An analog LTI system with impulse response function h(t) = u(2-t) + u(t-2) is presented with an input x(t) = t[u(t-2) -u(t-3)].a) Determine and plot the system output y(t) = x(t) * h(t).

The plot shows the output \( y(t) \) of the LTI system, where:

- \( y(t) = \frac{5}{2} \) for the interval \( 2 \leq t < 5 \).
- \( y(t) = 0 \) for \( t < 2 \) and \( t \geq 5 \).

This output reflects the result of convolving the input \( x(t) \) with the impulse response \( h(t) \).

Would you like more details on any specific part, or do you have any questions?

### Related Questions:

1. How would the output change if the impulse response function \( h(t) \) were different?
2. Can you explain the significance of convolution in LTI systems?
3. What is the effect of the unit step functions in defining the input and impulse response?
4. How would you compute the convolution if the input were a more complex function?
5. How does the convolution property relate to the frequency domain?

#### Tip:
Convolution in time domain corresponds to multiplication in the frequency domain, which can simplify the analysis of LTI systems.

Explore the analysis and plotting of the output \( y(t) \) of a Linear Time-Invariant (LTI) system with an impulse response function \( h(t) = u(2-t) + u(t-2) \) and input \( x(t) = t[u(t-2) - u(t-3)] \). Understand how convolution is applied to determine the system's response. Detailed step-by-step solution provided.

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