Matlab Low Pass filter using fft - matlab

I implemented a simple low pass filter in matlab using a forward and backward fft.
It works in principle, but the minimum and maximum values differ from the original.
signal = data;
%% fourier spectrum
% number of elements in fft
NFFT = 1024;
% fft of data
Y = fft(signal,NFFT)/L;
% plot(freq_spectrum)
%% apply filter
fullw = zeros(1, numel(Y));
fullw( 1 : 20 ) = 1;
filteredData = Y.*fullw;
%% invers fft
iY = ifft(filteredData,NFFT);
% amplitude is in abs part
fY = abs(iY);
% use only the length of the original data
fY = fY(1:numel(signal));
filteredSignal = fY * NFFT; % correct maximum
clf; hold on;
plot(signal, 'g-')
plot(filteredSignal ,'b-')
hold off;
the resulting image looks like this
What am I doing wrong? If I normalize both data the filtered signal looks correct.

Just to remind ourselves of how MATLAB stores frequency content for Y = fft(y,N):
Y(1) is the constant offset
Y(2:N/2 + 1) is the set of positive frequencies
Y(N/2 + 2:end) is the set of negative frequencies... (normally we would plot this left of the vertical axis)
In order to make a true low pass filter, we must preserve both the low positive frequencies and the low negative frequencies.
Here's an example of doing this with a multiplicative rectangle filter in the frequency domain, as you've done:
% make our noisy function
t = linspace(1,10,1024);
x = -(t-5).^2 + 2;
y = awgn(x,0.5);
Y = fft(y,1024);
r = 20; % range of frequencies we want to preserve
rectangle = zeros(size(Y));
rectangle(1:r+1) = 1; % preserve low +ve frequencies
y_half = ifft(Y.*rectangle,1024); % +ve low-pass filtered signal
rectangle(end-r+1:end) = 1; % preserve low -ve frequencies
y_rect = ifft(Y.*rectangle,1024); % full low-pass filtered signal
hold on;
plot(t,y,'g--'); plot(t,x,'k','LineWidth',2); plot(t,y_half,'b','LineWidth',2); plot(t,y_rect,'r','LineWidth',2);
legend('noisy signal','true signal','+ve low-pass','full low-pass','Location','southwest')
The full low-pass fitler does a better job but you'll notice that the reconstruction is a bit "wavy". This is because multiplication with a rectangle function in the frequency domain is the same as a convolution with a sinc function in the time domain. Convolution with a sinc fucntion replaces every point with a very uneven weighted average of its neighbours, hence the "wave" effect.
A gaussian filter has nicer low-pass filter properties because the fourier transform of a gaussian is a gaussian. A gaussian decays to zero nicely so it doesn't include far-off neighbours in the weighted average during convolution. Here is an example with a gaussian filter preserving the positive and negative frequencies:
gauss = zeros(size(Y));
sigma = 8; % just a guess for a range of ~20
gauss(1:r+1) = exp(-(1:r+1).^ 2 / (2 * sigma ^ 2)); % +ve frequencies
gauss(end-r+1:end) = fliplr(gauss(2:r+1)); % -ve frequencies
y_gauss = ifft(Y.*gauss,1024);
hold on;
plot(t,x,'k','LineWidth',2); plot(t,y_rect,'r','LineWidth',2); plot(t,y_gauss,'c','LineWidth',2);
legend('true signal','full low-pass','gaussian','Location','southwest')
As you can see, the reconstruction is much better this way.

Related

MATLAB: How to apply ifft correctly to bring a "filtered" signal back to the time doamin?

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.

