- 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 - Time-Invariant Systems
For a time-invariant system, the output and input should be delayed by some time unit. Any delay provided in the input must be reflected in the output for a time invariant system.
![Time Invariant](/digital_signal_processing/images/time_invariant.jpg)
Examples
a) $y(T) = x(2T)$
If the above expression, it is first passed through the system and then through the time delay (as shown in the upper part of the figure); then the output will become $x(2T-2t)$. Now, the same expression is passed through a time delay first and then through the system (as shown in the lower part of the figure). The output will become $x(2T-t)$.
Hence, the system is not a time-invariant system.
b) $y(T) = sin [x(T)]$
If the signal is first passed through the system and then through the time delay process, the output be $sin x(T-t)$. Similarly, if the system is passed through the time delay first then through the system then output will be $sin x(T-t)$. We can see clearly that both the outputs are same. Hence, the system is time invariant.
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