Re: Z-Transform
- To: mathgroup at smc.vnet.net
- Subject: [mg21212] Re: Z-Transform
- From: cheng lee lung <eeacheng at cityu.edu.hk>
- Date: Fri, 17 Dec 1999 01:29:11 -0500 (EST)
- Organization: City University of Hong Kong
- References: <831un6$g1v@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Angela Faessler wrote: > Hello > > How can I get the solution of > > InverseZTransform[(z^2+z)/(z^2-6z+13),z,k] > > into its form with trigonometric functions instead of a complex > expression? > > Thanks for an answer! > > Angela Faessler Try ComplexExpand[InverseZTransform[(z^2+z)/(z^2-6z+13),z,k]] you should get (13^(k/2) Cos[k ArcTan[2/3]] + 2*13^(k/2) Sin[k ArcTan[2/3]]). Hope this can help. Best regards, LL Cheng (City Univ. of Hong Kong EE Dept.)