# Documentation

### This is machine translation

Translated by
Mouseover text to see original. Click the button below to return to the English verison of the page.

Note: This page has been translated by MathWorks. Please click here
To view all translated materals including this page, select Japan from the country navigator on the bottom of this page.

## Measuring Signal Similarities

This example shows how to measure signal similarities. It will help you answer questions such as: How do I compare signals with different lengths or different sampling rates? How do I find if there is a signal or just noise in a measurement? Are two signals related? How to measure a delay between two signals (and how do I align them)? How do I compare the frequency content of two signals? Similarities can also be found in different sections of a signal to determine if a signal is periodic.

### Comparing Signals with Different Sampling Rates

Consider a database of audio signals and a pattern matching application where you need to identify a song as it is playing. Data is commonly stored at a low sampling rate to occupy less memory.

% Load data
load relatedsig.mat

figure
ax(1) = subplot(3,1,1);
plot((0:numel(T1)-1)/Fs1,T1,'k')
ylabel('Template 1')
grid on
ax(2) = subplot(3,1,2);
plot((0:numel(T2)-1)/Fs2,T2,'r')
ylabel('Template 2')
grid on
ax(3) = subplot(3,1,3);
plot((0:numel(S)-1)/Fs,S)
ylabel('Signal')
grid on
xlabel('Time (secs)')
linkaxes(ax(1:3),'x')
axis([0 1.61 -4 4])

The first and the second subplot show the template signals from the database. The third subplot shows the signal which we want to search for in our database. Just by looking at the time series, the signal does not seem to match to any of the two templates. A closer inspection reveals that the signals actually have different lengths and sampling rates.

[Fs1 Fs2 Fs]
ans =

4096        4096        8192

Different lengths prevent you from calculating the difference between two signals but this can easily be remedied by extracting the common part of signals. Furthermore, it is not always necessary to equalize lengths. Cross-correlation can be performed between signals with different lengths, but it is essential to ensure that they have identical sampling rates. The safest way to do this is to resample the signal with a lower sampling rate. The resample function applies an anti-aliasing(low-pass) FIR filter to the signal during the resampling process.

[P1,Q1] = rat(Fs/Fs1);          % Rational fraction approximation
[P2,Q2] = rat(Fs/Fs2);          % Rational fraction approximation
T1 = resample(T1,P1,Q1);        % Change sampling rate by rational factor
T2 = resample(T2,P2,Q2);        % Change sampling rate by rational factor

### Finding a Signal in a Measurement

We can now cross-correlate signal S to templates T1 and T2 with the xcorr function to determine if there is a match.

[C1,lag1] = xcorr(T1,S);
[C2,lag2] = xcorr(T2,S);

figure
ax(1) = subplot(2,1,1);
plot(lag1/Fs,C1,'k')
ylabel('Amplitude')
grid on
title('Cross-correlation between Template 1 and Signal')
ax(2) = subplot(2,1,2);
plot(lag2/Fs,C2,'r')
ylabel('Amplitude')
grid on
title('Cross-correlation between Template 2 and Signal')
xlabel('Time(secs)')
axis(ax(1:2),[-1.5 1.5 -700 700 ])

The first subplot indicates that the signal and template 1 are less correlated while the high peak in the second subplot indicates that signal is present in the second template.

[~,I] = max(abs(C2));
SampleDiff = lag2(I)
SampleDiff = 499
timeDiff = SampleDiff/Fs
timeDiff = 0.0609

The peak of the cross correlation implies that the signal is present in template T2 starting after 61 ms. In other words, signal T2 leads signal S by 499 samples as indicated by SampleDiff. This information can be used to align the signals.

### Measuring Delay Between Signals and Aligning Them

Consider a situation where you are collecting data from different sensors, recording vibrations caused by cars on both sides of a bridge. When you analyze the signals, you may need to align them. Assume you have 3 sensors working at same sampling rates and they are measuring signals caused by the same event.

figure,
ax(1) = subplot(3,1,1);
plot(s1)
ylabel('s1')
grid on
ax(2) = subplot(3,1,2);
plot(s2,'k')
ylabel('s2')
grid on
ax(3) = subplot(3,1,3);
plot(s3,'r')
ylabel('s3')
grid on
xlabel('Samples')
linkaxes(ax,'xy')

We can also use the finddelay function to find the delay between two signals.

t21 = finddelay(s1,s2)
t21 = -350
t31 = finddelay(s1,s3)
t31 = 150

t21 indicates that s2 lags s1 by 350 samples, and t31 indicates that s3 leads s1 by 150 samples. This information can now used to align the 3 signals by time shifting the signals. We can also use the alignsignals function directly to align the signals which aligns two signals by delaying the earliest signal.

s1 = alignsignals(s1,s3);
s2 = alignsignals(s2,s3);

figure
ax(1) = subplot(3,1,1);
plot(s1)
grid on
title('s1')
axis tight
ax(2) = subplot(3,1,2);
plot(s2)
grid on
title('s2')
axis tight
ax(3) = subplot(3,1,3);
plot(s3)
grid on
title('s3')
axis tight
linkaxes(ax,'xy')

### Comparing the Frequency Content of Signals

A power spectrum displays the power present in each frequency. Spectral coherence identifies frequency-domain correlation between signals. Coherence values tending towards 0 indicate that the corresponding frequency components are uncorrelated while values tending towards 1 indicate that the corresponding frequency components are correlated. Consider two signals and their respective power spectra.

