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How to Compare Two Groups of Signals?

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Rightia Rollmann
Rightia Rollmann on 27 Mar 2017
Commented: the cyclist on 27 Mar 2017
Imagine I have two groups of signals A and B having four and three samples, respectively.
A1 = [1 3 -1 2];
A2 = [1 3 -2 3];
A3 = [1 4 -3 4];
A4 = [2 4 -2 3];
B1 = [2 3 -1 5];
B2 = [2 4 -2 6];
B3 = [2 4 -4 5];
How can I find out whether there is a statistically significant difference between groups A and B?

Answers (2)

the cyclist
the cyclist on 27 Mar 2017
Edited: the cyclist on 27 Mar 2017
To determine whether these signals are significantly different, you first need to define a null hypothesis that will define the range of typical differences. This generally involves an underlying theoretical model, such as "these samples are all drawn from the a normal distribution with the same parameters".
What is your null hypothesis? If your answer is, "I don't have one", then you cannot do hypothesis testing, or determine statistically significant differences.
  4 Comments
Rightia Rollmann
Rightia Rollmann on 27 Mar 2017
Okay, here’s the more detailed description of my example. Imagine there are speakers A and B and I want to see whether they produce the same output audio or not. In other words, I want to test whether these two speakers are identical or not. I play a certain sound with speaker A for 4 times and record it and therefore I create samples A1, A2, A3, and A4. I also play that certain sound with speaker B for 3 times and create samples B1, B2, and B3. Each sample consists of 4 samples.
A1 = [1 3 -1 2];
A2 = [1 3 -2 3];
A3 = [1 4 -3 4];
A4 = [2 4 -2 3];
B1 = [2 3 -1 5];
B2 = [2 4 -2 6];
B3 = [2 4 -4 5];
the cyclist
the cyclist on 27 Mar 2017
And what do the four numbers in A1 represent? Four different amplitudes, at four different time points? Why are they always integers, rather than values like 1.63?

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John BG
John BG on 27 Mar 2017
When comparing signals this way you could start averaging both signals amplitude this way:
A=[A1;A2;A3;A4];B=[B1;B2;B3];
mean(A)
=
1.25 3.50 -2.00 3.00
mean(B)
=
2.00 3.67 -2.33 5.33
then could check time average
mean(mean(A))
=
1.44
mean(mean(B))
=
2.17
that was 1st order statistical all-weather very useful mean, now 2nd order, variance:
var(mean(A))
ans =
6.18
var(mean(B))
ans =
10.85
the more samples the better, the longer the signals the more accurate the measurements
You may also want to see if the signals match a known statistical probability distributions that you may think would fit, for instance normal
fitdist(A,'Normal')
this in turn will give the probability distrition, then you can generate the pdf curve and measure how far the points of the signals are from the probability. The distance of the samples to an expected curve is an error measurement you may find useful
Error Vector Measurements are common in for instance DVB-T or for the case, any IQ signals.
if you find these lines useful would you please mark my answer as Accepted Answer?
To any other reader, if you find this answer of any help, would you please click on the thumbs-up vote link,
thanks in advance for time and attention
John BG

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