Fft graphs exhibit symmetry on the x axis, but not around 0

I'm pretty inexperienced with MatLab and Fourier transform. I'm trying to convert a matrix into a Fourier transform graph using the following code:

Y = fft(ppt1_female_dob_10_10_2020_20min_normal_colon_vid_05_10_2021matrx);

n = length(Y);

pwr = abs(Y(1:floor(n/2))).^2;

freq = (1:n/2)/(n/2) * 0.5;

plot(freq,pwr)

figure,plot(freq*100,pwr)

When I did this, I obtained bimodal data for every figure generated, and they were all symmetrical around an x value not equal to 0. (See attached files). Why might I be getting graphs that look like this?

Answer 1

[Star Strider]

It appears that you are plotting part of a two-sided Fourier transform.

I generally use this approach to plot a one-sided Fourier transform —

t = linspace(0, 20, 250);                           % Time Vector
s = sum(sin([1:5]'*2*pi*t));                        % Signal Vector
Fs = 1/(t(2)-t(1));                                 % Sampling Frequency

Fs = 12.4500

Fn = Fs/2;                                          % Nyquist Frequency

figure
plot(t, s)
grid
xlabel('t')
ylabel('s')

L = numel(t);
NFFT = 2^nextpow2(L);
FTs = fft(s, NFFT)/L;
Fv = linspace(0, 1, NFFT/2+1)*Fn;
Iv = 1:numel(Fv);

figure
plot(Fv, abs(FTs(Iv))*2)
grid
xlabel('Frequency')
ylabel('Amplitude')

There are other approaches as well that work. There is nothing particularly special about this approach, other than it generally works for me.

.

[Sophie S]

Thanks for getting back to me! Excuse my ignorance, but how did you determine the signal vector?

[Star Strider]

My pleasure!

No need to apologise!

I just created it to demonstrate the code. It can be anything.

[Sophie S]

Thanks!! & what exactly does the signal vector represent?

[Star Strider]

It is a vector that is a function of time. Here, it is simply an arbitrary vector designed to demonstrate the code.

[Sophie S]

Understood! Thank you!

[Star Strider]

My pleasure!

If my Answer helped you solve your problem, please Accept it!

.

[Sophie S]

What I am still a bit confused about is how to translate the concept of the signal vector into the context of matrix data. I have a matrix that's 1200x1920 and am trying to determine the frequencies of the signals within each column; how could I go about doing this and plotting the amplitude of these determined frequencies?

[Star Strider]

The fft function operates column-wise, so assuming that they all have a common time vector, one call to the fft function will work for the entire matrix:

t = linspace(0, 12, 1200);                          % Time Vector
fc = linspace(30, 1, 1920).';                       % Frequency Vector
SM = sin(fc*2*pi*t).';                              % Create Signal Vector Matrix

figure
subplot(3,1,1)
plot(t, SM(:,1))
title(sprintf('Column %4d',1))
grid
subplot(3,1,2)
plot(t, SM(:,640))
title(sprintf('Column %4d',640))
grid
subplot(3,1,3)
plot(t, SM(:,1280))
title(sprintf('Column %4d',1280))
grid
xlabel('Time')

Fs = 1/(t(2)-t(1));
Fn = Fs/2;
L = size(SM,1);
NFFT = 2^nextpow2(L);
FT_SM = fft(SM, NFFT)/L;
Fv = linspace(0, 1, NFFT/2+1)*Fn;
Iv = 1:numel(Fv);

figure
subplot(3,1,1)
plot(Fv, abs(FT_SM(Iv,1))*2)
title(sprintf('Column %4d, Freq = %.1f',1,fc(1)))
grid
subplot(3,1,2)
plot(Fv, abs(FT_SM(Iv,640))*2)
title(sprintf('Column %4d, Freq = %.1f',640,fc(640)))
grid
subplot(3,1,3)
plot(Fv, abs(FT_SM(Iv,1280))*2)
title(sprintf('Column %4d, Freq = %.1f',1280,fc(1280)))
grid
xlabel('Frequency')

Plotting them or analysing them would require looping over the columns of ‘FT_SM’ (or whatever it would be named in your code). You would use the findpeaks or similar functions (such as medfreq) to determine the frequencies for each column.

This code is simply an example.

.

[Sophie S]

This is exceedingly helpful; thanks! One additional follow-up question: I attached a matrix as a .mat file as an example. Where would I put this variable name within the code to yield the signal vector for each column?

[Star Strider]

My pleasure!

The provided .mat file is a doouble matrix, not a table, so there are no variable names associated with any of the columns. There is also no time vector or sampling frequency provided.

My previous code would change to:

LD = load('vid00232_05212022_Colon_3cm_matrx.mat');

Mouse2 = LD.vid00232_Mouse2_05212022_Colon_3cm_matrx;

L = size(Mouse2,1);

t = linspace(0, L-1, L); % Create Time Vector

figure

subplot(3,1,1)

plot(t, Mouse2(:,1))

title(sprintf('Column %4d',1))

grid

subplot(3,1,2)

plot(t, Mouse2(:,640))

title(sprintf('Column %4d',640))

grid

subplot(3,1,3)

plot(t, Mouse2(:,1280))

title(sprintf('Column %4d',1280))

grid

xlabel('Time')

Fs = 1/(t(2)-t(1));

Fn = Fs/2;

NFFT = 2^nextpow2(L);

FT_Mouse2 = fft(Mouse2-mean(Mouse2), NFFT)/L;

Fv = linspace(0, 1, NFFT/2+1)*Fn;

Iv = 1:numel(Fv);

[maxpk,maxloc] = max(abs(FT_Mouse2)*2); % Maximum Peak Amplitude % Index For Each Column

PksLocs = table((1:size(Mouse2,2)).',maxpk(:),maxloc(:),Fv(maxloc(:)).', 'VariableNames',{'Column','Amplitude','Index','Frequency'});

figure

subplot(3,1,1)

plot(Fv, abs(FT_Mouse2(Iv,1))*2)

title(sprintf('Column %4d, Freq = %.6f',1,Fv(maxloc(1))))

grid

subplot(3,1,2)

plot(Fv, abs(FT_Mouse2(Iv,640))*2)

title(sprintf('Column %4d, Freq = %.6f',1,Fv(maxloc(640))))

grid

subplot(3,1,3)

plot(Fv, abs(FT_Mouse2(Iv,1280))*2)

title(sprintf('Column %4d, Freq = %.6f',1,Fv(maxloc(1280))))

grid

xlabel('Frequency')

This plots only a sample of the data, since there are too many to plot all of them. It calculates the Fourier transforms for each column and returns a table of the column number, maximum amplitude, row index of the maximum, and the corresponding frequency for each column. You can add the column names as a separate variable if you want. I did not because there were no column names provided.

If a common sampling frequency ‘Fs’ is provided, the time vector would change to:

t = linspace(0, L-1, L)/Fs; % Create Time Vector

This will also automatically change the frequencies for the peaks, and the frequency vector for the plots.

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