I'm trying to apply a bandpass around freq 0 without luck. I'd be happy to receive some help please
x=scan11(1,:)*1e-3/3e8; y=scan11(2,:);
plot(x,y) % my function
[XX,ff]=trans_fourier(y,mean(diff(x)));
plot(ff,abs(XX)) % gives the Fourier transform
I want to choose the freq around 0. let's suppose -1e13 till 1e13 and than to make ifft and to plot the signal after this filer.
How should I start doing this? the command
YY=bandpass(y,[-1e13 1e13],1/mean(diff(x)))
didn't help here unfortunately.
Since, i can't upload here files, here is also my question on matlab forum with all the files
matlab link
I am not sure what the contents of the trans_fourier function exactly are, but in 'plain matlab functions', you could attempt something along the lines of the following.
Nt = 1024; % Number of samples
Fs = 10; % Sampling frequency (samples / second)
t = (0:Nt-1)/Fs; % Time array
x = sin(t/10); % Low-frequency signal
x = x + 0.25*randn(1,Nt); % add some noise
X = fftshift(fft(x)); % FFT of signal with 0 Hz centered
fr = (-Nt/2 : Nt/2-1)/(Nt/Fs); % Frequency axis
% Filter: squared cosine (edit as desired)
fsl = 10; % Length of filter slope, in samples
filt = zeros(size(X));
filt(Nt/2+1+(-fsl:fsl)) = cos( linspace(-pi/2,pi/2,2*fsl+1) ).^2;
x_filt = real(ifft(ifftshift( filt.*X ))); % Filtered x
figure();
subplot(2,2,1); plot(t,x); ax=axis; title('original signal');
subplot(2,2,4); plot(t,x_filt); axis(ax); title('Low-pass filtered signal');
subplot(2,2,2); plot(fr,abs(X)); ax=axis; title('original amplitude spectrum');
subplot(2,2,3); plot(fr,abs(X).*filt); axis(ax); title('Filtered amplitude spectrum');
I am not sure what exactly you meant when you said
let's suppose -1e13 till 1e13
, but keep in mind that extracting a single fourier component (essentially setting all values of the spectrum to zero, except the one you are interested in) acts as a brick-wall filter, and you will get considerable artefacts if you take the inverse transform. Refer to this topic or this page if you're interested.
Related
I have a matlab code to Frequency modulation and demodulation a signal. My code is work well for modulation part. My message signal is m and modulated signal is u, code plot the message signal and its integral in one graph for plotting 1.
Then signal modulated with carrier and program plots the modulated signal in time domain for plotting 2.
After that, by the help of some code blocks program find the frequency spectrum of modulated signal and message signal to plot graph of them for plotting 3.
In demodulation part program make some fundamental calculation for FM detection, then to obtain message signal it uses filter.
Last part program plots the graph of recovered signal with message signal to compare them.
I summarized all code because ı do not know whre is the problem.
My problem about plotting 3 when I make zoom graph 3 I see some phase foldings or like it.Graph is not symmetric according to y-axis.
I did not solve this problem, I research about them and I decided to use unwrap(). Although I tried a lot, I could not be successful. How can I get rid of this phase folding with unwrap() function. Thank you.