MATLAB fft vs Mathematical fourier

I am using the following code to generate a fft and mathematical Fourier transform of a signal. I want to then mathematically recreate the original signal of the fft. This works on the mathematical signal but not on the fft since it is a Discrete Transform. Does anyone know what change I can make to my inverse transform equation that will make it work for fft?
clear all; clc;
N = 1024;
N2 = 1023;
SNR = -10;
fs = 1024;
Ts = 1/fs;
t = (0:(N-1))*Ts;
x = 0.5*sawtooth(2*2*pi*t);
x1 = fft(x);
Magnitude1 = abs(x1);
Phase1 = angle(x1)*360/(2*pi);
for m = 1:1024
f(m) = m; % Sinusoidal frequencies
a = (2/N)*sum(x.*cos(2*pi*f(m)*t)); % Cosine coeff.
b = (2/N)*sum(x.*sin(2*pi*f(m)*t)); % Sine coeff
Magnitude(m) = sqrt(a^2 + b^2); % Magnitude spectrum
Phase(m) = -atan2(b,a); % Phase spectrum
end
subplot(2,1,1);
plot(f,Magnitude1./512); % Plot magnitude spectrum
......Labels and title.......
subplot(2,1,2);
plot(f,Magnitude,'k'); % Plot phase spectrum
ylabel('Phase (deg)','FontSize',14);
pause();
x2 = zeros(1,1024); % Waveform vector
for m = 1:24
f(m) = m; % Sinusoidal frequencies
x2 = (1/m)*(x2 + Magnitude1(m)*cos(2*pi*f(m)*t + Phase1(m)));
end
x3 = zeros(1,1024); % Waveform vector
for m = 1:24
f(m) = m; % Sinusoidal frequencies
x3 = (x3 + Magnitude(m)*cos(2*pi*f(m)*t + Phase(m)));
end
plot(t,x,'--k'); hold on;
plot(t,x2,'k');
plot(t,x3,'b');```
There are a few comments about the Fourier Transform, and I hope I can explain everything for you. Also, I don't know what you mean by "Mathematical Fourier transform", as none of the expressions in your code is resembles the Fourier series of the sawtooth wave.
To understand exactly what the fft function does, we can do things step by step.
First, following your code, we create and plot one period of the sawtooth wave.
n = 1024;
fs = 1024;
dt = 1/fs;
t = (0:(n-1))*dt;
x = 0.5*sawtooth(2*pi*t);
figure; plot(t,x); xlabel('t [s]'); ylabel('x');
We can now calculate a few things.
First, the Nyquist frequency, the maximum detectable frequency from the samples.
f_max = 0.5*fs
f_max =
512
Also, the minimum detectable frequency,
f_min = 1/t(end)
f_min =
1.000977517106549
Calculate now the discrete Fourier transform with MATLAB function:
X = fft(x)/n;
This function obtains the complex coefficients of each term of the discrete Fourier transform. Notice it calculates the coefficients using the exp notation, not in terms of sines and cosines. The division by n is to guarantee that the first coefficient is equal to the arithmetic mean of the samples
If you want to plot the magnitude/phase of the transformed signal, you can type:
f = linspace(f_min,f_max,n/2); % frequency vector
a0 = X(1); % constant amplitude
X(1)=[]; % we don't have to plot the first component, as it is the constant amplitude term
XP = X(1:n/2); % we get only the first half of the array, as the second half is the reflection along the y-axis
figure
subplot(2,1,1)
plot(f,abs(XP)); ylabel('Amplitude');
subplot(2,1,2)
plot(f,angle(XP)); ylabel('Phase');
xlabel('Frequency [Hz]')
What does this plot means? It shows in a figure the amplitude and phase of the complex coefficients of the terms in the Fourier series that represent the original signal (the sawtooth wave). You can use this coefficients to obtain the signal approximation in terms of a (truncated) Fourier series. Of course, to do that, we need the whole transform (not only the first half, as it is usual to plot it).
X = fft(x)/n;
amplitude = abs(X);
phase = angle(X);
f = fs*[(0:(n/2)-1)/n (-n/2:-1)/n]; % frequency vector with all components
% we calculate the value of x for each time step
for j=1:n
x_approx(j) = 0;
for k=1:n % summation done using a for
x_approx(j) = x_approx(j)+X(k)*exp(2*pi*1i/n*(j-1)*(k-1));
end
x_approx(j) = x_approx(j);
end
Notice: The code above is for clarification and does not intend to be well coded. The summation can be done in MATLAB in a much better way than using a for loop, and some warnings will pop up in the code, warning the user to preallocate each variable for speed.
The above code calculates the x(ti) for each time ti, using the terms of the truncated Fourier series. If we plot both the original signal and the approximated one, we get:
figure
plot(t,x,t,x_approx)
legend('original signal','signal from fft','location','best')
The original signal and the approximated one are nearly equal. As a matter of fact,
norm(x-x_approx)
ans =
1.997566360514140e-12
Is almost zero, but not exactly zero.
Also, the plot above will issue a warning, due to the use of complex coefficients when calculating the approximated signal:
Warning: Imaginary parts of complex X and/or Y arguments ignored
But you can check that the imaginary term is very close to zero. It is not exactly zero due to roundoff errors in the computations.
norm(imag(x_approx))
ans =
1.402648396024229e-12
Notice in the codes above how to interpret and use the results from the fft function and how they are represented in the exp form, not on terms of sines and cosines, as you coded.

