
From:  Brian Kaczynski 
Subject:  Re: Any FFT expert around? 
Date:  Thu, 13 Mar 2014 17:51:03 +0100 
Hi Brian,
> freq=linspace(Fs1*(N1)/(2*N), Fs1*(N1)/(2*N), Fs1*at)';
ok, i will rewrite it with an example:
If i sample 11 points (N=11) at the rate Fs=4 and fft it,
my first point on the frequency vector will be zero.
The next will be df=Fs/N=0.36364.
The 3rd will be 2*df=Fs/N=0.72727
...
...
In other words:
point k, frequency
point 1: 0 * 0.36364 = 0.00000
point 2: 1 * 0.36364 = 0.36364
point 2: 2 * 0.36364 = 0.72727
...
point 6: 5 * 0.36364 = 1.81818
point 7: 1 * 0.36364 = 0.36364
point 8: 2 * 0.36364 = 0.72727
...
point 11: 5 * 0.36364 = 1.81818
Again in other words, here is the code:
Fs=4 % sampling rate
N=11 % nr. of points
df=Fs/N % freq. increment per point
MaxFreq=df*(N1)/2 % max measurable freq
freqIdx=[(0:N/2)'; (1:N/2)'] % index vector to get the freqs
freq=freqIdx*df; % frequency vector
[freqIdx freq] % index beside frequency
fftshift(freq) % shifted freq vector
Thanks very much again for your very helpful wisdom!!!
All the best ...
On 3/13/14, oxy <address@hidden> wrote:
> Hi Brian,
>
>> freq=linspace(Fs1*(N1)/(2*N), Fs1*(N1)/(2*N), Fs1*at)';
>
> ok, i will rewrite it with an example:
>
> If i sample 11 points (N=11) at the rate Fs=4 and fft it,
> my first point on the frequency vector will be zero.
> The next will be df=Fs/N=0.36364.
> The 3rd will be 2*df=Fs/N=0.72727
> ...
> ...
>
> In other words:
>
> point k, frequency
> point 1: 0 * 0.36364 = 0.00000
> point 2: 1 * 0.36364 = 0.36364
> point 2: 2 * 0.36364 = 0.72727
> ...
> point 6: 5 * 0.36364 = 1.81818
> point 7: 1 * 0.36364 = 0.36364
> point 8: 2 * 0.36364 = 0.72727
> ...
> point 11: 5 * 0.36364 = 1.81818
>
>
> Again in other words, here is the code:
>
>
> Fs=4 % sampling rate
> N=11 % nr. of points
> df=Fs/N % freq. increment per point
> MaxFreq=df*(N1)/2 % max measurable freq
> freqIdx=[(0:N/2)'; (1:N/2)'] % index vector to get the freqs
> freq=freqIdx*df; % frequency vector
> [freqIdx freq] % index beside frequency
> fftshift(freq) % shifted freq vector
>
>
> Thanks very much again for your very helpful wisdom!!!
> All the best ...
>
>
>> 20140312 11:45 GMT+01:00 oxy <address@hidden>:
>>
>>> hi all,
>>>
>>> again in the question about FFT frequency axis.
>>>
>>> Pls confirm if I'm doing it right here:
>>>
>>> If I do fftshift(fft(time_domain_signal))
>>> of a certain time_domain_signal with even number of points,
>>> then the frequency axis becomes
>>>
>>> freq=linspace(Fs1/2, Fs1/2Fs1/(Fs1*at), Fs1*at)';
>>>
>>> where Fs= sampling rate, at=acquisition time.
>>>
>>> If however my time domain signal has an odd number of points, then
>>>
>>> freq=linspace(Fs1/2, Fs1/2, Fs1*at)';
>>>
>>> Am id doing it right?
>>>
>>> We had a recent thread on that. See some msg below.
>>>
>>> Thx for your wisdom!
