4.2.3. Pulse and Shift

\(\newcommand{\op}[1]{\mathsf #1}\) \(\newcommand{\ztarrow}{\stackrel{\op Z}{\longrightarrow}}\)

Let’s start with a simple signal \(x[n]=\delta[n]\). Then we have:

\[\begin{split}X(z) &= \sum_{n=-\infty}^{\infty} x[n] z^{-n} \\ &= \sum_{n=-\infty}^{\infty} \delta[n] z^{-n} \\ &= z^0 = 1\end{split}\]

The last step uses the sifting property of the delta pulse. A pulse in the time domain corresponds with a constant value 1 in the Z-domain:

\[\delta[n]\ztarrow 1\]

In case we take a shifted pulse \(\delta[n-n_0]\) we get:

\[X(z) = \sum_{n=\infty}^{\infty} x[n] z^{-n} = \sum_{n=\infty}^{\infty} \delta[n-n_0] z^{-n} = z^{-n_0}\]

i.e.:

\[\delta[n-n_o]\ztarrow z^{-n_0}\]

Shifting over \(n_0\) in the time domain thus corresponds with multiplication with \(z^{-n_0}\) in the Z-domain. We have seen that in the above equation for the pulse but we can prove it for any signal \(x[n]\).

Consider the shifted signal \(x[n-n_0]\), then by definition the Z-transform of the shifted signal is:

\[\sum_{n=\infty}^{\infty} x[n-n_0] z^{-n}\]

changing from variable \(m=n-n_0\) we get

\[\sum_{m=\infty}^{\infty} x[m] z^{-(m+n_0)} = z^{-n_0} \sum_{m=\infty}^{\infty} x[m] z^{-m)} = z^{-n_0} X(z)\]

where \(X(z)\) is the Z-transform of the original signal \(x[n]\).

Summarizing, given the Zt{} pair:

\[x[n]\stackrel{\op Z}{\longrightarrow}X(z)\]

we have that:

\[x[n-n_0]\stackrel{\op Z}{\longrightarrow}z^{-n_0}X(z)\]