Question 4.12: Modeling a more complicated feedback system with excitation ......
Modeling a more complicated feedback system with excitation
Find the zero-state response of the system in Figure 4.43, for times n ≥ 0 to x[n] = 1 applied at time n = 0, by assuming all the signals in the system are zero before time n = 0 for a = 1, b = −1.5 and three different values of c, 0.8, 0.6 and 0.5.
The difference equation for this system is
y[n] = a(x[n]− b y[n −1]− c y[n − 2]) = x[n] + 1.5 y[n −1] − c y[n − 2] (4.18)
The response is the total solution of the difference equation with initial conditions. We can find a closed-form solution by finding the total solution of the difference equation. The homogeneous solution is y_h[n]= K_{h1}z_1^n+ K_{h2}z_2^n where z_{1,2}= 0.75 ±\sqrt{0.5625 − c}. The particular solution is in the form of a linear combination of the input signal and all its unique differences. The input signal is a constant. So all its differences are zero. Therefore the particular solution is simply a constant K_p . Substituting into the difference equation,
K_p – 1.5 K_p+ cK_p= 1⇒ K_p= \frac{1}{c-0.5}.
Using (4.18) we can find the initial two values of y[n] needed to solve for the remaining two unknown constants K_{h1} and K_{h2}. They are y[0] = 1 and y[1] = 2.5.
In Chapter 1 three responses were illustrated for a = 1, b = −1.5 and c = 0.8, 0.6 and 0.5.
Those responses are replicated in Figure 4.44.
