MATLAB Central Newsreader - FFT -- sum( v(t)) /= sum (v(t))

FFT -- sum( v(t)) /= sum (v(t))

Travis Bland — Mon, 02 Nov 2009 23:39:02 -0500

Hi,
I'm taking a course using Matlab, and i'm also doing data analysis for a job. Since i started using Matlab, i thought it might be easier to use than my program written in C++, and i was hoping to check the numbers against eachother.
I have a file data.dat with two columns, time and voltage. The sum v(t)^2 should equal the sum of v(f)^2, correct? I should have the same total power in the frequency domain as the time domain?
Well i can't get this to prove true. Here's what i tried

volt_t = data(:,2)

sum ( v(f)^2 ) --------->
>> sum( (fft(volt_t)) .* (fft(volt_t)) )
ans = 7.6598

sum( ( v(f)/ N )^2) ,where N = length volt_t-------->
>> sum( abs( (fft(volt_t) / N) .* (fft(volt_t) / N) ))
ans = 8.6741e-08

sum ( (v(t))^2)----------->
>> sum(volt_t .* volt_t)
ans = 0.0035

I notice when i read in the file to volt_t, the precision isn't very high. My values are #*e-06 , and matlab gives 0.0001 or 0.0000. Could this be my issue; if so, how do i fix it? If not, any suggestions on what to do?

Thanks,
Travis

Re: FFT -- sum( v(t)) /= sum (v(t))

dpb — Mon, 02 Nov 2009 23:56:20 -0500

Travis Bland wrote:
> Hi,
...
> I notice when i read in the file to volt_t, the precision isn't very
> high. My values are #*e-06 , and matlab gives 0.0001 or 0.0000.
> Could this be my issue; if so, how do i fix it? If not, any
> suggestions on what to do?
>
ML uses default double for internal representation irrespective of what
the command window display format is.

Try

format long

or

format long e

to see full internal precision.

doc format % for more info

As for the fft normalization, w/ correct normalization you should find
all the energy, yes.

--

Re: FFT -- sum( v(t)) /

TideMan — Tue, 03 Nov 2009 00:13:54 -0500

On Nov 3, 12:39 pm, "Travis Bland" <travisblan...@yahoo.com> wrote:
> Hi,
> I'm taking a course using Matlab, and i'm also doing data analysis for a job. Since i started using Matlab, i thought it might be easier to use than my program written in C++, and i was hoping to check the numbers against eachother.
> I have a file data.dat with two columns, time and voltage. The sum v(t)^2 should equal the sum of v(f)^2, correct? I should have the same total power in the frequency domain as the time domain?
> Well i can't get this to prove true. Here's what i tried
>
> volt_t = data(:,2)
>
> sum ( v(f)^2 ) --------->>> sum( (fft(volt_t)) .* (fft(volt_t)) )
>
> ans = 7.6598
>
> sum( ( v(f)/ N )^2) ,where N = length volt_t-------->>> sum( abs( (fft(volt_t) / N) .* (fft(volt_t) / N) ))
>
> ans = 8.6741e-08
>
> sum ( (v(t))^2)----------->>> sum(volt_t .* volt_t)
>
> ans = 0.0035
>
> I notice when i read in the file to volt_t, the precision isn't very high. My values are #*e-06 , and matlab gives 0.0001 or 0.0000. Could this be my issue; if so, how do i fix it? If not, any suggestions on what to do?
>
> Thanks,
> Travis

Try this:
N=10000;
y=randn(y,1);
[var(y) sum(abs(fft(y)/N).^2)]

Re: FFT -- sum( v(t)) /

Travis Bland — Tue, 03 Nov 2009 03:35:04 -0500

TideMan <mulgor@gmail.com> wrote in message <18ca1852-949a-41c5-b2dd-0564cad82341@z4g2000prh.googlegroups.com>...
> On Nov 3, 12:39?pm, "Travis Bland" <travisblan...@yahoo.com> wrote:
> > Hi,
> > ?I'm taking a course using Matlab, and i'm also doing data analysis for a job. Since i started using Matlab, i thought it might be easier to use than my program written in C++, and i was hoping to check the numbers against eachother.
> > I have a file data.dat with two columns, time and voltage. The sum ?v(t)^2 ?should equal the sum of v(f)^2, correct? I should have the same total power in the frequency domain as the time domain?
> > Well i can't get this to prove true. Here's what i tried
> >
> > volt_t = data(:,2)
> >
> > sum ( v(f)^2 ) ?--------->>> sum( ? (fft(volt_t)) ?.* (fft(volt_t)) ? )
> >
> > ans = 7.6598
> >
> > sum( ?( v(f)/ N )^2) ,where N = length volt_t-------->>> sum( ?abs( (fft(volt_t) / N) ?.* (fft(volt_t) / N) ))
> >
> > ans = 8.6741e-08
> >
> > sum ( (v(t))^2)----------->>> sum(volt_t .* volt_t)
> >
> > ans = 0.0035
> >
> > I notice when i read in the file to volt_t, the precision isn't very high. My values are #*e-06 , and matlab gives ? 0.0001 or 0.0000. Could this be my issue; if so, how do i fix it? If not, any suggestions on what to do?
> >
> > ?Thanks,
> > ?Travis
>
> Try this:
> N=10000;
> y=randn(y,1);
> [var(y) sum(abs(fft(y)/N).^2)]
>

