Thread Subject: frequency array of fft

hi, can someone please tell me how i can find the frequency array of some fft data.
please thanks.

> hi, can someone please tell me how i can find the
> frequency array of some fft data.
> please thanks.

ive done fft(data), to get the fft, but i need to find the frequency array of that fft. how is that done?

vishan <vishan.sondhi@guidance.eu.com> wrote in message <18773913.16072.1245748422329.JavaMail.jakarta@nitrogen.mathforum.org>...
> > hi, can someone please tell me how i can find the
> > frequency array of some fft data.
> > please thanks.
>
> ive done fft(data), to get the fft, but i need to find the frequency array of that fft. how is that done?

First you will notice that your FFT comes out as an array with the same number of elements as your data. What you need to know is how far apart each adjacent element is in frequency space. This is easy it is 1/T where T is the total length of time you took to gather your data. Thus if your data array took 1 second to gather, your FFT array elements ar 1Hz apart, if it only took 0.1 seconds then they are 10Hz apart.

Next the confusing part is what frequency each element represents. Well the first element represents 0Hz or DC as we electrical types like to consider it, infact it is related to the mean value of your initial data. From there in steps of 1/T Hz each element represents a measure of the signal occuring at that frequency, thus the 4 element represents 3/T Hz etc. This continues until you reach 'halfway' along the array when all of a sudden the rules change. From here on out you are in the 'negative frequency' zone, in which the frequency represents the negative Nyquist frequency (effectively the sample rate you were taking your data at /2) from there the negative frequencies decrease by 1/T Hz until you reach the end of your array.

Note that if you apply the function fftshift() to your FFT'ed data, then it rearranges it nicely so you start from the negative frequencies, DC (0 Hz) is in the centre and the positive frequencies follow that. This is the conventional way that most text books illustrate a spectrum.

If your input data is real (meaning it is not complex) then the negative frequencies have a symettrical relationship to the positive frequency. and if you are measuring power or amplitude, you can forget all about the negative frequency components.

Also one thing you will find that the data, once transformed contains complex numbers. Here the amplitude of the complex numbers represent how big the signal is at that frequency, and the ratio of the real to imaginary part gives you the phase of that signal. Often if you are only interested in power, this information is not needed. However if you want to do some processing in the frequency domain, and then retrieve your time domain data, it is essential to hang on to this phase data, because without it the inverse transform cannot reconstruct the waveform you are after.

Hope that helps

Regards

Dave Robinson

vishan <vishan.sondhi@guidance.eu.com> wrote in message <25735227.15743.1245743293465.JavaMail.jakarta@nitrogen.mathforum.org>...
> hi, can someone please tell me how i can find the frequency array of some fft data.
> please thanks.

The Frequency axis (after fftshifting) is given by

FrequencyAxis=( -ceil((N-1)/2):N-1-ceil((N-1)/2) )/(N*dt)

where N is the number of points in your array and dt is your time sampling interval. The above holds regardless of whether N is even or odd.

hi, thanks for your reply.
is there a formuale that i can input into matlab that will calcuate teh frequency?

vishan <vishan.sondhi@guidance.eu.com> wrote in message <7213463.16289.1245754480280.JavaMail.jakarta@nitrogen.mathforum.org>...
> hi, thanks for your reply.
> is there a formuale that i can input into matlab that will calcuate teh frequency?

FrequencyAxis=( -ceil((N-1)/2):N-1-ceil((N-1)/2) )/(N*dt)

thanks for that matt. is there a way to calculate N?

is it n=length(data)
or n=length(X) ?

vishan <vishan.sondhi@guidance.eu.com> wrote in message <27823741.16379.1245755410959.JavaMail.jakarta@nitrogen.mathforum.org>...
> thanks for that matt. is there a way to calculate N?

