- 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
DSP - Z-Transform Inverse
If we want to analyze a system, which is already represented in frequency domain, as discrete time signal then we go for Inverse Z-transformation.
Mathematically, it can be represented as;
$$x(n) = Z^{-1}X(Z)$$where x(n) is the signal in time domain and X(Z) is the signal in frequency domain.
If we want to represent the above equation in integral format then we can write it as
$$x(n) = (frac{1}{2Pi j})oint X(Z)Z^{-1}dz$$Here, the integral is over a closed path C. This path is within the ROC of the x(z) and it does contain the origin.
Methods to Find Inverse Z-Transform
When the analysis is needed in discrete format, we convert the frequency domain signal back into discrete format through inverse Z-transformation. We follow the following four ways to determine the inverse Z-transformation.
Long Division Method
Partial Fraction expansion method
Residue or Contour integral method
Long Division Method
In this method, the Z-transform of the signal x (z) can be represented as the ratio of polynomial as shown below;
$$x(z)=N(Z)/D(Z)$$Now, if we go on spaniding the numerator by denominator, then we will get a series as shown below
$$X(z) = x(0)+x(1)Z^{-1}+x(2)Z^{-2}+...quad...quad...$$The above sequence represents the series of inverse Z-transform of the given signal (for n≥0) and the above system is causal.
However for n<0 the series can be written as;
$$x(z) = x(-1)Z^1+x(-2)Z^2+x(-3)Z^3+...quad...quad...$$Partial Fraction Expansion Method
Here also the signal is expressed first in N (z)/D (z) form.
If it is a rational fraction it will be represented as follows;
$x(z) = b_0+b_1Z^{-1}+b_2Z^{-2}+...quad...quad...+b_mZ^{-m})/(a_0+a_1Z^{-1}+a_2Z^{-2}+...quad...quad...+a_nZ^{-N})$
The above one is improper when m<n and an≠0
If the ratio is not proper (i.e. Improper), then we have to convert it to the proper form to solve it.
Residue or Contour Integral Method
In this method, we obtain inverse Z-transform x(n) by summing residues of $[x(z)Z^{n-1}]$ at all poles. Mathematically, this may be expressed as
$$x(n) = displaystylesumpmits_{allquad polesquad X(z)}residuesquad of[x(z)Z^{n-1}]$$Here, the residue for any pole of order m at $z = eta$ is
$$Residues = frac{1}{(m-1)!}pm_{Z ightarrow eta}lbrace frac{d^{m-1}}{dZ^{m-1}}lbrace (z-eta)^mX(z)Z^{n-1} brace$$ Advertisements