PSD vs. Fourier transform

On Mar 2, 8:08 am, Rune Allnor <all...@tele.ntnu.no> wrote:
> On 1 Mar, 19:05, HardySpicer <gyansor...@gmail.com> wrote:
>
>
>
> > On Mar 2, 5:10 am, Oli Charlesworth <ca...@olifilth.co.uk> wrote:
>
> > > m26k9 wrote:
> > > > Hello,
>
> > > > I have a confusion regarding Fourier transform of a time signal and the
> > > > PSD.
> > > > By definition, power spectral density (PSD) is the Fourier transform of
> > > > the autocorrelation function.
>
> > > > Suppose my time-domain signal is a voltage signal.
> > > > I sample this at the correct rate, and taking the Fourier transform gives
> > > > me the maginitudes of of various frequency components present in the signal
> > > > (within the context of sampling rate). Taking the modulus squared of this
> > > > quantity will give me power/Hz.
>
> > > > On the other hand, literature says, taking the Fourier transform of the
> > > > autocorrelation function (rather than the direct transform as discussed
> > > > earlier) gives the PSD. And confusingly gives the units as power/Hz.
>
> > > > I am confused as what is the difference between these two? if both have
> > > > units of power/Hz, why/when should I do one over the other?
>
> > > One difference is that the Fourier transform of a signal may not exist,
> > > for instance, a stochastic (i.e. random) process doesn't have a Fourier
> > > transform, but it will have an autocorrelation function.  Therefore, the
> > > PSD will exist even if you can't compute the Fourier transform of the
> > > signal itself.
>
> > > --
> > > Oli
>
> > You can still compute the periodogram (PSD) of a random signal (say
> > speech) using the FFT directly of course
> > and not via the autocorrelation.
>
> How would you do this if you want to use some window function?
>
> Rune

I normally just use magnitude squared/N for periodogram. I average of
course over successive periodograms if possible.
The other way is to usea forgetting factor

S(i) =S(i-1)*beta + (1-beta)*X(i)*X(i)

beta is a forgetting factor <1.

The conventional Welch method  is a WOSA (weighted opverlapped
spectral averaging method).


Hardy

Thank a lot guys.

What I gather from the discussion is that:

1) Both methods give essentially the same amount of information.

2) Direct FFT method is easier, but the FFT{autocorrelation} gives a
averaged value, which can better characterize a stochastic process.


What I am confused/interested in now is;

1) How to prove that the FFT{autcorrelation} method gives better results
over the FFT{time-signal} ?

2) What is the advantage of doing FFT{autcorrelation} over averaging
several FFT{time-signal} calculations ?

3) How is it possible for me to evaluate the PSD at f= F Hz, and get the
value with units of power/Hz? this is like a rate/slope?  If I am actually
measuring the energy content at F Hz, I should get a value with units
power(watts), not power/Hz. What is this dilemma?

Thank you a lot for all the help.

m26k9 wrote:
> Thank a lot guys.
> 
> What I gather from the discussion is that:
> 
> 1) Both methods give essentially the same amount of information.
> 
> 2) Direct FFT method is easier, but the FFT{autocorrelation} gives a
> averaged value, which can better characterize a stochastic process.
> 
> 
> What I am confused/interested in now is;
> 
> 1) How to prove that the FFT{autcorrelation} method gives better results
> over the FFT{time-signal} ?
> 
> 2) What is the advantage of doing FFT{autcorrelation} over averaging
> several FFT{time-signal} calculations ?
> 
> 3) How is it possible for me to evaluate the PSD at f= F Hz, and get the
> value with units of power/Hz? this is like a rate/slope?  If I am actually
> measuring the energy content at F Hz, I should get a value with units
> power(watts), not power/Hz. What is this dilemma?

You are measuring the total power in a band ("bin"), not at a point.

Jerry
-- 
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������

Jerry Avins  <jya@ieee.org> wrote:

>m26k9 wrote:

>> What I gather from the discussion is that:
>> 
>> 1) Both methods give essentially the same amount of information.
>> 
>> 2) Direct FFT method is easier, but the FFT{autocorrelation} gives a
>> averaged value, which can better characterize a stochastic process.

>> What I am confused/interested in now is;

>> 1) How to prove that the FFT{autcorrelation} method gives better results
>> over the FFT{time-signal} ?