Morlet wavelet transformation function returns nonsensical plot

I have written a matlab function (Version 7.10.0.499 (R2010a)) to evaluate incoming FT signal and calculate the morlet wavelet for the signal. I have a similar program, but I needed to make it more readable and closer to mathematical lingo. The output plot is supposed to be a 2D plot with colour showing the intensity of the frequencies. My plot seems to have all frequencies the same per time. The program does make an fft per row of time for each frequency, so I suppose another way to look at it is that the same line repeats itself per step in my for loop. The issue is I have checked with the original program, which does return the correct plot, and I cannot locate any difference beyond what I named the values and how I organized the code.
function[msg] = mile01_wlt(FT_y, f_mn, f_mx, K, N, F_s)
%{
Fucntion to perform a full wlt of a morlet wavelett.
optimization of the number of frequencies to be included.
FT_y satisfies the FT(x) of 1 envelope and is our ft signal.
f min and max enter into the analysis and are decided from
the f-image for optimal values.
While performing the transformation there are different scalings
on the resulting "intensity".
Plot is made with a 2D array and a colour code for intensity.
version 05.05.2016
%}
%--------------------------------------------------------------%
%{
tableofcontents:
1: determining nr. of analysis f, prints and readies f's to be used.
2: ensuring correct orientation of FT_y
3:defining arrays
4: declaring waveletdiagram and storage of frequencies
5: for-loop over all frequencies:
6: reducing file to manageable size by truncating time.
7: marking plot to highlight ("randproblemer")
8: plotting waveletdiagram
%}
%--------------------------------------------------------------%
%1: determining nr. of analysis f, prints and readies f's to be used.
DF = floor( log(f_mx/f_mn) / log(1+( 1/(8*K) ) ) ) + 1;% f-spectre analysed
nr_f_analysed = DF %output to commandline
f_step = (f_mx/f_mn)^(1/(DF-1)); % multiplicative step for new f_a
f_a = f_mn; %[Hz] frequency of analysis
T = N/F_s; %[s] total time sampled
C = 2.0; % factor to scale Psi
%--------------------------------------------------------------%
%2: ensuring correct orientation of FT_y
siz = size(FT_y);
if (siz(2)>siz(1))
FT_y = transpose(FT_y);
end;
%--------------------------------------------------------------%
%3:defining arrays
t = linspace(0, T*(N-1)/N, N); %[s] timespan
f = linspace(0, F_s*(N-1)/N, N); %[Hz] f-specter
%--------------------------------------------------------------%
%4: declaring waveletdiagram and storage of frequencies
WLd = zeros(DF,N); % matrix of DF rows and N columns for storing our wlt
f_store = zeros(1,DF); % horizontal array for storing DF frequencies
%--------------------------------------------------------------%
%5: for-loop over all frequencies:
for jj = 1:DF
o = (K/f_a)*(K/f_a); %factor sigma
Psi = exp(- 0*(f-f_a).*(f-f_a)); % FT(\psi) for 1 envelope
Psi = Psi - exp(-K*K)*exp(- o*(f.*f)); % correctional element
Psi = C*Psi; %factor. not set in stone
%next step fits 1 row in the WLd (3 alternatives)
%WLd(jj,:) = abs(ifft(Psi.*transpose(FT_y)));
WLd(jj,:) = sqrt(abs(ifft(Psi.*transpose(FT_y))));
%WLd(jj,:) = sqrt(abs(ifft(Psi.*FT_y))); %for different array sizes
%and emphasizes weaker parts.
%prep for next round
f_store (jj) = f_a; % storing used frequencies
f_a = f_a*f_step; % determines the next step
end;
%--------------------------------------------------------------%
%6: reducing file to manageable size by truncating time.
