Power spectrum question -- seek verification

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Could some electronics guru here please verify this ?
Suppose that I calculate the power spectrum of an  
oscillator output. In the time domain the output is a  
sine wave across a 50 Ohm resistor, with units of Volts.  
When the Discrete Fourier Transform(DFT) is computed,  
followed by the power spectrum, what would be units of  
the vertical axis. The horizontal axis unit is Hz.
As the raw DFT values are complex numbers, the power
spectrum must contain the magnitudes of these DFT values,
so that the units for the vertical axis must be Volt/Hz.  
Is this correct ? Thanks in advance.

Re: Power spectrum question -- seek verification
On 4/22/19 7:01 AM, snipped-for-privacy@gmail.com wrote:
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Nope.  The vertical axis is in volts squared, because the frequency bin  
width is 1/T, where T is the total measurement time.  So the vertical  
scaling in volts is the power spectral density (PSD) in V**2/Hz times  
1/T Hz.

The horizontal axis is also 1/T per bin.


Cheers

Phil Hobbs

--  
Dr Philip C D Hobbs
Principal Consultant
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Re: Power spectrum question -- seek verification
On Monday, April 22, 2019 at 8:10:51 AM UTC-4, Phil Hobbs wrote:
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Considering the math, can you explain how this comes out of analysis of the
 units?  

In a DFT is the sinewave the inputs are multiplied by unitless or the same  
units as the input?  If unitless, the resultant values are just volts or vo
lt seconds depending on how you regard the input units.  The magnitudes wou
ld then be volts^2 or volts^2*seconds^2 which can be related to power throu
gh a constant impedance so watts or joule seconds.  

I remember in college being able to come up with the units for virtually an
y calculation.  Sometimes it took some odd paths to simplify the units, but
 it always worked.  

--  

  Rick C.

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Re: Power spectrum question -- seek verification
On 4/22/19 9:59 AM, snipped-for-privacy@gmail.com wrote:
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Parseval and Nyquist.  The total power is the same in both domains, and  
the bin width in each domain is the reciprocal of the total measurement  
time in the other domain.  (You have to leave the power spectrum  
two-sided for this to work.)

Cheers

Phil Hobbs


--  
Dr Philip C D Hobbs
Principal Consultant
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Re: Power spectrum question -- seek verification
On Monday, April 22, 2019 at 7:24:57 PM UTC-4, Phil Hobbs wrote:
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How does that relate to the units of the bins?  

--  

  Rick C.

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Re: Power spectrum question -- seek verification
On 4/22/19 10:40 PM, snipped-for-privacy@gmail.com wrote:
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The amplitude is always going to be in volts.  It's just the  
bandwidth/time duration that you have to worry about.

The energy is the same in the time and frequency domains.  Let's say we  
take N samples, a[0]...a[N-1], in a time T. Parseval says

     sum     |a[i]|**2  =      sum        |A[i]|**2.
i=0 to N-1                 i = 1 to N-1

where is the transform of .  An A/D converter implicitly divides  
its input by the reference voltage, so to convert to V**2, we need to  
multiply both sides by Vref**2.

It's natural to want the RHS to be normalized in terms of frequency.  
The frequency bin width is 1/T Hz, and the two-sided frequency range is  
N/T.  Thus each frequency bin has the total energy in a band of 1/T Hz.

To convert the raw FFT power spectrum to PSD, besides multiplying by  
Vref**2, you have to multiply the frequency-domain samples by T, i.e.  
divide by the bin width in frequency.

PSD_2sided[i] =  Vref**2 * T * |A[i]|**2, i = 0...N-1

Since the negative frequencies have the same power spectrum as the  
positive ones, we usually switch to the analytic signal convention, where

PSD_1sided[i] = { Vref**2 * T * |A[0]|**2   , i = 0 }
                 { 2 Vref**2 * T * |A[i]|**2 , 0 < i < (N+1)//2 }

where // is integer division.  What to do about the peculiar sample at  
N/2 for even N is left as an exercise for the reader. ;)  (It shouldn't  
have anything much in it anyway, provided that you've done your  
antialiasing filter correctly.)

Scope and spectrum analyzer FFT power spectra are always quoted in  
single-sided terms.

Cheers

Phil Hobbs

--  
Dr Philip C D Hobbs
Principal Consultant
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Re: Power spectrum question -- seek verification
On Tuesday, April 23, 2019 at 11:21:47 PM UTC+5:30, Phil Hobbs wrote:
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Thanks for the detailed explanation. I was only referring to the  
raw DFT power spectrum. My simple C program does a somewhat
brute force implementation of DFT, and the raw power spectrum,
when plotted with gnuplot shows the peaks at the correct,
expected frequencies(i.e.,fundamental, first harmonic etc.,)
I will add the PSD feature soon.

Re: Power spectrum question -- seek verification
On 22/04/2019 13:10, Phil Hobbs wrote:
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Although there are some versions of "power spectrum" display that  
actually display the amplitude of the signal at each frequency bin.

That is amplitude A determined from A.exp(i.theta) = x + iy

Rather than the true power spectrum A^2 = x^2 + y^2

Always worth checking which display your spectrum analyser is using.

Displaying A, theta plots is quite common in VLBI interferometry.

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--  
Regards,
Martin Brown

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