Find the LTI system response if its impulse response h[n] = ( 1
2
)
2u[n] and input x[n] is,

(a) 2nu[n] (b) 2nu[n − 2] (c) 2n−2u[n]

To find the response of a Linear Time-Invariant (LTI) system, we use the **convolution** of the input signal \( x[n] \) and the impulse response \( h[n] \). The convolution of two discrete-time signals \( x[n] \) and \( h[n] \) is given by:

\[
y[n] = (x * h)[n] = \sum_{k=-\infty}^{\infty} x[k] h[n-k]
\]

### Given:
- Impulse response: 
  \[
  h[n] = \left(\frac{1}{2}\right)^2 u[n] = \frac{1}{4} u[n]
  \]
  where \( u[n] \) is the unit step function.

We'll now calculate the output for each input signal.

### (a) Input: \( x[n] = 2^n u[n] \)

1. Express \( x[n] \) and \( h[n] \):
   \[
   x[n] = 2^n u[n], \quad h[n] = \frac{1}{4} u[n]
   \]
   
2. Convolution:
   \[
   y[n] = \sum_{k=-\infty}^{\infty} x[k] h[n-k]
   \]
   For \( x[k] = 2^k u[k] \) and \( h[n-k] = \frac{1}{4} u[n-k] \), the convolution becomes:
   \[
   y[n] = \sum_{k=0}^{n} 2^k \cdot \frac{1}{4} = \frac{1}{4} \sum_{k=0}^{n} 2^k
   \]

3. Sum of a geometric series:
   \[
   \sum_{k=0}^{n} 2^k = \frac{2^{n+1} - 1}{2 - 1} = 2^{n+1} - 1
   \]

4. Multiply by \( \frac{1}{4} \):
   \[
   y[n] = \frac{1}{4} (2^{n+1} - 1) = \frac{1}{4} \cdot 2^{n+1} - \frac{1}{4}
   \]
   Thus:
   \[
   y[n] = \frac{2^{n+1}}{4} - \frac{1}{4}
   \]

### (b) Input: \( x[n] = 2^n u[n-2] \)

1. Shift \( x[n] \):
   \[
   x[n] = 2^n u[n-2]
   \]
   This is just a shifted version of the input in part (a).

2. Convolution for this shifted input:
   \[
   y[n] = \sum_{k=-\infty}^{\infty} x[k] h[n-k]
   \]
   Applying the same steps as in part (a), but now accounting for the shift by 2:
   \[
   y[n] = \frac{2^{n-1}}{4} - \frac{1}{4}
   \]
   Thus, the output will also be shifted by 2 compared to the output of part (a):
   \[
   y[n] = \frac{2^{n+1}}{4} u[n-2] - \frac{1}{4}
   \]

### (c) Input: \( x[n] = 2^{n-2} u[n] \)

1. Modify \( x[n] \):
   \[
   x[n] = 2^{n-2} u[n]
   \]
   
2. Convolution:
   \[
   y[n] = \sum_{k=-\infty}^{\infty} x[k] h[n-k]
   \]
   Using the same process:
   \[
   y[n] = \frac{1}{4} \sum_{k=0}^{n} 2^{k-2}
   \]
   Sum of a geometric series:
   \[
   \sum_{k=0}^{n} 2^{k-2} = \frac{1}{4} \left( 2^{n+1} - 1 \right)
   \]

3. Final output:
   \[
   y[n] = \frac{2^{n-1}}{4} - \frac{1}{4}
   \]

### Summary of Responses:

1. \( y_a[n] = \frac{2^{n+1}}{4} - \frac{1}{4} \)
2. \( y_b[n] = \frac{2^{n+1}}{4} u[n-2] - \frac{1}{4} \)
3. \( y_c[n] = \frac{2^{n-1}}{4} - \frac{1}{4} \)

Let me know if you'd like more detailed steps or explanations!

#### Relative Questions:
1. How do you compute the convolution for discrete signals?
2. What properties of an LTI system make convolution the appropriate method for determining system output?
3. How does a time shift in the input signal affect the output in LTI systems?
4. What is the significance of the unit step function \( u[n] \) in discrete-time systems?
5. Can you apply the convolution sum formula to non-causal systems?

**Tip**: When working with convolution in discrete-time signals, it helps to visualize the signals using simple plots or sketches for clarity.

Find the LTI system response if its impulse response h[n] = (1/2)^2u[n] and input x[n] is, (a) 2^n u[n] (b) 2^n u[n−2] (c) 2^(n−2) u[n]

Explore how to compute the response of a Linear Time-Invariant (LTI) system with impulse response h[n] = (1/2)^2 u[n] and inputs 2^n u[n], 2^n u[n-2], and 2^(n-2) u[n]. Learn key concepts in discrete-time signal processing, convolution, and the impact of time-shifting on system response.

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