Re: Z-Transform
- To: mathgroup at smc.vnet.net
- Subject: [mg21148] Re: [mg21069] Z-Transform
- From: "Tomas Garza" <tgarza at mail.internet.com.mx>
- Date: Fri, 17 Dec 1999 01:21:54 -0500 (EST)
- Sender: owner-wri-mathgroup at wolfram.com
Angela Faessler [angela.faessler at isg.ch] wrote:
> 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?
Try ComplexExpand:
In[1]:=
InverseZTransform[(z^2 + z)/(z^2 - 6z + 13), z, k]
Out[1]=
(1/2)*((1 +
2[ImaginaryI])(3 - 2[ImaginaryI])^k +(1 -
2[ImaginaryI])(3 + 2[ImaginaryI])^k)
In[2]:=
ComplexExpand[%]
Out[2]=
13^(k/2) Cos[k ArcTan[2/3]] + 2 13^(k/2) Sin[k ArcTan[2/3]]
Tomas Garza
Mexico City