Proving the Relation between Angular and Frequency Power Spectral Density

In summary, pseudophonist's example shows that the power in any frequency interval is the same as measured using both functions, provided that they are of the same functional form. However, pseudophonist's example is flawed because he assumes that P(ν) and P(ω) are of the same functional form, which is not the case.
  • #1
KFC
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Suppose the power spectral density denoted by [tex]P(\omega)[/tex] where $\omega$ is the angular frequency and [tex]\omega = 2\pi \nu[/tex], I wonder how to prove that

[tex]2\pi P(\omega) = P(\nu)[/tex]

without know the functional form of [tex]P(\omega)[/tex]. I saw this relation in some book, but I don't know how to prove that.
 
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  • #2
It looks like a scale factor needed to insure that the total power (integral) is the same whatever the function argument (integral differential) is.
 
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  • #3
Disproof by Counterexample:

Suppose [tex]P(\omega) = 1/\omega^{2}[/tex] then [tex]P(\nu) = 1/\nu^{2} = 4\pi^{2}/\omega^{2}[/tex]
Thus
[tex]P(\nu) = 4\pi^{2} P(\omega)[/tex]
 
  • #4
pseudophonist said:
Disproof by Counterexample:

Suppose [tex]P(\omega) = 1/\omega^{2}[/tex] then [tex]P(\nu) = 1/\nu^{2} = 4\pi^{2}/\omega^{2}[/tex]
Thus
[tex]P(\nu) = 4\pi^{2} P(\omega)[/tex]

But I saw the statement about this somewhere else, like mathman said, when you integrate the power spectral density, no matter on [tex]\omega[/tex] or [tex]\nu[/tex] you should get the same energy, so you will find a relation between [tex]P(\omega)[/tex] and [tex]P(\nu)[/tex]. But as to your example, I don't know why it leads to this ...
 
  • #5
What I was trying to say was in essence:

[tex]d\omega = 2\pi d\nu[/tex]
 
  • #6
mathman said:
What I was trying to say was in essence:

[tex]d\omega = 2\pi d\nu[/tex]

Yes, that's what I mean. But what I am thinking to pseudophonist's example is: [tex]
2\pi P(\omega) = P(\nu)
[/tex] is only valid in the sense of integral but not the expression alone, right? Otherwise, how do you explain the counter example?
 
  • #7
The fallacy in pseudophonist's example is that he assumes that P(ν) and P(ω) are of the same functional form, and that going from one to the other is merely a matter of changing what symbol is used for the argument. However, in order to satisfy the condition that the power in any frequency interval is the same as measured using both functions, it's pretty clear that they can't be of the same functional form (even though we are confusingly using the same symbol, P, for both). So let's not use the same symbol for them. Let's call the power per unit frequency interval Pν(ν) and the power per unit angular frequency interval Pω(ω) (by adding the subscripts, we've used a different symbol for each spectrum, making it clear that these are in fact two different functions).

First of all, the sloppy physicist's derivation of the result is to talk about the power in any "small" (i.e. infinitesimal) frequency interval being the same regardless of what kind of frequency you're talking about. Hence:

Pν(ν) dν = Pω(ω)dω​

and since

2πdν = dω​

The result follows immediately:

Pν(ν) = 2πPω(ω)​

A more mathematically sensible version of the derivation is to talk about the power over some *finite* frequency interval (which is calculated by integrating the power spectral densities over that interval). It shouldn't matter whether you're using angular frequency or just plain old frequency: you should get the same answer for the total power in that interval:

[tex] \int_{\omega_1}^{\omega_2} P_{\omega}(\omega)\, d\omega = \int_{\nu_1}^{\nu_2} P_{\nu}(\nu)\, d\nu [/tex]​

Again, it's pretty clear that, for this condition to be satisfied, there's no way that Pν(ν) and Pω(ω) can be of the same functional form.
 
  • #8
cepheid said:
The fallacy in pseudophonist's example is that he assumes that P(ν) and P(ω) are of the same functional form, and that going from one to the other is merely a matter of changing what symbol is used for the argument. However, in order to satisfy the condition that the power in any frequency interval is the same as measured using both functions, it's pretty clear that they can't be of the same functional form (even though we are confusingly using the same symbol, P, for both). So let's not use the same symbol for them. Let's call the power per unit frequency interval Pν(ν) and the power per unit angular frequency interval Pω(ω) (by adding the subscripts, we've used a different symbol for each spectrum, making it clear that these are in fact two different functions).

First of all, the sloppy physicist's derivation of the result is to talk about the power in any "small" (i.e. infinitesimal) frequency interval being the same regardless of what kind of frequency you're talking about. Hence:

Pν(ν) dν = Pω(ω)dω​

and since

2πdν = dω​

The result follows immediately:

Pν(ν) = 2πPω(ω)​

A more mathematically sensible version of the derivation is to talk about the power over some *finite* frequency interval (which is calculated by integrating the power spectral densities over that interval). It shouldn't matter whether you're using angular frequency or just plain old frequency: you should get the same answer for the total power in that interval:

[tex] \int_{\omega_1}^{\omega_2} P_{\omega}(\omega)\, d\omega = \int_{\nu_1}^{\nu_2} P_{\nu}(\nu)\, d\nu [/tex]​

Again, it's pretty clear that, for this condition to be satisfied, there's no way that Pν(ν) and Pω(ω) can be of the same functional form.

Thanks. Make sense
 

1. What is power spectral density?

Power spectral density (PSD) is a measure of the distribution of power into frequency components of a signal. It is commonly used in signal processing and engineering to analyze and characterize signals in the frequency domain.

2. How is power spectral density calculated?

Power spectral density is calculated by taking the Fourier transform of the autocorrelation function of a signal. This converts the signal from the time domain to the frequency domain, allowing for the analysis of power at different frequencies.

3. What is the difference between power spectral density and energy spectral density?

Power spectral density measures the distribution of power in a signal, while energy spectral density measures the distribution of energy. As power is proportional to the square of the signal amplitude, PSD tends to emphasize higher amplitude components, while ESD gives equal weight to all frequency components.

4. How is power spectral density used in practical applications?

Power spectral density is used in a wide range of applications, including noise analysis, vibration analysis, and signal processing. It can be used to identify the dominant frequencies in a signal, detect periodicity, and analyze the effects of noise on a signal.

5. What are the limitations of power spectral density?

One limitation of power spectral density is that it assumes stationary signals, meaning that the statistical properties of the signal do not change over time. It also does not take into account phase information, which can be important in certain applications. Additionally, PSD may not accurately represent non-linear or non-stationary signals.

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