Prev: Interaction of MEX and MATLAB or mexCallMATLAB and Co.
Next: the distance between two time series
From: Steve Amphlett on 27 Apr 2010 12:03
Greg Heath <heath(a)alumni.brown.edu> wrote in message <0bb0996b-f5fc-455e-8101-b4ff3c7651d1(a)j15g2000vbq.googlegroups.com>...
> On Apr 27, 2:50 am, "Steve Amphlett" <Firstname.Lastn...(a)Where-I-
> Work.com> wrote:
> > Greg Heath <he...(a)alumni.brown.edu> wrote in message <c17b6883-f695-4934-877c-838b7ff5f...(a)n5g2000yqh.googlegroups.com>...
> > > On Apr 26, 11:05 am, Luis <luisthearc...(a)gmail.com> wrote:
> > > > Hi.
> >
> > > > In the example on how to use fft, once they have the signal, they do this:
> >
> > > > NFFT = 2^nextpow2(L); % Next power of 2 from length of y
> > > > Y = fft(y,NFFT)/L;
> >
> > > > Meaning that they get the next power of 2 from length of y (the signal), and then they calculate the NFFT-points fft, to then divide it by L (length of y).
> > > > ¿Can you tell me why they do that?
> >
> > > > What I'm trying to do is capture a sound signal and then with the fft get the frequency of that signal.
> >
> > > > Thank you!
> >
> > > 1. You don't have to do either.
> >
> > > a. If L is huge and/or you have to process many cases,
> > > you can reduce running time by a factor of log(NFFT)/L .
> > > If this doesn't concern you, just use
> >
> > > Y = fft(y)/L.
> >
> > > b. The normalizing factor L will yield an amplitude of 1
> > > for exp(i*2*f0*t) at f = f0.
> >
> > > Hope this helps.
> >
> > > Greg
> >
> > ... except at DC and Nyquist, where you need 2/L
>
> No.
>
> >> Fs = 1, N = 6, dt = 1/Fs
> t = dt*(0:N-1);
> x = 1 + cos(2*pi*(Fs/2)*t);
> X = fft(x)/N
>
> Fs = 1
> N = 6
> dt = 1
>
> X = 1 0 0 1 0 0
>
> Hope this helps.
>
> Greg

Is this a recent change?
From: Greg Heath on 27 Apr 2010 20:40
On Apr 27, 12:03 pm, "Steve Amphlett" <Firstname.Lastn...(a)Where-I-
Work.com> wrote:
> Greg Heath <he...(a)alumni.brown.edu> wrote in message <0bb0996b-f5fc-455e-8101-b4ff3c765...(a)j15g2000vbq.googlegroups.com>...
> > On Apr 27, 2:50 am, "Steve Amphlett" <Firstname.Lastn...(a)Where-I-
> > Work.com> wrote:
> > > Greg Heath <he...(a)alumni.brown.edu> wrote in message <c17b6883-f695-4934-877c-838b7ff5f...(a)n5g2000yqh.googlegroups.com>...
----SNIP

> > > > Y = fft(y)/L.
>
> > > > b. The normalizing factor L will yield an amplitude of 1
> > > > for exp(i*2*f0*t) at f = f0.
>
> > > ... except at DC and Nyquist, where you need 2/L
>
> > No.
>
> > >> Fs = 1, N = 6, dt = 1/Fs
> > t = dt*(0:N-1);
> > x = 1 + cos(2*pi*(Fs/2)*t);
> > X = fft(x)/N
>
> > Fs = 1
> > N = 6
> > dt = 1
>
> > X = 1 0 0 1 0 0
>
>
> Is this a recent change?

No.

Too make things more clear:

>> x1 = 1 + exp(i*2*pi*(Fs/2)*t');
x2 = 1 + exp(-i*2*pi*(Fs/2)*t');
x = [x1 x2]
x0 = mean(x)
X = fft(x)/N

x =
2 2
0 +1.2246e-016i 0 -1.2246e-016i
2 -2.4493e-016i 2 +2.4493e-016i
0 +3.6739e-016i 0 -3.6739e-016i
2 -4.8986e-016i 2 +4.8986e-016i
0 +6.1232e-016i 0 -6.1232e-016i

x0 = 1 +6.1232e-017i 1 -6.1232e-017i

X =
1 +6.1232e-017i 1 -6.1232e-017i
-3.5353e-017 +6.1232e-017i 3.5353e-017 -6.1232e-017i
-1.0606e-016 +6.1232e-017i 1.0606e-016 -6.1232e-017i
1 -3.0616e-016i 1 +3.0616e-016i
1.0606e-016 +6.1232e-017i -1.0606e-016 -6.1232e-017i
3.5353e-017 +6.1232e-017i -3.5353e-017 -6.1232e-017i

or, after thresholding

x =
2 2
0 0
2 2
0 0
2 2
0 0

x0 = 1 1

X =
1 1
0 0
0 0
1 1
0 0
0 0

Hope this helps.

