I have a problem in my MATLAB program. I'm trying to find a cutoff frequency to create a low pass filter for compass data. I'm trying to go from the time domain to the frequency domain and find an Fc, so I used the FFT but it seems that's it's nor working.
This is what i have done:
dataset=xlsread('data.xlsx','Feuil1','A1:A751');
t=1:length(dataset);
z=abs(fft(dataset));
subplot(2,2,3)
plot(dataset)
title('dataNonFiltrer')
subplot(2,2,4)
plot(z)
title('frequenciel')
And i get this wish seems to be not correct:
You are just not plotting the data right.
To plot the fft of a signal X, do (from the docs):
Fs = 1000; % Sampling frequency of your data. YOU NEED TO KNOW THIS, change
L = length(X); % Length of signal
Y = fft(X);
P2 = abs(Y/L);
P1 = P2(1:L/2+1);
P1(2:end-1) = 2*P1(2:end-1);
f = Fs*(0:(L/2))/L;
plot(f,P1)
title('frequenciel X(t)')
xlabel('f (Hz)')
ylabel('|P1(f)|')
Related
I'm trying to calculate the inverse Fourier transform of some data using Matlab. I start with raw data in the frequency domain and want to visualise the data in the time domain. Here is my MWE:
a = 1.056;
% frequency of data (I cannot change this)
w = linspace(-100,100,1e6);
L = length(w); % no. sample points
ts = L/1000; % time sampling
Ts = ts/L; % sampling rate
Fs = 1/Ts; % sampling freq
t = (-L/2:L/2-1)*ts/L; % time
Y = sqrt(pi/a)*exp(-w.^2/(4*a)); % my data
yn = Fs*ifftshift(ifft(fftshift(Y(end:-1:1)))) % numerical soln
ya = exp(-a*t.^2); % analytic solution
figure; hold on
plot(t,yn,'.')
plot(t,ya,'-')
xlabel('time, t')
legend('numerical','analytic')
xlim([-5,5])
I have adapted the code from this question however the amplitude is too large:
Can you please tell me what I'm doing wrong?
There are three issues with your code:
You define ts = L/1000 and then compute Fs, which gives you 1000. But Fs is given by the w array you've set up: the full range of w is 2*pi*Fs:
Fs = -w(1)/pi; % sampling freq
Ts = 1/Fs; % sampling rate
Or, equivalently, set Fs = mean(diff(w))*L / (2*pi)
w is defined, but does not include 0. Just like you define t carefully to include the 0 in just the right place, so should you define w to include 0 in just the right place. One simple way to do this is to define it with one more value, then delete the last value:
w = linspace(-100,100,1e6+1);
w(end) = [];
If your input data does not include the 0 frequency, you should resample it so that it does. The DFT (FFT) expects a 0 frequency bin.
You're using ifftshift and fftshift reversed: fftshift shifts the origin from the leftmost array element to the middle, and ifftshift shifts it from the middle to the left. You define your signal with the origin in the middle, so you need to use ifftshift on it to move the origin where the fft and ifft functions expect it. Use fftshift on the output of these two functions to center the origin for display. Because your data is even sized, these two functions do exactly the same thing, and you will not notice the difference. But if the data were odd sized, you'd see the difference.
The following code gives a perfect match:
a = 1.056;
% frequency of data (I cannot change this)
w = linspace(-100,100,1e6+1); w(end) = [];
L = length(w); % no. sample points
Fs = -w(1)/pi; % sampling freq
Ts = 1/Fs; % sampling rate
t = (-L/2:L/2-1)*Ts; % time
Y = sqrt(pi/a)*exp(-w.^2/(4*a)); % my data
yn = Fs*fftshift(ifft(ifftshift(Y(end:-1:1)))); % numerical soln
ya = exp(-a*t.^2); % analytic solution
figure; hold on
plot(t,yn,'.')
plot(t,ya,'-')
xlabel('time, t')
legend('numerical','analytic')
xlim([-5,5])
I am trying to implement a basic DFT algorithm on Matlab.
