Z-transform of an impulse signal in discrete time

Question

am trying to compute the Z-transform of the following signal \begin{equation*} x\left[n\right]\:=\:\sum_{k=-\infty \:}^{\infty \:}\:\delta \:\left[n-k\right] \end{equation*} so I thought it would be \begin{equation*} X\left(z\right)\:=\:\sum _{n=-\infty \:}^{\infty }\left(\sum_{k=-\infty }^{\infty}\delta \:\left[n-k\right]\right)\:z^{-n} \end{equation*} exchanging the order of summation and including the z inside the parenthesis \begin{equation*} X\left(z\right)\:=\:\sum _{k=-\infty \:}^{\infty }\left(\sum_{n=-\infty }^{\infty}\delta \:\left[n-k\right]z^{-n}\right)\: \end{equation*} then we get \begin{equation*} X\left(z\right)\:=\:\sum _{k=-\infty \:}^{\infty }z^{-k}\: \end{equation*} which would be \begin{equation*} X\left(z\right)\:=\:\frac{1}{\left(1-z^{-1}\right)} \end{equation*}

Is this correct?? can somebody help me please. Thanks.

Answer 1

There's a ROC (region of convergence) problem in your derivation.

Consider a decomposition of the signal as $x[n] = x_1[n] + x_2[n]$. Then in order for the $Z$ transform of $x[n]$ to exist, the corresponding $Z$-transforms of both $x_1[n]$ and $x_2[n]$ must exist for a set of $z$ which is the intersection of the individual ROCs of the components $x_1[n]$ and $x_2[n]$. If the intersection set is empty then the Z transform does not exist. Applying this to your signal yields.

$$x[n] = \sum_{k=-\infty}^{\infty} \delta[n-k] = u[n] + u[-n-1] = 1 ~~~~,~~~\text{for all n}$$

Where the individual $Z$-transform and their corresponding ROCs are $$ u[n] \longleftrightarrow \frac{1}{1-z^{-1}} ~~~,~~~ ROC_1 ~~~ |z| > 1 $$ $$ u[-n-1] \longleftrightarrow \frac{-1}{1-z^{-1}} ~~~,~~~ ROC_2 ~~~ |z| < 1 $$

As you can see the intersection of $ROC_1$ and $ROC_2$ is empty; i.e; there is no set of values $z$ for which both of the components has convergent $Z$ transforms. Hence we conclude that the $Z$ transform $X(z)$ of the signal $x[n]=1$ for all $n$ does not exist.