Digital Signal Processing Tutorial
Operations on Signals
Basic System Properties
Z-Transform
Discrete Fourier Transform
Fast Fourier Transform
Digital Signal Processing Resources
Selected Reading
- DSP - Miscellaneous Signals
- DSP - Classification of DT Signals
- DSP - Classification of CT Signals
- DSP - Basic DT Signals
- DSP - Basic CT Signals
- DSP - Signals-Definition
- DSP - Home
Operations on Signals
- Operations Signals - Convolution
- Operations Signals - Integration
- Operations Signals - Differentiation
- Operations Signals - Reversal
- Operations Signals - Scaling
- Operations Signals - Shifting
Basic System Properties
- DSP - Solved Examples
- DSP - Unstable Systems
- DSP - Stable Systems
- DSP - Time-Variant Systems
- DSP - Time-Invariant Systems
- DSP - Non-Linear Systems
- DSP - Linear Systems
- DSP - Anti-Causal Systems
- DSP - Non-Causal Systems
- DSP - Causal Systems
- DSP - Dynamic Systems
- DSP - Static Systems
Z-Transform
- Z-Transform - Solved Examples
- Z-Transform - Inverse
- Z-Transform - Existence
- Z-Transform - Properties
- Z-Transform - Introduction
Discrete Fourier Transform
- DFT - Solved Examples
- DFT - Discrete Cosine Transform
- DFT - Sectional Convolution
- DFT - Linear Filtering
- DTF - Circular Convolution
- DFT - Time Frequency Transform
- DFT - Introduction
Fast Fourier Transform
Digital Signal Processing Resources
Selected Reading
- Who is Who
- Computer Glossary
- HR Interview Questions
- Effective Resume Writing
- Questions and Answers
- UPSC IAS Exams Notes
Z-Transform - Introduction
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 stabipty, 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 $x(n) = lbrace 7,3,4,9,5 brace$, 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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