Re: freqz problems

[Robert A. Macy]

octave uses indices starting at 1 going up 512

you forced 0 to 511
different

that's my guess.

               - Robert -

On Wed, 28 Sep 2005 12:08:09 -0500
 "Raman Venkataramani" <address@hidden> wrote:
> Hello,
> 
> I was wondering if this is a known bug. I am running
> octave version 2.1.50.
> In this example, I am computing the frequency response of
> an FIR filter "h".
> 
> octave:96>
> octave:96> h=[1, 0.5]
> h =
> 
>   1.00000  0.50000
> 
> octave:97> H=freqz(h, 1, 512, "whole");
> octave:98> H2=freqz(h, 1, 2*pi*[0:511]/512);
> octave:99> norm(H-H2)
> ans = 27.713
> octave:100> norm(abs(H)-abs(H2))
> ans =  5.3545e-15
> octave:101>
> 
> I would have expected that H = H2, but only their
> magnitudes are equal. It
> turns out that H is the correct answer and H2 is the
> reponse of a shifter
> version of h with h(2) being the sample at the origin and
> h(1) is the
> sampele at time -1.
> 
> Thanks,
> Raman
> 
> 



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