Wow, i didn't know you guys would respond so quick!!
Thanks dpb, that outputs the full length.
TideMan
how can i use y to declare y (y=randn(y,1)? It gives an error. I would normally just figure out what you meant to say... but since i'm new to matlab i can't quite figure out what that code is trying to do.
Also, my data.dat file is 40,004 long. I tried N = 40004 instead of length(volt) but it didn't help any.

Thanks for the quick reply!

Re: FFT -- sum( v(t)) /

TideMan — Tue, 03 Nov 2009 06:21:32 -0500

On Nov 3, 4:35 pm, "Travis Bland" <travisblan...@yahoo.com> wrote:
> TideMan <mul...@gmail.com> wrote in message <18ca1852-949a-41c5-b2dd-0564cad82...@z4g2000prh.googlegroups.com>...
> > On Nov 3, 12:39?pm, "Travis Bland" <travisblan...@yahoo.com> wrote:
> > > Hi,
> > > ?I'm taking a course using Matlab, and i'm also doing data analysis for a job. Since i started using Matlab, i thought it might be easier to use than my program written in C++, and i was hoping to check the numbers against eachother.
> > > I have a file data.dat with two columns, time and voltage. The sum ?v(t)^2 ?should equal the sum of v(f)^2, correct? I should have the same total power in the frequency domain as the time domain?
> > > Well i can't get this to prove true. Here's what i tried
>
> > > volt_t = data(:,2)
>
> > > sum ( v(f)^2 ) ?--------->>> sum( ? (fft(volt_t)) ?.* (fft(volt_t)) ? )
>
> > > ans = 7.6598
>
> > > sum( ?( v(f)/ N )^2) ,where N = length volt_t-------->>> sum( ?abs( (fft(volt_t) / N) ?.* (fft(volt_t) / N) ))
>
> > > ans = 8.6741e-08
>
> > > sum ( (v(t))^2)----------->>> sum(volt_t .* volt_t)
>
> > > ans = 0.0035
>
> > > I notice when i read in the file to volt_t, the precision isn't very high. My values are #*e-06 , and matlab gives ? 0.0001 or 0.0000. Could this be my issue; if so, how do i fix it? If not, any suggestions on what to do?
>
> > > ?Thanks,
> > > ?Travis
>
> > Try this:
> > N=10000;
> > y=randn(y,1);
> > [var(y) sum(abs(fft(y)/N).^2)]
>
> Wow, i didn't know you guys would respond so quick!!
> Thanks dpb, that outputs the full length.
> TideMan
> how can i use y to declare y (y=randn(y,1)? It gives an error. I would normally just figure out what you meant to say... but since i'm new to matlab i can't quite figure out what that code is trying to do.
> Also, my data.dat file is 40,004 long. I tried N = 40004 instead of length(volt) but it didn't help any.
>
> Thanks for the quick reply!

OOps
Typo. It should have read:
y=randn(N,1);

