# Z-Transform Identity

I've come across an identity and would like to know if it has some sort of formal name or derivation or explanation or something! Also, I'm curious as to whether others are aware of such an identity.

The identity is as follows: $$h[n]*c^n = H(c) \cdot c^n$$

Also, I know that if $|c| = 1$, $c$ represents half a cosine wave (from Euler's formula) and you can use conjugate symmetry and superposition to show that the above is true for cosine and sine function (replace $c$ with a cosine or sine function). However, what happens if $|c|\not= 1$? What would that be? Would it be a complex geometric series or can you create something similar to sine/cosine by letting $|c| \not= 1$?

I know this is not a Laplace Transform, but there is no Z-Transform tag so I just added Laplace because Z-Transform is basically the same thing, but in the discrete domain.

Edit 1: I'm thinking that if $|c| > 1$, then you could use superpostion and conjugate symmetry to make whatever the opposite of decaying cosine/sine function, and if $|c| < 1$ then you could make a decaying cosine/sine function.

Edit 2:

$h[n]$ is a series (or sequence or whatever it's called). An example of $h[n]$ could be $h[n] = \{3, 3, 7, 4\}$.

$y[n]$ didn't need to be there so it has been removed.

$c$ is a constant complex number.

$H(c)$ is the Z-Transform of $h[c]$.

$n$ is a discrete variable.

$\mathscr{Z}\{\}$ is the Z-Transform operator. For anyone who doesn't know, the Z-Transform is the discrete version of the Laplace Transform. Here's it's definition: $\mathscr{Z}\{x[n]\} = \sum\limits_{n=-\infty}^\infty x[n]z^{-n}$

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I think an answer is more likely to emerge if you define what $y$, $h$, $H$, $Z$, $c$, and $n$ are, and how they are related one to another. –  user31373 Jun 13 '12 at 14:50
Okay, added it into the edit. –  user968243 Jun 14 '12 at 9:09

Note that convolution operation is exclusive entity of Linear Time-Invariant systems. Also from the convolution summation given above it can be proved that any function of the form $x[n] = c^n$ is a eigen function of LTI system, by which I mean, the output of the system is just scaled version of the input, i.e., $$y[n] = Kc^n$$ where $$K = H(c) = \displaystyle\sum_{k}h[k]c^{-k}$$ is some complex constant called eigenvalue. Here $h[k]$ is called impulse response of the LTI system.
I don't think that $h[n]∗c^n=H(c)⋅c^n$ is the convolution theorem. Either I'm not understanding, or you've missed the fact the in the equation I've given, the Z-Transform of $c$ is not taken on both sides of the equality; on the right hand side, it isn't $\mathcal{Z}^-1\{ H(c) \cdot \mathcal{Z}\{ c^n \}$. –  user968243 Jun 17 '12 at 10:22