My matlab code is ;
ts = 0.0001;% Sampling interval
t0 = 0.15; % Duration
t = 0:ts:t0;% define time vector
%% OTHER PARAMETERS
fc = 200; % Carrier signal frequency
kf =50; % Frequency deviation constant
fs = 1/ts; % Sampling frequency
%% MESSAGE SIGNAL SIMPLY
m = 1*(t<t0/3)-2*(t<2*t0/3).*(t>=t0/3);
%% Integration of m
int_m(1) = 0;
for k =1:length(m)-1
int_m(k+1) = int_m(k) + m(k)*ts;
end
%% PLOTTING 1
figure; subplot(211); % Message signal
plot(t,m);grid on;xlabel('time');ylabel('Amplitude');
title('m(t)');ylim([-3 2]);
subplot(212);plot(t,int_m);% Integral of message signal
grid on; xlabel('time');ylabel('Amplitude');title('integral of m(t)');
ylim([-0.07 0.07]);
%% FM MODULATED SIGNAL
u = cos(2*pi*fc*t + 2*pi*kf*int_m);
%% PLOTTING 2
figure; plot(t,u); % Modulated signal in time domain
grid on;xlabel('time');
ylabel('Amplitude');title('FM :u(t)');
ylim([-1.2 1.2]);
%% FINDING FREQUENCY SPECTRUM AND PLOTTING 3
% Frequency spectrum of m(t)
f=linspace(-1/(2*ts),1/(2*ts),length(t));
M=fftshift(fft(m))./length(t); % Taking fourier transform for m(t)
U=fftshift(fft(u))./length(t); % Taking fourier transform for u(t)
figure;subplot(211); % Frequence spectrum of m(t)
plot(f,abs(M)); grid;
xlabel('Frequency in Hz');xlim([-500 500]);
ylabel('Amplitude');title('Double sided Magnitude spectrum of m(t)');
subplot(212);plot(f,abs(U)); % Frequency spectrum of u(t)
grid;xlabel('Frequency in Hz');xlim([-500 500]);
ylabel('Amplitude');title('Double sided Magnitude spectrum of u(t)');
%% DEMODULATION (Using Differentiator)
dem = diff(u);
dem = [0,dem];
rect_dem = abs(dem);
%% Filtering out High Frequencies
N = 80; % Order of Filter
Wn = 1.e-2; % Pass Band Edge Frequency.
a = fir1(N,Wn);% Return Numerator of Low Pass FIR filter
b = 1; % Denominator of Low Pass FIR Filter
rec = filter(a,b,rect_dem);
%% Finding frequency Response of Signals
fl = length(t);
fl = 2^ceil(log2(fl));
f = (-fl/2:fl/2-1)/(fl*1.e-4);
mF = fftshift(fft(m,fl)); % Frequency Response of Message Signal
fmF = fftshift(fft(u,fl)); % Frequency Response of FM Signal
rect_demF = fftshift(fft(rect_dem,fl));% Frequency Response of Rectified FM Signal
recF = fftshift(fft(rec,fl)); % Frequency Response of Recovered Message Signal
%% PLOTTING 4
figure;subplot(211);plot(t,m);grid on;
xlabel('time');ylabel('Amplitude');
title('m(t)');ylim([-3 2]);
subplot(212);plot(t,rec);
title('Recovered Signal');xlabel('{\it t} (sec)');
ylabel('m(t)');grid;
My problem is in this third graph to show well I put big picture
k = -(length(X)-1)/2:1:length(X)/2;
Your k is not symmetric.
If you work with symmetric k is it working fine?
I am trying to get the output of a Gaussian pulse going through a coax cable. I made a vector that represents a coax cable; I got attenuation and phase delay information online and used Euler's equation to create a complex array.
I FFTed my Gaussian vector and convoluted it with my cable. The issue is, I can't figure out how to properly iFFT the convolution. I read about iFFt in MathWorks and looked at other people's questions. Someone had a similar problem and in the answers, someone suggested to remove n = 2^nextpow2(L) and FFT over length(t) instead. I was able to get more reasonable plot from that and it made sense to why that is the case. I am confused about whether or not I should be using the symmetry option in iFFt. It is making a big difference in my plots. The main reason I added the symmetry it is because I was getting complex numbers in the iFFTed convolution (timeHF). I would truly appreciate some help, thanks!
