Matlab truth

EDAboard.com Forum Index -> Digital Signal Processing -> Matlab truth

Author

Message

Aya2002

Joined: 12 Dec 2006
Posts: 1409
Helped: 254
Location: Iraq

12 Mar 2009 0:20 ifft axis scaling matlab

Hi experts, i have a quistion here, please read the following paragragh as it is in a pdf document on the web:

The Fourier transform can be computed using the Matlab code
Fk = fft(fn)/N; % The FFT of the signal fn
fn = ifft(Fk)*N; % The Inverse FFT
ifft(fft(fn)); % This result = fn (within machine precision)

We must divide by the number of grid points N if we want to use the results of the FFT, because Matlab puts the normalization only in the inverse transform. As a result we must then multiply the IFFT by N to counter Matlab's dividing by N.

my doubt is: is the above is correct?

Thank you in advance.

Added after 3 hours:

any reply please

Aya2002

Joined: 12 Dec 2006
Posts: 1409
Helped: 254
Location: Iraq

13 Mar 2009 16:14 centeredfft

any help please?

rramya

Joined: 14 Dec 2008
Posts: 76
Helped: 19

13 Mar 2009 20:35 normalizing data matlab

Hi Aya2002,
I will wrtie some code & then i will conclude.

Tested without using normalisation factor (1/N):

time_signal=[ 1 2 3 4 3 2 ] ;
FFT_time_signal = fft(time_signal);
INVERSE_FFT_time_signal = ifft(FFT_time_signal );

o/p i got:
FFT_time_signal = 15 -4 0 -1 0 -4
INVERSE_FFT_time_signal = 1 2 3 4 3 2

Tested with using normalisation factor (1/N):

time_signal=[ 1 2 3 4 3 2 ] ;
FFT_time_signal = fft(time_signal)./length(time_signal);
INVERSE_FFT_time_signal = ifft(FFT_time_signal ).*length(time_signal);

FFT_time_signal = 2.5000 -0.6667 0 -0.1667 0 -0.6667
INVERSE_FFT_time_signal = 1.0000 2.0000 3.0000 4.0000 3.0000
2.0000

conclusion:
As per formula DFT does not use normalisation.
Instead IDFT uses normalisation factor (1/N) in the formula.
hence the matlab functon "fft" also operates as per formula does n't use normalisation factor (1/N) & "ifft" also operates as per formula uses normalisation factor (1/N) .

even without scaling we can get the same o/p.
what is the adv of using a scaling factor in fft and ifft???
i think one can think and get the answer for that.

I hope we will find it soon......

happy learning.

Aya2002

Joined: 12 Dec 2006
Posts: 1409
Helped: 254
Location: Iraq

13 Mar 2009 21:48 normalize data matlab

My Friend,

see my experement below:

x=ones(1,5)

x =

1 1 1 1 1

>> s=fft(x)

s =

5 0 0 0 0

>> ifft(s)

ans =

1 1 1 1 1

as shown above, Matlab did a complete conversion and inverse conversion with out need to do any normalization, I mean , that i do not need any normalization, while the Pdf file said that we must use the normalization. From which i conclude that this document is not correct.

Now, for your quiation "what is the adv of using a scaling factor in fft and ifft???", please see my answere below;

when we represent a signal within matlab, we usually use two vectors, one for the x-data and one for the y-data. The FFT command only operates on the y-data (converting the y-data from time domain to the frequency domain). So it is up to the user to determine what the x-data in the frequency domain will be!

Now let us to take an example:

y(t)=2cos( 2 Π fo t )
-------------------------

fo=4;
Fs=100;
Ts=1/Fs;
t=0:Ts:1-Ts;
n=length(t) ;
y=2*sin(2*pi*fo*t) ;
plot(t,y) ;

when we take the fft of this curve, we would ideally expect to get the following spectrum in the frequency domain , we expect to see one peak of amplitude 1 at -4 Hz, and another peak of amplitude 1 at +4 Hz. So, let us use matlab's built in function fft to see the results.

yfr=fft(y) ;
stem(abs(yfr ) ) ;

from the figure above, it does not quite look like what we predicted above. If we notice there are acouple of things that are missing:
1. The x-axis gives us no information on the frequency. How can we tell that the peaks are in the right place?

2. The amplitude is all the way up to 100.

3. The spectrum is not centered around zero.

To solve these problems, we must normalize the data and use the matlab function FFTshift after the function FFT as follows;

function [X freq]=centeredFFT(x,Fs)

N=length(x) ;
if mod(N,2) ==0
k=-N/2:N/2-1;
else
k=-(N/2)/2: (N-1)/2;
end
T=N/Fs;
freq=k/T; % The Frequency axis.
X=fft(x)/N; % Normalize the data.
X=fftshift(X); % shift the fft data so that it is centered.

the o/p of the above function will be correct frequency range and the transformed signal.

let's use this function in example:

[yfDom,frange]=centeredFFt(y,Fs);
stem(frange,abs(yfDom ) ) ;

as we can see from the plot above, the information within the frequency spectrum is entirely symmetric.

Hope this is clear.

Montadar

Google
AdSense
Google Adsense

13 Mar 2009 21:48 Ads

rramya

Joined: 14 Dec 2008
Posts: 76
Helped: 19

14 Mar 2009 19:34 division by n affter fft in matlab

please see this site.

normally when we do fft(x) we get huge ampitude . we do scaling by dividing by N and while calculating inverse fft we also do scaling . if we do so, we cant satisfy parsevals theorem which maintains unitary between FFT & IFFT.
Also, the reason for why we scale fft basically????

all this answer were being discussed in the below mentioned site

pl see dicussion (between Matt and ramya )
http://www.mathworks.fr/matlabcentral/newsreader/view_thread/174066

also mainly this site:
http://www.mathworks.fr/matlabcentral/newsreader/view_thread/246628

See finally we got the answer.

Happy learning