Re: Z-Transform
- To: mathgroup at smc.vnet.net
- Subject: [mg21205] Re: Z-Transform
- From: Richard Mercer <richard.mercer at wright.edu>
- Date: Fri, 17 Dec 1999 01:25:00 -0500 (EST)
- Organization: Wright State University
- References: <831un6$g1v@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Angela, InverseZTransform[(z^2+z)/(z^2-6z+13),z,k] //ComplexExpand gives 13^(k/2)*Cos[k*ArcTan[2/3]] + 2*13^(k/2)*Sin[k*ArcTan[2/3]] Is that what you were hoping for? Richard Mercer 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