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
2018-11-30
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 个评论
Sandip Kumar
2014-2-14
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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