Fs = FsSig;         % Sampling Rate

[P1,f1] = periodogram(sig1,[],[],Fs,'power');
[P2,f2] = periodogram(sig2,[],[],Fs,'power');

figure
t = (0:numel(sig1)-1)/Fs;
subplot(2,2,1)
plot(t,sig1,'k')
ylabel('s1')
grid on
title('Time Series')
subplot(2,2,3)
plot(t,sig2)
ylabel('s2')
grid on
xlabel('Time (secs)')
subplot(2,2,2)
plot(f1,P1,'k')
ylabel('P1')
grid on
axis tight
title('Power Spectrum')
subplot(2,2,4)
plot(f2,P2)
ylabel('P2')
grid on
axis tight
xlabel('Frequency (Hz)')

The mscohere function calculates the spectral coherence between the two signals. It confirms that sig1 and sig2 have two correlated components around 35 Hz and 165 Hz. In frequencies where spectral coherence is high, the relative phase between the correlated components can be estimated with the cross-spectrum phase.

[Cxy,f] = mscohere(sig1,sig2,[],[],[],Fs);
Pxy     = cpsd(sig1,sig2,[],[],[],Fs);
phase   = -angle(Pxy)/pi*180;
[pks,locs] = findpeaks(Cxy,'MinPeakHeight',0.75);

figure
subplot(2,1,1)
plot(f,Cxy)
title('Coherence Estimate')
grid on
hgca = gca;
hgca.XTick = f(locs);
hgca.YTick = 0.75;
axis([0 200 0 1])
subplot(2,1,2)
plot(f,phase)
title('Cross-spectrum Phase (deg)')
grid on
hgca = gca;
hgca.XTick = f(locs);
hgca.YTick = round(phase(locs));
xlabel('Frequency (Hz)')
axis([0 200 -180 180])

The phase lag between the 35 Hz components is close to -90 degrees, and the phase lag between the 165 Hz components is close to -60 degrees.

### Finding Periodicities in a Signal

Consider a set of temperature measurements in an office building during the winter season. Measurements were taken every 30 minutes for about 16.5 weeks.

% Load Temperature Data
load officetemp.mat
Fs = 1/(60*30);                 % Sample rate is 1 sample every 30 minutes
days = (0:length(temp)-1)/(Fs*60*60*24);

figure
plot(days,temp)
title('Temperature Data')
xlabel('Time (days)')
ylabel('Temperature (Fahrenheit)')
grid on

With the temperatures in the low 70s, you need to remove the mean to analyze small fluctuations in the signal. The xcov function removes the mean of the signal before computing the cross-correlation. It returns the cross-covariance. Limit the maximum lag to 50% of the signal to get a good estimate of the cross-covariance.

maxlags = numel(temp)*0.5;
[xc,lag] = xcov(temp,maxlags);

[~,df] = findpeaks(xc,'MinPeakDistance',5*2*24);
[~,mf] = findpeaks(xc);

figure
plot(lag/(2*24),xc,'k',...
lag(df)/(2*24),xc(df),'kv','MarkerFaceColor','r')
grid on
xlim([-15 15])
xlabel('Time (days)')
title('Auto-covariance')

Observe dominant and minor fluctuations in the auto-covariance. Dominant and minor peaks appear equidistant. To verify if they are, compute and plot the difference between the locations of subsequent peaks.

cycle1 = diff(df)/(2*24);
cycle2 = diff(mf)/(2*24);

subplot(2,1,1)
plot(cycle1)
ylabel('Days')
grid on
title('Dominant peak distance')
subplot(2,1,2)
plot(cycle2,'r')
ylabel('Days')
grid on
title('Minor peak distance')

mean(cycle1)
ans = 7
mean(cycle2)
ans = 1.0000

The minor peaks indicate 7 cycles/week and the dominant peaks indicate 1 cycle per week. This makes sense given that the data comes from a temperature-controlled building on a 7 day calendar. The first 7-day cycle indicates that there is a weekly cyclic behavior of the building temperature where temperatures lower during the weekends and go back to normal during the week days. The 1-day cycle behavior indicates that there is also a daily cyclic behavior - temperatures lower during the night and increase during the day.

Was this topic helpful?

Download ebook