calculation norm divided by square root of data length
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Amanda Kamphoff
am 10 Okt. 2021
Kommentiert: Jan
am 11 Okt. 2021
I'm trying to calculate noise level which the formula is already given by book
this is the formula
here I have:
r = [-33.1869365513747 - 7.19942077700726i 9.95754431609273 - 9.07572996519880i 5.11935460162031 - 9.45713125754552i 17.2406749451800 + 7.50738378923969i 22.3786796564404 + 12.6012073987034i -29.1193546016203 + 0.557636320933856i 15.3381897144724 - 3.13336628613266i 30.2746887309409 - 7.75771048053827i];
n = 16;
I tried calculate sd by:
sd=norm(r)/ sqrt(n);
the result of sd = 7.406256475699323
but if I tried calculate sd by
g = norm(r);
h = sqrt(n);
sd = g/ h;
the result would be sd = 67.332259630053100
why is that happend? what's the diffence calculating in those two?
I'm new in Matlab
3 Kommentare
DGM
am 10 Okt. 2021
I'm going to assume that something happened to the value of r or n in the intervening code. I don't immediately see how this could happen through any version-dependent behavior or anything.
David Goodmanson
am 10 Okt. 2021
Bearbeitet: David Goodmanson
am 10 Okt. 2021
Hi Amanda,
The number of elements of r is length(r) which is 8, not 16. So this consideration will be part of the calculation as well.
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Jan
am 10 Okt. 2021
Bearbeitet: Jan
am 10 Okt. 2021
r = [-33.1869365513747 - 7.19942077700726i 9.95754431609273 - 9.07572996519880i 5.11935460162031 - 9.45713125754552i 17.2406749451800 + 7.50738378923969i 22.3786796564404 + 12.6012073987034i -29.1193546016203 + 0.557636320933856i 15.3381897144724 - 3.13336628613266i 30.2746887309409 - 7.75771048053827i];
n = 16;
sd1 = norm(r) / sqrt(n)
g = norm(r);
h = sqrt(n);
sd2 = g / h
For me it works: both give the same result, as expected, but neither 7.4 nor 67.3 .
Did you create a script called "norm" or "sqrt"?
which norm
which sqrt
2 Kommentare
Jan
am 11 Okt. 2021
Fine. Then the solution is to rename your script, because it shadows the builtin function with the same name.
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