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From: Erwin on 4 Jun 2010 18:18
hi,

i'm generating a power law noise spectrum, F(w), generating the real and imaginary part of a signal.
so, having these, i plug them in a ifft to get the function in time, f(t).
next i want to do the following:

i) test Parseval's theorem: integral [f(t)]^2 = (1/N) * integral [F(w)]^2
ii) plot the time function, f(t).

the way i'm doing the item i) is with the trapz() function, but it's not right.
my imaginary function F(w), is not symmetric, so do i have to put it symmetric?

to plot the time function i plot it with time, but only the real part, because the ifft() gives a complex valued function (shouldn't it be only real?).

so, being x the time function, and Y the complex generated function, here's my code:

x = ifft(Y,length(f));
Power = abs(Y).^2;
areaTime = trapz(t,abs(x).^2);
areaFreq = (1/length(f))*trapz(f,Power);
areaRatio = areaTime/areaFreq; %check Parseval's Theorem

thank you!
From: Matt J on 5 Jun 2010 04:54
"Erwin " <carvas(a)gmail.com> wrote in message <hubu2s$63p$1(a)fred.mathworks.com>...
> hi,
>
> i'm generating a power law noise spectrum, F(w), generating the real and imaginary part of a signal.
> so, having these, i plug them in a ifft to get the function in time, f(t).
> next i want to do the following:
>
> i) test Parseval's theorem: integral [f(t)]^2 = (1/N) * integral [F(w)]^2
===============

Your equation is a confused combination of continuous and discrete space concepts.

If you're trying to test the discrete space version of Parseval, you should have sums in the above expression instead of integrals. Therefore, in that case also, trapz() would not be the relevant tool to use.

If you're trying to test the continuous space version of Parseval, it's questionable why you have a factor of (1/N) in there. Since N is a discrete space parameter, it shouldn't appear in any textbook's expression of continuous space Parseval.

Realize also that

(1) A test of continuous space Parseval will have an accuracy that depends on how finely you discretize the Fourier transform integrals. A test of discrete space Parseval, meanwhile, will be good up to the numerical precision of the sums.

(2) MATLAB's fft function works in Hz rather than radians.
From: Greg Heath on 5 Jun 2010 09:24
On Jun 4, 6:18 pm, "Erwin " <car...(a)gmail.com> wrote:
> hi,
>
> i'm generating a power law noise spectrum, F(w), generating the
real and imaginary part of a signal.
> so, having these, i plug them in a ifft to get the function in time, f(t)..
> next i want to do the following:
>
> i) test Parseval's theorem: integral [f(t)]^2 = (1/N) * integral [F(w)]^2

It's confusing to combine the discrete time length N with otherwise
continuous time notation.

It is more fruitful to work backwards: if

N = length(x);
X = fft(x);

then

sum(abs(x).^2) = sum((abs(X).^2)/N

This is easy to demonstrate with

x = randn(N,1) + i * rand(N,1);

for arbitrary positive integers N.

The continuous time integral connection follows
by defining the discrete time differentials

dt = 1/Fs;
df = Fs/N;

and using the continuous time scaling

Xc = dt*fft(x);

to obtain

dt*sum(abs(x).^2) = df*sum((abs(Xc).^2)

which, of course, does not contain N.

> ii) plot the time function, f(t).
>
> the way i'm doing the item i) is with the trapz() function,

I think trapz weights the first and last components of x by
a factor of 2 so CT/DT comparisons using trapz require
the periodic extension x(N+1) = x(1). Then X(N+1) will
automatically be X(1) and the CT/DT comparisons follow
easily..

Hope this helps.

Greg

From: Erwin on 6 Jun 2010 22:22
hi,
first of all, thank you matt and greg.
i think it would be more helpful for me and for you if i explain what i am trying to do.

i am generating a power law function, which describes noise in an experiment, with an algorithm i saw in a paper, and the main parameter that defines this kind of noise is the slope of the log-log curve, which i called "beta".
(the paper is this: http://adsabs.harvard.edu/abs/1995A&A...300..707T )

so what i will have is a function in the frequency space defined by the following for loop:

beta=0.5; %slope of the log-log power law
a=5; %amplitude

N = 500; %number of samples
Fs = 1/5; %time between samples
f = Fs:Fs:N; %generated frequency space

for i = 1:length(f),
S(i) = (1/f(i))^(beta); %power function S(w)=(1/w)^beta
r(i) = a.*normrnd(0,(1/2)*S(i))*sqrt((1/2)*S(i)); %real part
im(i) = a.*normrnd(0,(1/2)*S(i))*sqrt((1/2)*S(i)); %imaginary part
Y(i) = complex(r(i),im(i)); %complex function which gives power law if Y.^2
end

now i have a function Y(i) that has real and complex values, and plotting abs(Y).^2 versus de frequency it gives me what i want...

my doubt here, is: now how to get the equivalent function in time?!

thanks a lot!

PS: sorry i'm kind of new to MATLAB and to Applied Fourier Transforms
PS2: should i repost this regarding what i posted here?
From: TideMan on 7 Jun 2010 01:51
On Jun 7, 2:22 pm, "Erwin " <car...(a)gmail.com> wrote:
> hi,
> first of all, thank you matt and greg.
> i think it would be more helpful for me and for you if i explain what i am trying to do.
>
> i am generating a power law function, which describes noise in an experiment, with an algorithm i saw in a paper, and the main parameter that defines this kind of noise is the slope of the log-log curve, which i called "beta".
> (the paper is this:http://adsabs.harvard.edu/abs/1995A&A...300..707T)
>
> so what i will have is a function in the frequency space defined by the following for loop:
>
> beta=0.5; %slope of the log-log power law
> a=5; %amplitude
>
> N = 500; %number of samples
> Fs = 1/5;  %time between samples
> f = Fs:Fs:N; %generated frequency space
>
> for i = 1:length(f),
>     S(i) = (1/f(i))^(beta);      %power function S(w)=(1/w)^beta
>     r(i) = a.*normrnd(0,(1/2)*S(i))*sqrt((1/2)*S(i));       %real part
>     im(i) = a.*normrnd(0,(1/2)*S(i))*sqrt((1/2)*S(i));     %imaginary part
>     Y(i) = complex(r(i),im(i));      %complex function which gives power law if Y.^2
> end
>
> now i have a function Y(i) that has real and complex values, and plotting abs(Y).^2 versus de frequency it gives me what i want...
>
> my doubt here, is: now how to get the equivalent function in time?!
>
> thanks a lot!
>
> PS: sorry i'm kind of new to MATLAB and to Applied Fourier Transforms
> PS2: should i repost this regarding what i posted here?

This is NOT the way to do it.
Search on "phase randomisation" in this forum and you'll get an
algorithm that does exactly what you want, but in the proper
(numerically robust) way.
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