In the next phase of the course we are going to introduce some new transforms in order to handle a class of problems that we could not handle previously with the Fourier transform. Recall some of the Fourier transform issues. We talked about


(a) {\bf absolutely summable signals}


These signals have fourier transforms that are continuously infinitely differentiable and well-defined. Let's say that we have $x(n) = \alpha^n u(n)$ where $\abs{\alpha} \lt 1$. What was the frequency response? The transform of this particular signal is $\frac{1}{1-\alpha e^{-i\omega}}$. If you were to write this in terms of $e^{i\omega}$, this expression is differentiable and smooth when you plot it. As a complex function it is a perfectly smooth function and infinitely differentiable. All finite-length signals also have this property. For example, $x(n) = \delta(n+1) + \delta(n) + \delta(n-1)$. With finite length signals we never have an issue the convergence of the fourier transform. \\

Absolutely summable signals can be defined as 

\begin{nicebox}
$$l^1 : \{ x: \Z \to \C \st \sum_{n \in \Z} \abs{x(n)} \lt \infty \}$$
\end{nicebox}

Another class of signals that we have discussed are


(b) {\bf square summable (finite-energy signals)}

\begin{nicebox}
$$l^2 : \{ x: \Z \to \C \st \sum_{n \in \Z} \abs{x(n)}^2 \lt \infty \}$$
\end{nicebox}

Consider a function that is square summable but not absolutely summable, that is, $x \in l^2 \setminus l^1$. An example of this is $x(n) = \frac{1}{\pi n} \sin(A n)$. $X(\omega)$ has discontinuities. We have also discussed signals that are


(c) not absolutely or square summable signals

If $x \notin l^1 \cup l^2 \Rightarrow$ ``All bets are off''. There are only two cases:

        i) DTFT Defined $\Rightarrow$ DTFT will have diracs. 
An example of this type of signal is $x(n) = e^{i\omega_0 n}$. $X(\omega) = \delta(\omega - \omega_0)$ for $\abs{\omega} \lt \pi$.

		ii) DTFT not defined.
An example of this type of signal is $x(n) = \alpha^n u(n)$ where $\abs{\alpha} \gt 1$.
    

There is another class of signals called bounded signals, which are given by

\begin{nicebox}
$$l^\infty : \{ x: \Z \to \C \st \abs{x(n)} \lt M \lt \infty \mbox{ } \forall n \}$$
\end{nicebox}


Why use Z-Transforms