>>>
>>> oxy
>>>
>>>
>>> > % TTF Demo: this code shows how to calculate the
>>> > % frequency domain when doing FFT of tdsignals.
>>> >
>>> > % parameter setup: same signal, 2 sampling rates 
>>> > clear all
>>> > nu=10; % signal frequency
>>> > at=5; % acquisition time
>>> > T1=2 % signal decay constant in time domain
>>> > FsFactor1=4; % sampling rate factor 1, Fs1/nu
>>> > FsFactor2=16; % sampling rate factor 2, Fs2/nu
>>> >
>>> > % calculating 
>>> > Fs1 = FsFactor1*nu; % Sampling rate 1
>>> > Fs2 = FsFactor2*nu; % Sampling rate 2
>>> >
>>> > t1 = ((0:(at*Fs11))/Fs1)'; % Time vector 1
>>> > t2 = ((0:(at*Fs21))/Fs2)'; % Time vector 2
>>> >
>>> > tdsig1=exp(i*2*pi*nu*t1).*exp(t1./(T1)); % time domain signal 1
>>> > tdsig2=exp(i*2*pi*nu*t2).*exp(t2./(T1)); % time domain signal 2
>>> >
>>> > % I realized, the signal was appearing negative, thus minus...
>>> > freq1=linspace(Fs1/2,Fs1/2Fs1/(Fs1*at),Fs1*at)'; % freq. vector 1
>>> > freq2=linspace(Fs2/2,Fs2/2Fs2/(Fs2*at),Fs2*at)'; % freq. vector 2
>>> >
>>> > freqsig1=fftshift(fft(tdsig1)); % signal vector 1
>>> > freqsig2=fftshift(fft(tdsig2)); % signal vector 2
>>> >
>>> >
>>> > % ploting 
>>> > clf
>>> > subplot(1,2,1)
>>> > plot(freq1, freqsig1)
>>> > axis([20 0]), title('FsFactor=4')
>>> >
>>> > subplot(1,2,2)
>>> > plot(freq2, freqsig2)
>>> > axis([20 0]), title('FsFactor=16')
>>> > %  end 
>>> >
>>> >>>================================================
>>> >>> Just one question more: you are calculating the frequency
>>> >>> vector as:
>>> >>>
>>> >>> freq1=linspace(Fs1/2, Fs1/21/at, Fs1*at)';
>>> >>>
>>> >>> that makes it nonsymetric. Isn't it better ...
>>> >>>
>>> >>> freq1=linspace(Fs1/2+1/at, Fs1/21/at, Fs1*at)';
>>> >>>
>>> >>> ...? Then it's symetric. Hey, thanks a lot! Great help!!!!!
>>> >>
>>> >> That would be "better" if you really want to insist on having a
>>> >> symmetrical frequency vector at the expense of accuracy.
>>> >> Unfortunately, it's inaccurate. For one thing, your symmetrical
>>> >> frequency vector does not contain a DC term if you have an even
>>> >> number
>>> >> of FFT points (which you should based on your examples).
>>> >>
>>> >> If you want the FFT value at the nth point to really correspond to
>>> >> freq1(n) you should use the slightly offset frequency vector that I
>>> >> provided.
>>> >>
>>> >> If it is more important to whatever you're doing that the frequency
>>> >> vector be exactly symmetrical and you can live with sacrificing the
>>> >> true 1to1 correspondence between frequency points and FFT points,
>>> >> you can use your symmetrical frequency vector.
