How do I calculate the amplitude ratio and phase lag for two sinusoidal signals in MATLAB?

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How do I calculate the amplitude ratio and phase lag for two sinusoidal signals in MATLAB?

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MathWorks Support Team
MathWorks Support Team el 30 de Nov. de 2018
The amplitude ratio and phase lag for two sinusoidal signals can be determined by taking the Fourier Transform and comparing the magnitude and phase of the signals.
For example:
% clear
clear all
% set the lag
lag = pi/2;
% amplitude for sinusodial 1
amp_x = 2;
% amplitude for sinusodial 2
amp_y = 4;
% Sampling frequency
Fs = 1024;
% Time vector of 0.5 second
t = 0:1/Fs:0.5*(Fs-1)/Fs;
% number of points
npts = length(t);
% Create a sine wave of 20 Hz.
x = amp_x * (sin(2*pi*t*20) + ...
0.25*randn(1,npts)) + 5;
% Create a sine wave of 20 Hz
% with a phase of pi/2.
y = amp_y * (sin(2*pi*t*20 + lag)...
+ 0.25*randn(1,npts));
% remove bias
x = x - mean(x);
y = y - mean(y);
% plot the signal
figure(1)
plot(t,x,t,y);
xlabel('Time (s)');
ylabel('Amplitude');
legend('Signal x(t):sin(2*pi*t*20)',...
'Signal y(t):sin(2*pi*t*20+lag)');
% take the FFT
X=fft(x);
Y=fft(y);
% Calculate the numberof unique points
NumUniquePts = ceil((npts+1)/2);
figure(2)
subplot(211);
f = (0:NumUniquePts-1)*Fs/npts;
plot(f,abs(X(1:NumUniquePts)));
title('X(f) : Magnitude response');
ylabel('|X(f)|')
subplot(212)
plot(f,abs(Y(1:NumUniquePts)));
title('Y(f) : Magnitude response')
xlabel('Frequency (Hz)');
ylabel('|Y(f)|')
figure(3)
subplot(211)
plot(f,angle(X(1:NumUniquePts)));
title('X(f) : Phase response');
ylabel('Phase (rad)');
subplot(212)
plot(f,angle(Y(1:NumUniquePts)));
title('Y(f) : Phase response');
xlabel('Frequency (Hz)');
ylabel('Phase (rad)');
% Determine the max value and max point.
% This is where the sinusoidal
% is located. See Figure 2.
[mag_x idx_x] = max(abs(X));
[mag_y idx_y] = max(abs(Y));
% determine the phase difference
% at the maximum point.
px = angle(X(idx_x));
py = angle(Y(idx_y));
phase_lag = py - px
% determine the amplitude scaling
amplitude_ratio = mag_y/mag_x
For additional background information, see
  1 comentario
Sandip Kumar
Sandip Kumar el 14 de Feb. de 2014
Sorry for the delay in response. This is done to ensure 512 samples of time and not 513 samples.
Fs = 1024;
t1 = 0:1/Fs:0.5;
t2 = 0:1/Fs:0.5*(Fs-1)/Fs;
Now t1 is 513 samples (with last sample as 0.5) whereas t2 is 512 samples.

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