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**posted on**

- dakupoto

April 22, 2019, 11:01 am

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.

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

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

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:

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.

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.

- Get a 1,000 miles of free Supercharging

Rick C.

- Get a 1,000 miles of free Supercharging

We've slightly trimmed the long signature. Click to see the full one.

Re: Power spectrum question -- seek verification

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

Dr Philip C D Hobbs

Principal Consultant

We've slightly trimmed the long signature. Click to see the full one.

Re: Power spectrum question -- seek verification

On 4/22/19 10:40 PM, snipped-for-privacy@gmail.com wrote:

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]|

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

multiply both sides by Vref

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

divide by the bin width in frequency.

PSD_2sided[i] = Vref

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 Vref

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

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 tomultiply 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-1Since 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

Dr Philip C D Hobbs

Principal Consultant

We've slightly trimmed the long signature. Click to see the full one.

Re: Power spectrum question -- seek verification

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:

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.

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.

--

Regards,

Martin Brown

Regards,

Martin Brown

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