Properties of Z-Transform

The z-transform has a set of properties in parallel with that of the Fourier transform (and Laplace transform). The difference is that we need to pay special attention to the ROCs. In the following, we always assume

$${\cal Z}[x[n]]=X(z)\;\;\;\;ROC=R_x$$

and

$${\cal Z}[y[n]]=Y(z)\;\;\;\;ROC=R_y$$

• Linearity
  
  $${\cal Z}[a x[n]+b y[n]]=aX(z)+bY(z), \;\;\;\;ROC \supseteq (R_x \cap R_y)$$
  
  
  While it is obvious that the ROC of the linear combination of $x[n]$ and $y[n]$ should be the intersection of the their individual ROCs $R_x \cap R_y$ in which both $X(z)$ and $Y(z)$ exist, note that in some cases the ROC of the linear combination could be larger than $R_x \cap R_y$. For example, for both $x[n]=a^n u[n]$ and $y[n]=a^n u[n-1]$, the ROC is $\vert z\vert > \vert a\vert$, but the ROC of their difference $a^n u[n]-a^n u[n-1]=\delta[n]$ is the entire z-plane.

• Time Shifting
  
  $${\cal Z}[x[n-n_0]]=z^{-n_0} X(z),\;\;\;\;ROC=R_x$$
  
  
  Proof:
  
  $${\cal Z}[x[n-n_0]]=\sum_{n=-\infty}^\infty x[n-n_0]z^{-n}$$
  
  
  Define $m=n-n_0$, we have $n=m+n_0$ and
  
  $$\sum_{m=-\infty}^\infty x[m]z^{-m}z^{-n_0}=z^{-n_0} X(z)$$
  
  
  The new ROC is the same as the old one except the possible addition/deletion of the origin or infinity as the shift may change the duration of the signal.

• Time Expansion (Scaling)

  
  

  $${\cal Z}[x[n/k]]=X(z^k),\;\;\;\;ROC=R_x^{1/k}$$
  
  
  The discrete signal $x[n]$ cannot be continuously scaled in time as $n$ has to be an integer (for a non-integer $n$ $x[n]$ is zero). Therefore $x[n/k]$ is defined as
  
  $$x[n/k]\stackrel{\triangle}{=}\left\{ \begin{array}{ll} x[n/k... ... is a multiple of $k$} 0 & \mbox{else} \end{array} \right.$$
  
  
  Example: If $x[n]$ is ramp

  $n$	1	2	3	4	5	6
  $x[n]$	1	2	3	4	5	6

  then the expanded version $x[n/2]$ is

  $n$	1	2	3	4	5	6
  $n/2$	0.5	1	1.5	2	2.5	3
  $m$		1		2		3
  $x[n/2]$	0	1	0	2	0	3

  where $m$ is the integer part of $n/k$.

  Proof: The z-transform of such an expanded signal is
  

  $${\cal Z}[x[n/k]]=\sum_{n=-\infty}^\infty x[n/k]z^{-n} =\sum_{m=-\infty}^\infty x[m]z^{-km}=X(z^k)$$
  
  
  Note that the change of the summation index from $n$ to $m$ has no effect as the terms skipped are all zeros.

• Convolution
  
  $${\cal Z}[x[n]*y[n]]=X(z) Y(z), \;\;\;\;ROC \supseteq (R_x \cap R_y)$$
  
  
  The ROC of the convolution could be larger than the intersection of $R_x$ and $R_y$, due to the possible pole-zero cancellation caused by the convolution.

• Time Difference
  
  $${\cal Z}[x[n]-x[n-1]]=(1-z^{-1})X(z), \;\;\;\;\;ROC=R_x$$
  
  
  Proof:
  
  $${\cal Z}[x[n]-x[n-1]]=X(z)-z^{-1}X(z)=(1-z^{-1})X(z)=\frac{z-1}{z}X(z)$$
  
  
  Note that due to the additional zero $z=1$ and pole $z=0$, the resulting ROC is the same as $R_x$ except the possible deletion of $z=0$ caused by the added pole and/or addition of $z=1$ caused by the added zero which may cancel an existing pole.

