calculation norm divided by square root of data length

Question

Amanda Kamphoff 2021-10-10

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1549561-calculation-norm-divided-by-square-root-of-data-length

评论： Jan 2021-10-11

采纳的回答： Jan

在 MATLAB Online 中打开

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 个评论
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DGM 2021-10-10

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 2021-10-10

编辑：David Goodmanson 2021-10-10

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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Answer 1

Jan 2021-10-10

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1549561-calculation-norm-divided-by-square-root-of-data-length#answer_805591

编辑：Jan 2021-10-10

在 MATLAB Online 中打开

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)
sd1 = 16.8331
g = norm(r);
h = sqrt(n);
sd2 = g / h
sd2 = 16.8331