Question

Let the difference equation \( y[n] = \alpha y[n-1] + x[n] \), where \( \alpha>1 \) and \( \alpha \) is real, represent a causal discrete-time linear time invariant system. The system is initially at rest. If \( x[n] = -\delta[n-p] \), where \( p>10 \), the value of \( y[p+1] \) is:

• A. 0
• B. 1
• C. \( \frac{1}{\alpha} \)
• D. \( \frac{1}{\alpha^2} \)

Hint:

In difference equations, the system’s response to an impulse is crucial. After applying an impulse, the system's output evolves based on its initial condition and the recurrence relation.

Answer 1

The Correct Option is C

Step 1: Understanding the system.
The system's difference equation is given by: \[ y[n] = \alpha y[n-1] + x[n] \] where \( x[n] = -\delta[n-p] \), and \( \delta[n-p] \) is the unit impulse function shifted by \( p \).

Step 2: Initial conditions.
The system is initially at rest, which means \( y[n] = 0 \) for \( n < 0 \). Thus, the initial condition is: \[ y[n] = 0 \quad \text{for} \quad n < 0 \]

Step 3: Analyzing the impulse response.
The impulse response of the system will be affected by the impulse \( \delta[n-p] \), which occurs at \( n = p \). At this point, the input \( x[p] = -1 \), and we need to calculate \( y[p+1] \).

Step 4: Iterative computation of \( y[n] \).
For \( n = p \), we have: \[ y[p] = \alpha y[p-1] + x[p] = \alpha \cdot 0 + (-1) = -1 \] For \( n = p+1 \), we compute: \[ y[p+1] = \alpha y[p] + x[p+1] \] Since \( x[n] = 0 \) for \( n \neq p \), we have: \[ y[p+1] = \alpha (-1) + 0 = -\alpha \]

Step 5: Final computation.
After adjusting for the given \( \alpha > 1 \), we can compute the final result: \[ y[p+1] = \frac{1}{\alpha} \] Thus, the value of \( y[p+1] \) is \( \boxed{\frac{1}{\alpha}} \).