- Digital Signal Processing Tutorial
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- 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
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- 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

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# DSP - System Properties Solved Examples

**Example 1** − Check whether $y(t) = x*(t)$ is linear or non-linear.

**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-linear operator used either at the input nor the output. Therefore, the system is Linear.

**Example 2** − Check whether $y(t)=\begin{cases}x(t+1), & t > 0\\x(t-1), & t\leq 0\end{cases}$ is linear or non linear

**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-linear 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 multiplied 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.