https://file.aiquickdraw.com/mathbot/file/1730441133653n0zjujqs.jpg

To solve this problem using the graphical method, we need to perform the convolution of the input sequence \( x(n) = \{0.5, 2\} \) with the impulse response \( h(n) = \{1, 1, 1\} \) of the LTI (Linear Time-Invariant) system. The output sequence \( y(n) \) is given by the convolution:

\[
y(n) = x(n) * h(n)
\]

### Steps for Graphical Convolution

1. **Identify Lengths of Sequences**:
   - The input sequence \( x(n) = \{0.5, 2\} \) has a length of 2.
   - The impulse response \( h(n) = \{1, 1, 1\} \) has a length of 3.
   - The length of the output sequence \( y(n) \) will be \( (2 + 3 - 1) = 4 \).

2. **Reverse the Sequence \( h(n) \)**:
   - Since we are using the graphical convolution method, we need to flip \( h(n) \) (even though for this finite sequence it doesn’t change much visually).
   - Flipped \( h(n) \) is still \( \{1, 1, 1\} \).

3. **Shift and Multiply**:
   - Shift \( h(n) \) for each time step \( n \), align it with \( x(n) \), multiply element-wise, and sum the products for each shift.

### Calculating \( y(n) \) for Each Shift

1. **\( y(0) \)**:
   - Align the first element of \( x(n) \) with the first element of \( h(n) \).
   - Only the first element of \( x(n) \) and the first element of \( h(n) \) overlap.
   - \( y(0) = (0.5 \times 1) = 0.5 \).

2. **\( y(1) \)**:
   - Shift \( h(n) \) one position to the right.
   - Now \( x(0) \) aligns with \( h(1) \) and \( x(1) \) aligns with \( h(0) \).
   - \( y(1) = (0.5 \times 1) + (2 \times 1) = 0.5 + 2 = 2.5 \).

3. **\( y(2) \)**:
   - Shift \( h(n) \) another position to the right.
   - Now \( x(1) \) aligns with \( h(1) \), and \( h(0) \) is out of range.
   - \( y(2) = (2 \times 1) + (0.5 \times 1) = 2 + 0.5 = 2.5 \).

4. **\( y(3) \)**:
   - Shift \( h(n) \) one more position to the right.
   - Now only \( x(1) \) overlaps with \( h(2) \).
   - \( y(3) = (2 \times 1) = 2 \).

### Final Output Sequence

The output sequence \( y(n) \) is:
\[
y(n) = \{0.5, 2.5, 2.5, 2\}
\]

Would you like a more detailed explanation of each step, or do you have any questions?

### Related Questions
1. How does the graphical method differ from the analytical method of convolution?
2. What is the significance of convolution in LTI systems?
3. How would the output sequence change if the impulse response \( h(n) \) had a different length?
4. Can we use the convolution property in the frequency domain instead of the time domain?
5. How do properties like causality and stability affect the convolution process?

### Tip
When using the graphical method, visualize each shift and overlap to simplify the calculations, especially for longer sequences.

Using graphical method, find the output sequence y(n) of the LTI system whose response h(n) is given and input x(n) is given as follows. X(n)={0.5,2}; h(n)={1,1,1}.

Learn how to find the output sequence y(n) for a Linear Time-Invariant (LTI) system by performing graphical convolution of the input sequence x(n)={0.5,2} and the impulse response h(n)={1,1,1}. This problem covers key concepts in signal processing, including convolution in discrete-time systems.

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