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
DSP - Stable Systems
Digital Signal Processing - Stable Systems
A stable system satisfies the BIBO (bounded input for bounded output) condition. Here, bounded means finite in ampptude. For a stable system, output should be bounded or finite, for finite or bounded input, at every instant of time.
Some examples of bounded inputs are functions of sine, cosine, DC, signum and unit step.
Examples
a) $y(t) = x(t)+10$
Here, for a definite bounded input, we can get definite bounded output i.e. if we put $x(t) = 2, y(t) = 12$ which is bounded in nature. Therefore, the system is stable.
b) $y(t) = sin [x(t)]$
In the given expression, we know that sine functions have a definite boundary of values, which pes between -1 to +1. So, whatever values we will substitute at x(t), we will get the values within our boundary. Therefore, the system is stable.
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