N is the number of samples, including any zero-padding, operated upon by the fft

% MODIFY
DT = 0.2; % sampling time interval
N = 46; % series length
KFFT = 10;

% Parameters
NFFT = KFFT*N; % spectral points (zero pading)
DF = 1/(NFFT*DT); % Sampling frequency interval

% Series and its transform:
X = rand(N,1);
CXa = fft(X,NFFT)*DT;


% FREQUEANCIES:

% a) two-side (MATLAB default)
fa = (0:NFFT-1)'*DF;
figure, plot(fa,[real(CXa) imag(CXa)]), title('two-side')

% b) two-side centered
fb = (-floor(NFFT/2):(NFFT-1)/2)'*DF;
CXb = fftshift(CXa);
figure, plot(fb,[real(CXb) imag(CXb)]),title('two-side centered')

% c) one-side (positive)
fc = (0:floor(NFFT/2))'*DF;
CXc = CXa(1:floor(NFFT/2)+1);
figure, plot(fc,[real(CXc) imag(CXc)]),title('one-side (positive)')


% Note:
% * NFFT "increase" the spectral resolution
% (by KFFT in this case) but do not change the
% Nyquist frequency: 1/(2*DT).
% * The use of DT.

% Carlos Vargas

On Jun 23, 3:47 am, vishan <vishan.son...@guidance.eu.com> wrote:
> hi, can someone please tell me how i can find the frequency array of some fft data.
> please thanks.

Consider the following relationships.

dt = 1/Fs
T = N*dt
t = dt*(0:N-1);
t = (0:dt:T-dt);
t = linspace(0,T-dt,N);

df = 1/T
Fs = N*df
f = df*(0:N-1);
f = (0:df:Fs-df);
f = linspace(0,Fs-df,N);

Hope this helps.

Greg

On Jun 23, 6:37 am, "Dave Robinson" <dave.robin...@somewhere.biz>
wrote:
> vishan <vishan.son...@guidance.eu.com> wrote in message <18773913.16072.1245748422329.JavaMail.jaka...@nitrogen.mathforum.org>...
> > > hi, can someone please tell me how i can find the
> > > frequency array of some fft data.
> > > please thanks.
>
> > ive done fft(data), to get the fft, but i need to find the frequency array of that fft. how is that done?
>
> First you will notice that your FFT comes out as an array with the same number of elements as your data. What you need to know is how far apart each adjacent element is in frequency space. This is easy it is 1/T where T is the total length of time you took to gather your data.

Not quite:

t = dt*(0:N-1) = 0:dt:T-dt

Therefore, df = 1/T but

T = max(t)+ dt

>Thus if your data array took 1 second to gather, your FFT array elements ar 1Hz apart, if it only took 0.1 seconds then they are 10Hz apart.

> Next the confusing part is what frequency each element represents. Well the first element represents 0Hz or DC as we electrical types like to consider it, infact it is related to the mean value of your initial data. From there in steps of 1/T Hz each element represents a measure of the signal occuring at that frequency, It represents the interval over which the signal represented by thus >the 4 element represents 3/T Hz etc.

Not quite:

The value f is the center of a df Hz frequency range (f-df/2,f+df/2)
over which the energy corresponding to X(f) is spread.

> This continues until you reach 'halfway' along the array when all of a sudden the rules change. From here on out you are in the 'negative frequency' zone, in which the frequency represents the negative Nyquist frequency (effectively the sample rate you were taking your data at /2) from there the negative frequencies decrease by 1/T Hz until you reach the end of your array.

Specifically:

T = N*dt
df = 1/T
Fs = 1/dt
Fs = N*df
f = df*(0:N-1) = 0:df:Fs-df
fNyquist = Fs/2 = (N/2)*df

N even ==> fNyquist is on a grid point
N odd ==> fNyqist is half way between two grid points

fft generates a spectrum over [0,Fs-df] whos extension
outside of that interval is periodic with period Fs.

The spectrum for frequencies f >= FNyquist should be
interpreted as the periodic extension of the negative
frequency part of the spectrum.

> Note that if you apply the function fftshift() to your FFT'ed data, then it rearranges it nicely so you start from the negative frequencies, DC (0 Hz) is in the centre and the positive frequencies follow that. This is the conventional way that most text books illustrate a spectrum.

Thus Xshift = fftshift(X) where X = fft(x), should be used
with the frequency scales

fshift = df* (-N/2:N/2-1) for N even
fshift = df* (-(N-1)/2:(N-1)/2) for N odd

> If your input data is real (meaning it is not complex) then the negative frequencies have a symettrical relationship to the positive frequency. and if you are measuring power or amplitude, you can forget all about the negative frequency components.

Not quite.

