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.


1 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.

  • $\begingroup$ Thank you very much for your answer. I see your logic is valid. But in my derivation when I exchange the order of summation, then evaluate it at Z. Is that wrong? Because I think that is property of the impulse signal $\endgroup$
    – JordenSH
    Oct 7, 2017 at 13:16
  • $\begingroup$ What is wrong in your derivation? Well it's pretty easy... The double infinite series sum $\sum_{k=-\infty}^{\infty} z^{-k}$ simply cannot converge for any $z$. It would converge for example if it were $\sum_{k=0}^{\infty} z^{-k}$ ... see the lower limit change? $\endgroup$
    – Fat32
    Oct 7, 2017 at 14:41
  • $\begingroup$ @Raykh If this answer was what you were looking for, then you can select it and upvote so that it can be closed by the system. Otherwise it will hang indefinite and block the system. Thanks for your understanding. $\endgroup$
    – Fat32
    Oct 7, 2017 at 15:57

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