MATLAB: I have removed frequency from (3-200)HZ from FFT signal.Now i want to remove noise from last figure which is in time domain so that i can transmit signal.Thanks in advance.

remove noise

clc,close all,clear all

codn=70;

% fc=6e+3;

fs=3600;

bode=10;

code=round(rand(1,codn));

code_len=round(1/bode/(1/fs))

for ii=1:codn

x((ii-1)*code_len+1:code_len*ii)=code(ii);

end

x2 = x-(1/2) % get rid of most of the dc peak

% set up time and frequency arrays

N = length(x);

delt = 1/fs;

delf = fs/N

tvec = (1:N)*delt

fvec = (-N/2:N/2-1)*delf ; % shifted frequency array

figure(1)

plot(tvec,x2(1,:)+0.5)

title('orignal baseband')

xlabel('time');

ylabel('amplitude')

ylim([-1 1.5]);

y = fftshift(fft(x2)/N);

z=abs(y);

figure(2)

plot(fvec,abs(y))

title('FFT')

xlabel('frequency')

ylabel('amplitude')

figure(3)

z=y;

z(abs(fvec)>=3 & abs(fvec)<=200)=0

plot(fvec,abs(z))

xlabel('frequency removed from 3 to 200 HZ');

ylabel('amplitude')

figure(4)

zf=fftshift(z)*N;

zifft=ifft(zf)+0.5

plot(tvec,abs(zifft))

ylim([-1 1.5])

title('recovered signal')

xlabel('time');

ylabel('amplitude')

Best Answer

For your signal a median filter may work fine enough.

clc,close all,clear all
codn=70;
%  fc=6e+3;
fs=3600;
bode=10;
code=round(rand(1,codn));
code_len=round(1/bode/(1/fs));
for ii=1:codn
    x((ii-1)*code_len+1:code_len*ii)=code(ii);
end
% set up time and frequency arrays
N = length(x);
delt = 1/fs;
delf = fs/N;
tvec = (1:N)*delt;
fvec = [0:N-1]*delf ;   % frequency array
subplot(3,1,1)
plot(tvec,x(1,:))
title('orignal baseband')
xlabel('time');
ylabel('amplitude')
ylim([-1 1.5]);
% Filter x
xFFT = fft(x-mean(x)); % In this way you remove all the dc component
xFFT(abs(fvec)>=3 & abs(fvec)<=200) = 0;
x2 = ifft(xFFT,'symmetric');
subplot(3,1,2)
plot(tvec,x2(1,:))
title('Noise Signal')
xlabel('time');
ylabel('amplitude')
ylim([-1 1.5]);
x2 = medfilt1(x2,50); % Filter with median filter
subplot(3,1,3)
plot(tvec,x2(1,:))
title('Filtered Signal with median filter')
xlabel('time');
ylabel('amplitude')
ylim([-1 1.5]);

Related Solutions

MATLAB: FFT of ON OFF signal

Hi imran,

you appear to be pretty close on this. The code below uses ftshift to put f = 0 at the middle of the freq array, and uses a freq array with the same span as yours but shifted appropriately.

Another modoification concerns dc offset in the time domain. The pulse train has an average value of approximately 1/2 (not exactly 1/2 due to the random nature of the pulses) which leads to a large uninteresting spike at f = 0 that dominates the freq domain data. Subtracting 1/2 off of the data before the fft gives a reasonable peak at f = 0.

Also the fft is divided by the number of points N to give the corrrect amplitude in the freq domain.

codn=70;
%  fc=6e+3;
fs=36000;
bode=1000;
code=round(rand(1,codn));
code_len=round(1/bode/(1/fs))
 for ii=1:codn   
     x((ii-1)*code_len+1:code_len*ii)=code(ii);
 end
 
 x2 = x-(1/2);    % get rid of most of the dc peak
% set up time and frequency arrays 
fs = 36000;
N = length(x);
delt = 1/fs;
delf = fs/N;
tvec = (1:N)*delt;
fvec = (-N/2:N/2-1)*delf;    % shifted frequency array
figure(1)
plot(tvec,x2)
ylim([-1 1.5])
y = fftshift(fft(x2)/N);    
figure(2)
plot(fvec,abs(y))

MATLAB: I get the peaks at the wrong frequency and with the wrong amplitude.

