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Let's solve the problem based on the provided first-order difference equation:

### Given:
\[
y[n] - a y[n-1] = x[n]
\]
where \( 0 < a < 1 \).

#### Part (a): Verify that the impulse response \( h[n] \) for this system is \( h[n] = a^n u[n] \).

1. The difference equation given is for a linear time-invariant (LTI) system. To find the impulse response, we solve the system's response to the input \( x[n] = \delta[n] \), where \( \delta[n] \) is the unit impulse.

2. Substituting \( x[n] = \delta[n] \) into the equation:
   \[
   y[n] - a y[n-1] = \delta[n]
   \]
   
3. Assume \( h[n] \) is the impulse response, so we are solving:
   \[
   h[n] - a h[n-1] = \delta[n]
   \]
   
4. Solving this recursively:
   - For \( n = 0 \): \( h[0] - a h[-1] = 1 \). Since \( h[-1] = 0 \) for causality, we get \( h[0] = 1 \).
   - For \( n = 1 \): \( h[1] - a h[0] = 0 \Rightarrow h[1] = a \).
   - For \( n = 2 \): \( h[2] - a h[1] = 0 \Rightarrow h[2] = a^2 \).
   
   This pattern continues for all \( n \geq 0 \), giving the solution:
   \[
   h[n] = a^n u[n]
   \]
   where \( u[n] \) is the unit step function.

Thus, the impulse response is indeed \( h[n] = a^n u[n] \).

---

#### Part (b): Is the system memoryless, stable, and causal?

- **Memoryless:** A system is memoryless if the output at time \( n \) depends only on the input at time \( n \). Since \( y[n] \) depends on \( y[n-1] \), the system is not memoryless.

- **Stable:** For BIBO (bounded-input bounded-output) stability, the impulse response \( h[n] \) must be absolutely summable, i.e., \( \sum_{n=-\infty}^{\infty} |h[n]| < \infty \).
  
  The impulse response is \( h[n] = a^n u[n] \), which sums as:
  \[
  \sum_{n=0}^{\infty} |a^n| = \frac{1}{1 - a} \quad \text{(since \( 0 < a < 1 \))}.
  \]
  Since this sum is finite, the system is stable.

- **Causal:** A system is causal if the output at any time \( n \) depends only on the current and past inputs, not future inputs. From the given equation \( y[n] - a y[n-1] = x[n] \), we see that \( y[n] \) depends on \( y[n-1] \), meaning the system is causal.

---

#### Part (c): Is this stable for \( |a| > 1 \)?

- For \( |a| > 1 \), the impulse response would still be \( h[n] = a^n u[n] \), but now the sum:
  \[
  \sum_{n=0}^{\infty} |a^n|
  \]
  will diverge, since \( a^n \) grows exponentially for \( |a| > 1 \).

Thus, the system would **not** be stable for \( |a| > 1 \), because the impulse response is not absolutely summable.

---

Do you have any questions or need further clarifications? 

Here are 5 related questions you might find useful:
1. What is the general solution of a first-order difference equation with arbitrary coefficients?
2. How does causality relate to the structure of difference equations?
3. Can you describe the conditions under which an LTI system is time-invariant?
4. What happens to the system's stability if the input is periodic instead of an impulse?
5. How does the value of \( a \) affect the system's response for real-world signals?

**Tip:** When working with difference equations, always check for key properties like memorylessness, stability, and causality by analyzing the system's impulse response and behavior over time.

The first order difference equation y[n] - ay[n - 1] = x[n], 0 < a < 1, describes a particular discrete time system.
(a) Verify that impulse response h[n] for this system is h[n] = a^n u[n].
(b) Is this system memoryless, stable and causal?
(c) Is this stable for |a| > 1?

Learn how to solve a first-order difference equation y[n] - ay[n - 1] = x[n] and verify the impulse response h[n] = a^n u[n]. Explore the system's properties, including memorylessness, stability, and causality, with step-by-step solutions.

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