How do I use a for-loop to do fft and plot the power?

5 visualizaciones (últimos 30 días)
Emma
Emma el 16 de Dic. de 2014
Comentada: Emma el 17 de Dic. de 2014
Hello,
I have a data set (time and variable Be) and I want to do a fft on one part of the data at a time (1000 years). To do this, I would like to use a for-loop that plots the power of the period for each of the different ffts. How do I do this? I'm new to Matlab but I have tried the code below:
hold on
for ts=(0:1000:5000)'; % I have data between 0 and 5000 years ago.
A=fft(Be); % Do FFT on the Be data.
A(1)=[];
n=length(A);
power=abs(A(1:floor(n/2))).^2;
nyquist=1/2;
freq=(1:n/2)/(n/2)*nyquist;
period=1./freq;
plot(period,power);
end
When I do this, I get only one plot. What am I doing wrong?
Thanks!

Respuesta aceptada

Thorsten
Thorsten el 17 de Dic. de 2014
Editada: Thorsten el 17 de Dic. de 2014
May be that's because they are the same? I get different curves using random input.
BTW: note that n is always 1000 in your case, so you can compute "period" outside the loop; further, as you have an offset of 350 now, a could start from 0; just be(0) is wrong syntax in Matlab because matrix indices start with 1.
figure;
be_i_dt = rand(1, 20000);
hAxes = gca;
hold( hAxes, 'on' )
offset = 305;
n = 1000;
n=length(A);
nyquist=1/2;
freq=(1:n/2)/(n/2)*nyquist;
period=1./freq;
for a = 1:9
A = fft(be_i_dt(offset+a*n:offset+(a+1)*n));
A(1)=[];
power=abs(A(1:floor(n/2))).^2;
plot(hAxes,period,power)
end
  3 comentarios
Thorsten
Thorsten el 17 de Dic. de 2014
Yes, you probably have to few data points. The last value
offset+(a+1)*n
for the highest a (9 in your case) in your for loop must always be lower or equal to
numel(be_i_dt)
Emma
Emma el 17 de Dic. de 2014
Thank you!

Iniciar sesión para comentar.

Más respuestas (2)

David Young
David Young el 16 de Dic. de 2014
Editada: David Young el 16 de Dic. de 2014
The variable Be is not changed between iterations, so A will also be the same each time, so each plot will be the same as the previous one. If Be is a vector with all 5000 data points, then you could replace the first line in the loop with
A = fft(Be(ts+1:ts+999));
to select 1000 data points starting from ts.
  4 comentarios
Emma
Emma el 17 de Dic. de 2014
Thanks for the tips!
There is only one plot. I did plot the ffts for each period separately (without the loop) and they look different, so the 5 plots should not look the same.
I also tried to add:
thisfig = figure();
to try to get 5 different plots, but then I get the same one as before plus an empty figure.
How come it has 6 iterations?
Thorsten
Thorsten el 17 de Dic. de 2014
Editada: Thorsten el 17 de Dic. de 2014
Could you please show your present code, and provide the Be matrix so that we can try it and see what went wrong?

Iniciar sesión para comentar.


Sudharsana Iyengar
Sudharsana Iyengar el 17 de Dic. de 2014
HI,
Try this and let me know.
for a= 0:4
A=fft(Be(a*1000:(a+1)*1000));
A(1)=[];
n=length(A);
power=abs(A(1:floor(n/2))).^2;
nyquist=1/2;
freq=(1:n/2)/(n/2)*nyquist;
period=1./freq;
figure()
plot(period,power)
end
  1 comentario
Emma
Emma el 17 de Dic. de 2014
Hi!
Thanks! I tried it but got an error message (Subscript indices must either be real positive integers or log) so I changed a to
a=1:5
and then it plots 5 different figures (no error message=good) but it's the same plot in each figure. So the plot doesn't change.
I have changed some things (the period I'm lookin at is now from year 305 to 9305 before present, be is now called be_i_dt) and the code looks like this now:
for a= 1:9
A=fft(be_i_dt(305+a*1000:305+(a+1)*1000));
A(1)=[];
n=length(A);
power=abs(A(1:floor(n/2))).^2;
nyquist=1/2;
freq=(1:n/2)/(n/2)*nyquist;
period=1./freq;
figure()
plot(period,power)
end
And I still get 9 figures that look the same.

Iniciar sesión para comentar.

Categorías

Más información sobre Fourier Analysis and Filtering en Help Center y File Exchange.

Etiquetas

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by