Question

6. Consider the second-order difference equation, 114 pt 16 a. Find the characteristie polynomial and characteristic roots of this system. b. Is this a stable system? Justify your answer mathematically. c. Draw a system diagram for implementing this difference equation using adders, multipliers and time-delays, as needed. d. Find the zero-input solution for initial conditions: l-11-0 and yI-2]--1 in closed- form. e. Find y[l iteratively using the initial conditions given in part f. Find yIl] using the closed form solution found in part-d.Compare the value with the one found in part-e. Find the impulse response values, HO) and MlJ iteratively. g.

Answer 1

Given data y (n) + 16/15 y (n - 1) Y (n - 2) = -x (n) rightarrow equation (a) finding characteristic polynomial and characteristic roots: - Taking 'z' transform on both sides z = y (z)/x (z) y (z) = -1 x (z) = 1 + 16/5 z^- 1 - z^- 2 Given values are apply in above equation = -1/1 + 16/5 z^- 1 - z^- 2 = -z^2/z^2 + 16/5 z - 1 The characteristic polynomial is z^2 + 16/5 z - 1 = 0 The characteristic roots are 0.2867, - 3.4867 \r\n (b) It is not a stable system because modules are one of the roots is greater than 1 have Vertical-bar z Vertical-bar < 1 (c) y (n) = -16/5 y (n - 1) + y (n - 2) - x (n) It has 3 multiplication 2 adders and 2 delays (d) in zero input solution closed firm is y (- 1) = 0 y (- 2) = -1