>> 2) What is the advantage of doing FFT{autcorrelation} over averaging
>> several FFT{time-signal} calculations ?

>> 3) How is it possible for me to evaluate the PSD at f= F Hz, and get the
>> value with units of power/Hz? this is like a rate/slope?  If I am actually
>> measuring the energy content at F Hz, I should get a value with units
>> power(watts), not power/Hz. What is this dilemma?

>You are measuring the total power in a band ("bin"), not at a point.

Which goes back to my earlier comment -- you cannot compute
the PSD with a discrete fourier transform.  The PSD is defined
at any frequency, whereas the DFT only gives you bins.

You might get something that is PSD-like, it may serve your
purposes, but it is not the real deal.

Steve

Thank you.


>Which goes back to my earlier comment -- you cannot compute
>the PSD with a discrete fourier transform.  The PSD is defined
>at any frequency, whereas the DFT only gives you bins.
>
>You might get something that is PSD-like, it may serve your
>purposes, but it is not the real deal.
>
>Steve

If PSD is defined at any frequency, what is the meaning of getting a value
with units power/Hz?

And how to calculate PSD of discrete signals (when working in digital
domain) if DFT cannot be used?

Thank you.

On Mar 1, 8:17 pm, "m26k9" <maduranga.liyan...@gmail.com> wrote:
> Thank a lot guys.
>
. What I gather from the discussion is that:
.
. 1) Both methods give essentially the same amount of information.
True

. 2) Direct FFT method is easier, but the FFT{autocorrelation} gives a
. averaged value, which can better characterize a stochastic process.
Still confused. See Bob Adams post with his correction.

. What I am confused/interested in now is;
.
. 1) How to prove that the FFT{autcorrelation} method gives better
results
. over the FFT{time-signal} ?
The FFT{autcorrelation} method is equivalent to the FFT{time-signal
zero extended to 2x}

. 2) What is the advantage of doing FFT{autcorrelation} over averaging
. several FFT{time-signal} calculations ?
See above. Average either as you choose. Averaging decreases the
variance of estimation of signals that span the averaging interval

. 3) How is it possible for me to evaluate the PSD at f= F Hz, and get
the
. value with units of power/Hz? this is like a rate/slope?  If I am
actually
. measuring the energy content at F Hz, I should get a value with
units
. power(watts), not power/Hz. What is this dilemma?
Appropriate units and scaling depend on the type of signal. See:
http://www.bksv.com/doc/bo0438.pdf

Good luck,
Dale B. Dalrymple

Thank you Dale.
Loos very informative. Will have a read.

m26k9 <maduranga.liyanage@gmail.com> wrote:

>spope33 writes,

>>Which goes back to my earlier comment -- you cannot compute
>>the PSD with a discrete fourier transform.  The PSD is defined
>>at any frequency, whereas the DFT only gives you bins.

>>You might get something that is PSD-like, it may serve your
>>purposes, but it is not the real deal.

> If PSD is defined at any frequency, what is the meaning of getting 
> a value with units power/Hz?

It means the power in a suitably small region around that point is
equal to the PSD multiplied by the width (in Hz) of the region.

>And how to calculate PSD of discrete signals (when working in digital
>domain) if DFT cannot be used?

The sampled-time signal gives you a sampled-time autocorrelation
function, but then when you take its continuous time cosine 
transform you get a continuous-time PSD.

Steve

Thank you Steve.

But what did you mean exactly when you said DFT cannot be used to
calculate the PSD? Suppose if we only want to find the frequency content of
discrete bins, will CT-cosine transform give anything better than thr plain
DFT?

m26k9 <maduranga.liyanage@gmail.com> wrote:

> But what did you mean exactly when you said DFT cannot be used to
> calculate the PSD? Suppose if we only want to find the frequency 
> content of discrete bins, will CT-cosine transform give anything 
> better than the plain DFT?

I think it might.  Let's say you have a billion samples at 1 MHz,
and there is a frequency component at 400.13 Hz.  The PSD can
be evaluated at exactly this frequency, whereas if you choose
a reasonable-sized DFT to use instead of the continuous-time 
transform, it might not have a bin at this frequency, so the energy 
associated with this component bleeds into adjacent bins.  You do 
not get the same information, at least not very directly.  

Steve