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 - Unstable Systems
Digital Signal Processing - Unstable Systems
Unstable systems do not satisfy the BIBO conditions. Therefore, for a bounded input, we cannot expect a bounded output in case of unstable systems.
Examples
a) $y(t) = tx(t)$
Here, for a finite input, we cannot expect a finite output. For example, if we will put $x(t) = 2 Rightarrow y(t) = 2t$. This is not a finite value because we do not know the value of t. So, it can be ranged from anywhere. Therefore, this system is not stable. It is an unstable system.
b) $y(t) = frac{x(t)}{sin t}$
We have discussed earper, that the sine function has a definite range from -1 to +1; but here, it is present in the denominator. So, in worst case scenario, if we put t = 0 and sine function becomes zero, then the whole system will tend to infinity. Therefore, this type of system is not at all stable. Obviously, this is an unstable system.
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