clc, clear
Fs = 14E12; %1 sample per pico seconds
tlim = 4000E-12;
t = -tlim:1/Fs:tlim; %in pico seconds
ag = 0.5; %peak of guassian
bg = 0; %peak location
wg = 50E-12; %FWHM
x = ag.*exp(-4 .* log(2) .* (t-bg).^2 / (wg).^2); %Gauss. in terms of FWHM
Ly = x;
L = length(t);
%n = 2^nextpow2(L); %test output in time domain with and without as suggested online
fNum = fft(Ly,L);
frange = Fs/L*(0:(L/2)); %half of the spectrum
fNumMag = abs(fNum/L); %divide by n to normalize
% COAX modulation ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
%phase data
mu = 4*pi*1E-7;
sigma_a = 2.9*1E7;
sigma_b = 5.8*1E6;
a = 0.42E-3;
b = 1.75E-3;
er = 1.508;
vf = 0.66;
c = 3E8;
l = 1;
Lso = sqrt(mu) /(4*pi^3/2) * (1/(sqrt(sigma_a)*a) + 1/(b*sqrt(sigma_b)));
Lo = mu/(2*pi) * log(b/a);
%to = l/(vf*c);
to = 12E-9; %measured
phase = -pi*to*(frange + 1/2 * Lso/Lo * sqrt(frange));
%attenuation Data
k1 = 0.34190;
k2 = 0.00377;
len = 1;
mldb = (k1 .* sqrt(frange) + k2 .* frange) ./ 100 .* len ./1E6;
mldb1 = mldb ./ 0.3048; %original eqaution is in inch
tfMag = 10.^(mldb1./-10);
% combine to make in complex form
tfC = [];
for ii = 1: L/2 + 1
tfC(ii) = tfMag(ii) * (cosd(phase(ii)) + 1j*sind(phase(ii)));
end
%END ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
%convolute both h and signal
fNum = fNum(1:L/2+1);
convHF = tfC.*fNum;
convHFMag = abs(convHF/L);
timeHF = ifft(convHF, length(t), 'symmetric'); %this is the part im confused about
% Ignore,
% tfC(numel(fNum)) = 0;
% convHF = tfC.*fNum;
% convHFMag = abs(convHF/n);
% timeHF = ifft(convHF);
%% plotting
% subplot(2, 2, 1);
% plot(t, Ly)
% title('Gaussian input');
% xlabel('time in seconds')
% ylabel('V')
% grid
subplot(2, 2, 1)
plot(frange, abs(tfC(1: L/2 + 1)));
set(gca, 'Xscale', 'log')
title('coax cable model')
xlabel('Hz')
ylabel('|H(s)|V/V')
grid
ylim([0 1.1])
subplot(2, 2, 2);
plot(frange, convHFMag(1:L/2+1), '.-', frange, fNumMag(1:L/2+1)) %make both range and function the same lenght
title('The input signal Vs its convolution with coax');
xlabel('Hz')
ylabel('V')
legend('Convolution','Lorentzian in frequecuency domain');
xlim([0, 5E10])
grid
subplot(2, 2, [3, 4]);
plot(t, Ly, t, timeHF)
% plot(t, real(timeHF(1:length(t)))) %make both range and function the same lenght
legend('Input', 'Output')
title('Signal at the output')
xlabel('time in seconds')
ylabel('V')
grid
It's important to understand deeply the principles of the FFT to use it correctly.
When you apply Fourier transform to a real signal, the coefficients at negative frequencies are the conjugate of the ones at positive frequencies. When you apply FFT to a real numerical signal, you can show mathematically that the conjugates of the coefficients that should be at negative frequencies (-f) will now appear at (Fsampling-f) where Fsampling=1/dt is the sampling frequency and dt the sampling period. This behavior is called aliasing and is present when you apply fft to a discrete time signal and the sampling period should be chosen small enaough for those two spectra not to overlap Shannon criteria.
When you want to apply a frequency filter to a signal, we say that we keep the first half of the spectrum because the high frequencies (>Fsampling/2) are due to aliasing and are not characteristics of the original signal. To do so, we put zeros on the second half of the spectra before multiplying by the filter. However, by doing so you also lose half of the amplitude of the original signal that you will not recover with ifft. The option 'symmetric' enable to recover it by adding in high frequencis (>Fsampling/2) the conjugate of the coefficients at lower ones (<Fsampling/2).
I simplified the code to explain briefly what's happening and implemented for you at line 20 a hand-made symmetrisation. Note that I reduced the sampling period from one to 100 picoseconds for the spectrum to display correctly:
close all
clc, clear
Fs = 14E10; %1 sample per pico seconds % CHANGED to 100ps
tlim = 4000E-12;
t = -tlim:1/Fs:tlim; %in pico seconds
ag = 0.5; %peak of guassian
bg = 0; %peak location
wg = 50E-12; %FWHM
NT = length(t);
x_i = ag.*exp(-4 .* log(2) .* (t-bg).^2 / (wg).^2); %Gauss. in terms of FWHM
fftx_i = fft(x_i);
f = 1/(2*tlim)*(0:NT-1);
fftx_r = fftx_i;
fftx_r(floor(NT/2):end) = 0; % The removal of high frequencies due to aliasing leads to losing half the amplitude
% HER YOU APPLY FILTER
x_r1 = ifft(fftx_r); % without symmetrisation (half the amplitude lost)
x_r2 = ifft(fftx_r, 'symmetric'); % with symmetrisation
x_r3 = ifft(fftx_r+[0, conj(fftx_r(end:-1:2))]); % hand-made symmetrisation
figure();
subplot(211)
hold on
plot(t, x_i, 'r')
plot(t, x_r2, 'r-+')
plot(t, x_r3, 'r-o')
plot(t, x_r1, 'k--')
hold off
legend('Initial', 'Matlab sym', 'Hand made sym', 'No sym')
title('Time signals')
xlabel('time in seconds')
ylabel('V')
grid
subplot(212)
hold on
plot(f, abs(fft(x_i)), 'r')
plot(f, abs(fft(x_r2)), 'r-+')
plot(f, abs(fft(x_r3)), 'r-o')
plot(f, abs(fft(x_r1)), 'k--')
hold off
legend('Initial', 'Matlab sym', 'Hand made sym', 'No sym')
title('Power spectra')
xlabel('frequency in hertz')
ylabel('V')
grid
Plots the result:
Do not hesitate if you have further questions. Good luck!
