[Scilab-users] Frequency response

[Claus Futtrup]

Hi Tim

 >So, this is complicated.

I admitted from the very beginning, it's probably just me that doesn't 
know how to read the Scilab manual. It cannot be complicated for someone 
who knows and/or understand the manual.

BTW, I also notice that nobody answered the question what ZPK means? ... 
ZPK = Zagrepa Plivaci Klub ... at least the Scilab could spell out what 
it stands for.

/Claus

On 15.09.2018 22:50, Tim Wescott wrote:
> So, this is complicated.
>
> First, if you have an impulse response, in the form of a vector of
> samples, then you can turn it into a FIR filter, and you can find the
> frequency response of that with any of the available Scilab tools.  I'm
> not sure if there's a more direct way, but if you had an impulse
> response h = [1, 1/2, 1/4, 1/8], then you can turn that into H = 1 +
> 0.5*%z^(-1) + 0.25*%z^(-2) + 0.125*%z^(-3) (note that %z is a built-in:
> %z = poly(0, 'z')).  Then you can use techniques surrounding repfreq to
> get the frequency response.
>
> Assign a sampling rate to it (i.e., by setting H.dt = 0.001, for a 1kHz
> sampling rate) and you can just use the Bode plot function: bode(H).
>
> Second, you can take your impulse response, pad it with zeros, and then
> take the FFT of it -- i.e. H = fft([h zeros(1, 252)]).  You need to pad
> it because the FFT is only exact for a periodic function in time --
> padding it gives you an approximation of an actual impulse response
> (and before someone jumps in and says you need to window it -- no, you
> don't, windowing is for sample vectors of continuous signals, not
> impulse responses).  But if you don't know WHY the method is
> approximate, you're going to have trouble correctly massaging the data
> before it's presented to the FFT, and with interpreting the results.
>
> On Sat, 2018-09-15 at 22:15 +0200, Claus Futtrup wrote:
>> Hi Samuel and Scilabers
>>
>> My problem with freq and repfreq is that they require a "sys" input,
>> which implies I need to describe a response function of some sort.
>> For example (from the Scilab web-help):
>>
>> https://help.scilab.org/docs/6.0.1/en_US/freq.html (example code)
>> s=poly(0,'s');
>> sys=(s+1)/(s^3-5*s+4)
>> rep=freq(sys("num"),sys("den"),[0,0.9,1.1,2,3,10,20])
>>
>> https://help.scilab.org/docs/6.0.1/en_US/repfreq.html (syntax)
>> [ [frq,] repf]=repfreq(sys,fmin,fmax [,step])
>> [ [frq,] repf]=repfreq(sys [,frq])
>> [ frq,repf,splitf]=repfreq(sys,fmin,fmax [,step])
>> [ frq,repf,splitf]=repfreq(sys [,frq])
>>
>> ... So, the thing with the matlab freqz (as the example repeated
>> below shows) is just a basic FFT with sampling, etc:
>>
>> h = rand(1,64); // impulse response (Matlab source code)
>> fs = 1000;
>> Nfft = 128;
>> [H,F] = freqz(h,1,Nfft,fs);
>> semilogx(F,mag2db(abs(H))); grid on;
>> xlabel('Frequency (Hz)'); ylabel('Magnitude (dB)');
>>
>> It may very well be that I just don't understand the help page of
>> repfreq. For example it says about sys: "A siso or simo linear
>> dynamical system, in state space, transfer function or zpk
>> representations, in continuous or discrete time." - al right then, it
>> seems pretty capable, but so, what's "zpk" for example? ... my
>> apologies for finding the Scilab help to be cryptic and the examples
>> insufficient for me to solve my problem, hence I ask if someone can
>> take above matlab code and make it work in Scilab.
>>
>> Best regards,
>> Claus
>>
>> On 15.09.2018 00:32, Samuel Gougeon wrote:
>>> Le 14/09/2018 à 20:57, Claus Futtrup a écrit :
>>>> Dear Scilabers
>>>> I have calculated an impulse response and wish to do an FFT to
>>>> achieve the frequency response. I know what to expect. In the
>>>> matlab forum someone asked the same question and was recommended
>>>> to use freqz ... I wonder what would be the equivalent function
>>>> in Scilab?
>>>> https://www.mathworks.com/matlabcentral/answers/350350-how-to-plo
>>>> t-loudspeaker-frequency-response-from-its-impulse-response
>>>> For example, to replicate the code snippet (second answer in
>>>> above link), how to do this in Scilab?
>>>> h = rand(1,64); // impulse response (Matlab source code)
>>>> fs = 1000;
>>>> Nfft = 128;
>>>> [H,F] = freqz(h,1,Nfft,fs);
>>>   
>>> Did you have a look around freq() or repfreq()?
>>>
>>> We have somewhat the equivalence invfreqz(H,F,m,n,W)  <=>
>>> frfit(F*2*%pi, H, n, W) // Scilab
>>>
>>> So you may look for the reciprocal of Scilab's frfit()
>>>
>>> HTH
>>> Samuel
>>>
>>>
>>>
>>> _______________________________________________
>>> users mailing list
>>> users at lists.scilab.org
>>> http://lists.scilab.org/mailman/listinfo/users
>> 	Virus-free. www.avast.com
>>   _______________________________________________
>> users mailing list
>> users at lists.scilab.org
>> http://lists.scilab.org/mailman/listinfo/users



---
This email has been checked for viruses by Avast antivirus software.
https://www.avast.com/antivirus