Skip to main content
Signal Processing

Return to Answer

added 1 character in body
Source Link
David Jonsson
David Jonsson
  • 221
  • 1
  • 8

Here is a better answer (than the one I gave abovebefore) based on properly scaled Fourier transformations where basis vectors are normalized to length 1.

Noise, a normal distribution of a random variable, has the following variance, based on https://en.wikipedia.org/wiki/Variance#Basic_properties for variance of a linear combination with no correlation, applies to both the Fourier transformation and it's inverse

$$ {Var}\left(\sum _{{i=1}}^{{N}}a_{i}X_{i}\right) = \sum _{{i=1}}^{{N}}a_{i}^{2}\operatorname {Var}(X_{i}) $$

The same variance for all Xi means $$ {Var}\left(\sum _{{i=1}}^{{N}}a_{i}X_{i}\right) = \sum _{{i=1}}^{{N}}a_{i}^{2}\operatorname {Var}(X) = \left(\sum _{{i=1}}^{{N}}a_{i}^{2}\right)\operatorname {Var}(X) $$

Example 1

The first row DC basis vector in a even Fourier transform
$$ \left(\frac{1}{\sqrt N } , \frac{1}{\sqrt N } , \frac{1}{\sqrt N } , ... \right) $$ gives a variance of

$$ Var(Transform) = N \left( \frac{1}{\sqrt N } \right)^2 Var(X) = Var(X) $$

Example 2

A basis vector in an even fast Fourier transform with a wavelength of 4 samples, a repeating series of

$$ \left( \sqrt{\frac{2}{N}} , 0 , -\sqrt{\frac{2}{N}} , 0 , ... \right) $$

This is the basis vector with most zero elements, every second. The variance becomes

$$ Var(Transform) = \frac{N}{2} \left( \sqrt{\frac{2}{N}} \right)^2 Var(X) = Var(X) $$

So variance, and thus standard deviation, is independent of basis vectors and zero elements. But what happens to the signal, the expected value? Maybe someone else can show that?

Here is a better answer (than the one I gave above) based on properly scaled Fourier transformations where basis vectors are normalized to length 1.

Noise, a normal distribution of a random variable, has the following variance, based on https://en.wikipedia.org/wiki/Variance#Basic_properties for variance of a linear combination with no correlation, applies to both the Fourier transformation and it's inverse

$$ {Var}\left(\sum _{{i=1}}^{{N}}a_{i}X_{i}\right) = \sum _{{i=1}}^{{N}}a_{i}^{2}\operatorname {Var}(X_{i}) $$

The same variance for all Xi means $$ {Var}\left(\sum _{{i=1}}^{{N}}a_{i}X_{i}\right) = \sum _{{i=1}}^{{N}}a_{i}^{2}\operatorname {Var}(X) = \left(\sum _{{i=1}}^{{N}}a_{i}^{2}\right)\operatorname {Var}(X) $$

Example 1

The first row DC basis vector in a even Fourier transform
$$ \left(\frac{1}{\sqrt N } , \frac{1}{\sqrt N } , \frac{1}{\sqrt N } , ... \right) $$ gives a variance of

$$ Var(Transform) = N \left( \frac{1}{\sqrt N } \right)^2 Var(X) = Var(X) $$

Example 2

A basis vector in an even fast Fourier transform with a wavelength of 4 samples, a repeating series of

$$ \left( \sqrt{\frac{2}{N}} , 0 , -\sqrt{\frac{2}{N}} , 0 , ... \right) $$

This is the basis vector with most zero elements, every second. The variance becomes

$$ Var(Transform) = \frac{N}{2} \left( \sqrt{\frac{2}{N}} \right)^2 Var(X) = Var(X) $$

So variance, and thus standard deviation, is independent of basis vectors and zero elements. But what happens to the signal, the expected value? Maybe someone else can show that?

Here is a better answer (than the one I gave before) based on properly scaled Fourier transformations where basis vectors are normalized to length 1.