Re: FFT -- sum( v(t)) /

Travis Bland — Wed, 04 Nov 2009 16:24:03 -0500

TideMan <mulgor@gmail.com> wrote in message <ab1631de-4e0f-4f32-88ef-7afb8935a075@f1g2000prf.googlegroups.com>...
> On Nov 3, 4:35?pm, "Travis Bland" <travisblan...@yahoo.com> wrote:
> > TideMan <mul...@gmail.com> wrote in message <18ca1852-949a-41c5-b2dd-0564cad82...@z4g2000prh.googlegroups.com>...
> > > On Nov 3, 12:39?pm, "Travis Bland" <travisblan...@yahoo.com> wrote:
> > > > Hi,
> > > > ?I'm taking a course using Matlab, and i'm also doing data analysis for a job. Since i started using Matlab, i thought it might be easier to use than my program written in C++, and i was hoping to check the numbers against eachother.
> > > > I have a file data.dat with two columns, time and voltage. The sum ?v(t)^2 ?should equal the sum of v(f)^2, correct? I should have the same total power in the frequency domain as the time domain?
> > > > Well i can't get this to prove true. Here's what i tried
> >
> > > > volt_t = data(:,2)
> >
> > > > sum ( v(f)^2 ) ?--------->>> sum( ? (fft(volt_t)) ?.* (fft(volt_t)) ? )
> >
> > > > ans = 7.6598
> >
> > > > sum( ?( v(f)/ N )^2) ,where N = length volt_t-------->>> sum( ?abs( (fft(volt_t) / N) ?.* (fft(volt_t) / N) ))
> >
> > > > ans = 8.6741e-08
> >
> > > > sum ( (v(t))^2)----------->>> sum(volt_t .* volt_t)
> >
> > > > ans = 0.0035
> >
> > > > I notice when i read in the file to volt_t, the precision isn't very high. My values are #*e-06 , and matlab gives ? 0.0001 or 0.0000. Could this be my issue; if so, how do i fix it? If not, any suggestions on what to do?
> >
> > > > ?Thanks,
> > > > ?Travis
> >
> > > Try this:
> > > N=10000;
> > > y=randn(y,1);
> > > [var(y) sum(abs(fft(y)/N).^2)]
> >
> > Wow, i didn't know you guys would respond so quick!!
> > Thanks dpb, that outputs the full length.
> > TideMan
> > how can i use y to declare y (y=randn(y,1)? It gives an error. ?I would normally just figure out what you meant to say... but since i'm new to matlab i can't quite figure out what that code is trying to do.
> > Also, my data.dat file is 40,004 long. I tried N = 40004 instead of length(volt) but it didn't help any.
> >
> > Thanks for the quick reply!
>
> OOps
> Typo. It should have read:
> y=randn(N,1);

alright, so that gives me the same value each time. But I'm not quite sure what it's doing. I fed my data into y
N= 40004;
y=data(:,2);
[var(y) sum(abs(fft(y)/N).^2)]

but that doesn't give me the same values.

ans =
1.0e-007 *
0.3966 0.4485

What did i do wrong?

Re: FFT -- sum( v(t)) /

TideMan — Wed, 04 Nov 2009 18:51:31 -0500

On Nov 5, 5:24 am, "Travis Bland" <travisblan...@yahoo.com> wrote:
> TideMan <mul...@gmail.com> wrote in message <ab1631de-4e0f-4f32-88ef-7afb8935a...@f1g2000prf.googlegroups.com>...
> > On Nov 3, 4:35?pm, "Travis Bland" <travisblan...@yahoo.com> wrote:
> > > TideMan <mul...@gmail.com> wrote in message <18ca1852-949a-41c5-b2dd-0564cad82...@z4g2000prh.googlegroups.com>...
> > > > On Nov 3, 12:39?pm, "Travis Bland" <travisblan...@yahoo.com> wrote:
> > > > > Hi,
> > > > > ?I'm taking a course using Matlab, and i'm also doing data analysis for a job. Since i started using Matlab, i thought it might be easier to use than my program written in C++, and i was hoping to check the numbers against eachother.
> > > > > I have a file data.dat with two columns, time and voltage. The sum ?v(t)^2 ?should equal the sum of v(f)^2, correct? I should have the same total power in the frequency domain as the time domain?
> > > > > Well i can't get this to prove true. Here's what i tried
>
> > > > > volt_t = data(:,2)
>
> > > > > sum ( v(f)^2 ) ?--------->>> sum( ? (fft(volt_t)) ?.* (fft(volt_t)) ? )
>
> > > > > ans = 7.6598
>
> > > > > sum( ?( v(f)/ N )^2) ,where N = length volt_t-------->>> sum( ?abs( (fft(volt_t) / N) ?.* (fft(volt_t) / N) ))
>
> > > > > ans = 8.6741e-08
>
> > > > > sum ( (v(t))^2)----------->>> sum(volt_t .* volt_t)
>
> > > > > ans = 0.0035
>
> > > > > I notice when i read in the file to volt_t, the precision isn't very high. My values are #*e-06 , and matlab gives ? 0.0001 or 0.0000. Could this be my issue; if so, how do i fix it? If not, any suggestions on what to do?
>
> > > > > ?Thanks,
> > > > > ?Travis
>
> > > > Try this:
> > > > N=10000;
> > > > y=randn(y,1);
> > > > [var(y) sum(abs(fft(y)/N).^2)]
>
> > > Wow, i didn't know you guys would respond so quick!!
> > > Thanks dpb, that outputs the full length.
> > > TideMan
> > > how can i use y to declare y (y=randn(y,1)? It gives an error. ?I would normally just figure out what you meant to say... but since i'm new to matlab i can't quite figure out what that code is trying to do.
> > > Also, my data.dat file is 40,004 long. I tried N = 40004 instead of length(volt) but it didn't help any.
>
> > > Thanks for the quick reply!
>
> > OOps
> > Typo. It should have read:
> > y=randn(N,1);
>
> alright, so that gives me the same value each time. But I'm not quite sure what it's doing. I fed my data into y
> N= 40004;
> y=data(:,2);
> [var(y) sum(abs(fft(y)/N).^2)]
>
> but that doesn't give me the same values.
>
> ans =
> 1.0e-007 *
> 0.3966 0.4485
>
> What did i do wrong?