If x is real, then X is congugate symmetric:

Xshift(-f) = Xshift(f)'

By convention, when N is even. the Nyquist frequency
was put in the negative part of the spectrum. However,
to use only a single-sided positive frequency spectrum,
the Nyquist component should be included for mathematical
accuracy. Then, for 0 < f < fNyquist. X should be multiplied
by sqrt(2) to account for the energy in the negative
frequency part of the spectrum. Whereas the factor for
the DC and Nyquist frequencies is unity.

> Also one thing you will find that the data, once transformed contains complex numbers. Here the amplitude of the complex numbers represent how big the signal is at that frequency, and the ratio of the real to imaginary part gives you the phase of that signal. Often if you are only interested in power, this information is not needed. However if you want to do some processing in the frequency domain, and then retrieve your time domain data, it is essential to hang on to this phase data, because without it the inverse transform cannot reconstruct the waveform you are after.

Hope this helps.

Greg

On Jun 23, 12:24 pm, "Carlos Adrian Vargas Aguilera"
<nubeobsc...@hotmail.com> wrote:
> % MODIFY
> DT = 0.2; % sampling time interval
> N = 46; % series length
> KFFT = 10;
>
> % Parameters
> NFFT = KFFT*N; % spectral points (zero pading)
> DF = 1/(NFFT*DT);     % Sampling frequency interval
>
> % Series and its transform:
> X   = rand(N,1);
> CXa = fft(X,NFFT)*DT;
>
> % FREQUEANCIES:
>
> % a) two-side (MATLAB default)
> fa = (0:NFFT-1)'*DF;
> figure, plot(fa,[real(CXa) imag(CXa)]), title('two-side')
>
> % b) two-side centered
> fb = (-floor(NFFT/2):(NFFT-1)/2)'*DF;

fb = (-floor(NFFT/2):NFFT/2-1)'*DF;

> CXb = fftshift(CXa);
> figure, plot(fb,[real(CXb) imag(CXb)]),title('two-side centered')
>
> % c) one-side (positive)
> fc = (0:floor(NFFT/2))'*DF;
> CXc = CXa(1:floor(NFFT/2)+1);
> figure, plot(fc,[real(CXc) imag(CXc)]),title('one-side (positive)')
>
> % Note:
> % * NFFT "increase" the spectral resolution
> % (by KFFT in this case) but do not change the
> % Nyquist frequency: 1/(2*DT).
> % * The use of DT.

Not quite:

Although the spacing between frequency gridpoints
is dectreased by the factor KFFT, the result is
just interpolation which adds no additional information
with respect to separating the effects of closely spaced
frequencies.

Therefore, the frequency resolution, 1/T is unchanged.

Hope this helps.

Greg

confused.com

im a beginner in all this lol

On Jun 24, 8:26 am, vishan <vishan.son...@guidance.eu.com> wrote:
> confused.com
>
> im a beginner in all this lol

I have posted code in previous threads. Search on

greg-heath close fft
greg-heath clear fft

Hopr this helps.

Greg

Yes Greg, I know, that's way the quotes in "increase". In fact, it is not an interpolation, but a convolution with the SINC window because of the RECTWIN in time space, that's is way the "peaks" looks like SINCs. I'll submit a function about this latter.

Besides, my centered frequency is correct for any NFFT
>> fb = (-floor(NFFT/2):(NFFT-1)/2)'*DF;
but yours
>> fb = (-floor(NFFT/2):NFFT/2-1)'*DF;
is incorrect for odd NFFT, CHECK IT OUT!!

Carlos Vargas

vishan <vishan.sondhi@guidance.eu.com> wrote in message <33454626.22783.1245846433940.JavaMail.jakarta@nitrogen.mathforum.org>...
> confused.com
>
> im a beginner in all this lol

vishan - I reiterate.

Once you've chosen a number of samples N and time sampling interval dt, your frequency axis grid values are as folllows (after you've centered your spectrum using fftshift).