Noise, a normal distribution of a random variable, has the following variance, based on https://en.wikipedia.org/wiki/Variance#Basic_properties for variance of a linear combination with no correlation, applies to both the Fourier transformation and it's inverse

$$ {Var}\left(\sum _{{i=1}}^{{N}}a_{i}X_{i}\right) = \sum _{{i=1}}^{{N}}a_{i}^{2}\operatorname {Var}(X_{i}) $$

The same variance for all Xi means $$ {Var}\left(\sum _{{i=1}}^{{N}}a_{i}X_{i}\right) = \sum _{{i=1}}^{{N}}a_{i}^{2}\operatorname {Var}(X) = \left(\sum _{{i=1}}^{{N}}a_{i}^{2}\right)\operatorname {Var}(X) $$

Example 1

The first row DC basis vector in a even Fourier transform
$$ \left(\frac{1}{\sqrt N } , \frac{1}{\sqrt N } , \frac{1}{\sqrt N } , ... \right) $$ gives a variance of

$$ Var(Transform) = N \left( \frac{1}{\sqrt N } \right)^2 Var(X) = Var(X) $$

Example 2

A basis vector in an even fast Fourier transform with a wavelength of 4 samples, a repeating series of

$$ \left( \sqrt{\frac{2}{N}} , 0 , -\sqrt{\frac{2}{N}} , 0 , ... \right) $$

This is the basis vector with most zero elements, every second. The variance becomes

$$ Var(Transform) = \frac{N}{2} \left( \sqrt{\frac{2}{N}} \right)^2 Var(X) = Var(X) $$

So variance, and thus standard deviation, is independent of basis vectors and zero elements. But what happens to the signal, the expected value? Maybe someone else can show that?

deleted 39 characters in body
Source Link
David Jonsson
David Jonsson
  • 221
  • 1
  • 8

Here is a better answer (than the one I gave above) based on properly scaled Fourier transformations where basis vectors are normalized to length 1.

Noise, a normal distribution of a random variable, has the following variance, based on https://en.wikipedia.org/wiki/Variance#Basic_properties for variance of a linear combination with no correlation, applies to both the Fourier transformation and it's inverse

$$ {Var}\left(\sum _{{i=1}}^{{N}}a_{i}X_{i}\right) = \sum _{{i=1}}^{{N}}a_{i}^{2}\operatorname {Var}(X_{i}) $$

The same variance for all Xi means $$ {Var}\left(\sum _{{i=1}}^{{N}}a_{i}X_{i}\right) = \sum _{{i=1}}^{{N}}a_{i}^{2}\operatorname {Var}(X) = \left(\sum _{{i=1}}^{{N}}a_{i}^{2}\right)\operatorname {Var}(X) $$

Example 1 TheExample 1

The first row DC basis vector in a even Fourier transform
$$ \left(\frac{1}{\sqrt N } , \frac{1}{\sqrt N } , \frac{1}{\sqrt N } , ... \right) $$ gives a variance of

$$ Var(Transform) = N \left( \frac{1}{\sqrt N } \right)^2 Var(X) = Var(X) $$

Example 2 With the uppper middle rowExample 2

A basis vector in an even fast Fourier transform with a wavelength of 4 samples in the basis vector, a repeating series of

$$ \left( \sqrt{\frac{2}{N}} , 0 , -\sqrt{\frac{2}{N}} , 0 , ... \right) $$

This is the basis vector with most zero elements, every second. The variance becomes

$$ Var(Transform) = \frac{N}{2} \left( \sqrt{\frac{2}{N}} \right)^2 Var(X) = Var(X) $$

So variance, and thus standard deviation, is independent of basis vectors and zero elements. But what happens to the signal, the expected value? Maybe someone else can show that?

Here is a better answer (than the one I gave above) based on properly scaled Fourier transformations where basis vectors are normalized to length 1.

Noise, a normal distribution of a random variable, has the following variance, based on https://en.wikipedia.org/wiki/Variance#Basic_properties for variance of a linear combination with no correlation, applies to both the Fourier transformation and it's inverse

$$ {Var}\left(\sum _{{i=1}}^{{N}}a_{i}X_{i}\right) = \sum _{{i=1}}^{{N}}a_{i}^{2}\operatorname {Var}(X_{i}) $$

The same variance for all Xi means $$ {Var}\left(\sum _{{i=1}}^{{N}}a_{i}X_{i}\right) = \sum _{{i=1}}^{{N}}a_{i}^{2}\operatorname {Var}(X) = \left(\sum _{{i=1}}^{{N}}a_{i}^{2}\right)\operatorname {Var}(X) $$

