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.