I simply use in phase and quadrature components of a sine wave with phase modulation(increasing frequency a.k.a chirp). I do compare my results with fft command of Matlab. My code gives the same results whenever there is no phase modulation(pure sine). Whenever I add chirp modulation, results differ. For example, when I use a chirp with some bandwidth around a carrier, the expected results should be a frequency distribution of chirp bandwidth starting from carrier frequency. However, I get a copy of that result backwards starting from carrier frequency as well. You can use my code below without modifying anything. Figure 5 is my result and figure 6 is the expected result. Carrier is 256 Hz with a 10Hz bandwidth of chirp. You can see the code below. The important part is for loop where I take dft of my signal. Also uou can see my dft result below.
close all;
clear all;
%% signal generation
t = (0:0.0001:1); % 1 second window
f = 256; %freq of input signal in hertz
bw = 10; % bandwidth sweep of signal
phaseInput = 2*pi*t*bw.*t;
signalInput = sin(2*pi*f*t + phaseInput); %input signal
inphase = sin(2*pi*f*t).*cos(phaseInput); %inphase component
quadrature = cos(2*pi*f*t).*sin(phaseInput); %quadrature component
figure
plot(t,signalInput,'b',t,inphase,'g',t,quadrature,'r');
title('Input Signal');
xlabel('Time in seconds');
ylabel('Amplitude');
%% sampling signal previously generated
Fs = 1024; %sampling freq
Ts = (0:1/Fs:1);%sample times for 1 second window
sPhase = 2*pi*Ts*bw.*Ts;
sI = sin(2*pi*f*Ts).*cos(sPhase);
sQ = cos(2*pi*f*Ts).*sin(sPhase);
hold on;
plot(Ts,sI+sQ,'b*',Ts,sI,'g*',Ts,sQ,'r*');
fftSize = Fs; %Using all samples in dft
sampleIdx = (0:1:fftSize-1)';
sampledI = sI(1:fftSize)';
sampledQ = sQ(1:fftSize)';
figure;
plot(sampleIdx,sampledI,sampleIdx,sampledQ);
title('Sampled IQ Components');
%% DFT Calculation
dftI = zeros(fftSize,1);
dftQ = zeros(fftSize,1);
for w = 0:fftSize-1
%exp(-2*pi*w*t) = cos(2*pi*w*t) - i*sin(2*pi*w*t)
cI = cos(2*pi*w*sampleIdx/fftSize); %correlation cos
cQ = -sin(2*pi*w*sampleIdx/fftSize); %correlation sin
dftI(w+1) = sum(sampledI.*cI - sampledQ.*cQ); %
dftQ(w+1) = sum(sampledI.*cQ + sampledQ.*cI);
end;
figure;
plot(Fs*sampleIdx/fftSize,dftI);
title('DFT Inphase');
xlabel('Hertz');
figure
plot(Fs*sampleIdx/fftSize,dftQ);
title('DFT Quadrature');
xlabel('Hertz');
figure;
plot(Fs*sampleIdx/fftSize,sqrt(dftQ.^2+dftI.^2));
%% For comparison
sampledInput = sin(2*pi*f*Ts + sPhase);
Y = fft(sampledInput(1:1024),1024);
Pyy = Y.*conj(Y)/1024;
f = (0:1023);
figure;
plot(f,Pyy)
title('Power spectral density')
xlabel('Frequency (Hz)')
the reason lies in the fact that two different signals will definitely give your two different frequency spectrums. check out the code below, you will find that the input of the dft algorithm you actually gave is sampledI+jsampledQ. as a result, what you are doing here is NOT merely decomposing your original signal into In-phase and quadrature components, instead, you are doing Hilbert transform here -- to change a real signal into a complex one.
cI = cos(2*pi*w*sampleIdx/fftSize); %correlation cos
cQ = -sin(2*pi*w*sampleIdx/fftSize); %correlation sin
dftI(w+1) = sum(sampledI.*cI - sampledQ.*cQ); %
dftQ(w+1) = sum(sampledI.*cQ + sampledQ.*cI);
so the sampledInput for comparison should be sampledInput = sI+1i*sQ;.
