MATLAB: How to calculate Amplitude and phase out of FFT transform of the signal

amplitudefft

Hi All ! How are you people ? I have a question in matlab: I tried to write a code that finds an Amplitude and Phase_shift of the Periodic signal(cosine wave) BUT it doesn't give the correct answer(instead of Amplitude=10,i am getting answer Amplitude=5)!!! Could someone to see the code and to tell me what i'am doing the wrong way ?

%THE CODE function ask_portal_fft per=24;%number of periods w=2*pi*per;% points=1000; t=linspace(0,1,points); a=10;%the amplitude so the answer of Amplitude suppose to be 10 y=a*cos(w*t-1);%Phase shift suppose to be 1 %plot(y)%for debug %————-

Y=fft(y); N=length(Y); Psd=Y.*conj(Y); mx=find(Psd==max(Psd),1); Amplitude=sqrt(Psd(mx)); Amplitude=Amplitude/N;%=4.99,why not 10 ?

Phase=angle(Y(mx));%suppose to be 1 end

Best Answer

This reminds me of the same issure here:

http://www.mathworks.co.uk/matlabcentral/answers/25479-sdof-frf-fft-magnitude-discrepancy

Basically your maximum amplitude should be at 24Hz, right? Here is the Fourier transform of your signal at 24Hz.

>> ff = 24;
>> G_ff = quad(@(t)10*cos(2*pi*ff*t-1).*exp(-1i*2*pi*ff*t),0,1)
G_ff =
       2.7015 -     4.2074i
>> sqrt(G_ff*conj(G_ff))
ans =
            5

It's 5 !!!

Related Solutions

MATLAB: Specter of a square signal(10Khz) at 10MHz frequency

hello

there is a much faster method to generate a square signal (using square )

see demo below for signal generation and fft analysis

NB your sampling freq is 100 kHz in the code, not 10 MHz as stated in the post

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%



% load signal
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% dummy data
Fs = 100e3;     % sampling freq
f0 = 10e3;     % signal freq
duration = 1;   % signal duration (s)
samples = Fs*duration;
t = (0:samples-1)*1/Fs;
% signal = square(2*pi*f0*t); % neg / pos symetrical square wave ( -1 / +1)
signal = 0.5*(square(2*pi*f0*t)+1); % 0 / pos  square wave ( 0 / +1)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% FFT parameters
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
NFFT = 10000;    % 
Overlap = 0.75;
w = hanning(NFFT);     % Hanning window / Use the HANN function to get a Hanning window which has the first and last zero-weighted samples.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% display : averaged FFT spectrum 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
[freq,fft_spectrum] = myfft_peak(signal, Fs, NFFT, Overlap);
sensor_spectrum_dB = 20*log10(fft_spectrum);% convert to dB scale (ref = 1)
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(' dB')
text(locs+.02,pks,num2str(freq(locs)))
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

MATLAB: What is the relation between DFT and PSD of a signal

It can be shown that PSD can alternatively be estimated by the square of magnitude of FFT. You should be able to find it in most spectral analysis text books, e.g., Kay's Modern Spectral Estimation, equation 4.2, pp 65

Best Answer

Related Solutions

MATLAB: Specter of a square signal(10Khz) at 10MHz frequency

MATLAB: What is the relation between DFT and PSD of a signal

Related Question