Create an input so that poles show up as outputs?

Question

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 $A$.

Answer 1

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.