---------- EDIT ----------
The amplitude of discrete Fourier transform is not the same as the continuous one. If you are interested in showing signal in frequency domain, you will need to apply a normalization based on the convention you have chosen. In general, you use the convention that the amplitude of the Fourier transform of a Dirac delta function has amplitude one everywhere.
A numerical Dirac delta function has an amplitude of one at an index and zeros elsewhere and leads to a power spectrum equal to one everywhere. However in your case, the time axis has sample period dt, the integral over time of a numerical Dirac in that case is not 1 but dt. You must normalize your frequency domain signal by multiplying it by a factor dt (=1picoseceond in your case) to respect the convention. You can also note that this makes the frequency domain signal homogeneous to [unit of the original multiplied by a time] which is the correct unit of a Fourier transform.
Here I am plotting a spectrogram twice, once using imagesc and once using the spectrogram automatic plotting. I don't know why I get different scaling results, probably some filtering by the automatic plotting function but I would like to know what exactly and how to transform it so that it matches.
fs = 44100;
% Frequency sweep signal
sw = logspace(log10(500),log10(5000),fs*5);
x = 0.95 * sin(cumsum((2*pi*sw)/fs));
N = 128;
win_size = N;
noverlap = N/2;
win = window(#blackman,win_size);
[s,f,t] = spectrogram(x,win,noverlap,N,fs,'yaxis');
%% IMAGESC --- FIGURE 1
figure(1)
imagesc(t,f/1000,20*log(abs(s)));
title('Spectrogram');
set(gca,'Ydir','Normal');
xlabel('Time (secs)');
ylabel('Frequency (kHz)');
hcb=colorbar;
title(hcb,'Spectral Magnitude (dB)');
%% test with automatic spectrogram plot ---FIGURE2
figure(2)
spectrogram(x,win,noverlap,N,fs,'yaxis');
end
Help is appreciated :)
Change your code "imagesc(t,f/1000,20*log(abs(s)))" as imagesc(t,f/1000,20*log10(abs(s))).
log() is natural logarithm.
When you calculate dB scale, you should use log10(), not log().
In the book "Computational Fourier Optics, A Matlab Tutorial" by David Voelz, it is written that a call to fftshift is needed before a call to fft or ifft, but in the MATLAB documentation of fftshift it's only written that this command
rearranges the outputs of fft, fft2, and fftn by moving the zero-frequency component to the center of the array.
There is no mention in documentation that this command should be called before the call to fft, and I saw some examples that call fft without a call to fftshift beforehand.
My question is: Whether or not fftshift needed to be called before a call to fft or ifft?
If fftshift doesn't need to be called before a call to fft, and only after a call to fft, then when should we use (if at all) the command ifftshift with relation to a calculation of the fft of a data set?
The matlab fft only computes half of the frequency spectrum (positive frequencies and the zero frequency if your number of samples is odd) in order to save computation time.
Then, the second half of the frequency spectrum (which is the complex conjugate of the first half) is just added at the end of this vector.
So what you get after a fft is the following vector:
0 1 2 3 ... Fs/2 -Fs/2 ... -3 -2 -1
<----------------> <------------------>
positive freq negative freq
where Fs is the frequency sample.
Now, what fftshift does is just shifting the negative frequency bins (2nd part of the frequency spectrum) at the beginning of your vector so that you can display a nice frequency spectrum starting at -Fs/2 and ending at +Fs/2.
