- 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 - Time-Variant Systems
For a time variant system, also, output and input should be delayed by some time constant but the delay at the input should not reflect at the output. All time scaling cases are examples of time variant system. Similarly, when coefficient in the system relationship is a function of time, then also, the system is time variant.
Examples
a) $y(t) = x[\cos T]$
If the above signal is first passed through the system and then through the time delay, the output will be $x\cos (T-t)$. If it is passed through the time delay first and then through the system, it will be $x(\cos T-t)$. As the outputs are not same, the system is time variant.
b) $y(T) = \cos T.x(T)$
If the above expression is first passed through the system and then through the time delay, then the output will be $\cos(T-t)x(T-t)$. However, if the expression is passed through the time delay first and then through the system, the output will be $\cos T.x(T-t)$. As the outputs are not same, clearly the system is time variant.