Question 4.10: Modeling a feedback system without excitation Find the outpu......

The system in Figure 4.40 is described by the difference equation

y[n] = 1.97 y[n −1]− y[n − 2]   (4.16)

This equation, along with initial conditions y[0] = 1 and y[−1] = 0, completely determines the response y[n] , which is the zero-input response. The zero-input response can be found by iterating on (4.16). This yields a correct solution, but it is in the form of an infinite sequence of values of the response. The zero-input response can be found in closed form by solving the difference equation (see Web Appendix D). Since there is no input signal exciting the system, the equation is homogeneous. The functional form of the homogeneous solution is the complex exponential Kz^n . Substituting that into the difference equation we get Kz^n = 1.97 Kz^{n−1}− Kz^{n−2}. Dividing through by Kz^{n−2} we get the characteristic equation and solving it for z, we get

The fact that there are two eigenvalues means that the homogeneous solution is in the form

\mathrm{y}[n]=K_{h 1} z_1^n+K_{h 2} z_2^n  (4.17)

We have initial conditions y[0] = 1 and y[−1] = 0 and we know from (4.17) that y[0]=K_{h1}+K_{h2} and \mathrm{y}[-1]=K_{h 1} z_1^{-1}+K_{h 2} z_2^{-1}. Therefore

Solving for the two constants, K_{h1} = 0.5 − j2.853  and  K_{h2} = 0.5 + j 2.853. So the complete solution is

This is a correct solution but it is not in a very convenient form. We can rewrite it in the form

or

or

y[n] = cos(0.1734n) + 5.706sin(0.1734n).

The first 50 values of the signal produced by this system are illustrated in Figure 4.41.