It assumes the mean is zero.
Do this first:
y=detrend(y,0);

Re: FFT -- sum( v(t)) /= sum (v(t))

Matt — Wed, 04 Nov 2009 19:14:03 -0500

"Travis Bland" <travisbland88@yahoo.com> wrote in message <hcnqim$l20$1@fred.mathworks.com>...
> Hi,
> I'm taking a course using Matlab, and i'm also doing data analysis for a job. Since i started using Matlab, i thought it might be easier to use than my program written in C++, and i was hoping to check the numbers against eachother.
> I have a file data.dat with two columns, time and voltage. The sum v(t)^2 should equal the sum of v(f)^2, correct? I should have the same total power in the frequency domain as the time domain?
> Well i can't get this to prove true. Here's what i tried
>
> volt_t = data(:,2)
>
> sum ( v(f)^2 ) --------->
> >> sum( (fft(volt_t)) .* (fft(volt_t)) )
> ans = 7.6598
>
> sum( ( v(f)/ N )^2) ,where N = length volt_t-------->
> >> sum( abs( (fft(volt_t) / N) .* (fft(volt_t) / N) ))
> ans = 8.6741e-08

No. To properly normalize the fft to obtain equivalent energies, you must divide by
sqrt(N), not by N. Try the following for any N:

N=100;
a=rand(1,N);

norm(a),
norm(fft(a)/sqrt(N)),

Re: FFT -- sum( v(t)) /

Greg Heath — Thu, 05 Nov 2009 01:12:09 -0500

On Nov 2, 6:39 pm, "Travis Bland" <travisblan...@yahoo.com> wrote:
> Hi,
> I'm taking a course using Matlab, and i'm also doing data analysis for a job. Since i started using Matlab, i thought it might be easier to use than my program written in C++, and i was hoping to check the numbers against eachother.
> I have a file data.dat with two columns, time and voltage. The sum v(t)^2 should equal the sum of v(f)^2, correct? I should have the same total power in the frequency domain as the time domain?
> Well i can't get this to prove true. Here's what i tried
>
> volt_t = data(:,2)
>
> sum ( v(f)^2 ) --------->>> sum( (fft(volt_t)) .* (fft(volt_t)) )
>
> ans = 7.6598
>
> sum( ( v(f)/ N )^2) ,where N = length volt_t-------->>> sum( abs( (fft(volt_t) / N) .* (fft(volt_t) / N) ))
>
> ans = 8.6741e-08
>
> sum ( (v(t))^2)----------->>> sum(volt_t .* volt_t)
>
> ans = 0.0035
>
> I notice when i read in the file to volt_t, the precision isn't very high. My values are #*e-06 , and matlab gives 0.0001 or 0.0000. Could this be my issue; if so, how do i fix it? If not, any suggestions on what to do?
>
> Thanks,
> Travis

Take a look at

http://groups.google.com/group/comp.soft-sys.matlab/msg/2222327db2ea7f51?hl=en

Hope this helps.

Greg

Re: FFT -- sum( v(t)) /= sum (v(t))

Travis Bland — Thu, 05 Nov 2009 03:05:19 -0500

Matt " <xys@whatever.com> wrote in message
> No. To properly normalize the fft to obtain equivalent energies, you must divide by
> sqrt(N), not by N. Try the following for any N:
>
>
> N=100;
> a=rand(1,N);
>
> norm(a),
> norm(fft(a)/sqrt(N)),

THANKYOU MATT!! This is what i was looking for. I wondered if it just wasn't normalized correctly.
I had no clue what was going on in the other methods, i don't understand the need for all of the functions... i thought this was a simple thing... sum v(t) = sum v(f)

But interestingly enough, i wrote my own function to sum the two arrays, and when i used this code... i got the two values to be the same... and i normalized using N. There must be something going on internally somewhere that i don't understand.

load std000.txt
volt = std000(:,2);
v_s = 0;
for n = 1:40004
v_s = (volt(n).* volt(n)) + v_s ; %sum of v(t)
end

v_f = (fft(volt));
v_f_s = 0;

for b = 1:40004

v_f_s = v_f_s + (abs((v_f(b) .* v_f(b))/ 40004));
end
v_s %sum of v(t)
v_f_s % sum of v(f)

but thanks to everyone