Question

Analyze Unilateral Z transform to generate the output of a system which is characterized by the difference equation as y[n] + (1/2)y[n-1] + (1/4)y[n-2] = 0 with initial condition of y[-1]=1 and y[-2]=1.

a) Zy[n] = Y(z) is Y(z) + (1/2)z^(-1)Y(z) + (1/4)z^(-2)Y(z) = 0

b) Zy[n] = Y(z) is Y(z) + (1/2)z^(-1)Y(z) + (1/4)z^(-2)Y(z) = 1 + z^(-1) + z^(-2)

c) Zy[n] = Y(z) is Y(z) + (1/2)z^(-1)Y(z) + (1/4)z^(-2)Y(z) = z^(-1) + z^(-2)

d) Zy[n] = Y(z) is Y(z) + (1/2)z^(-1)Y(z) + (1/4)z^(-2)Y(z) = 1

Answer 1

The correct representation of the difference equation using Unilateral Z-Transform, with initial conditions factored in, is Y(z) + (1/2)z-1Y(z) + (1/4)z-2Y(z) = z-1 + z-2 which is option (b) in the Z-domain.

Step-by-step explanation:

The student is asking about solving a difference equation using the Unilateral Z-transform. This transformation is a mathematical tool used in signal processing and control theory to analyze and solve linear difference equations, often representing discrete-time systems.

The difference equation given is y[n] + (1/2)y[n-1] + (1/4)y[n-2] = 0. With initial conditions y[-1] = 1 and y[-2] = 1, we can use the Z-Transform properties to convert the time-domain difference equation into the Z-domain. The Z-Transform of y[n] is represented by Y(z).

Considering the non-zero initial conditions, option (b) is the correct representation in the Z-domain, which includes these conditions:

• Y(z) + (1/2)z-1Y(z) + (1/4)z-2Y(z) = z-1 + z-2

This equation now includes the initial conditions that were missing in the other choices (a, c, and d), which did not account for the non-zero starting values of y[n].