Greg
From: Steve Amphlett on 28 Apr 2010 04:43
Greg Heath <heath(a)alumni.brown.edu> wrote in message <2f45f49b-2586-4d1b-badf-8b3687382c2c(a)y12g2000vbr.googlegroups.com>...
> On Apr 27, 12:03 pm, "Steve Amphlett" <Firstname.Lastn...(a)Where-I-
> Work.com> wrote:
> > Greg Heath <he...(a)alumni.brown.edu> wrote in message <0bb0996b-f5fc-455e-8101-b4ff3c765...(a)j15g2000vbq.googlegroups.com>...
> > > On Apr 27, 2:50 am, "Steve Amphlett" <Firstname.Lastn...(a)Where-I-
> > > Work.com> wrote:
> > > > Greg Heath <he...(a)alumni.brown.edu> wrote in message <c17b6883-f695-4934-877c-838b7ff5f...(a)n5g2000yqh.googlegroups.com>...
> ----SNIP
>
> > > > > Y = fft(y)/L.
> >
> > > > > b. The normalizing factor L will yield an amplitude of 1
> > > > > for exp(i*2*f0*t) at f = f0.
> >
> > > > ... except at DC and Nyquist, where you need 2/L
> >
> > > No.
> >
> > > >> Fs = 1, N = 6, dt = 1/Fs
> > > t = dt*(0:N-1);
> > > x = 1 + cos(2*pi*(Fs/2)*t);
> > > X = fft(x)/N
> >
> > > Fs = 1
> > > N = 6
> > > dt = 1
> >
> > > X = 1 0 0 1 0 0
> >
> >
> > Is this a recent change?
>
> No.
>
> Too make things more clear:
>
> >> x1 = 1 + exp(i*2*pi*(Fs/2)*t');
> x2 = 1 + exp(-i*2*pi*(Fs/2)*t');
> x = [x1 x2]
> x0 = mean(x)
> X = fft(x)/N
>
> x =
> 2 2
> 0 +1.2246e-016i 0 -1.2246e-016i
> 2 -2.4493e-016i 2 +2.4493e-016i
> 0 +3.6739e-016i 0 -3.6739e-016i
> 2 -4.8986e-016i 2 +4.8986e-016i
> 0 +6.1232e-016i 0 -6.1232e-016i
>
> x0 = 1 +6.1232e-017i 1 -6.1232e-017i
>
> X =
> 1 +6.1232e-017i 1 -6.1232e-017i
> -3.5353e-017 +6.1232e-017i 3.5353e-017 -6.1232e-017i
> -1.0606e-016 +6.1232e-017i 1.0606e-016 -6.1232e-017i
> 1 -3.0616e-016i 1 +3.0616e-016i
> 1.0606e-016 +6.1232e-017i -1.0606e-016 -6.1232e-017i
> 3.5353e-017 +6.1232e-017i -3.5353e-017 -6.1232e-017i
>
> or, after thresholding
>
> x =
> 2 2
> 0 0
> 2 2
> 0 0
> 2 2
> 0 0
>
> x0 = 1 1
>
> X =
> 1 1
> 0 0
> 0 0
> 1 1
> 0 0
> 0 0
>
> Hope this helps.
>
> Greg

My example is:

>> t=(0:1023)/1024;
>> x=1+cos(2*pi*t);
>> fx=fft(x)/length(x);
>> fx(1:2)

