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I'm considering the following problem from some course notes.

Suppose the following is known about a discrete-time LTI system:

  1. Given the input $x[n]=(1/3)^n$ for all $n$, the system produces output $y[n]=(2/3)(1/3)^n$ for all $n$.

  2. Given the input $x[n]=(1/2)^nu[n]$, where $u[n]$ is the unit step, the output is $y[n]=\delta[n]+a(1/4)^nu[n]$.

The problem asks for the value of $a$ and for the response $y[n]$ given the input $x[n]=(1/6)^nu[n]$.

I thought from (1) I could determine the system function by taking a Fourier or Z transform, and then use that with (2) to deduce the value of $a$. For example, the Z transform of (2) says $Y(z)=H(z)X(z)$, where $Y$ and $X$ are easily computed (or obtainable from a table). If I could deduce $H$ from (1), I could get $a$.

But, as discussed here, the signals $(1/3)^n$ and $(2/3)(1/3)^n$ for all $n$ do not have Fourier or Z transforms, so I'm stuck. I could imagine there being a typo: perhaps $(1/3)^n$ should be $(1/3)^{|n|}$. But if that's not the case, I'm not sure how to proceed.

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    $\begingroup$ so $$x[n] = \left(\tfrac13 \right)^n \qquad \forall n \in \mathbb{Z} $$ ? even negative $n$? $\endgroup$ – robert bristow-johnson Aug 17 '17 at 3:45
  • $\begingroup$ I have quoted the problem verbatim. If the author intended only nonnegative $n$, why use the step function $u[n]$? $\endgroup$ – symplectomorphic Aug 17 '17 at 3:50
  • $\begingroup$ oh yeah, you're right. there are no values of $z$ that cover it for both. so #1 is a crappy problem in any case. so you need $u[n]$ in it. $\endgroup$ – robert bristow-johnson Aug 17 '17 at 5:01
  • $\begingroup$ @robertbristow-johnson: You don't need the Z-transform of the input signal to exist for the problem to make sense. Take $x[n]=\sin(n\omega_0)$, which doesn't have a Z-transform, but the output is perfectly defined as a scaled and phase shifted sinusoid with frequency $\omega_0$. Same is true for $x[n]=z_0^n$ as long as $z_0$ is inside the ROC of $H(z)$. $\endgroup$ – Matt L. Aug 17 '17 at 19:24
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This problem tests the understanding of two basic properties of (discrete-time) LTI systems. First, for an input signal of the form $x[n]=z_0^n$, $-\infty<n<\infty$, the output is given by

$$y[n]=z_0^nH(z_0)\tag{1}$$

where $H(z)$ is the system's transfer function, and where it is assumed that $z_0$ is inside the region of convergence (ROC) of $H(z)$. [Note that $(1)$ is not true for $z_0$ outside the ROC.]

The second property is simply the $\mathcal{Z}$-domain input-output relationship

$$Y(z)=X(z)H(z)\tag{2}$$

Applying $(1)$ to your example you can conclude that

$$H\left(\frac{1}{3}\right)=\frac23\tag{3}$$

and from $(2)$ you can get an expression for $H(z)$ in terms of the constant $a$. This constant can be determined from Eq. $(3)$.

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    $\begingroup$ "...where it is assumed that $z_0$ is inside the region of convergence (ROC) of $H(z)$." ---- so what value of $z_0$ would that be for question #1? $\endgroup$ – robert bristow-johnson Aug 17 '17 at 11:28
  • $\begingroup$ @Matt L.: ah, yes; I even noticed $(1/3)^n$ is an eigenfunction but didn't think of $H(1/3)$ as the eigenvalue, when of course that's the basic point. But there's a logical subtlety here: if we define $H(z_0)$ as the factor by which $z_0^n$ gets multiplied, for all input signals $z_0^n$, mustn't it be true that $H(z_0)$ exists (that is, converges) for all $z_0$? That seems like a bad assumption. For example, (2) gives an expression for $H(z)$ that could have ROC either $|z|>1/4$ or $|z|<1/4$. If we assume the former so that $z_0=1/3$ is included, what about $z_0=1/5$? Shouldn't there exist $\endgroup$ – symplectomorphic Aug 17 '17 at 15:12
  • $\begingroup$ exist a (finite) factor $H(1/5)$ by which the signal $(1/5)^n$ gets multiplied, according to our basic assumption that for all $z_0$ there exists $H(z_0)$ such that $y[n]=H(z_0)z_0^n$? $\endgroup$ – symplectomorphic Aug 17 '17 at 15:12
  • $\begingroup$ I suppose that to be precise, the basic principle is as follows: for any given $z_0$, there might exist a finite (complex) number $H(z_0)$ such that $y[n]=H(z_0)z_0^n$, depending on the region of convergence of the $Z$ transform of the impulse response $h[n]$. What happens when we then feed $(z_0)^n$ as input when $H(z_0)$ is undefined? The system still responds with some $y[n]$, but it isn't a constant times $(z_0)^n$. Is this right? $\endgroup$ – symplectomorphic Aug 17 '17 at 15:32
  • $\begingroup$ @robertbristow-johnson: $\frac13$, see Eq. (3). $\endgroup$ – Matt L. Aug 17 '17 at 19:08

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