>>> >>
>>> >>> oxy
>>> >>>
>>> >>> =================================================
>>> >>> On 2/19/14, oxy <address@hidden> wrote:
>>> >>>> On 2/12/14, Brian Kaczynski <address@hidden> wrote:
>>> >>>>> Hi Oxy,
>>> >>>>>
>>> >>>>> Since you're taking the FFT of a complex vector it will have
>>> >>>>> unique
>>> >>>>> terms
>>> >>>>> for negative vs. positive frequency. You are correct that it
>>> >>>>> makes
>>> >>>>> more
>>> >>>>> sense to think of the FFT from Fs/2 to +Fs/2. It's just that the
>>> >>>>> Octave
>>> >>>>> fft function doesn't compute the bins in that order. Due to
>>> aliasing,
>>> >>>>> any
>>> >>>>> frequency Fs/2 + df is equivalent to (Fs/2 + df) so it's more a
>>> >>>>> matter
>>> >>>>> of
>>> >>>>> choice how you want to display the data.
>>> >>>>>
>>> >>>>> If you prefer plotting from Fs/2 to +Fs/2 I would change these
>>> >>>>> two
>>> >>>>> lines
>>> >>>>> of your code as follows:
>>> >>>>>
>>> >>>>> freq=linspace(Fs/2,Fs/2Fs/(Fs*at),Fs*at)'
>>> >>>>> freqsig=fftshift(fft(tdsig);
>>> >>>>>
>>> >>>>> Let us know if that gives you what you want!
>>> >>>>>
>>> >>>>> Brian
>>> >>>>>
>>> >>>>>
>>> >>>>> 20140212 17:15 GMT+01:00 oxy <address@hidden>:
>>> >>>>>
>>> >>>>>> Hi Brian,
>>> >>>>>>
>>> >>>>>> > The frequency vector for FFT should go from DC as bin 1 to
>>> slightly
>>> >>>>>> > less
>>> >>>>>> > than Fs in the last bin (Fs  Fs/N where N is the number of FFT
>>> >>>>>> > points).
>>> >>>>>>
>>> >>>>>> If i understand u correctly, the frequency vector (freq) must be
>>> >>>>>> written
>>> >>>>>> as in this modified version of code below. However, according to
>>> >>>>>> Nyquist
>>> >>>>>> we
>>> >>>>>> cannot measure a frequency higher than Fs/2. Thus i do not see
>>> >>>>>> the meaning of plotting up to ~Fs.
>>> >>>>>>
>>> >>>>>> Also, if I rerun the code below changing FsFactor, I again see
>>> >>>>>> this
>>> >>>>>> dependency of the signal frequency on FsFactor.
>>> >>>>>> I cannot understand it. Looks like basics, yet not that obvious.
>>> >>>>>>
>>> >>>>>> # start code 
>>> >>>>>> clear all
>>> >>>>>> nu=10; % signal frequency
>>> >>>>>> at=5; % acquisition time
>>> >>>>>> T1=2 % signal decay constant in time
>>> >>>>>> domain
>>> >>>>>> FsFactor=16; % the ratio (sampling rate)/(signal
>>> >>>>>> frequency), or Fs/nu
>>> >>>>>>
>>> >>>>>> clf
>>> >>>>>> Fs = FsFactor*nu; % Sampling rate
>>> >>>>>> t = ((0:(at*Fs1))/Fs)'; % Time vector
>>> >>>>>> % freq=linspace(1,1,Fs*at)' * Fs/2; % frequency
>>> >>>>>> vector,
>>> >>>>>> first
>>> >>>>>> version
>>> >>>>>> freq=linspace(0,1,Fs*at)' * Fs Fs/(Fs*at); % frequency vector
>>> >>>>>> suggested by Brian
>>> >>>>>> tdsig=exp(i*2*pi*nu*t).*exp(t./(T1)); % time domain signal
>>> >>>>>> freqsig=fft(tdsig); % freq.
>>> >>>>>> domain
>>> >>>>>> signal
>>> >>>>>> subplot(1,2,1)
>>> >>>>>> plot(t,tdsig)
>>> >>>>>> axis([ 0 0.4]) % zooming time domain to see that
>>> >>>>>> period=1/nu
>>> >>>>>> subplot(1,2,2)
>>> >>>>>> plot(freq, freqsig)
>>> >>>>>> # end code 
>>> >>>>>>
>>> >>>>>> thx guys ...
