Question 4.12: Modeling a more complicated feedback system with excitation ......

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.