In analogy to how the Laplace Transform can be used to solve differential equations, then the Z transform can be used to solve difference equations. A difference equation with initial condition is shown below.

eg parameters (0.7, 0.1, 1.0, 1.0) real poles.

eg parameters (-1.0, 0.0, 1.0, 1.0) real poles.

eg parameters (-1.0, 0.1, 1.0, 0.03) real poles, one outside the unit circle.

eg parameters (-1.8, 0.8, 2.0, 0.2) real poles, one on the unit circle.

eg parameters (-2.0, 1.0, 2.0, 0.02) 2 real poles on the unit circle.

eg parameters (-1.98, 1.0, 2.0, 0.2) CC poles on the unit circle and close to 1.

eg parameters (-1.98, 1.035, 2.0, 0.1) CC poles outside the unit circle and close to 1.

eg parameters (2.0, 1.005, 5.0, 0.06) CC poles outside the unit circle and close to -1.

eg parameters (0.0, 1.03, 5.0, 0.05) CC poles outside the unit circle and close to ±j.

Unlike a linear 2nd order differential equation, a linear 2nd order difference eqn can have unstable solutions.

Copyright © 2006 Cuthbert A. Nyack.