Properties of the Z-transform

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This section first describes the main properties of the ZT. Finally, the so called one-sides ZT will be introduce, which is useful for solving Difference Equations with initial conditions.

Assume the ZT of the sequences $x[n]$ and $y[n]$ are given by $X(z)$ and $Y(z)$ and their ROC by $R_x$ and $R_y$ respectively.

$$ \begin{eqnarray*} x[n] \text{ } \circ \hspace{-1.7mm} - \hspace{-1.7mm} \circ \text{ } X(z) \text{ with ROC } R_x & \quad\text{and} \quad& y[n] \text{ } \circ \hspace{-1.7mm} - \hspace{-1.7mm} \circ \text{ } Y(z) \text{ with ROC } R_y \end{eqnarray*} $$

Property: Linearity

The ZT, as the FTD, is a linear operation:

$$ \boxed{w[n] = \alpha x[n] + \beta y[n] \quad \circ \hspace{-1.7mm} - \hspace{-1.7mm} \circ \quad W(z) = \alpha X(z) + \beta Y(z) \text{ with ROC } R_w=R_x \cap R_y} $$

The resulting ROC $R_w$ of the sequence $w[n]$ will include the intersection of $R_x$ and $R_y$. However the ROC $R_w$ may be larger.

Property: Shifting

Shifting a sequence, which can be delaying or advancing, multiplies the ZT by a power of $z$:

$$ \boxed{x[n-n_0] \quad \circ \hspace{-1.7mm} - \hspace{-1.7mm} \circ \quad z^{-n_0} X(z) \text{ with ROC } R_x} $$

Because shifting a sequence does not affect its absolute sumability, shifting does not change the ROC. Therefor, the ZT of $x[n]$ and $x[n-n_0]$ have the same ROC, with the possible exception of adding or deleting the points $z=0$ and $z=\infty$.

Property: Time reversal

The ZT of the time reversed sequence $x[-n]$ is given by $X(z^{-1}$.

$$ \boxed{x[-n] \quad \circ \hspace{-1.7mm} - \hspace{-1.7mm} \circ \quad X(z^{-1}) \text{ with ROC } 1/R_x} $$

In case the ROC $R_x$ is given by the annulus in between $r_{-}$ and $r_{+}$, the ROC of the time reversed sequence is given by the annulus in between $\frac{1}{r_{+}}$ and $\frac{1}{r_{-}}$, which is denoted by $1/R_x$.

Property: Multiply by $a^n$

$$ \boxed{a^n x[n] \quad \circ \hspace{-1.7mm} - \hspace{-1.7mm} \circ \quad X(a^{-1} z) \text{ with ROC } R_y= |a|R_x} $$

The proof of this property is as follows: If a sequence $x[n]$ is multiplied by a complex exponential $a^n$ the resulting ZT $Y(z)$ can be obtained by first combining the parameter $a^{-1}$ and the variable $z$ resulting in a scaled version of $X(z)$:

$$ \begin{eqnarray*} y[n] = a^n x[n] \text{ } \circ \hspace{-1.7mm} - \hspace{-1.7mm} \circ \text{ } &Y(z)& = \sum_{n=-\infty}^{\infty} a^n x[n] z^{-n} =\sum_{n=-\infty}^{\infty} x[n] (a^{-1} \cdot z)^{-n}\newline & & = X(a^{-1} z) \quad\text{ with ROC } R_y= |a|R_x \end{eqnarray*} $$

The resulting ROC is also scaled by the parameter $a$.

Property: Convolution Theorem

Perhaps the most important ZT property is the convolution theorem, which states that convolution in time domain is mapped to multiplication in $z$ domain.

$$ \boxed{x[n] \star y[n] \text{ } \circ \hspace{-1.7mm} - \hspace{-1.7mm} \circ \text{ } X(z) \cdot Y(z) \hspace{3mm} \text{with } R_w=R_x \cap R_y} $$

The proof of this property is shown in the following steps: First write out the convolution sum and transform to the $z$-domain.

$$ \begin{eqnarray*} w[n]= x[n] \ast y[n] & \circ \hspace{-1.7mm} - \hspace{-1.7mm} \circ & W(z) = \sum_{n=-\infty}^{\infty}\left ( \sum_{p=-\infty}^{\infty} x[p] y[n-p] \right ) z^{-n} \end{eqnarray*} $$

In the next step change the order of the two summations and then add an extra exponent $z^{-p}$ in the first summation which has to be compensated this in the second summation:

$$ \begin{eqnarray*} W(z) &=& \sum_{p=-\infty}^{\infty} x[p] \left\{ \sum_{n=-\infty}^{\infty} y[n-p] z^{-n} \right\} \newline &=& \sum_{p=-\infty}^{\infty} x[p] z^{-p} \cdot \left\{ \sum_{n=-\infty}^{\infty} y[n-p] z^{-(n-p)} \right\} \end{eqnarray*} $$

The first summation is the ZT of the sequence $x[n]$ and the second summation between large brackets is the ZT of the sequence $y[n]$ , which finalizes the proof.

$$ \begin{eqnarray*} W(z)&=& X(z) \cdot Y(z) \hspace{3mm} \text{with } R_w=R_x \cap R_y \end{eqnarray*} $$

The ROC of the result $W(z)$ of the convolution property includes the intersection of the ROC's of $X(z)$ and $Y(z)$. However, the ROC $R_w$ may be larger, if there is a pole-zero cancellation in the product $X(z) \cdot Y(z)$.