P = floor( (K*F_s) / (24*f_mx) );%24 not set in stone
using_every_P_point = P %printout to cmdline for monitoring
N_P = floor(N/P);
points_in_time = N_P %printout to cmdline for monitoring
% truncating WLd and time
WLd2 = zeros(DF,N_P);
for jj = 1:DF
for ii = 1:N_P
WLd2(jj,ii) = WLd(jj,ii*P);
end
end
t_P = zeros(1,N_P);
for ii = 1:N_P % set outside the initial loop to reduce redundancy
t_P(ii) = t(ii*P);
end
%--------------------------------------------------------------%
%7: marking plot to highlight boundary value problems
maxval = max(WLd2);%setting an intensity
mxv = max(maxval);
% marks in wl matrix
for jj= 1:DF
m = floor( K*F_s / (P*pi*f_store(jj)) ); %finding edges of envelope
WLd2(jj,m) = mxv/2; % lower limit
WLd2(jj,N_P-m) = mxv/2;% upper limit
end
%--------------------------------------------------------------%
%8: plotting waveletdiagram
figure;
imagesc(t_P, log10(f_store), WLd2, 'Ydata', [1 size(WLd2,1)]);
set(gca, 'Ydir', 'normal');
xlabel('Time [s]');
ylabel('log10(frequency [Hz])');
%title('wavelet power spectrum'); % for non-sqrt inensities
title('sqrt(wavelet power spectrum)'); %when calculating using sqrt
colorbar('location', 'southoutside');
msg = 'done.';
There are no error message, so I am uncertain what exactly I am doing wrong.
Hope I followed all the guidelines. Otherwise, I apologize.
edit:
my calling program:
% establishing parameters
N = 2^(16); % | number of points to sample
F_s = 3.2e6; % Hz | samplings frequency
T_t = N/F_s; % s | length in seconds of sample time
f_c = 2.0e5; % Hz | carrying wave frequency
f_m = 8./T_t; % Hz | modulating wave frequency
w_c = 2pif_c; % Hz | angular frequency("omega") of carrying wave
w_m = 2pif_m; % Hz | angular frequency("omega") of modulating wave
% establishing parameter arrays
t = linspace(0, T_t, N);
% function variables
T_h = 2*f_m.*t; % dimless | 1/2 of the period for square signal
% combined carry and modulated wave
% y(t) eq. 1):
y_t = 0.5.*cos(w_c.*t).*(1+cos(w_m.*t));
% y(t) eq. 2):
% y_t = 0.5.*cos(w_c.*t)+0.25*cos((w_c+w_m).*t)+0.25*cos((w_c-w_m).*t);
%square wave
sq_t = cos(w_c.*t).*(1 - mod(floor(t./T_h), 2)); % sq(t)
% the following can be exchanged between sq(t) and y(t)
plot(t, y_t)
% plot(t, sq_t)
xlabel('time [s]');
ylabel('signal amplitude');
title('plot of harmonically modulated signal with carrying wave');
% title('plot of square modulated signal with carrying wave');
figure()
hold on
% Fourier transform and plot of freq-image
FT_y = mile01_fftplot(y_t, N, F_s);
% FT_sq = mile01_fftplot(sq_t, N, F_s);
% Morlet wavelet transform and plot of WLdiagram
%determining K, check t-image
K_h = 57*4; % approximation based on 1/4 of an envelope, harmonious
%determining f min and max, from f-image
f_m = 1.995e5; % minimum frequency. chosen to showcase all relevant f
f_M = 2.005e5; % maximum frequency. chosen to showcase all relevant f
%calling wlt function.
name = 'mile'
msg = mile01_wlt(FT_y, f_m, f_M, K_h, N, F_s)
siz = size(FT_y);
if (siz(2)>siz(1))
FT_y = transpose(FT_y);
end;
name = 'arnt'
msg = arnt_wltransf(FT_y, f_m, f_M, K_h, N, F_s)
The time image has a constant frequency, but the amplitude oscillates resempling a gaussian curve. My code returns a sharply segmented image over time, where each point in time holds only 1 frequency. It should reflect a change in intensity across the spectra over time.
hope that helps and thanks!
I found the error. There is a 0 rather than an o in the first instance of Psi. Thinking I'll maybe rename the value as sig or something. besides this the code works. sorry for the trouble there

