# single plot, dft graph?

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david hanna on 8 Mar 2016
Commented: Star Strider on 10 May 2016
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.
hope this makes sense
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david hanna on 9 Mar 2016
Star Strider
i need an activation license to see the document you mention. I'm using matlab at my college,i.e. their computers, when i type license into the matlab prompt for the code i get: 347765 when i type this in for access onto the document,it doesnt allow me. any ideas? thanks

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.
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
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david hanna on 14 Mar 2016
llham Hardy
not exactly, but it does do the trick i suppose for my case, and just working through matlab. am having difficulty regarding doing a fourier transform or short-time fourier transform, on this signal. having problems relating to the tutorials compared to my own signal :(

david hanna on 14 Mar 2016
Fs = 1000; T = 1/Fs; L = 1000; t = (0:L-1)*T; x = 0:5:120; y = [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(Fs*t(1:50),y(1:50)) xlabel('time (seconds)')
Index exceeds matrix dimensions.
Getting that index error, maybe i'm not on the right track here for getting the fft even?
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david hanna on 21 Mar 2016
BUMP

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?

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?