Example 1 The first row DC basis vector in a even Fourier transform
$$ \left(\frac{1}{\sqrt N } , \frac{1}{\sqrt N } , \frac{1}{\sqrt N } , ... \right) $$ gives a variance of

$$ Var(Transform) = N \left( \frac{1}{\sqrt N } \right)^2 Var(X) = Var(X) $$

Example 2 With the uppper middle row vector in an even fast Fourier transform with a wavelength of 4 samples in the basis vector, a repeating series of

$$ \left( \sqrt{\frac{2}{N}} , 0 , -\sqrt{\frac{2}{N}} , 0 , ... \right) $$

This is the basis vector with most zero elements. The variance becomes

$$ Var(Transform) = \frac{N}{2} \left( \sqrt{\frac{2}{N}} \right)^2 Var(X) = Var(X) $$

So variance, and thus standard deviation, is independent of basis vectors and zero elements. But what happens to the signal, the expected value? Maybe someone else can show that?

Here is a better answer (than the one I gave above) based on properly scaled Fourier transformations where basis vectors are normalized to length 1.

Noise, a normal distribution of a random variable, has the following variance, based on https://en.wikipedia.org/wiki/Variance#Basic_properties for variance of a linear combination with no correlation, applies to both the Fourier transformation and it's inverse

$$ {Var}\left(\sum _{{i=1}}^{{N}}a_{i}X_{i}\right) = \sum _{{i=1}}^{{N}}a_{i}^{2}\operatorname {Var}(X_{i}) $$

The same variance for all Xi means $$ {Var}\left(\sum _{{i=1}}^{{N}}a_{i}X_{i}\right) = \sum _{{i=1}}^{{N}}a_{i}^{2}\operatorname {Var}(X) = \left(\sum _{{i=1}}^{{N}}a_{i}^{2}\right)\operatorname {Var}(X) $$

Example 1

The first row DC basis vector in a even Fourier transform
$$ \left(\frac{1}{\sqrt N } , \frac{1}{\sqrt N } , \frac{1}{\sqrt N } , ... \right) $$ gives a variance of

$$ Var(Transform) = N \left( \frac{1}{\sqrt N } \right)^2 Var(X) = Var(X) $$

Example 2

A basis vector in an even fast Fourier transform with a wavelength of 4 samples, a repeating series of

$$ \left( \sqrt{\frac{2}{N}} , 0 , -\sqrt{\frac{2}{N}} , 0 , ... \right) $$

This is the basis vector with most zero elements, every second. The variance becomes

$$ Var(Transform) = \frac{N}{2} \left( \sqrt{\frac{2}{N}} \right)^2 Var(X) = Var(X) $$

So variance, and thus standard deviation, is independent of basis vectors and zero elements. But what happens to the signal, the expected value? Maybe someone else can show that?

Source Link
David Jonsson
David Jonsson
  • 221
  • 1
  • 8

Here is a better answer (than the one I gave above) based on properly scaled Fourier transformations where basis vectors are normalized to length 1.

Noise, a normal distribution of a random variable, has the following variance, based on https://en.wikipedia.org/wiki/Variance#Basic_properties for variance of a linear combination with no correlation, applies to both the Fourier transformation and it's inverse

$$ {Var}\left(\sum _{{i=1}}^{{N}}a_{i}X_{i}\right) = \sum _{{i=1}}^{{N}}a_{i}^{2}\operatorname {Var}(X_{i}) $$

The same variance for all Xi means $$ {Var}\left(\sum _{{i=1}}^{{N}}a_{i}X_{i}\right) = \sum _{{i=1}}^{{N}}a_{i}^{2}\operatorname {Var}(X) = \left(\sum _{{i=1}}^{{N}}a_{i}^{2}\right)\operatorname {Var}(X) $$

Example 1 The first row DC basis vector in a even Fourier transform
$$ \left(\frac{1}{\sqrt N } , \frac{1}{\sqrt N } , \frac{1}{\sqrt N } , ... \right) $$ gives a variance of

$$ Var(Transform) = N \left( \frac{1}{\sqrt N } \right)^2 Var(X) = Var(X) $$

Example 2 With the uppper middle row vector in an even fast Fourier transform with a wavelength of 4 samples in the basis vector, a repeating series of

$$ \left( \sqrt{\frac{2}{N}} , 0 , -\sqrt{\frac{2}{N}} , 0 , ... \right) $$

This is the basis vector with most zero elements. The variance becomes

$$ Var(Transform) = \frac{N}{2} \left( \sqrt{\frac{2}{N}} \right)^2 Var(X) = Var(X) $$

So variance, and thus standard deviation, is independent of basis vectors and zero elements. But what happens to the signal, the expected value? Maybe someone else can show that?