• Time Accumulation
  
  $${\cal Z}[\sum_{k=-\infty}^nx[k]]=\frac{1}{1-z^{-1}}X(z), \;\;\;\;ROC \supseteq [R_x \cap (\vert z\vert>1)]$$
  
  
  Proof: The accumulation of $x[n]$ can be written as its convolution with $u[n]$:
  
  $$u[n]*x[n]=\sum_{k=-\infty}^\infty u[n-k]x[k]=\sum_{k=-\infty}^n x[k]$$
  
  
  Applying the convolution property, we get
  
  $${\cal Z}[\sum_{k=-\infty}^nx[k]]={\cal Z}[u[n]*x[n]]=\frac{1}{1-z^{-1}}X(z)$$
  
  
  as ${\cal Z}[u[n]]=1/(1-z^{-1})$.

• Time Reversal
  
  $${\cal Z}[x[-n]]=X(1/z)\;\;\;\;ROC=1/R_x$$
  
  
  Proof:
  
  $${\cal Z}[x[-n]]=\sum_{n=-\infty}^\infty x[-n]z^{-n} =\sum_{m=-\infty}^\infty x[m](\frac{1}{z})^{-m}=X(1/z)$$
  
  
  where $m=-n$.

• Scaling in Z-domain

  
  

  $${\cal Z}[a^n x[n]]=X\left(\frac{z}{a}\right),\;\;\;\;ROC=\vert a\vert R_x$$
  
  
  Proof:
  
  $${\cal Z}[a^n x[n]]=\sum_{n=-\infty}^\infty x[n]\left(\frac{z}{a} \right)^{-n} =X\left(\frac{z}{a}\right)$$
  
  
  In particular, if $a=e^{j\omega_0}$, the above becomes
  
  $${\cal Z}[e^{jn\omega_0} x[n]]=X(e^{-j\omega_0}z)\;\;\;\;ROC=R_x$$
  
  
  The multiplication by $e^{-j\omega_0}$ to $z$ corresponds to a rotation by angle $\omega_0$ in the z-plane, i.e., a frequency shift by $\omega_0$. The rotation is either clockwise ($\omega_0>0$) or counter clockwise ($\omega_0<0$) corresponding to, respectively, either a left-shift or a right shift in frequency domain. The property is essentially the same as the frequency shifting property of discrete Fourier transform.

• Conjugation
  
  $${\cal Z}[x^*[n]]=X^*(z^*), \;\;\;\;\;ROC=R_x$$
  
  
  Proof: Complex conjugate of the z-transform of $x[n]$ is
  
  $$X^*(z)=[\sum_{n=-\infty}^\infty x[n]z^{-n}]^* =\sum_{n=-\infty}^\infty x^*[n](z^*)^{-n}$$
  
  
  Replacing $z$ by $z^*$, we get the desired result.

• Differentiation in z-Domain
  
  $${\cal Z}[n x[n]]=-\frac{d}{dz}X(z), \;\;\;\; ROC=R_x$$
  
  
  Proof:
  
  $$\frac{d}{dz}X(z)=\sum_{n=-\infty}^\infty x[n] \frac{d}{dz} (z... ...n] z^{-n-1} =\frac{-1}{z}\sum_{n=-\infty}^\infty n x[n] z^{-n}$$
  
  
  i.e.,
  
  $${\cal Z}[n x[n]]=-z\frac{d}{dz}X(z)$$
  
  

  Example: Taking derivative with respect to $z$ of the right side of
  

  $${\cal Z}[a^nu[n]]=\frac{1}{1-az^{-1}}\;\;\;\;\vert z\vert>\vert a\vert$$
  
  
  we get
  
  $$\frac{d}{dz}\left[ \frac{1}{1-az^{-1}}\right]=\frac{-az^{-2}}{(1-az^{-1})^2}$$
  
  
  Due to the property of differentiation in z-domain, we have
  
  $${\cal Z}[n a^n u[n]]=\frac{az^{-1}}{(1-az^{-1})^2}\;\;\;\;\vert z\vert>\vert a\vert$$
  
  
  Note that for a different ROC $\vert z\vert<\vert a\vert$, we have
  
  $${\cal Z}[-n a^n u[-n-1]]=\frac{az^{-1}}{(1-az^{-1})^2}\;\;\;\;\vert z\vert<\vert a\vert$$