JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech III DIGITAL SIGNAL PROCESSING Previous Question Paper (August/September - 2021 - R18)

The given difference equation is y(n) - 4y(n-1) - 12y(n-2) = x(n). For the total response, we consider the zero-input response (ZIR) as x(n) = 0. The characteristic equation is r² - 4r - 12 = 0. Factoring this gives (r - 6)(r + 2) = 0, yielding roots r₁ = 6 and r₂ = -2. The homogeneous solution is y_h(n) = C₁(6)ⁿ + C₂(-2)ⁿ.

Using initial conditions y(-1) = 1 and y(-2) = 2:

For n = 0: y(0) = 4y(-1) + 12y(-2) = 4(1) + 12(2) = 28.

For n = 1: y(1) = 4y(0) + 12y(-1) = 4(28) + 12(1) = 112 + 12 = 124.

Applying these to the homogeneous solution:

Solving the system: 2C₁ + 2C₂ = 56 and 6C₁ - 2C₂ = 124. Adding them gives 8C₁ = 180, so C₁ = 22.5. Then C₂ = 28 - 22.5 = 5.5. The total response is y(n) = 22.5(6)ⁿ + 5.5(-2)ⁿ for n ≥ 0.

Given the difference equation y(n) - 2y(n-1) = x(n), apply the Z-transform to both sides assuming zero initial conditions:

Factor out Y(z):

The transfer function H(z) is defined as the ratio of the output Z-transform to the input Z-transform:

This can also be expressed as H(z) = z / (z - 2), with a region of convergence |z| > 2.