Property: Conjugation

$$ \boxed{x^\ast[n] \text{ } \circ \hspace{-1.7mm} - \hspace{-1.7mm} \circ \text{ } X^\ast(z^\ast) \text{ ; } R_x} $$ As a corollary, note that if $x[n]$ is real-valued, $y[n]=x^\ast[n]=x[n]$, then $X(z)=X^\ast(z^\ast)$ is such as case.

The proof of the conjugation property goes as follows:

The ZT of the complex conjugate of the sequence $x[n]$ can be rewritten by taking the complex conjugation operation outside the brackets and interchange in the exponents of the variable $z$ which leads to the final result. $$ \begin{eqnarray*} y[n] = x^\ast[n] \text{ } \circ \hspace{-1.7mm} - \hspace{-1.7mm} \circ \text{ } & Y(z)& = \sum_{n=-\infty}^{\infty} x^\ast[n] z^{-n} = \left ( \sum_{n=-\infty}^{\infty} x[n] {z^{-n}}^{\ast} \right )^\ast\newline &&= \left ( \sum_{n=-\infty}^{\infty} x[n] {z^{\ast}}^{-n} \right )^\ast = X^\ast(z^\ast) \text{ ; } R_y=R_x \end{eqnarray*} $$ The ROC is not influenced by the complex operation.

Property: Derivative

$$ \boxed{n \cdot x[n] \text{ } \circ \hspace{-1.7mm} - \hspace{-1.7mm} \circ \text{ } -z \frac{\text{d}}{\text{d} z} X(z) \text{ ; } R_x} $$

The proof of this property goes as follows:

The ZT of the sequence $n \cdot x[n]$ can be pre-multiplied with $z$, which has to be compensated inside the summation: $$ \begin{eqnarray*} y[n] = n \cdot x[n] &\circ \hspace{-1.7mm} - \hspace{-1.7mm} \circ & Y(z)=\sum_{n=-\infty}^{\infty} n \cdot x[n] z^{-n} =z \sum_{n=-\infty}^{\infty} \left\{ n \cdot x[n] z^{-n-1} \right\} \end{eqnarray*} $$ The function in between brackets can be rewritten as a differentiation, which can be taken outside the summation, because of linearity, which leads to the final result: $$ \begin{eqnarray*} Y(z)& =& =z \sum_{n=-\infty}^{\infty}\left\{ -\frac{\text{d}}{\text{d} z} x[n] z^{-n} \right\} =-z \frac{\text{d}}{\text{d} z} \left( \sum_{n=-\infty}^{\infty}x[n] z^{-n} \right) = -z \frac{\text{d}}{\text{d} z} X(z) \end{eqnarray*} $$ The ROC is not influenced.

Property: Initial value

When the sequence $x[n]=0$ for $n<0$, the initial value $x[0]$ may be found from the ZT $X(z)$ by taking the limit for $z$ to infinity: $$ \boxed{x[0] = \lim_{z \rightarrow \infty} \left \{ X(z) \right \}} $$ The proof goes as follows with $x[n]=0$ for $n<0$: $$ \begin{eqnarray*} \lim_{z \rightarrow \infty} \left\{ X(z) \right\} &=& \lim_{z \rightarrow \infty} \left\{ \sum_{n=-\infty}^{\infty} x[n] z^{-n} \right\} \newline &=&\lim_{z \rightarrow \infty} \left\{ \cdots x[-1] z + x[0] + x[1] z^{-1} + \cdots \right\}\newline &=& \lim_{z \rightarrow \infty} \left\{ 0 + x[0] + x[1] z^{-1} + \cdots \right\} =x[0] \end{eqnarray*} $$ This initial value property can be shown in the following example:

One sided Z-Transform

The ZT $X(z)$ of the sequence $x[n]$, as discussed before, is the two-sided ZT. The one-sided ZT $\tilde{X}(z)$ is defined by the given equation in which the summation index $n$ does not start from $n=-\infty$ but from $n=0$. \begin{equation*} \boxed{\tilde{X}(z)=\sum_{{\color{red}{n=0}}}^{\infty} x[n] z^{-n}} \end{equation*} Most of the properties of the one-sided ZT are the same as the properties of the two-sided ZT. One that is different, however, is the shift property. Specifically, if the sequence $x[n]$ has a one-sided ZT $\tilde{X}(z)$, the difference of the one-sided ZT of the shifted sequence $x[n-k]$ is represented in red in the following shift property of the one-sided ZT: $$ \boxed{x[n-k] \text{ } \circ \hspace{-1.7mm} - \hspace{-1.7mm} \circ \text{ } {\color{red}{\sum_{p=-k}^{-1} x[p] z^{-(p+k)}}} + z^{-k} \cdot \tilde{X}(z)} $$ For the proof of this property we apply the one-sided ZT to the shifted sequence $x[n-k]$ and then substitute a new variable $p$ for $n-k$, which finalizes the proof: $$ \begin{eqnarray*} x[n-k] & \circ \hspace{-1.7mm} - \hspace{-1.7mm} \circ & \sum_{n=0}^{\infty} x[n-k] z^{-n} \overset{p=n-k}{=\!=\!=\!=} \sum_{p=-k}^{\infty} x[p] z^{-(p+k)} \newline & & = {\color{red}{\sum_{p=-k}^{-1} x[p] z^{-(p+k)}}} + z^{-k} \cdot \left( \sum_{p=0}^{\infty} x[p] z^{-p} \right) \end{eqnarray*} $$ It is this shifting property that makes the one-sided ZT useful for solving Difference Equations with initial conditions, as can be shown by the following example.