FrequencyAxis=( -ceil((N-1)/2):N-1-ceil((N-1)/2) )/(N*dt)

It's as simple as that. Additionally, your time axis grid values are


TimeAxis=( -ceil((N-1)/2):N-1-ceil((N-1)/2) ) * dt

 

"Matt " <xys@whatever.com> wrote in message <h1u52h$ut$1@fred.mathworks.com>...
> vishan <vishan.sondhi@guidance.eu.com> wrote in message <33454626.22783.1245846433940.JavaMail.jakarta@nitrogen.mathforum.org>...
> > confused.com
> >
> > im a beginner in all this lol
>
> vishan - I reiterate.
>
> Once you've chosen a number of samples N and time sampling interval dt, your frequency axis grid values are as folllows (after you've centered your spectrum using fftshift).
>
> FrequencyAxis=( -ceil((N-1)/2):N-1-ceil((N-1)/2) )/(N*dt)
>
> It's as simple as that. Additionally, your time axis grid values are
>
>
> TimeAxis=( -ceil((N-1)/2):N-1-ceil((N-1)/2) ) * dt









--------------


I dont think NFFT Stuff works, ghosh i spended last three month on the stuff which dont work

Well i am trying to construct the frequency axis for N =odd but it doesnot seem working
freq=1/dt/N/2*[-N:2:N-1]

is its fine for N = odd

when i give plot(freq,s)....my spectrum doesnot seem to be at 30HZ .....

"Dave Robinson" <dave.robinson@somewhere.biz> wrote in message <h1qb8d$air$1@fred.mathworks.com>...
> vishan <vishan.sondhi@guidance.eu.com> wrote in message <18773913.16072.1245748422329.JavaMail.jakarta@nitrogen.mathforum.org>...
> > > hi, can someone please tell me how i can find the
> > > frequency array of some fft data.
> > > please thanks.
> >
> > ive done fft(data), to get the fft, but i need to find the frequency array of that fft. how is that done?
>
> First you will notice that your FFT comes out as an array with the same number of elements as your data. What you need to know is how far apart each adjacent element is in frequency space. This is easy it is 1/T where T is the total length of time you took to gather your data. Thus if your data array took 1 second to gather, your FFT array elements ar 1Hz apart, if it only took 0.1 seconds then they are 10Hz apart.
>
> Next the confusing part is what frequency each element represents. Well the first element represents 0Hz or DC as we electrical types like to consider it, infact it is related to the mean value of your initial data. From there in steps of 1/T Hz each element represents a measure of the signal occuring at that frequency, thus the 4 element represents 3/T Hz etc. This continues until you reach 'halfway' along the array when all of a sudden the rules change. From here on out you are in the 'negative frequency' zone, in which the frequency represents the negative Nyquist frequency (effectively the sample rate you were taking your data at /2) from there the negative frequencies decrease by 1/T Hz until you reach the end of your array.
>
> Note that if you apply the function fftshift() to your FFT'ed data, then it rearranges it nicely so you start from the negative frequencies, DC (0 Hz) is in the centre and the positive frequencies follow that. This is the conventional way that most text books illustrate a spectrum.
>
> If your input data is real (meaning it is not complex) then the negative frequencies have a symettrical relationship to the positive frequency. and if you are measuring power or amplitude, you can forget all about the negative frequency components.
>
> Also one thing you will find that the data, once transformed contains complex numbers. Here the amplitude of the complex numbers represent how big the signal is at that frequency, and the ratio of the real to imaginary part gives you the phase of that signal. Often if you are only interested in power, this information is not needed. However if you want to do some processing in the frequency domain, and then retrieve your time domain data, it is essential to hang on to this phase data, because without it the inverse transform cannot reconstruct the waveform you are after.
>
> Hope that helps
>
> Regards
>
> Dave Robinson