The shifted vector becomes:
-Fs/2 ... -3 -2 -1 0 1 2 3 ... Fs/2
<------------------> <---------------->
negative freq positive freq
So, to answer your question, no you don't need to use fftshift after or before a call to fft or ifft. But if you used a fftshift on your vector, you should undo it by applying an ifftshift or fftshift. (I think both calls are equivalent.)
If you read onwards in the documentation on fftshift:
"It is useful for visualizing a Fourier transform with the zero-frequency component in the middle of the spectrum."
The shift is not necessary to perform the fft, but it is handy to visualise the Fourier transform. Whether to use fftshift or not is thus dependent upon whether you want to visualise your transform or not.
Note that ifftshift is different than fftshift because it shifts the negative back to the positive. Assume a simple 3 bin unit impulse in frequency domain before fftshift
[0, exp(-jw1), exp(-(jw1-pi)=exp(-jw1+pi)];
fftshift gives
[exp(-jw1+pi)], 0, exp(-jw1)];
you can see the discontinuity in phase. If you perform ifftshift, negative freq is shifted back to positive:
[0, exp(-jw1), exp(-jw1+pi)];
whereas, fftshift again gives:
[exp(-jw1), exp(-jw1+pi), 0];
one can see the phase monotonicity is not the same, (aka, fftshift-ifftshift case gives [decrease, increase] and fftshift-fftshift case gives [increase, decrease] in phase).
Given the title of the book you are studying, I assume you are working with images. Then the proper way is to call both fftshift and ifftshift before and after you call [i]fft.
Your code should look like
spectrum = fftshift(fft2(ifftshift(myimage))
It is quite the same when applying the inverse Fourier transform
myimage = fftshift(ifft2(ifftshift(spectrum))
The simple answer is call to fftshift not needed before a call to fft. fftshift does not influence the calculation of Fast fourier transform. fftshift rearranges values inside a matrix. For eg:
cameraman = imread('cameraman.tif');
fftshifted_cameraman = fftshift(cameraman);
subplot(1,2,1); imshow(cameraman); title('Cameraman');
subplot(1,2,2); imshow(fftshifted_cameraman); title('FFTShifted Cameraman');
It depends on what you are going to do with the transformed data. If you don't perform an fftshift before transforming, the fft result will have every other value multiplied by -1. This doesn't matter if you plan to view the magnitude or magnitude squared of the result. However, if you plan to compare adjacent spectral values and phase is important, you'll need to apply fftshift before transforming to avoid the alternating sign.
Here is a MATLAB script I use to test basic sound analysis functions of MATLAB including fftshift() in displaying the output of fft().
if ~exist('inputFile', 'var')
inputFile = 'vibe.wav';
end
[inputBuffer, Fs] = audioread(inputFile);
fileSize = length(inputBuffer);
numSamples = 2.^(ceil(log2(fileSize))); % round up to nearest power of 2
x = zeros(numSamples, 1); % zero pad if necessary
x(1:fileSize) = inputBuffer(:,1); % if multi-channel, use left channel only
clear inputBuffer; % free this memory
clear fileSize;
t = linspace(0, (numSamples-1)/Fs, numSamples)';
f = linspace(-Fs/2, Fs/2 - Fs/numSamples, numSamples)';
X = fft(x);
plot(t, x);
xlabel('time (seconds)');
ylabel('amplitude');
title(['time-domain plot of ' inputFile]);
sound(x, Fs); % play the sound
pause;
% display both positive and negative frequency spectrum
plot(f, real(fftshift(X)));
xlabel('frequency (Hz)');
ylabel('real part');
title(['real part frequency-domain plot of ' inputFile]);
pause;
plot(f, imag(fftshift(X)));
xlabel('frequency (Hz)');
ylabel('imag part');
title(['imag part frequency-domain plot of ' inputFile]);
pause;
plot(f, abs(fftshift(X))); % linear amplitude by linear freq plot
xlabel('frequency (Hz)');
ylabel('amplitude');
