MATLAB Central Newsreader - fft sign?

fft sign?

h xu — Tue, 29 Apr 2008 15:57:02 -0400

i've written a small and simple script to test the
fft-function of matlab that transformates exp(i*n*pi/xmax *
x) from xmin=-xmax to xmax. that should result in a
kronecker delta with positive amplitude with value 1... but
for some reason the amplitude is negative for odd n,... why
is that?

xmin=-5;
xmax = 5;

N= 2^8;
dx = (xmax-xmin)/N;
x = linspace(xmin,xmax,N);

y= exp(3* i*pi/xmax *x);
NFFT = 2^nextpow2(N);
Y_ = fft(y,NFFT)*dx;
q = [0:NFFT-1]/NFFT/dx*xmax*2;
Y = Y_(1:NFFT)/dx/NFFT;
plot(q,real(Y),'.b')

Re: fft sign?

David — Tue, 29 Apr 2008 16:30:03 -0400

"h xu" <quabla123@gmx.de> wrote in message
<fv7gge$a1d$1@fred.mathworks.com>...
> i've written a small and simple script to test the
> fft-function of matlab that transformates exp
(i*n*pi/xmax *
> x) from xmin=-xmax to xmax. that should result in a
> kronecker delta with positive amplitude with value 1...
but
> for some reason the amplitude is negative for odd n,...
why
> is that?
>
> xmin=-5;
> xmax = 5;
>
> N= 2^8;
> dx = (xmax-xmin)/N;
> x = linspace(xmin,xmax,N);
>
> y= exp(3* i*pi/xmax *x);
> NFFT = 2^nextpow2(N);
> Y_ = fft(y,NFFT)*dx;
> q = [0:NFFT-1]/NFFT/dx*xmax*2;
> Y = Y_(1:NFFT)/dx/NFFT;
> plot(q,real(Y),'.b')
>
>

try
y= -1*exp(3* i*(pi/xmax) *x);

Re: fft sign?

NZTideMan — Wed, 30 Apr 2008 00:55:11 -0400

On Apr 30, 4:30=A0am, "David " <d...@bigcompany.com> wrote:
> "h xu" <quabla...@gmx.de> wrote in message
>
> <fv7gge$a1...@fred.mathworks.com>...
>
>
>
>
>
> > i've written a small and simple script to test the
> > fft-function of matlab that transformates exp
> (i*n*pi/xmax *
> > x) from xmin=3D-xmax to xmax. that should result in a
> > kronecker delta with positive amplitude with value 1...
> but
> > for some reason the amplitude is negative for odd n,...
> why
> > is that?
>
> > xmin=3D-5;
> > xmax =3D 5;
>
> > N=3D 2^8;
> > dx =3D (xmax-xmin)/N;
> > x =3D linspace(xmin,xmax,N);
>
> > y=3D exp(3* i*pi/xmax *x);
> > NFFT =3D 2^nextpow2(N);
> > Y_ =3D fft(y,NFFT)*dx;
> > q =3D [0:NFFT-1]/NFFT/dx*xmax*2;
> > Y =3D Y_(1:NFFT)/dx/NFFT;
> > plot(q,real(Y),'.b')
>
> try
> y=3D -1*exp(3* i*(pi/xmax) *x);- Hide quoted text -
>
> - Show quoted text -

Try:
plot(q,abs(Y),'.b')
The sign of the real component (assuming the imaginary component is
zero) is just telling you the phase.
phase=3Dangle(Y)*180/pi;

Re: fft sign?

h xu — Wed, 30 Apr 2008 12:07:01 -0400

"h xu" <quabla123@gmx.de> wrote in message
<fv7gge$a1d$1@fred.mathworks.com>...
> i've written a small and simple script to test the
> fft-function of matlab that transformates exp(i*n*pi/xmax *
> x) from xmin=-xmax to xmax. that should result in a
> kronecker delta with positive amplitude with value 1... but
> for some reason the amplitude is negative for odd n,... why
> is that?
>
> xmin=-5;
> xmax = 5;
>
> N= 2^8;
> dx = (xmax-xmin)/N;
> x = linspace(xmin,xmax,N);
>
> y= exp(3* i*pi/xmax *x);
> NFFT = 2^nextpow2(N);
> Y_ = fft(y,NFFT)*dx;
> q = [0:NFFT-1]/NFFT/dx*xmax*2;
> Y = Y_(1:NFFT)/dx/NFFT;
> plot(q,real(Y),'.b')
>
>

hm but i want to transformate functions, that transformated
also have a imaginary part, so what do i need to do, to have
the sign right for every value?

Re: fft sign?

David — Wed, 30 Apr 2008 18:31:03 -0400

"h xu" <quabla123@gmx.de> wrote in message <fv9nd5
$9oe$1@fred.mathworks.com>...
> "h xu" <quabla123@gmx.de> wrote in message
> <fv7gge$a1d$1@fred.mathworks.com>...
> > i've written a small and simple script to test the
> > fft-function of matlab that transformates exp
(i*n*pi/xmax *
> > x) from xmin=-xmax to xmax. that should result in a
> > kronecker delta with positive amplitude with value
1... but
> > for some reason the amplitude is negative for odd
n,... why
> > is that?
> >
> > xmin=-5;
> > xmax = 5;
> >
> > N= 2^8;
> > dx = (xmax-xmin)/N;
> > x = linspace(xmin,xmax,N);
> >
> > y= exp(3* i*pi/xmax *x);
> > NFFT = 2^nextpow2(N);
> > Y_ = fft(y,NFFT)*dx;
> > q = [0:NFFT-1]/NFFT/dx*xmax*2;
> > Y = Y_(1:NFFT)/dx/NFFT;
> > plot(q,real(Y),'.b')
> >
> >
>
>
> hm but i want to transformate functions, that
transformated
> also have a imaginary part, so what do i need to do, to
have
> the sign right for every value?

try:
y= -1*exp(3* i*(pi/xmax) *x);

yes, it looks the same on the surface... but it REALLY
isn't.

