- 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 - System Properties Solved Examples
Example 1 − Check whether $y(t) = x*(t)$ is pnear or non-pnear.
Solution − The function represents the conjugate of input. It can be verified by either first law of homogeneity and law of additivity or by the two rules. However, verifying through rules is lot easier, so we will go by that.
If the input to the system is zero, the output also tends to zero. Therefore, our first condition is satisfied. There is no non-pnear operator used either at the input nor the output. Therefore, the system is Linear.
Example 2 − Check whether $y(t)=egin{cases}x(t+1), & t > 0\x(t-1), & tleq 0end{cases}$ is pnear or non pnear
Solution − Clearly, we can see that when time becomes less than or equal to zero the input becomes zero. So, we can say that at zero input the output is also zero and our first condition is satisfied.
Again, there is no non-pnear operator used at the input nor at the output. Therefore, the system is Linear.
Example 3 − Check whether $y(t) = sin t.x(t)$ is stable or not.
Solution − Suppose, we have taken the value of x(t) as 3. Here, sine function has been multipped with it and maximum and minimum value of sine function varies between -1 to +1.
Therefore, the maximum and minimum value of the whole function will also vary between -3 and +3. Thus, the system is stable because here we are getting a bounded input for a bounded output.
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