should FFT{x-mean(x)} = FFT{x}-FFT{mean(x)} ?

Hi guys,

    The Linearity property of Fourier transform says:

z(t)=af1(t)+bf2(t)

the linear combination of the terms is unaffected by the transform.
Z(ω)=aF1(ω)+bF2(ω)

and 

Z(ω) = FFT{ z(t) }.

I synthesised a signal, sn as follows. I want to remove the mean values in
the signal. According to the Fourier transform property,

FFT{x-mean(x)} = FFT{x}-FFT{mean(x)}

the left hand side should equal to the right hand side. But my simulation
suggested they are different.

Can anyone sports any mistakes I've made?

Cheers,

 ---------------------------- Code Starts Here ----------------------------


%% Load FMCW data
clear all; close all; clc;
fs = 1; % We use scaleed samling rate here
NFFT = 2048;
scale = @(y) 10*log10(abs(y));
n = 0:149;
s = 2*sin(2*pi*0.2*n');
r = 10*sin(2*pi*0.15*n');

%  Generate Data and Visulize it
for jj = 1:37
    v = 0.3*randn(150,1);
    sn(:,jj) = s + v;
    rn(:,jj) = r + v;
end
sn (:,1:10) = rn(:,1:10);
sn (:,11:end) = sn(:,11:end) + rn(:,11:end);

% Unless you want to visualize the signal :-)
% for jj = 1:37
%     [Sn(:,jj),f] = periodogram(sn(:,jj),[],NFFT,fs);
% end
% figure;
% subplot(121); plot(f,scale(Sn(:,2)))
% subplot(122); imagesc(scale(Sn)); colorbar

ii = 2;
%% Time Domain 
sn_bg = sn-repmat(mean(sn,2),1,37);
[P_bg_1, f] = periodogram(sn_bg(:,ii),hamming(size(sn_bg,1)),NFFT,fs);

%% Freq Domain
for jj = 1:size(sn,2)
    [Psn_0(:,jj), f] = periodogram(sn(:,jj),hamming(size(sn,1)),NFFT,fs);
end
P_bg_2 = Psn_0-repmat(mean(Psn_0,2),1,37);
P_bg_2 = P_bg_2(:,ii);

%% Comparisons
figure; plot(f,scale(P_bg_1),'r-'); % title('Per-Ham, Mean Removal');
hold on;
plot(f,scale(P_bg_2),'b--')
legend('FFT(x-mean(x))','FFT(x) - FFT(mean(x))')

Yes.  Time to check your code.

Also, complicated code like that is pretty lazy dude!  I wouldn't even 
read it.  Easier to write something simpler like this which makes the point:

x=rand(64,1);
y(1:64)=mean(x);  %did you do this?  Easy mistake to not do this.....
y=y';
x1=x-y;
xdiff=x-y-x1
fx=fft(x);
fy=fft(y);
fx1=fft(x1);
fdiff=fx-fy-fx1

Shows that the functions are equal in time and equal in frequency 
because their differences are zero (or as close as one can get numerically).

Fred

Hi,

x-mean(x) ??

If the x is a sample value, then mean(x) is x :-)
This means mean(x) = x
Then x-mean(x) = ZERO --> FFT(x) = FFT(mean(x)) ;-)

BR,
SPHiNX


>Yes.  Time to check your code.
>
>Also, complicated code like that is pretty lazy dude!  I wouldn't even 
>read it.  Easier to write something simpler like this which makes the
point:
>
>x=rand(64,1);
>y(1:64)=mean(x);  %did you do this?  Easy mistake to not do this.....
>y=y';
>x1=x-y;
>xdiff=x-y-x1
>fx=fft(x);
>fy=fft(y);
>fx1=fft(x1);
>fdiff=fx-fy-fx1
>
>Shows that the functions are equal in time and equal in frequency 
>because their differences are zero (or as close as one can get
numerically).
>
>Fred
>

On 08/28/2010 11:03 AM, thedean515 wrote:
> Hi guys,
>
>      The Linearity property of Fourier transform says:
>
> z(t)=af1(t)+bf2(t)
>
> the linear combination of the terms is unaffected by the transform.
> Z(ω)=aF1(ω)+bF2(ω)
>
> and
>
> Z(ω) = FFT{ z(t) }.
>
> I synthesised a signal, sn as follows. I want to remove the mean values in
> the signal. According to the Fourier transform property,
>
> FFT{x-mean(x)} = FFT{x}-FFT{mean(x)}
>
> the left hand side should equal to the right hand side. But my simulation
> suggested they are different.
>
> Can anyone sports any mistakes I've made?
>
> Cheers,
>
>   ---------------------------- Code Starts Here ----------------------------
(and gets snipped)

I see windowing in the code -- if you're windowing x you need to 
subtract the mean _after_ windowing; subtracting the mean, then 
windowing, means that instead of finding

fft(x - mean(x)) you're finding fft((x - mean(x)) .* window).  I think 
you'll find that fft(mean(x) * window) has energy at a lot more places 
than fft(mean(x)) does.

Try _just_ what you're talking about above, without windowing.  Then go 
from there.

If there's still issues, then make sure they're not numeric.

-- 

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" was written for you.
See details at http://www.wescottdesign.com/actfes/actfes.html

Tim Wescott wrote:

> 
> Try _just_ what you're talking about above, without windowing.  Then go 
> from there.
> 

Good advice.  Or, just window the sequence before doing anything else 
and proceed from there just as intended without further consideration of 
the window - it's just a new sequence to deal with.

Fred

On Aug 28, 2:03&nbsp;pm, "thedean515" <jiangwei0515@n_o_s_p_a_m.gmail.com>
wrote:
> Hi guys,
>
> &nbsp; &nbsp; The Linearity property of Fourier transform says:
>
> z(t)=af1(t)+bf2(t)
>
> the linear combination of the terms is unaffected by the transform.
> Z(&omega;)=aF1(&omega;)+bF2(&omega;)
>
> and
>
> Z(&omega;) = FFT{ z(t) }.
>
> I synthesised a signal, sn as follows. I want to remove the mean values in
> the signal. According to the Fourier transform property,
>
> FFT{x-mean(x)} = FFT{x}-FFT{mean(x)}
>
> the left hand side should equal to the right hand side. But my simulation
> suggested they are different.
>
> Can anyone sports any mistakes I've made?
>
> Cheers,

The most egregious mistake you made was wasting helpers
time by separately sending identical posts to comp.dsp and
comp.soft-sys.matlab.

Next time use a single submission with a comma separated
distribution list; e.g.,

Newsgroups: comp.dsp, comp.soft-sys.matlab.

Greg Heath

On 06/09/10 9:43 PM, Greg Heath wrote:

> Next time use a single submission with a comma separated
> distribution list; e.g.,
>
> Newsgroups: comp.dsp, comp.soft-sys.matlab.

Technical note: there should not be any spaces between the list of 
newsgroups.

Newsgroups: comp.dsp,comp.soft-sys.matlab

Your posting software might be forgiving and fix this for you, but the 
technical standards say that the spaces must not be there.