Re: fft sign?

h xu — Sun, 04 May 2008 13:09:03 -0400

"David " <dave@bigcompany.com> wrote in message
<fvadt7$e6u$1@fred.mathworks.com>...
> "h xu" <quabla123@gmx.de> wrote in message <fv9nd5
> $9oe$1@fred.mathworks.com>...
> > "h xu" <quabla123@gmx.de> wrote in message
> > <fv7gge$a1d$1@fred.mathworks.com>...
> > > i've written a small and simple script to test the
> > > fft-function of matlab that transformates exp
> (i*n*pi/xmax *
> > > x) from xmin=-xmax to xmax. that should result in a
> > > kronecker delta with positive amplitude with value
> 1... but
> > > for some reason the amplitude is negative for odd
> n,... why
> > > is that?
> > >
> > > xmin=-5;
> > > xmax = 5;
> > >
> > > N= 2^8;
> > > dx = (xmax-xmin)/N;
> > > x = linspace(xmin,xmax,N);
> > >
> > > y= exp(3* i*pi/xmax *x);
> > > NFFT = 2^nextpow2(N);
> > > Y_ = fft(y,NFFT)*dx;
> > > q = [0:NFFT-1]/NFFT/dx*xmax*2;
> > > Y = Y_(1:NFFT)/dx/NFFT;
> > > plot(q,real(Y),'.b')
> > >
> > >
> >
> >
> > hm but i want to transformate functions, that
> transformated
> > also have a imaginary part, so what do i need to do, to
> have
> > the sign right for every value?
>
> try:
> y= -1*exp(3* i*(pi/xmax) *x);
>
> yes, it looks the same on the surface... but it REALLY
> isn't.

mhm, sure for y=exp(n* i*(pi/xmax) *x) with n=3 or any other
odd number, the fft of y'=-y would change the amplitude from
negative to positive, but then the sign would be wrong if I
had even n.

well the exp() thing was just to make my problem with wrong
amplitude-sign clear.

my question is what i need to do to get the sign of the
ft-amplitude right for a abritary (also complex) function.

and also why sometimes the sign is false.

Re: fft sign?

Greg Heath — Sun, 04 May 2008 13:18:41 -0400

On Apr 29, 11:57=A0am, "h xu" <quabla...@gmx.de> wrote:
> i've written a small and simple script to test the
> fft-function of matlab that transformates exp(i*n*pi/xmax *
> x) from xmin=3D-xmax to xmax. that should result in a
> kronecker delta with positive amplitude with value 1... but
> for some reason the amplitude is negative for odd n,... why
> is that?
>
> xmin=3D-5;
> xmax =3D 5;
>
> N=3D 2^8;
> dx =3D (xmax-xmin)/N;

dx =3D (xmax-xmin)/(N-1)
x =3D xmin + dx*(0:N-1);

Hope this helps.

Greg

Re: fft sign?

Greg Heath — Mon, 05 May 2008 13:32:22 -0400

On Apr 29, 11:57 am, "h xu" <quabla...@gmx.de> wrote:
> i've written a small and simple script to test the
> fft-function of matlab that transformates exp(i*n*pi/xmax *
> x) from xmin=-xmax to xmax. that should result in a
> kronecker delta with positive amplitude with value 1... but
> for some reason the amplitude is negative for odd n,... why
> is that?
>
> xmin=-5;

xmin~=0 makes phase interpretation difficult

> xmax = 5;
>
> N= 2^8;

Too high; Try the 10 separate trials N = 3:12

> dx = (xmax-xmin)/N;

Incorrect. See below

> x = linspace(xmin,xmax,N);
>
> y= exp(3* i*pi/xmax *x);

Weird function. Try an FFT basis function. See below

-----SNIP

close all, clear, clc, j=0

N = 8 % Instructive to look at ALL of N=3:12
xmin = -5
xmax = 5
dx = (xmax-xmin)/(N-1)
X = N*dx % FFT imposed period = 10+dx

x = linspace(xmin,xmax,N)'; % OK (just changed to column)
x = xmin + dx*(0:N-1)' % Alternate form

df = 1/X % frequency spacing
f = df*(0:N-1)';

% The FFT basis functions are exp(2*pi*i*f(n)*x)

f0 = (N/4)*df % Half of the Nyquist frequency

y = exp(2*pi*i*f0*x);

z = zeros(size(x));
j=j+1,figure(j)

subplot(2,2,1), hold on
plot(x,z,'k')
plot(x,real(y))
subplot(2,2,2), hold on
plot(x,z,'k')
plot(x,imag(y))
subplot(2,2,3), hold on
plot(x,abs(y))
axis([xmin xmax 0 2])
subplot(2,2,4), hold on
plot(x,z,'k')
plot(x,angle(y))

Y = dx*fft(y);
y0 = df*Y; % Rescaling to the size of y

j=j+1,figure(j)
subplot(2,2,1), hold on
plot(f,z,'k')
plot(f,real(y0))
subplot(2,2,2), hold on
plot(f,z,'k')
plot(f,imag(y0))
subplot(2,2,3), hold on
plot(f,abs(y0))
% axis([0 max(f) 0 1.1*max(abs(Y))])
% axis([0 max(f) 0 1.1*max(abs(Y))])
subplot(2,2,4), hold on
plot(f,z,'k')
plot(f,angle(y0))

Hope this helps.

Greg