ans =

1.0000 0.5000 - 0.0000i


So the first and second bins require different scaling. They should both have the same amplitude. I thought it was because the cosine's amplitude is shared between the positive and negative parts of the spectrum (requiring 2/L scaling), whereas the DC's isn't.
From: Greg Heath on 28 Apr 2010 07:06
On Apr 28, 4:43 am, "Steve Amphlett" <Firstname.Lastn...(a)Where-I-
Work.com> wrote:
> Greg Heath <he...(a)alumni.brown.edu> wrote in message <2f45f49b-2586-4d1b-badf-8b3687382...(a)y12g2000vbr.googlegroups.com>...
> > On Apr 27, 12:03 pm, "Steve Amphlett" <Firstname.Lastn...(a)Where-I-
> > Work.com> wrote:
> > > Greg Heath <he...(a)alumni.brown.edu> wrote in message <0bb0996b-f5fc-455e-8101-b4ff3c765...(a)j15g2000vbq.googlegroups.com>...
> > > > On Apr 27, 2:50 am, "Steve Amphlett" <Firstname.Lastn...(a)Where-I-
> > > > Work.com> wrote:
> > > > > Greg Heath <he...(a)alumni.brown.edu> wrote in message <c17b6883-f695-4934-877c-838b7ff5f...(a)n5g2000yqh.googlegroups.com>...
> > ----SNIP
>
> > > > > > Y = fft(y)/L.
>
> > > > > > b. The normalizing factor L will yield an amplitude of 1
> > > > > > for exp(i*2*f0*t) at f = f0.
>
> > > > > ... except at DC and Nyquist, where you need 2/L
>
> > > > No.
>
> > > > >> Fs = 1, N = 6, dt = 1/Fs
> > > > t = dt*(0:N-1);
> > > > x = 1 + cos(2*pi*(Fs/2)*t);
> > > > X = fft(x)/N
>
> > > > Fs =     1
> > > > N =     6
> > > > dt =     1
>
> > > > X =     1     0     0     1     0     0
>
> > > Is this a recent change?
>
> > No.
>
> > Too make things more clear:
>
> > >> x1  = 1 + exp(i*2*pi*(Fs/2)*t');
> > x2 = 1 + exp(-i*2*pi*(Fs/2)*t');
> > x    = [x1 x2]
> > x0   = mean(x)
> > X    = fft(x)/N
>
> > x =
> >     2                         2
> >     0 +1.2246e-016i           0 -1.2246e-016i
> >     2 -2.4493e-016i           2 +2.4493e-016i
> >     0 +3.6739e-016i           0 -3.6739e-016i
> >     2 -4.8986e-016i           2 +4.8986e-016i
> >     0 +6.1232e-016i           0 -6.1232e-016i
>
> > x0 =  1 +6.1232e-017i    1 -6.1232e-017i
>
> > X =
> >            1 +6.1232e-017i            1 -6.1232e-017i
> > -3.5353e-017 +6.1232e-017i  3.5353e-017 -6.1232e-017i
> > -1.0606e-016 +6.1232e-017i  1.0606e-016 -6.1232e-017i
> >            1 -3.0616e-016i            1 +3.0616e-016i
> >  1.0606e-016 +6.1232e-017i -1.0606e-016 -6.1232e-017i
> >  3.5353e-017 +6.1232e-017i -3.5353e-017 -6.1232e-017i
>
> > or, after thresholding
>
> > x =
> >     2            2
> >     0            0
> >     2            2
> >     0            0
> >     2            2
> >     0            0
>
> > x0 =  1     1
>
> > X =
> >     1             1
> >     0             0
> >     0             0
> >     1             1
> >     0             0
> >     0             0
>
> > Hope this helps.
>
> > Greg
>
> My example is:
>
> >> t=(0:1023)/1024;
> >> x=1+cos(2*pi*t);
> >> fx=fft(x)/length(x);
> >> fx(1:2)
>
> ans =
>
>    1.0000             0.5000 - 0.0000i
>
> So the first and second bins require different scaling.  They should both have the same amplitude.  I thought it was because the cosine's amplitude is shared between the positive and negative parts of the spectrum (requiring 2/L scaling), whereas the DC's
> isn't.

Sorry, I don't get your point. (I hope you are agreeing with me).
Staying with your example (f0 = df = 1):

>> Fs = 1024; N = 1024; dt = 1/Fs; t = dt*(0:N-1)';
df = Fs/N; f0 = df;
x = 1 + 0.5*exp(i*2*pi*f0*t)+ 0.5*exp(-i*2*pi*f0*t);

f = (0:N-1)';
X = fft(x)/N;
tab = [1 2 N/2-1 N/2 N/2+1 N-1 N];
[f(tab) X(tab)]

ans =
0 1
1 0.5 -6.2224e-017i
510 -2.5434e-018 +9.1992e-019i
511 0 +6.3875e-018i
512 0
1022 1.4135e-017 -1.6012e-018i
1023 0.5 +6.2224e-017i

% thresholding:

ans =
0 1
1 0.5
510 0
511 0
512 0
1022 0
1023 0.5

Hope this helps.

Greg
First  |  Prev  | 
Pages: 1 2
Prev: Interaction of MEX and MATLAB or mexCallMATLAB and Co.
Next: the distance between two time series