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Let's say we have a rational, causal, stable LTI system with transfer function $$H(z) = \frac{A(z)}{B(z)}$$ If $H(z)$ has $N$ poles, we can in theory have only 1 of those poles, $p_i$, show up at the output. So $y[n]$ is $p_i^n$ after some $n>A$. To do this we need to come up with an input $x[n]$ that is causal and has finite length, and $A$, so that we get that output $y[n]$.

How would we construct such a finite $x[n]$ to get this behavior?

Edit: We can assume also that we know the most general form of $h[n]$ since $H(z)$ is causal, LTI and stable. $$h[n] = a_0\delta[n] + a_1\delta[n-1]+a_2\delta[n-2]...a_A\delta[n-A]\\+b_1p_1^nu[n]+b_2\cdot n \cdot p_1^nu[n] + b_3 \cdot n^2 \cdot p_1^nu[n] + ...\\+c_1p_2^nu[n]+c_2 \cdot n \cdot p_2^nu[n] + c_3 \cdot n^2 \cdot p_2^nu[n] + ...\\.\\.\\.$$ So would there be a clever construction of a finite $x[n]$ to cancel all other poles except $p_i$? The first part of $h[n]$ gives us the finite first part of $y[n]$ until integer $A$.

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  • $\begingroup$ so you want find out an input x[n] which supresses all the remaining poles other than the chosen one ? $\endgroup$
    – Fat32
    Commented Nov 20, 2017 at 21:23
  • $\begingroup$ Yes, but this x[n] is finite. And y[n] can be anything until n>A, where y[n] becomes p^n. $\endgroup$
    – ItM
    Commented Nov 20, 2017 at 21:25
  • $\begingroup$ After n>A, y[n] is exactly p^n. $\endgroup$
    – ItM
    Commented Nov 20, 2017 at 21:28

1 Answer 1

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You should always be able to achieve this after $N-1$ samples. For this I will denote the given transfer function as

$$ H(z) = \frac{A(z)}{\prod_{k=1}^N (z - p_k)}. \tag{1} $$

Now by starting with an impulse and filter it with

$$ F(z) = \frac{\prod_{k\neq i} (z - p_k)}{z^{N-1}} = \sum_{k=0}^{N-1} \alpha_k\,z^{-k} \tag{2} $$

and use the output of this filter as the input to the system. The combined transfer function looks like

$$ H(z)\,F(z) = \frac{A(z)}{(z - p_i)\,z^{N-1}}. \tag{3} $$

Therefore this filtered signal should only excite the mode of the system associated with the pole $p_i$. And the $z^{N-1}$ term only adds a delay of $N-1$ samples, so it only takes a finite amount of time after which only $p_i^n$ is visible in the output.

However it can be noted that this does require you to have perfect knowledge of the poles you cancel.

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  • $\begingroup$ So what does the finite x[n] look like? $\endgroup$
    – ItM
    Commented Nov 21, 2017 at 0:10
  • $\begingroup$ @ItM By expanding the products of all the term in the numerator of $F(z)$ and dividing each term by its denominator, then then the expression for $F(z)$ just becomes a FIR filter. From this it is easy to derive what $x[n]$ should be, given that the input to this filter is an impulse. $\endgroup$
    – fibonatic
    Commented Nov 21, 2017 at 4:44
  • $\begingroup$ I'm not familiar with FIR filters. The question was specifically what x[n] is. But thanks anyways! $\endgroup$
    – ItM
    Commented Nov 21, 2017 at 8:40
  • $\begingroup$ ItM: Note that, in the general case, exact pole cancellation with cascaded systems is practically impossible with finite precision arithmetic. So if using a computer, the answer to your question is that there is none. However @fibonatic's answer should work analytically. $\endgroup$
    – Andy Walls
    Commented Nov 21, 2017 at 11:35
  • $\begingroup$ @AndyWalls You can also filter the impulse twice with $F(z)$. This should set both the magnitude and the slope of the Bode diagram at the (estimated) pole locations to zero, which should give you a bit more robustness in the actual values of these poles. $\endgroup$
    – fibonatic
    Commented Nov 21, 2017 at 23:01

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