I have a question about the PSD estimation and where exactly does the 'Hz' term come from on the unit denominator. After taking the 1024pt FFT of a signal (let's say discrete vibration data sampled at 5kHz), we will get a plot of G vs Frequency (Hz)) with frequency resolution of 4.88Hz per data point. From my understanding of the PSD estimation (without windowing), we will square the magnitude of 'G' to obtain 'G^2', but then step of dividing 'G^2' by the "frequency bandwidth" to normalize is confusing... I've seen different methods of this, some of which say: df = Sampling Frequency (Matlab script?) df = User Defined Resolution? (LMS?) df = Fs/Spectral lines? What exactly is the 'G^2' divided by to get 'G^2/Hz'??? If we divide out the discrete points of the 'G^2' by different values of 'df', it would change the total energy of the signal. How exactly are the points divided by a frequency width without changing the signal energy and values? Help would be greatly appreciated. Kevin
PSD-Frequency Bin?
Started by ●October 14, 2008
Reply by ●October 14, 20082008-10-14
Kevin W wrote: ...> What exactly is the 'G^2' divided by to get 'G^2/Hz'???Hz? ... Jerry -- Engineering is the art of making what you want from things you can get. ����������������������������������������������������������������������� ** Posted from http://www.teranews.com **
Reply by ●October 14, 20082008-10-14
Hi Kevin, Suppose you sample a signal at 2,048 time per second, and then take a 256 point FFT. The frequency domain will have 129 samples (ignoring negative frequencies), with sample 0 representing DC, sample 1 representing 8 Hz, sample 2 representing 16 Hz, and so on up to sample 129 representing 1024 Hz. In other words, each of the samples is the signal strength of an 8 hz bandwidth. So instead of saying you have a certain "power per 8 Hz", you divide all of the values by 8 Hz, so you can cite the "power per Hz." There are more details here, but this is the simple answer: divide by the spacing between the frequency domain samples. In turn, this spacing is equal to the sample rate divided by the number of points in the FFT. Regards, Steve
Reply by ●October 15, 20082008-10-15
On Oct 15, 7:31�am, "Kevin W" <homesa...@gmail.com> wrote:> I have a question about the PSD estimation and where exactly does the 'Hz' > term come from on the unit denominator. > > After taking the 1024pt FFT of a signal (let's say discrete vibration data > sampled at 5kHz), we will get a plot of G vs Frequency (Hz)) with frequency > resolution of 4.88Hz per data point. > > From my understanding of the PSD estimation (without windowing), we will > square the magnitude of 'G' to obtain 'G^2', but then step of dividing > 'G^2' by the "frequency bandwidth" to normalize is confusing... > > I've seen different methods of this, some of which say: > df = Sampling Frequency (Matlab script?) > df = User Defined Resolution? (LMS?) > df = Fs/Spectral lines? > > What exactly is the 'G^2' divided by to get 'G^2/Hz'??? > > If we divide out the discrete points of the 'G^2' by different values of > 'df', it would change the total energy of the signal. �How exactly are the > points divided by a frequency width without changing the signal energy and > values? > > Help would be greatly appreciated. > > KevinThe area under the PSD (Periodogram) is the total average power. (or variance of the signal has zero dc value). Hardy
Reply by ●October 15, 20082008-10-15
HardySpicer wrote:> On Oct 15, 7:31 am, "Kevin W" <homesa...@gmail.com> wrote: >> I have a question about the PSD estimation and where exactly does the 'Hz' >> term come from on the unit denominator. >> >> After taking the 1024pt FFT of a signal (let's say discrete vibration data >> sampled at 5kHz), we will get a plot of G vs Frequency (Hz)) with frequency >> resolution of 4.88Hz per data point. >> >> From my understanding of the PSD estimation (without windowing), we will >> square the magnitude of 'G' to obtain 'G^2', but then step of dividing >> 'G^2' by the "frequency bandwidth" to normalize is confusing... >> >> I've seen different methods of this, some of which say: >> df = Sampling Frequency (Matlab script?) >> df = User Defined Resolution? (LMS?) >> df = Fs/Spectral lines? >> >> What exactly is the 'G^2' divided by to get 'G^2/Hz'??? >> >> If we divide out the discrete points of the 'G^2' by different values of >> 'df', it would change the total energy of the signal. How exactly are the >> points divided by a frequency width without changing the signal energy and >> values? >> >> Help would be greatly appreciated. >> >> Kevin > > The area under the PSD (Periodogram) is the total average power. (or > variance of the signal has zero dc value).Total average? Jerry -- Engineering is the art of making what you want from things you can get. ����������������������������������������������������������������������� ** Posted from http://www.teranews.com **
Reply by ●October 15, 20082008-10-15
>Hi Kevin, >Suppose you sample a signal at 2,048 time per second, and then take a256>point FFT. The frequency domain will have 129 samples (ignoringnegative>frequencies), with sample 0 representing DC, sample 1 representing 8 Hz, >sample 2 representing 16 Hz, and so on up to sample 129 representing1024>Hz. In other words, each of the samples is the signal strength of an 8hz>bandwidth. So instead of saying you have a certain "power per 8 Hz", you >divide all of the values by 8 Hz, so you can cite the "power per Hz." >There are more details here, but this is the simple answer: divide bythe>spacing between the frequency domain samples. In turn, this spacing is >equal to the sample rate divided by the number of points in the FFT. >Regards, >Steve >Steve, Thanks for your help. In your example, here's the thing that confuses me. Let's say that we took the same exact signal but instead sampled it at 1024Hz rather than 2048Hz. Now, your frequency width would be 4Hz. Assuming that the FFT of the signal gives the same or similar amplitudes at specific frequencies, now you are either dividing each discrete point by either 8Hz or 4Hz. This results in completely different signal energy! How can we take the same signal with different sampling frequencies and post-process them to have completely different energy? Kevin
