issue for any finite $a$.

However, the closed form representation includes a pole, and implements a pole zero cancellation. Eventhough theoretically (with infinite precision arithmetic) this would not pose a problem, for finite length arithmetic and for large values of $N$, and for $|a| > 1$, you could end up in numerical issues.

Specifcally, the instantaneous dynamic range (the mantissa length of the floating point representation) would be large enough to be able to accurately represent the arithmetic difference indicated in the numerator. Indeed this could also be aproblem with the finite length implementation when $a^N$ is too large to be added to $ax[n-1]$, and that could happen when $|a|>1$. So which one would be more sensitive to this numerical issue? You better test by yourself. Furthermore, the closed-from version could impose an additional problem for those $ a \approx 1$ which would not be the case for the FIR filter.

As an advantage of te closed form, note that the FIR form and IIR form requires the same amount of memory (to hold past values of input up to a delay of $N$ samples), however the closed form (truncated IIR) requires much less computation MACS compared to FIR version for large N.

Fat32
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  • In time series modelling, most models like the one discussed above ( not exactly that one but similar ones ) require < 1 in order to have a model that is not explosive. In this case, is this not true ? If so, how are you able to know that ? Thanks. – mark leeds May 05 '19 at 04:11
  • Does the initial representation not have the same restriction that $a$ should be less than 1.0 ? It looks like it will also explode for large $n$ if $a$ is greater than 1.0. But I might be mistaken. It's happened before !!!!! Thanks. – mark leeds Oct 02 '19 at 23:49
  • No the initial form is a truncated IIR; though it seems to have an exponential $a^n$ weight multiplier (that could go to unbounded values as $n$ goes to inf) , actually it's finite term sum. And a finite term sum is always finite as long as each term is finite. Practically however, the finite term sum also gets very large; so large may not even be fitting to the register size... – Fat32 Oct 03 '19 at 08:52
  • Ok. Thanks. I don't see how that's a finite term sum as $n \rightarrow \infty$ but I agree then that it's not explosive. – mark leeds Oct 03 '19 at 14:48
  • @markleeds this impulse response $$h[n] = a^n(u[n] - u[n-N]) $$ results in an output sum that includes only $N$ terms. Because the impulse response itself is of length $N$ from $n=0$ to $n=N-1$... That's how the convolution sum requires only $N$ terms to be added. This is independent from how $n$ goes to infinity because the impulse response $h[n]$ is fixed. But when $N$ goes to infinity, then it will be unstable if $a>1$. So $N \to \infty$ and $n \to \infty$ are different cases. – Fat32 Oct 03 '19 at 21:36
  • gotcha. I didn't see ( until you explained it ) that it was only N terms even as $n$ gets large. Definitely an interesting impulse response. Thanks for clarification. – mark leeds Oct 05 '19 at 00:45
  • @markleeds u[n]-u[n-N] is a rectangular (box) window of length N, from n=0 to n=N-1. – Fat32 Oct 05 '19 at 10:18
  • Nice. So, it's the equivalent of a moving sum except for a decay factor multiplying the sum. Thanks. – mark leeds Oct 06 '19 at 13:23