Low pass filter implementation Correct or wrong?

I've been working on 2 sensors signals measuring vibrations of a rotating shaft. Since there is residual noise in the signal. I tried to filter it by detrending, zero padding and applying low pass filter. Below I'm attaching the graphs of the signal before and after filtering. There is a huge variation in the signal after filtering that makes me think if I'm really doing it in the right way.
My Matlab code is
X = xlsread(filename,'F:F');
Y = xlsread(filename,'G:G');
%Calculate frequency axis
fs = 1e6 ; % Sampling frequency (Hz)
NFFT = 2^nextpow2(length(X)); % Zero padding to nearest N power 2
df = fs/NFFT;
dt = 1/df;
%Frequency Axis defintion
f = (-(fs-df)/2:df:(fs-df)/2)';
X(2^ceil(log2(length(X))))=0;
Y(2^ceil(log2(length(Y))))=0;
%calculate time axis
T = (dt:dt:(length(X)*dt))';
subplot(2,2,1)
plot(T,X);
xlabel('Time(s)')
ylabel('X amplitude')
title('X signal before filtering')
subplot(2,2,2)
plot(T,Y);
xlabel('Time(s)')
ylabel('Y amplitude')
title('Y signal before filtering')
X = detrend(X,0); % Removing DC Offset
Y = detrend(Y,0); % Removing DC Offset
% Filter parameters:
M = length(X); % signal length
L = M; % filter length
fc = 2*(38000/60); % cutoff frequency
% Design the filter using the window method:
hsupp = (-(L-1)/2:(L-1)/2);
hideal = (2*fc/fs)*sinc(2*fc*hsupp/fs);
h = hamming(L)' .* hideal; % h is our filter
% Zero pad the signal and impulse response:
X(2^ceil(log2(M)))=0;
xzp = X;
hzp = [ h zeros(1,NFFT-L) ];
% Transform the signal and the filter:
X = fft(xzp);
H = fft(hzp)';
X = X .* H;
X = ifft(X);
relrmserrX = norm(imag(X))/norm(X); % checked... this for zero
X = real(X)';
% Zero pad the signal and impulse response:
Y(2^ceil(log2(M)))=0;
xzp = Y;
hzp = [ h zeros(1,NFFT-L) ];
% Transform the signal and the filter:
Y = fft(xzp);
H = fft(hzp)';
Y = Y .* H;
Y = ifft(Y);
relrmserrY = norm(imag(Y))/norm(Y); % check... should be zero
Y = real(Y)';
I plotted the after filtering and as you can see in the picture there is a clear deviation. I want to only filter noise but the signal seem to loose other components and I'm little confused if thats the right way of doing filtering.
Any suggestion, hints or ideas would be helpful.
In the end I want to plot X vs Y to give orbits of the shaft vibration. Please also find below another picture of the unfiltered and filtered orbit. As you can see in the picture there is also change in the orbits from the original one (left image with lot of noise).
P.S.: I don't have DSP tool box
There is no problem with your FFT and IFFT.
You can test your code with a simple sine wave, with a very low frequency. The final output should be (almost) the same sine wave, since you are low-pass filtering it.
You've defined X (and Y) from 0 to some value. However, you've defined H from - (L-1)/2 to some positive value. Mathematically, this is fine, but you are simply taking an fft of H. Matlab thinks this has the same time-scale as X, when you multiply the ffts together!
So, in reality, you've taken the fft of XHfft(delta(t-d)), where d is the resulting time-shift. You can either undo this time-shift in the frequency domain, or the time-domain.

