Learning Z-Transform – Introduction work project make money

DSP – Z-Transform Introduction



Discrete Time Fourier Transform(DTFT) exists for energy and power signals. Z-transform also exists for neither energy nor Power (NENP) type signal, up to a certain extent only. The replacement $z=e^{jw}$ is used for Z-transform to DTFT conversion only for absolutely summable signal.

So, the Z-transform of the discrete time signal x(n) in a power series can be written as −

$$X(z) = sum_{n-infty}^infty x(n)Z^{-n}$$

The above equation represents a two-sided Z-transform equation.

Generally, when a signal is Z-transformed, it can be represented as −

$$X(Z) = Z[x(n)]$$

Or $x(n) longleftrightarrow X(Z)$

If it is a continuous time signal, then Z-transforms are not needed because Laplace transformations are used. However, Discrete time signals can be analyzed through Z-transforms only.

Region of Convergence

Region of Convergence is the range of complex variable Z in the Z-plane. The Z- transformation of the signal is finite or convergent. So, ROC represents those set of values of Z, for which X(Z) has a finite value.

Properties of ROC

  • ROC does not include any pole.
  • For right-sided signal, ROC will be outside the circle in Z-plane.
  • For left sided signal, ROC will be inside the circle in Z-plane.
  • For stability, ROC includes unit circle in Z-plane.
  • For Both sided signal, ROC is a ring in Z-plane.
  • For finite-duration signal, ROC is entire Z-plane.

The Z-transform is uniquely characterized by −

  • Expression of X(Z)
  • ROC of X(Z)

Signals and their ROC

x(n) X(Z) ROC
$delta(n)$ $1$ Entire Z plane
$U(n)$ $1/(1-Z^{-1})$ Mod(Z)>1
$a^nu(n)$ $1/(1-aZ^{-1})$ Mod(Z)>Mod(a)
$-a^nu(-n-1)$ $1/(1-aZ^{-1})$ Mod(Z)<Mod(a)
$na^nu(n)$ $aZ^{-1}/(1-aZ^{-1})^2$ Mod(Z)>Mod(a)
$-a^nu(-n-1)$ $aZ^{-1}/(1-aZ^{-1})^2$ Mod(Z)<Mod(a)
$U(n)cos omega n$ $(Z^2-Zcos omega)/(Z^2-2Z cos omega +1)$ Mod(Z)>1
$U(n)sin omega n$ $(Zsin omega)/(Z^2-2Z cos omega +1)$ Mod(Z)>1

Example

Let us find the Z-transform and the ROC of a signal given as $x(n) = lbrace 7,3,4,9,5rbrace$, where origin of the series is at 3.

Solution − Applying the formula we have −

$X(z) = sum_{n=-infty}^infty x(n)Z^{-n}$

$= sum_{n=-1}^3 x(n)Z^{-n}$

$= x(-1)Z+x(0)+x(1)Z^{-1}+x(2)Z^{-2}+x(3)Z^{-3}$

$= 7Z+3+4Z^{-1}+9Z^{-2}+5Z^{-3}$

ROC is the entire Z-plane excluding Z = 0, ∞, -∞

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