- Digital Signal Processing Tutorial
- DSP - Home
- DSP - Signals-Definition
- DSP - Basic CT Signals
- DSP - Basic DT Signals
- DSP - Classification of CT Signals
- DSP - Classification of DT Signals
- DSP - Miscellaneous Signals
- Operations on Signals
- Operations Signals - Shifting
- Operations Signals - Scaling
- Operations Signals - Reversal
- Operations Signals - Differentiation
- Operations Signals - Integration
- Operations Signals - Convolution
- Basic System Properties
- DSP - Static Systems
- DSP - Dynamic Systems
- DSP - Causal Systems
- DSP - Non-Causal Systems
- DSP - Anti-Causal Systems
- DSP - Linear Systems
- DSP - Non-Linear Systems
- DSP - Time-Invariant Systems
- DSP - Time-Variant Systems
- DSP - Stable Systems
- DSP - Unstable Systems
- DSP - Solved Examples
- Z-Transform
- Z-Transform - Introduction
- Z-Transform - Properties
- Z-Transform - Existence
- Z-Transform - Inverse
- Z-Transform - Solved Examples
- Discrete Fourier Transform
- DFT - Introduction
- DFT - Time Frequency Transform
- DTF - Circular Convolution
- DFT - Linear Filtering
- DFT - Sectional Convolution
- DFT - Discrete Cosine Transform
- DFT - Solved Examples
- Fast Fourier Transform
- DSP - Fast Fourier Transform
- DSP - In-Place Computation
- DSP - Computer Aided Design
- Digital Signal Processing Resources
- DSP - Quick Guide
- DSP - Useful Resources
- DSP - Discussion
DSP - Z-Transform Solved Examples
Example 1
Find the response of the system $s(n+2)-3s(n+1)+2s(n) = \delta (n)$, when all the initial conditions are zero.
Solution − Taking Z-transform on both the sides of the above equation, we get
$$S(z)Z^2-3S(z)Z^1+2S(z) = 1$$$\Rightarrow S(z)\lbrace Z^2-3Z+2\rbrace = 1$
$\Rightarrow S(z) = \frac{1}{\lbrace z^2-3z+2\rbrace}=\frac{1}{(z-2)(z-1)} = \frac{\alpha _1}{z-2}+\frac{\alpha _2}{z-1}$
$\Rightarrow S(z) = \frac{1}{z-2}-\frac{1}{z-1}$
Taking the inverse Z-transform of the above equation, we get
$S(n) = Z^{-1}[\frac{1}{Z-2}]-Z^{-1}[\frac{1}{Z-1}]$
$= 2^{n-1}-1^{n-1} = -1+2^{n-1}$
Example 2
Find the system function H(z) and unit sample response h(n) of the system whose difference equation is described as under
$y(n) = \frac{1}{2}y(n-1)+2x(n)$
where, y(n) and x(n) are the output and input of the system, respectively.
Solution − Taking the Z-transform of the above difference equation, we get
$y(z) = \frac{1}{2}Z^{-1}Y(Z)+2X(z)$
$= Y(Z)[1-\frac{1}{2}Z^{-1}] = 2X(Z)$
$= H(Z) = \frac{Y(Z)}{X(Z)} = \frac{2}{[1-\frac{1}{2}Z^{-1}]}$
This system has a pole at $Z = \frac{1}{2}$ and $Z = 0$ and $H(Z) = \frac{2}{[1-\frac{1}{2}Z^{-1}]}$
Hence, taking the inverse Z-transform of the above, we get
$h(n) = 2(\frac{1}{2})^nU(n)$
Example 3
Determine Y(z),n≥0 in the following case −
$y(n)+\frac{1}{2}y(n-1)-\frac{1}{4}y(n-2) = 0\quad given\quad y(-1) = y(-2) = 1$
Solution − Applying the Z-transform to the above equation, we get
$Y(Z)+\frac{1}{2}[Z^{-1}Y(Z)+Y(-1)]-\frac{1}{4}[Z^{-2}Y(Z)+Z^{-1}Y(-1)+4(-2)] = 0$
$\Rightarrow Y(Z)+\frac{1}{2Z}Y(Z)+\frac{1}{2}-\frac{1}{4Z^2}Y(Z)-\frac{1}{4Z}-\frac{1}{4} = 0$
$\Rightarrow Y(Z)[1+\frac{1}{2Z}-\frac{1}{4Z^2}] =\frac{1}{4Z}-\frac{1}{2}$
$\Rightarrow Y(Z)[\frac{4Z^2+2Z-1}{4Z^2}] = \frac{1-2Z}{4Z}$
$\Rightarrow Y(Z) = \frac{Z(1-2Z)}{4Z^2+2Z-1}$