Quote:
Originally Posted by TobbeL
Thanx a lot!
Note that my r1 and r2 are the reciprocal of your roots because they are solutions for z^-1 instead of z. Then since the roots are complex, you can reduce h[n] to sine functions multiplied by (sqrt(2)/2))^n using Euler's identities like
sin(x) = [exp(jx) - exp(-jx)] / 2j.
Depending on what is in your table of z-transform pairs, you could do this directly without a partial fraction expansion. For example, it may have the transform of a 2nd order transfer function like:
1 / [1 - 2r*cos(theta)*z^-1 + r^2*z^-2].
Then you can make the appropriate identifications for r and cos(theta), and recognize that the 1 - z^-1 in the numerator will transform to h[n] - h[n-1]. That comes from the partial fraction expansion though.
When you have 2 complex conjugate roots on a circle, the possible ROCs are actually |z| < |1/r1| for causal and |z| > |1/r1| for anti-causal using my notation, and the other root gives the same 2 possibilities, so they are either inside or outside the circle. In general, there will be a maximum of 3 ROCs for 2 roots, not 4, because it is either inside both, outside both, or the annulus in between. It can't be inside the inner one and outside the outer one (real roots) because the whole transfer function has to converge everywhere in the ROC.
Last edited by BruceZ; 05-15-2013 at 12:33 PM.