I'm new to Matlab for LTI signal processing and wondering if anyone can help with something that I'm sure is meant to be basic. I've spent hours and hours researching and obtaining background information and still cannot obtain a clear path to tackle these problems. So far, from scratch, I have generated a signal required and managed to use the fft function to produce the signal's DFT:
function x = fourier_rikki(A,t,O)
Fs = 1000;
t = 0:(1/Fs):1;
A = [0.5,0,0.5];
N = (length(A) - 1)/2;
x = zeros(size(t));
f1 = 85;
O1 = 2*pi*f1;
for k = 1:length(A)
x1 = x + A(k)*exp(1i*O1*t*(k-N-1));
end
f2 = 150;
O2 = 2*pi*f2;
for k = 1:length(A);
x2 = x + A(k)*exp(1i*O2*t*(k-N-1));
end
f3 = 330;
O3 = 2*pi*f3;
for k = 1:length(A);
x3 = x + A(k)*exp(1i*O3*t*(k-N-1));
end
signal = x1 + x2 + x3;
figure(1);
subplot(3,1,1);
plot(t, signal);
title('Signal x(t) in the Time Domain');
xlabel('Time (Seconds)');
ylabel('x(t)');
X = fft(signal); %DFT of the signal
subplot(3,1,2);
plot(t, X);
title('Power Spectrum of Discrete Fourier Transform of x(t)');
xlabel('Time (Seconds)');
ylabel('Power');
f = linspace(0, 1000, length(X)); %?
subplot(3,1,3);
plot(f, abs(X)); %Only want the positive values
title('Spectral Frequency');
xlabel('Frequency (Hz)'); ylabel('Power');
end
At this stage, I'm assuming this is correct for:
"Generate a signal with frequencies 85,150,330Hz using a sampling frequency of 1000Hz - plot 1seconds worth of the signal and its Discrete Fourier Transform."
The next step is to "Find the frequency response of an LTI system that filters out the higher and lower frequencies using the Fourier Transform". I'm stuck trying to create an LTI system that does that! I have to be left with the 150Hz signal, and I'm guessing I perform the filtering on the FFT, perhaps using conv.
My course is not a programming course - we are not assessed on our programming skills and I have minimal Matlab experience - basically we have been left to our own devices to struggle through, so any help would be greatly appreciated! I am sifting through tonnes of different examples and searching Matlab functions using 'help' etc, but since each one is different and does not have a break down of the variables used, explaining why certain parameters/values are chosen etc. it is just adding to the confusion.
Among many (many) others I have looked at:
http://www.ee.columbia.edu/~ronw/adst-spring2010/lectures/matlab/lecture1.html
http://gribblelab.org/scicomp/09_Signals_and_sampling.html section 10.4 especially.
As well as Matlab Geeks examples and Mathworks Matlab function explanations.
I guess the worst that can happen is that nobody answers and I continue burning my eyeballs out until I manage to come up with something :) Thanks in advance.
I found this bandpass filter code as a Mathworks example, which is exactly what needs to be applied to my fft signal, but I don't understand the attenuation values Ast or the amount of ripple Ap.
n = 0:159;
x = cos(pi/8*n)+cos(pi/2*n)+sin(3*pi/4*n);
d = fdesign.bandpass('Fst1,Fp1,Fp2,Fst2,Ast1,Ap,Ast2',1/4,3/8,5/8,6/8,60,1,60);
Hd = design(d,'equiripple');
y = filter(Hd,x);
freq = 0:(2*pi)/length(x):pi;
xdft = fft(x);
ydft = fft(y);
plot(freq,abs(xdft(1:length(x)/2+1)));
hold on;
plot(freq,abs(ydft(1:length(x)/2+1)),'r','linewidth',2);
legend('Original Signal','Bandpass Signal');
Here is something you can use as a reference. I think I got the gist of what you were trying to do. Let me know if you have any questions.
