Consider an LTI system with an impulse response $h(n)$:

$$ 
x(n) \to \bbox[5px,border:1px solid black]{h(n)} \to y(n) 
$$

What happens when $x(n) = z^n$, where $z = r e^{i\omega}$? Plug these two functions in to the convolution sum and solve:

\begin{align*}
y(n) &= \sum \limits_{k\in\Z} h(k) z^{n-k} \\
y(n) &= z^n \lr{\sum \limits_{k\in\Z} h(k) z^{-k}} \\
\end{align*}


We get $z^n \sum_{k\in\Z} h(k) z^{-k}$. We did exactly this in EE20 when we talked about the Eigen function properties of the complex exponential. It turns out $e^{i\omega n}$ does not define the entire space of complex exponentials that can have that property. \emph{Any} complex number raised to the $n^{th}$ power will come out at the other end scaled by some number. With that said, $\sum \limits_{k\in\Z} h(k) z^{-k}$ defines the transfer function of the system, $\hat{H}(z)$, a.k.a. the system function. This function is the \emph{Z-Transform} of the impulse response:

\begin{nicebox}
$$
\hat{X}(z) = \sum \limits_{n\in\Z} h(n) z^{-n}
$$
\end{nicebox}

Just as before we had a frequency response with fourier transforms, we have the transfer function with Z transforms.

\begin{nicebox}
\begin{align*}
h(n) &\ftp H(\omega) = \sum \limits_{n\in\Z} h(n) e^{-i\omega n} \\
h(n) &\ztp \hat{H}(z) = \sum \limits_{n\in\Z} h(n) z^{-n}
\end{align*}
\end{nicebox}

Lets expand this where $z = re^{i\omega}$:

$$
\hat{H}(z) = \sum \limits_{n\in\Z} h(n) r^{-n}e^{-i\omega n} = \ftrans{h(n)r^{-n}}
$$

This $r$ allows us to fly off the nest, which has been the unit circle. In the Z-transform we allow ourselves to go inside or outside the unit circle and gain flexibility in that way it is possible to deal with a function $h(n)$ that does not have fourier transform, because $h(n)r^{-n}$ does. We can potentially pick values of $r$, not on the unit circle, such that multiplying the function $h(n)$ with $r^{-n}$ tames whatever growth might occur.   

\begin{example}

Consider an impulse response given by $h(n) = 3^n u(n)$. $H(\omega)$ is undefined. Lets use the transfer function:

\begin{align*}
\hat{H}(z) &= \sum \limits_{n\in\Z} h(n)z^{-n} \\
\hat{H}(z) &= \sum \limits_{n\in\Z} 3^n u(n) z^{-n} \\
\hat{H}(z) &= \sum \limits_{n = 0}^{\infty} 3^n z^{-n} \\
\hat{H}(z) &= \sum \limits_{n = 0}^{\infty} \lr{3z^{-1}}^n \\
\hat{H}(z) &= \frac{1}{1 - 3z^{-1}} \mbox { if } \abs{3z^{-1}} \lt 1 \\
\end{align*}

The sum converges provided $\abs{3z^{-1}} \lt 1$, which is reduces to $\abs{z} \gt 3$. This makes sense because the radius is the magnitude of $z$. If we look at the sum and explicity solve with $r$, we get the same result:

\begin{align*}
\hat{H}(z) &= \sum \limits_{n = 0}^{\infty} 3^n \lr{re^{i\omega}}^{-n} \\
\hat{H}(z) &= \sum \limits_{n = 0}^{\infty} \lr{\frac{3}{z}}^n e^{-i\omega n} \\
\end{align*}

We found that the Z-transform was associated with a \emph{Region of Convergence}, an $RoC$. This regions contains the set of points in the complex plane which correspond to values of $z$ which provide convergence for the Z-transform. Note that when you break up $z$ into its polar form, $re^{i\omega}$, only the radius $r$ determines the range of convergence. If you find a particular radius for which an infinite series converges, any point on that circle will be perfectly fine as far as convergence is concerned.

\end{example}

{\bf Question}

1) For a function to have both a Z-transform and a DTFT, what must be true about the $RoC$ of the function? 

2) How can you get $H(\omega)$ from $\hat{H}(z)$?

{\bf Answer}


1) The $RoC$ must include the unit circle. That is, $\abs{z} = 1 \subset RoC(h)$. 

2) $H(\omega) = \left. \hat{H}(z) \right|_{z = e^{i\omega}}$


The Z-Transform