Getting a pole-zero diagram from a difference equation

Hi, I'm trying to figure out how to get a pole-zero map from a difference equation.

I have some difference equations and sketches of solutions but I can't figure out the relationship between them:

$y(n) = 1.34x(n) + 0.6y(n-1) + 0.9y(n-2)$

Gives a zero at (0,0), and poles at approx (0.2, 0.9) and (0.2, -0.9)

$y(n) = 2x(n) -x(n-1) - 0.3y(n-1)$

Gives a zero at (0, 0.5) and a pole at (0, -0.3)

$y(n) = x(n) - 0.8y(n-1) - 0.8y(n-2)$

Gives a zero at (0, 0) and complex conjugate poles at approximately (-0.4, 0.8) and (-0.4, -0.8)

Any help appreciated, even hints. Thanks

flag	asked Apr 22 2010 at 21:12 waitinforatrain 1

Please have a look at the first two questions in the faq (link on top of the page). – Harald Hanche-Olsen Apr 22 2010 at 21:41

when you say you have "sketches of solutions" and "I can't figure out the relationship between them", what are you actually asking for? Moreover, although this might not be homework per se, without a bit more background I fear that some people will take it as such. – Yemon Choi Apr 22 2010 at 21:44

I'm not convinced your examples are correct. In the first example, the transfer function that gives y from x is 1.34z^2/(z^2-0.6z-0.9) and this has poles at real values of z, not at z = 0.2+/-0.9i. BTW This looks like an ordinary homework question to me and these things are frowned upon on MO. – Dan Piponi Apr 22 2010 at 21:50