Z-Transform has following properties:
If $\,x (n) \stackrel{\mathrm{Z.T}}{\longleftrightarrow} X(Z)$
and $\,y(n) \stackrel{\mathrm{Z.T}}{\longleftrightarrow} Y(Z)$
Then linearity property states that
$a\, x (n) + b\, y (n) \stackrel{\mathrm{Z.T}}{\longleftrightarrow} a\, X(Z) + b\, Y(Z)$
If $\,x (n) \stackrel{\mathrm{Z.T}}{\longleftrightarrow} X(Z)$
Then Time shifting property states that
$x (n-m) \stackrel{\mathrm{Z.T}}{\longleftrightarrow} z^{-m} X(Z)$
If $\,x (n) \stackrel{\mathrm{Z.T}}{\longleftrightarrow} X(Z)$
Then multiplication by an exponential sequence property states that
$a^n\, . x(n) \stackrel{\mathrm{Z.T}}{\longleftrightarrow} X(Z/a)$
If $\, x (n) \stackrel{\mathrm{Z.T}}{\longleftrightarrow} X(Z)$
Then time reversal property states that
$x (-n) \stackrel{\mathrm{Z.T}}{\longleftrightarrow} X(1/Z)$
If $\, x (n) \stackrel{\mathrm{Z.T}}{\longleftrightarrow} X(Z)$
Then multiplication by n or differentiation in z-domain property states that
$ n^k x (n) \stackrel{\mathrm{Z.T}}{\longleftrightarrow} [-1]^k z^k{d^k X(Z) \over dZ^K} $
If $\,x (n) \stackrel{\mathrm{Z.T}}{\longleftrightarrow} X(Z)$
and $\,y(n) \stackrel{\mathrm{Z.T}}{\longleftrightarrow} Y(Z)$
Then convolution property states that
$x(n) * y(n) \stackrel{\mathrm{Z.T}}{\longleftrightarrow} X(Z).Y(Z)$
If $\,x (n) \stackrel{\mathrm{Z.T}}{\longleftrightarrow} X(Z)$
and $\,y(n) \stackrel{\mathrm{Z.T}}{\longleftrightarrow} Y(Z)$
Then correlation property states that
$x(n) \otimes y(n) \stackrel{\mathrm{Z.T}}{\longleftrightarrow} X(Z).Y(Z^{-1})$
Initial value and final value theorems of z-transform are defined for causal signal.
For a causal signal x(n), the initial value theorem states that
$ x (0) = \lim_{z \to \infty }X(z) $
This is used to find the initial value of the signal without taking inverse z-transform
For a causal signal x(n), the final value theorem states that
$ x ( \infty ) = \lim_{z \to 1} [z-1] X(z) $
This is used to find the final value of the signal without taking inverse z-transform.
The range of variation of z for which z-transform converges is called region of convergence of z-transform.
ROC of z-transform is indicated with circle in z-plane.
ROC does not contain any poles.
If x(n) is a finite duration causal sequence or right sided sequence, then the ROC is entire z-plane except at z = 0.
If x(n) is a finite duration anti-causal sequence or left sided sequence, then the ROC is entire z-plane except at z = ∞.
If x(n) is a infinite duration causal sequence, ROC is exterior of the circle with radius a. i.e. |z| > a.
If x(n) is a infinite duration anti-causal sequence, ROC is interior of the circle with radius a. i.e. |z| < a.
If x(n) is a finite duration two sided sequence, then the ROC is entire z-plane except at z = 0 & z = ∞.
The concept of ROC can be explained by the following example:
Example 1: Find z-transform and ROC of $a^n u[n] + a^{-}nu[-n-1]$
$Z.T[a^n u[n]] + Z.T[a^{-n}u[-n-1]] = {Z \over Z-a} + {Z \over Z {-1 \over a}}$
$$ ROC: |z| \gt a \quad\quad ROC: |z| \lt {1 \over a} $$
The plot of ROC has two conditions as a > 1 and a < 1, as you do not know a.
In this case, there is no combination ROC.
Here, the combination of ROC is from $a \lt |z| \lt {1 \over a}$
Hence for this problem, z-transform is possible when a < 1.
A discrete time LTI system is causal when
ROC is outside the outermost pole.
In The transfer function H[Z], the order of numerator cannot be grater than the order of denominator.
A discrete time LTI system is stable when
its system function H[Z] include unit circle |z|=1.
all poles of the transfer function lay inside the unit circle |z|=1.
x(t) | X[Z] |
---|---|
$\delta$ | 1 |
$u(n)$ | ${Z\over Z-1}$ |
$u(-n-1)$ | $ -{Z\over Z-1}$ |
$\delta(n-m)$ | $z^{-m}$ |
$a^n u[n]$ | ${Z \over Z-a}$ |
$a^n u[-n-1]$ | $- {Z \over Z-a}$ |
$n\,a^n u[n]$ | ${aZ \over |Z-a|^2}$ |
$n\,a^n u[-n-1] $ | $- {aZ \over |Z-a|^2}$ |
$a^n \cos \omega n u[n] $ | ${Z^2-aZ \cos \omega \over Z^2-2aZ \cos \omega +a^2}$ |
$a^n \sin \omega n u[n] $ | $ {aZ \sin \omega \over Z^2 -2aZ \cos \omega +a^2 } $ |