hello Anna

look at my fft function at the end of my code

clc
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%





% load signal
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% data
data = importdata('SFC5S_nov25_ST_1_1.txt');
dt = 1e-6; % 1 micro seconds
signal = data(:,1);
samples = length(signal);
Fs = 1/dt; % sampling frequency (Hz)
%% decimate (if needed)
% NB : decim = 1 will do nothing (output = input)
decim = 1;
if decim>1
    signal = decimate(signal,decim);
    Fs = Fs/decim;
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% FFT parameters
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
NFFT = 512;    % 
OVERLAP = 0.95;
% spectrogram dB scale
spectrogram_dB_scale = 80;  % dB range scale (means , the lowest displayed level is XX dB below the max level)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% options 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% if you are dealing with acoustics, you may wish to have A weighted
% spectrums 
% option_w = 0 : linear spectrum (no weighting dB (L) )
% option_w = 1 : A weighted spectrum (dB (A) )
option_w = 0;
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% display 1 : averaged FFT spectrum
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
[freq, sensor_spectrum] = myfft_peak(signal,Fs,NFFT,OVERLAP);
% convert to dB scale (ref = 1)
sensor_spectrum_dB = 20*log10(sensor_spectrum);
% apply A weigthing if needed

if option_w == 1
    pondA_dB = pondA_function(freq);
    sensor_spectrum_dB = sensor_spectrum_dB+pondA_dB;
    my_ylabel = ('Amplitude (dB (A))');
else
    my_ylabel = ('Amplitude (dB (L))');
end
figure(1),plot(freq,sensor_spectrum_dB,'b');grid
title(['Averaged FFT Spectrum  / Fs = ' num2str(Fs) ' Hz / Delta f = ' num2str(freq(2)-freq(1)) ' Hz ']);
xlabel('Frequency (Hz)');ylabel(my_ylabel);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% display 2 : time / frequency analysis : spectrogram demo
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
[sg,fsg,tsg] = specgram(signal,NFFT,Fs,hanning(NFFT),floor(NFFT*OVERLAP));  
% FFT normalisation and conversion amplitude from linear to dB (peak)
sg_dBpeak = 20*log10(abs(sg))+20*log10(2/length(fsg));     % NB : X=fft(x.*hanning(N))*4/N; % hanning only
% apply A weigthing if needed
if option_w == 1
    pondA_dB = pondA_function(fsg);
    sg_dBpeak = sg_dBpeak+(pondA_dB*ones(1,size(sg_dBpeak,2)));
    my_title = ('Spectrogram (dB (A))');
else
    my_title = ('Spectrogram (dB (L))');
end
% saturation of the dB range : 
% saturation_dB = 60;  % dB range scale (means , the lowest displayed level is XX dB below the max level)
min_disp_dB = round(max(max(sg_dBpeak))) - spectrogram_dB_scale;
sg_dBpeak(sg_dBpeak<min_disp_dB) = min_disp_dB;
% plots spectrogram
figure(2);
imagesc(tsg,fsg,sg_dBpeak);colormap('jet');
axis('xy');colorbar('vert');grid
title([my_title ' / Fs = ' num2str(Fs) ' Hz / Delta f = ' num2str(fsg(2)-fsg(1)) ' Hz ']);
xlabel('Time (s)');ylabel('Frequency (Hz)');
function pondA_dB = pondA_function(f)
	% dB (A) weighting curve
	n = ((12200^2*f.^4)./((f.^2+20.6^2).*(f.^2+12200^2).*sqrt(f.^2+107.7^2).*sqrt(f.^2+737.9^2)));
	r = ((12200^2*1000.^4)./((1000.^2+20.6^2).*(1000.^2+12200^2).*sqrt(1000.^2+107.7^2).*sqrt(1000.^2+737.9^2))) * ones(size(f));
	pondA = n./r;
	pondA_dB = 20*log10(pondA(:));
end
function  [freq_vector,fft_spectrum] = myfft_peak(signal, Fs, nfft, Overlap)
% FFT peak spectrum of signal  (example sinus amplitude 1   = 0 dB after fft).
% Linear averaging
%   signal - input signal, 
%   Fs - Sampling frequency (Hz).
%   nfft - FFT window size
%   Overlap - buffer overlap % (between 0 and 0.95)
signal = signal(:);
samples = length(signal);
% fill signal with zeros if its length is lower than nfft
if samples<nfft
    s_tmp = zeros(nfft,1);
    s_tmp((1:samples)) = signal;
    signal = s_tmp;
    samples = nfft;
end
% window : hanning
window = hanning(nfft);
window = window(:);
%    compute fft with overlap 
 offset = fix((1-Overlap)*nfft);
 spectnum = 1+ fix((samples-nfft)/offset); % Number of windows
%     % for info is equivalent to : 
%     noverlap = Overlap*nfft;
%     spectnum = fix((samples-noverlap)/(nfft-noverlap));	% Number of windows
    % main loop
    fft_spectrum = 0;
    for i=1:spectnum
        start = (i-1)*offset;
        sw = signal((1+start):(start+nfft)).*window;
        fft_spectrum = fft_spectrum + (abs(fft(sw))*4/nfft);     % X=fft(x.*hanning(N))*4/N; % hanning only 
    end
    fft_spectrum = fft_spectrum/spectnum; % to do linear averaging scaling
% one sidded fft spectrum  % Select first half 
    if rem(nfft,2)    % nfft odd
        select = (1:(nfft+1)/2)';
    else
        select = (1:nfft/2+1)';
    end
fft_spectrum = fft_spectrum(select);
freq_vector = (select - 1)*Fs/nfft;
end

Best Answer

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MATLAB: FFT of ON OFF signal

MATLAB: I get the peaks at the wrong frequency and with the wrong amplitude.

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