this
round(H.substitute(f=4).abs(),4)
give same error again...
then this:
round(RDF(H.substitute(f=4).abs()),4)
takes so much time, can't wait...

On Dec 9, 10:42 pm, Sand Wraith <[EMAIL PROTECTED]> wrote:
> you mean for f? this:
>
> round(H.substitute(f=1),4)
>
> gives type error again :-(
>
> On Dec 9, 9:51 pm, Tim Lahey <[EMAIL PROTECTED]> wrote:
>
> > On Dec 9, 2008, at 1:46 PM, Sand Wraith wrote:
>
> > > thank you! ))
>
> > > but I need to use my values, now just variables (Ts,Pole1 etc)
>
> > > I was trying this:
>
> > > r=2*2*pi
> > > Pole1=r*exp(I*2*pi/3)
> > > Pole2=-r
> > > Pole3=r*exp(-I*2*pi/3)
> > > Ts=0.1
>
> > > z = exp(I*2*pi*Ts*f)
> > > p = (2/Ts)*(z-1)/(z+1)
> > > H = (p^3)/((p-Pole1)*(p-Pole2)*(p-Pole3))
> > > show(H)
>
> > > show() works fine, but if i want to use H function now:
>
> > > round(H(4).abs(),5)
>
> > > a have got "Type error" message :-\
>
> > H as you've defined it isn't a function, it's an expression. Check
> > the type so H(4) doesn't mean anything. You could try subs and
> > substitute for p (I assume that's what you want evaluated at 4.
>
> > Cheers,
>
> > Tim.
>
> > ---
> > Tim Lahey
> > PhD Candidate, Systems Design Engineering
> > University of Waterloohttp://www.linkedin.com/in/timlahey
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