>>> >>>>>>
>>> >>>>>>
>>> >>>>>> > 20140212 14:22 GMT+01:00 oxy <address@hidden>:
>>> >>>>>> >
>>> >>>>>> >> hey guys,
>>> >>>>>> >>
>>> >>>>>> >> the (simple) code bellow is how i ve learned to do FFT
>>> >>>>>> >> according
>>> >>>>>> >> to
>>> >>>>>> >> several docs online. However i observe a dependency of the
>>> >>>>>> >> signal
>>> >>>>>> >> frequency in the spectrum on the constant FsFactor. In other
>>> >>>>>> >> words,
>>> >>>>>> >> the signal frequency depends on the sampling rate. I should'nt
>>> be,
>>> >>>>>> >> right? So what is wrong here?
>>> >>>>>> >>
>>> >>>>>> >> # start code 
>>> >>>>>> >> clear all
>>> >>>>>> >> nu=10; % signal frequency
>>> >>>>>> >> at=5; % acquisition time
>>> >>>>>> >> T1=2 % signal decay constant in time
>>> >>>>>> >> domain
>>> >>>>>> >> FsFactor=8; % the ratio (sampling rate)/(signal
>>> >>>>>> >> frequency), or Fs/nu
>>> >>>>>> >>
>>> >>>>>> >> clf
>>> >>>>>> >> Fs = FsFactor*nu; % Sampling
>>> >>>>>> >> rate
>>> >>>>>> >> t = ((0:(at*Fs1))/Fs)'; % Time
>>> >>>>>> >> vector
>>> >>>>>> >> freq=linspace(1,1,Fs*at)' * Fs/2; % frequency
>>> >>>>>> >> vector
>>> >>>>>> >> tdsig=exp(i*2*pi*nu*t).*exp(t./(T1)); % time domain
>>> >>>>>> >> signal
>>> >>>>>> >> freqsig=fft(tdsig); % freq.
>>> >>>>>> >> domain
>>> >>>>>> >> signal
>>> >>>>>> >> subplot(1,2,1)
>>> >>>>>> >> plot(t,tdsig)
>>> >>>>>> >> axis([ 0 0.4]) % zooming time domain to see that
>>> >>>>>> >> period=1/nu
>>> >>>>>> >> subplot(1,2,2)
>>> >>>>>> >> plot(freq, freqsig)
>>> >>>>>> >> # end code 
>>> >>>>>> >>
>>> >>>>>> >> Do it yourself. Just rerun the code trying different values of
>>> >>>>>> >> FsFactor (eg: 2, 4, 8).
>>> >>>>>> >> Thx a lot for any hint!!!
>>> >>>>>> >>
>>> >>>>>> >> oxy
>>> >>>>>> >>
>>> >>>>>> >> ps: cross post
>>> >>>>>> >>
>>> >>>>>>
>>> http://www.mathworks.com/matlabcentral/answers/115700fftwhysignalfrequencydependsonsamplingrate
>>> >>>>>> >> _______________________________________________
>>> >>>>>> >> Helpoctave mailing list
>>> >>>>>> >> address@hidden
>>> >>>>>> >> https://mailman.cae.wisc.edu/listinfo/helpoctave
>>> >>>>>> >>
>>> >>>>>> >
>>> >>>>>> _______________________________________________
>>> >>>>>> Helpoctave mailing list
>>> >>>>>> address@hidden
>>> >>>>>> https://mailman.cae.wisc.edu/listinfo/helpoctave
>>> >>>>>>
>>> >>>>>
>>> >>>>
>>> >>>
>>> >>
>>> >
>>>
>>
>
_______________________________________________
Helpoctave mailing list
address@hidden
https://mailman.cae.wisc.edu/listinfo/helpoctave
[Prev in Thread]  Current Thread  [Next in Thread] 