single plot, dft graph?
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Hi folks just asking how you can a single point graph plotted, so that its x against y. the x plot should be 0:120, every 5 seconds. and y plot should be these values: 62,57,56,55,57,56,58,62,62,63,61,61,62,62,62,61,60,60,60,61,65,68,67,66,67 These values represent a reading every 5 seconds
in the format of the link below is what i'm trying to get.
https://www.google.co.uk/search?q=matlab+dft+graph&source=lnms&tbm=isch&sa=X&ved=0ahUKEwjwoJnT77DLAhWCHxoKHQ6GArUQ_AUIBygB&biw=1920&bih=1033#imgrc=Npbkd6ifbaMW3M%3A
hope this makes sense
2 Comments
Star Strider
on 8 Mar 2016
See the R2015a documentation for fft. Pay special attention to the code between the top two figure images. Everything you need is there.
Answers (5)
Star Strider
on 13 Apr 2016
This is how I would code it:
xdat = 0:5:120;
ydat = [62,57,56,55,57,56,58,62,62,63,61,61,62,62,62,61,60,60,60,61,65,68,67,66,67];
Ts = mean(diff(xdat)); % Sampling Interval
Fs = 1/Ts; % Sampling Frequency
Fn = Fs/2; % Nyquist Frequency
L = length(ydat); % Signal Length (Obviously)
ft_y = fft(ydat)/L; % Fourier Transform (Normalised)
Fv = linspace(0, 1, fix(L/2)+1)*Fn; % Frequency Vector
Iv = 1:length(Fv); % Index Vector
figure(1)
plot(Fv, 2*abs(ft_y(Iv)))
grid
xlabel('Frequency (Arbitrary Units)')
ylabel('Amplitude (Arbitrary Units)')
There’s a relatively high d-c offset. You can eliminate that by subtracting the mean of ‘ydat’ before you take the fft. This will make the other frequencies more apparent.
6 Comments
davy hanna
on 10 May 2016
Star Strider
Are they though, the ydat for that example is pulse rates every 5 seconds, during walking. when the fft is plotted, the y-scale is from 0 to a maximum of 5? so am thinking it can't be the ydat of the original form?
Star Strider
on 10 May 2016
The Fourier transform is what it is, and I stand by my previous statements. It may not be appropriate for your analysis, since your data plotted as a function of time demonstrate a certain periodicity with respect to heart rate, with an increasing heart rate over time. This may be an artefact of your experimental conditions (for example, treadmill velocity, if you are studying heart rate over certain fixed periods of specific treadmill velocities). Since I don’t know what you’re doing, I can’t say with any certainty what analysis techniques are appropriate for your data.
If my ‘treadmill’ guess is correct, perhaps you should be regressing heart rate against treadmill velocity instead of doing a Fourier transform.
I have no idea.
Ilham Hardy
on 9 Mar 2016
Perhaps this is what you want,
xdat = 0:5:120;
ydat = [62,57,56,55,57,56,58,62,62,63,61,61,62,62,62,61,60,60,60,61,65,68,67,66,67];
scatter(xdat,ydat,'filled')
set(gca,'xtick',0:5:120)
grid on
davy hanna
on 12 Apr 2016
xdat = 0:5:120;
ydat = [62,57,56,55,57,56,58,62,62,63,61,61,62,62,62,61,60,60,60,61,65,68,67,66,67];
scatter(xdat,ydat,'filled')
set(gca,'xtick',0:5:120)
grid on
any idea on how to an fft on that plot above?
0 Comments
davy hanna
on 13 Apr 2016
Edited: davy hanna
on 13 Apr 2016
xdat= 0:5:120;
ydat=[62,57,56,55,57,56,58,62,62,63,61,61,62,62,62,61,60,60,60,61,65,68,67,66,67];
plot(xdat,ydat);
nfft=length(ydat);
nfft2=2.^nextpow2(nfft);
ffty=fftshift(fft(ydat,nfft2));
r=abs(ffty);
plot(r);
plot(r); ↑ Error: The input character is not valid in MATLAB statements or expressions.
can anyone tell me how/why im going wrong when trying to plot the FFT of this?
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