clear all
close all
Fs = 1000;
t = 0:(1/Fs):1;
N = length(t);
% 85, 150, and 330 Hz converted to radian frequency
w1 = 2*pi*85;
w2 = 2*pi*150;
w3 = 2*pi*330;
% amplitudes
a1 = 1;
a2 = 1.5;
a3 = .75;
% construct time-domain signals
x1 = a1*cos(w1*t);
x2 = a2*cos(w2*t);
x3 = a3*cos(w3*t);
% superposition of 85, 150, and 330 Hz component signals
x = x1 + x2 + x3;
figure
plot(t(1:100), x(1:100));
title('unfiltered time-domain signal, amplitude vs. time');
ylabel('amplitude');
xlabel('time (seconds)');
% compute discrete Fourier transform of time-domain signal
X = fft(x);
Xmag = 20*log10(abs(X)); % magnitude spectrum
Xphase = 180*unwrap(angle(X))./pi; % phase spectrum (degrees)
w = 2*pi*(0:N-1)./N; % normalized radian frequency
f = w./(2*pi)*Fs; % radian frequency to Hz
k = 1:N; % bin indices
% plot magnitude spectrum
figure
plot(f, Xmag)
title('frequency-domain signal, magnitude vs. frequency');
xlabel('frequency (Hz)');
ylabel('magnitude (dB)');
% frequency vector of the filter. attenuates undesired frequency components
% and keeps desired components.
H = 1e-3*ones(1, length(k));
H(97:223) = 1;
H((end-223):(end-97)) = 1;
% plot magnitude spectrum of signal and filter
figure
plot(k, Xmag)
hold on
plot(k, 20*log10(H), 'r')
title('frequency-domain signal (blue) and filter (red), magnitude vs. bin index');
xlabel('bin index');
ylabel('magnitude (dB)');
% filtering in frequency domain is just multiplication
Y = X.*H;
% plot magnitude spectrum of filtered signal
figure
plot(f, 20*log10(abs(Y)))
title('filtered frequency-domain signal, magnitude vs. frequency');
xlabel('frequency (Hz)');
ylabel('magnitude (dB)');
% use inverse discrete Fourier transform to obtain the filtered time-domain
% signal. This signal is complex due to imperfect symmetry in the
% frequency-domain, however the imaginary components are nearly zero.
y = ifft(Y);
% plot overlay of filtered signal and desired signal
figure
plot(t(1:100), x(1:100), 'r')
hold on
plot(t(1:100), x2(1:100), 'linewidth', 2)
plot(t(1:100), real(y(1:100)), 'g')
title('input signal (red), desired signal (blue), signal extracted via filtering (green)');
ylabel('amplitude');
xlabel('time (seconds)');
Here is the end result...
I am trying to extract a sinusoid which itself has a speed which changes sinusiodially. The form of this is approximately sin (a(sin(b*t))), a+b are constant.
This is what I'm currently trying, however it doesnt give me a nice sin graph as I hope for.
Fs = 100; % Sampling rate of signal
Fc = 2*pi; % Carrier frequency
t = [0:(20*(Fs-1))]'/Fs; % Sampling times
s1 = sin(11*sin(t)); % Channel 1, this generates the signal
x = [s1];
dev = 50; % Frequency deviation in modulated signal
z = fmdemod(x,Fc,Fs,fm); % Demodulate both channels.
plot(z);
Thank you for your help.
There is a bug in your code, instead of:
z = fmdemod(x,Fc,Fs,fm);
You should have:
z = fmdemod(x,Fc,Fs,dev);
Also to see a nice sine graph you need to plot s1.
It looks like you are not creating a FM signal that is modulated correctly, so you can not demodulate it correctly as well using fmdemod. Here is an example that does it correctly:
Fs = 8000; % Sampling rate of signal
Fc = 3000; % Carrier frequency
t = [0:Fs]'/Fs; % Sampling times
s1 = sin(2*pi*300*t)+2*sin(2*pi*600*t); % Channel 1
s2 = sin(2*pi*150*t)+2*sin(2*pi*900*t); % Channel 2
x = [s1,s2]; % Two-channel signal
dev = 50; % Frequency deviation in modulated signal
y = fmmod(x,Fc,Fs,dev); % Modulate both channels.
z = fmdemod(y,Fc,Fs,dev); % Demodulate both channels.
If you find thr answers useful you can both up-vote them and accept them, thanks.
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])