Question

Consider the system with difference equation y(n) – 0.64y(n − 2) = x(n).

When the input x(n) is the impulse signal ∂(n), what is the particular solution for this input?

Answer 1

The particular solution for the difference equation with the impulse signal input is y(n) = (1/1.28)N.

Step-by-step explanation:

The particular solution for the input ∂(n) can be found by considering the difference equation: y(n) - 0.64y(n-2) = x(n), where x(n) is the impulse signal. Since the impulse signal is defined as 1 at n=0 and 0 elsewhere, we can substitute x(n) = 1 in the difference equation. This gives us y(n) - 0.64y(n-2) = 1.

Let's solve this equation using a particular solution approach. Since the difference equation is linear, the particular solution can be assumed in the form of y(n) = AN, where A is a constant. Substituting this assumed form of the particular solution into the difference equation gives AN - 0.64A(N-2) = 1.

Simplifying the equation, we get 1.28A = 1, which implies A = 1/1.28. Therefore, the particular solution for the given difference equation with the impulse signal input is y(n) = (1/1.28)N.