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Re: Z-Transform

  • To: mathgroup at
  • Subject: [mg21212] Re: Z-Transform
  • From: cheng lee lung <eeacheng at>
  • Date: Fri, 17 Dec 1999 01:29:11 -0500 (EST)
  • Organization: City University of Hong Kong
  • References: <831un6$>
  • Sender: owner-wri-mathgroup at

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.)

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