Reply by ●October 15, 20082008-10-15
>>Hi Kevin, >>Suppose you sample a signal at 2,048 time per second, and then take a >256 >>point FFT. The frequency domain will have 129 samples (ignoring >negative >>frequencies), with sample 0 representing DC, sample 1 representing 8Hz,>>sample 2 representing 16 Hz, and so on up to sample 129 representing >1024 >>Hz. In other words, each of the samples is the signal strength of an 8 >hz >>bandwidth. So instead of saying you have a certain "power per 8 Hz",you>>divide all of the values by 8 Hz, so you can cite the "power per Hz." >>There are more details here, but this is the simple answer: divide by >the >>spacing between the frequency domain samples. In turn, this spacing is >>equal to the sample rate divided by the number of points in the FFT. >>Regards, >>Steve >> > >Steve, >Thanks for your help. In your example, here's the thing that confusesme.> Let's say that we took the same exact signal but instead sampled it at >1024Hz rather than 2048Hz. Now, your frequency width would be 4Hz. >Assuming that the FFT of the signal gives the same or similar amplitudesat>specific frequencies, now you are either dividing each discrete point by >either 8Hz or 4Hz. This results in completely different signal energy! >How can we take the same signal with different sampling frequencies and >post-process them to have completely different energy? > >Kevin >Another point- If we kept the same sampling frequency and changed the FFT size to 512, now your frequency width would be halved. That would also result in an energy change of the signal. Where am I going wrong in my thought process? Kevin
Reply by ●October 15, 20082008-10-15
Kevin W wrote:>>> Hi Kevin, >>> Suppose you sample a signal at 2,048 time per second, and then take a >> 256 >>> point FFT. The frequency domain will have 129 samples (ignoring >> negative >>> frequencies), with sample 0 representing DC, sample 1 representing 8 > Hz, >>> sample 2 representing 16 Hz, and so on up to sample 129 representing >> 1024 >>> Hz. In other words, each of the samples is the signal strength of an 8 >> hz >>> bandwidth. So instead of saying you have a certain "power per 8 Hz", > you >>> divide all of the values by 8 Hz, so you can cite the "power per Hz." >>> There are more details here, but this is the simple answer: divide by >> the >>> spacing between the frequency domain samples. In turn, this spacing is >>> equal to the sample rate divided by the number of points in the FFT. >>> Regards, >>> Steve >>> >> Steve, >> Thanks for your help. In your example, here's the thing that confuses > me. >> Let's say that we took the same exact signal but instead sampled it at >> 1024Hz rather than 2048Hz. Now, your frequency width would be 4Hz. >> Assuming that the FFT of the signal gives the same or similar amplitudes > at >> specific frequencies, now you are either dividing each discrete point by >> either 8Hz or 4Hz. This results in completely different signal energy! >> How can we take the same signal with different sampling frequencies and >> post-process them to have completely different energy?Energy that was distributed over 8 Hz would be nor divided into two 4-Hz bands. Of course the energy in each band would be half?> Another point- > If we kept the same sampling frequency and changed the FFT size to 512, > now your frequency width would be halved. That would also result in an > energy change of the signal. > > Where am I going wrong in my thought process?What is the whole band covered in each of the results? What is the total energy? Jerry -- Engineering is the art of making what you want from things you can get. ����������������������������������������������������������������������� ** Posted from http://www.teranews.com **
Reply by ●November 2, 20082008-11-02
On Oct 15, 7:04�am, Jerry Avins <j...@ieee.org> wrote:> > > The area under the PSD (Periodogram) is thetotal averagepower. (or > > variance of the signal has zero dc value). > > Total average? > > Jerry > -- > Engineering is the art of making what you want from things you can get. > �����������������������������������������������������������������������I think this term "total average" means the average power in the AC and the power in the DC components of the signal. The DC power is <x(t)>^2 and the average AC power is <x(t)^2>-<x(t)^>, then the total average power is <x(t)^>. (this is all for normalized powers, i.e. load is one ohm, and assuming ergodic signal so looking at the time averages here) Nasser
Reply by ●November 2, 20082008-11-02
On Oct 14, 10:31 am, "Kevin W" <homesa...@gmail.com> wrote:> I have a question about the PSD estimation and where exactly does the 'Hz' > term come from on the unit denominator. > > After taking the 1024pt FFT of a signal (let's say discrete vibration data > sampled at 5kHz), we will get a plot of G vs Frequency (Hz)) with frequency > resolution of 4.88Hz per data point. > > From my understanding of the PSD estimation (without windowing), we will > square the magnitude of 'G' to obtain 'G^2', but then step of dividing > 'G^2' by the "frequency bandwidth" to normalize is confusing... > > I've seen different methods of this, some of which say: > df = Sampling Frequency (Matlab script?) > df = User Defined Resolution? (LMS?) > df = Fs/Spectral lines? > > What exactly is the 'G^2' divided by to get 'G^2/Hz'??? > > If we divide out the discrete points of the 'G^2' by different values of > 'df', it would change the total energy of the signal. How exactly are the > points divided by a frequency width without changing the signal energy and > values? > > Help would be greatly appreciated. > > KevinKevin Take a look at: Application Note. Br�el & Kj�r. B. K. Choose your Units! by H. Konstantin-Hansen , J. Wismer, N. Thrane, S. Gade, Br�el & Kj�r, Denmark. available at: http://www.bksv.com/doc/bo0438.pdf for when and how. Dale B. Dalrymple http://dbdimages.com