Synthesize PSD in MatLab

I want to recover a time signals from a given power spectral density, assuming a normal distribution of the original signal:
PSD; % [(m/s)^2/Hz] given spectrum
T = 60; % [s] length of original signal
dt = 0.005; % [s] time step of original signal
N = T/dt; % [-] number of samples
NFFT = 2^nextpow2(N); % [-] number of bins for FFT
fs = 1/dt; % [Hz] sampling frequency
ASD = sqrt(PSD); % [(m/s)/sqrt(Hz)] get amplitude spectrum
omega = 2*pi*rand(NFFT/2,1); % [rad] generate phase vector
Z = ASD.*exp(1i*omega); % create complex amplitude vector
Z = [0;Z;flipud(conj(Z))]; % extend to satisfy symmetry
Y = real(ifft(Z)); % inverse FFT
[PSDY,f] = pwelch(Y,[],[],NFFT,fs); % generate PSD from Y to compare
The results show a power spectrum several orders of magnitude lower than the original, but the shape matches very good. I guess there is something wrong with the units or there might be a scaling factor missing. I'm not sure about the units of the time signal after ifft, since the amplitude has [(m/s)/sqrt(Hz)].
I believe there are two problems here.
First, I think that the PSD as you define it (or rather, as you use it) is in the wrong units.
When you define the signal as
Z = ASD.*exp(1i*omega);
then ASD should be in m/s and not in (m/s)/Hz.
So you should do something like that:
ASD = sqrt(PSD*fs/2)
Now, since PSD is in (m/s)^2/Hz, ASD is in units of m/s.
Next, the ifft should be normalised. That is, you should define the Y as
Y = ifft(Z)*sqrt(NFFT);
One more thing, I am not sure if this is on purpose, but the following line
[PSDY,f] = pwelch(Y,[],[],NFFT,fs);
results in Y being divided into 8 parts (with length <NFFT) with 50% overlap. Each part is zero padded to length of NFFT.
A better practice would be to use something like
[PSDY,f] = pwelch(Y,L,L/2,L,fs);
for some L or
[PSDY,f] = pwelch(Y,NFFT,[],NFFT,fs);
if you insist. To find out more, go to http://www.mathworks.com/help/signal/ref/pwelch.html
In conclusion, this is your (modified) code:
PSD = 5; % [(m/s)^2/Hz] given spectrum
T = 60; % [s] length of original signal
dt = 0.005; % [s] time step of original signal
N = T/dt; % [-] number of samples
NFFT = 2^nextpow2(N); % [-] number of bins for FFT
fs = 1/dt; % [Hz] sampling frequency
ASD = sqrt(PSD*fs/2); % [(m/s)] get amplitude spectrum
omega = 2*pi*rand(NFFT/2,1); % [rad] generate phase vector
Z = ASD.*exp(1i*omega); % create complex amplitude vector
Z = [0;Z;flipud(conj(Z))]; % extend to satisfy symmetry
Y = ifft(Z)*sqrt(NFFT); % inverse FFT
[PSDY,f] = pwelch(Y,256,128,256,fs); % generate PSD from Y to compare
which results in
where the blue line is the estimate PSD.