"Dave Robinson" <dave.robinson@somewhere.biz> wrote in message <h1qb8d$air$1@fred.mathworks.com>...
> vishan <vishan.sondhi@guidance.eu.com> wrote in message <18773913.16072.1245748422329.JavaMail.jakarta@nitrogen.mathforum.org>...
> > > hi, can someone please tell me how i can find the
> > > frequency array of some fft data.
> > > please thanks.
> >
> > ive done fft(data), to get the fft, but i need to find the frequency array of that fft. how is that done?
>
> First you will notice that your FFT comes out as an array with the same number of elements as your data. What you need to know is how far apart each adjacent element is in frequency space. This is easy it is 1/T where T is the total length of time you took to gather your data. Thus if your data array took 1 second to gather, your FFT array elements ar 1Hz apart, if it only took 0.1 seconds then they are 10Hz apart.
>
> Next the confusing part is what frequency each element represents. Well the first element represents 0Hz or DC as we electrical types like to consider it, infact it is related to the mean value of your initial data. From there in steps of 1/T Hz each element represents a measure of the signal occuring at that frequency, thus the 4 element represents 3/T Hz etc. This continues until you reach 'halfway' along the array when all of a sudden the rules change. From here on out you are in the 'negative frequency' zone, in which the frequency represents the negative Nyquist frequency (effectively the sample rate you were taking your data at /2) from there the negative frequencies decrease by 1/T Hz until you reach the end of your array.
>
> Note that if you apply the function fftshift() to your FFT'ed data, then it rearranges it nicely so you start from the negative frequencies, DC (0 Hz) is in the centre and the positive frequencies follow that. This is the conventional way that most text books illustrate a spectrum.
>
> If your input data is real (meaning it is not complex) then the negative frequencies have a symettrical relationship to the positive frequency. and if you are measuring power or amplitude, you can forget all about the negative frequency components.
>
> Also one thing you will find that the data, once transformed contains complex numbers. Here the amplitude of the complex numbers represent how big the signal is at that frequency, and the ratio of the real to imaginary part gives you the phase of that signal. Often if you are only interested in power, this information is not needed. However if you want to do some processing in the frequency domain, and then retrieve your time domain data, it is essential to hang on to this phase data, because without it the inverse transform cannot reconstruct the waveform you are after.
>
> Hope that helps
>
> Regards
>
> Dave Robinson

Hi Dave,

Speaking of frequency domain, I want to do some processing in the frequency domain (I just want to modify the amplitudes in the frequency domain.), and then retrieve the time domain data, how can it be done? I will appreciate your help.

Thanks, Monica

"Dave Robinson" <dave.robinson@somewhere.biz> wrote in message <h1qb8d$air$1@fred.mathworks.com>...
> vishan <vishan.sondhi@guidance.eu.com> wrote in message <18773913.16072.1245748422329.JavaMail.jakarta@nitrogen.mathforum.org>...
> > > hi, can someone please tell me how i can find the
> > > frequency array of some fft data.
> > > please thanks.
> >
> > ive done fft(data), to get the fft, but i need to find the frequency array of that fft. how is that done?
>
> First you will notice that your FFT comes out as an array with the same number of elements as your data. What you need to know is how far apart each adjacent element is in frequency space. This is easy it is 1/T where T is the total length of time you took to gather your data. Thus if your data array took 1 second to gather, your FFT array elements ar 1Hz apart, if it only took 0.1 seconds then they are 10Hz apart.
>
> Next the confusing part is what frequency each element represents. Well the first element represents 0Hz or DC as we electrical types like to consider it, infact it is related to the mean value of your initial data. From there in steps of 1/T Hz each element represents a measure of the signal occuring at that frequency, thus the 4 element represents 3/T Hz etc. This continues until you reach 'halfway' along the array when all of a sudden the rules change. From here on out you are in the 'negative frequency' zone, in which the frequency represents the negative Nyquist frequency (effectively the sample rate you were taking your data at /2) from there the negative frequencies decrease by 1/T Hz until you reach the end of your array.
>
> Note that if you apply the function fftshift() to your FFT'ed data, then it rearranges it nicely so you start from the negative frequencies, DC (0 Hz) is in the centre and the positive frequencies follow that. This is the conventional way that most text books illustrate a spectrum.
>
> If your input data is real (meaning it is not complex) then the negative frequencies have a symettrical relationship to the positive frequency. and if you are measuring power or amplitude, you can forget all about the negative frequency components.
>
> Also one thing you will find that the data, once transformed contains complex numbers. Here the amplitude of the complex numbers represent how big the signal is at that frequency, and the ratio of the real to imaginary part gives you the phase of that signal. Often if you are only interested in power, this information is not needed. However if you want to do some processing in the frequency domain, and then retrieve your time domain data, it is essential to hang on to this phase data, because without it the inverse transform cannot reconstruct the waveform you are after.
>
> Hope that helps
>
> Regards
>
> Dave Robinson

Hi Dave,

Speaking of frequency domain, I want to do some processing in the frequency domain (I just want to modify the amplitudes in the frequency domain.), and then retrieve the time domain data, how can it be done? I will appreciate your help.

Thanks, Monica