title(['abs frequency-domain plot of ' inputFile]);
pause;
plot(f, 20*log10(abs(fftshift(X))+1.0e-10)); % dB by linear freq plot
xlabel('frequency (Hz)');
ylabel('amplitude (dB)');
title(['dB frequency-domain plot of ' inputFile]);
pause;
% display only positive frequency spectrum for log frequency scale
semilogx(f(numSamples/2+2:numSamples), 20*log10(abs(X(2:numSamples/2)))); % dB by log freq plot
xlabel('frequency (Hz), log scale');
ylabel('amplitude (dB)');
title(['dB vs. log freq, frequency-domain plot of ' inputFile]);
pause;
semilogx(f(numSamples/2+2:numSamples), (180/pi)*angle(X(2:numSamples/2))); % phase by log freq plot
xlabel('frequency (Hz), log scale');
ylabel('phase (degrees)');
title(['phase vs. log freq, frequency-domain plot of ' inputFile]);
pause;
%
% this is an alternate method of unwrapping phase
%
% phase = cumsum([angle(X(1)); angle( X(2:numSamples/2) ./ X(1:numSamples/2-1) ) ]);
% semilogx(f(numSamples/2+2:numSamples), phase(2:numSamples/2)); % unwrapped phase by log freq plot
%
semilogx(f(numSamples/2+2:numSamples), unwrap(angle(X(2:numSamples/2)))); % unwrapped phase by log freq plot
xlabel('frequency (Hz), log scale');
ylabel('unwrapped phase (radians)');
title(['unwrapped phase vs. log freq, frequency-domain plot of ' inputFile]);
If you were windowing segments of audio and passing them to the FFT, then you should use fftshift() on the input to the FFT to define the center of the windowed segment as the t=0 point.
[x_input, Fs] = audioread('vibe.wav'); % load time-domain input
N = 2*floor(length(x_input)/2); % make sure N is even
x = x_input(1:N);
t = linspace(-N/2, N/2-1, N); % values of time in units of samples
omega = linspace(-pi, pi*(1-2/N), N); % values of (normalized) angular frequency
X = fftshift( fft( fftshift( x.*hamming(length(x)) ) ) );
[X_max k_max] = max( abs(X) );
figure(1);
plot(t, x, 'g');
figure(2);
plot(omega, abs(X), 'b');
hold on;
plot(omega(k_max), X_max, 'or');
hold off;
I had created a 3 three different frequency signal and filter out the signal i don't want. But when i using ifft in matlab, it shows a wrong graph.How to transform my frequency domain spectrum back into my 3 second time domain graph? Below my code is as below:
clc
clear all
Fs = 8192;
T = 1/Fs;
%create tones with different frequency
t=0:T:1;
t2=1:T:2;
t3=2:T:3;
y1 = sin(2*pi*220*t);
y2 = sin(2*pi*300*t2);
y3 = sin(2*pi*440*t3);
at=y1+y2+y3;
figure;
plot(t,y1,t2,y2,t3,y3),title('Tones with noise');
[b,a]=butter(2,[2*290/Fs,2*350/Fs],'stop');
e=filter(b,a,at);
et=(ifft(abs(e)));
figure,
plot(et)
As it is now, et is in the frequency domain, because of the fft. You don't need to fft. just plot(e) and you'll get the time domain filtered waveform. Yo can check the filter performance in the freq. domain by fft though, just
plot(abs(fftshift(fft(fftshift(e)))));
xlim([4000 5000])
Edit:
Your code as it is written on the question has the following bug: at has exactly 1 second of info in it (or 8192 elements). If you plot(at) you'll see the sum of frequencies alright, but they all happen in the same time. This is how to fix it:
clear all
Fs = 8192; % or multiply by 3 if needed
T = 1/Fs;
%create tones with different frequency
t=0:T:3;
y1 = sin(2*pi*220*t).*(t<1);
y2 = sin(2*pi*300*t).*(t<2 & t>=1);
y3 = sin(2*pi*440*t).*(t>=2);
at=y1+y2+y3;
[b,a]=butter(2,[2*290/Fs,2*350/Fs],'stop');
e=filter(b,a,at);
figure,
plot(t,e)
dt=t(2)-t(1);
N=length(at);
df=1/(N*dt); % the frequency resolution (df=1/max_T)
if mod(N,2)==0
f_vector= df*((1:N)-1-N/2); % frequency vector for EVEN length vectors: f =[-f_max,-f_max+df,...,0,...,f_max-df]
else
f_vector= df*((1:N)-0.5-N/2); % frequency vector for ODD length vectors f =[-f_max,-f_max+fw,...,0,...,f_max]
end
freq_vec=f_vector;
fft_vec=fftshift(fft(e));
plot(freq_vec,abs(fft_vec))
xlim([0 1000])