>On Mar 2, 5:05 pm, spop...@speedymail.org (Steve Pope) wrote:
>> I am wondering how exactly you would calculate
>> typical windowed PSD values without doing the autocorrelation
>> step. For example:
>>
>> autocorrelation -> Nuttall Window -> Cosine Transform
>>
>> What is the version of this that uses no autocorrelation?
>I said that windowing can be used for tradeoffs between mainlobe width
>and sidelobe rejection in both cases. The tradeoff need not be made by
>a window on the autocorrelation. My statement was that the input data
>was windowed.
Okay, but windowing the data and windowing the ACF are not
exactly mathematically interchageable.
>For long data sequences with multiple transforms performed on blocked
>subsets of data, Carter and Nuttall claim that the same estimate
>stability can be achieved by the two processes if they are constrained
>to the same frequency resolution and data size.
>On the weighted overlapped segment averaging method for power spectral
>estimation
>Carter, G.C. Nuttall, A.H. Yuen, C.K.
>Proceedings of the IEEE
>Oct. 1980, Vol: 68, Issue: 10
>page(s): 1352- 1354
Thanks. So, without looking up this article, I'm inferring it
says that to answer my question above, you would block and window
the data, then take DFT's, take the magnitudes, and combine.
Per the Carter/Nuttall/Yuen you get the same basic accuracy.
Seems reasonable. I haven't tried it, mostly because of
my instinctual feel that this underweights data that happens to
fall in the low-amplitude parts of an input window, whereas
the ACF method does not. The counterargument being that you
cannot resolve the corresponding low-frequency-resolution components
in either case.
Thanks for your replies, this has given me something to think about.
Steve
Reply by dbd●March 3, 20092009-03-03
On Mar 2, 5:05 pm, spop...@speedymail.org (Steve Pope) wrote:
> dbd <d...@ieee.org> wrote:
> >>> On Mar 2, 6:49 pm, spop...@speedymail.org (Steve Pope) wrote:
> >> >> 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.
.> >As has already been pointed out by others in this thread, for the
.> >discrete case, by zero extending the data to twice the (windowed)
data
.> >size and performing this larger DFT, the result is exactly the
same as
.> >the discrete version of the PSD-via-autocorrelation. You have
exactly
.> >the same information.
.>
.> Right. I agree with this. In the example I have given above this
.> requires a very large DFT.
Yes, the same size for either approach under the conditions of a
single data set transformed once.
.>
.> >The trade-offs in bin width and sidelobe rejection available by
.> >windowing are available in both approaches as well.
.>
.> I agree with this in principal (i.e. you can re-arrange the
.> linear and non-linear parts of a system by making suitable
.> changes); but I am wondering how exactly you would calculate
.> typical windowed PSD values without doing the autocorrelation
.> step. For example:
.>
.> autocorrelation -> Nuttall Window -> Cosine Transform
.>
.> What is the version of this that uses no autocorrelation?
.>
.> Steve
I said that windowing can be used for tradeoffs between mainlobe width
and sidelobe rejection in both cases. The tradeoff need not be made by
a window on the autocorrelation. My statement was that the input data
was windowed.
For long data sequences with multiple transforms performed on blocked
subsets of data, Carter and Nuttall claim that the same estimate
stability can be achieved by the two processes if they are constrained
to the same frequency resolution and data size.
On the weighted overlapped segment averaging method for power spectral
estimation
Carter, G.C. Nuttall, A.H. Yuen, C.K.
Proceedings of the IEEE
Oct. 1980, Vol: 68, Issue: 10
page(s): 1352- 1354
Dale B. Dalrymple
Reply by Rune Allnor●March 3, 20092009-03-03
On 2 Mar, 03:18, HardySpicer <gyansor...@gmail.com> wrote:
> 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.
But how would you modify your FFT algorithm to handle explicit
window functions in lag domain? That is, if you use something
like
rxx[k] = w[k]*E[x[n]x[n+k]]
where w[k] is an arbitrary window function, what will the
equivalent expression be in terms of FFTs?
The recipes I have seen for Welch's method [*] compute an
average of window-weighted periodograms.
Rune
[*] I might have seen more, but I remember the recipe
from Kay's 1988 book.
Reply by Rune Allnor●March 3, 20092009-03-03
On 1 Mar, 20:50, dbd <d...@ieee.org> wrote:
> On Mar 1, 11:08 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
>
> How could you do this without using some window function?
>
> Dale B. Dalrymple
Funny. I'll get back to you on that in the very moment your
Norwegian is at the level where it's worth splittng hairs
on exact phrasings.
Rune
Reply by Steve Pope●March 2, 20092009-03-02
dbd <dbd@ieee.org> wrote:
>>> On Mar 2, 6:49 pm, spop...@speedymail.org (Steve Pope) wrote:
>> >> 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.
>As has already been pointed out by others in this thread, for the
>discrete case, by zero extending the data to twice the (windowed) data
>size and performing this larger DFT, the result is exactly the same as
>the discrete version of the PSD-via-autocorrelation. You have exactly
>the same information.
Right. I agree with this. In the example I have given above this
requires a very large DFT.
>The trade-offs in bin width and sidelobe rejection available by
>windowing are available in both approaches as well.
I agree with this in principal (i.e. you can re-arrange the
linear and non-linear parts of a system by making suitable
changes); but I am wondering how exactly you would calculate
typical windowed PSD values without doing the autocorrelation
step. For example:
autocorrelation -> Nuttall Window -> Cosine Transform
What is the version of this that uses no autocorrelation?
Steve
Reply by dbd●March 2, 20092009-03-02
On Mar 2, 10:22 am, spop...@speedymail.org (Steve Pope) wrote:
> HardySpicer <gyansor...@gmail.com> wrote:
> >On Mar 2, 6:49 pm, spop...@speedymail.org (Steve Pope) wrote:
> >> 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.
> >Surely the cc is also an approximation? It is found from samples and
> >must be?
>
. I think for the PSD calculation, your frequency resolution/accuracy
. is on the order of the inverse of the original sample length (in
this
. case, 0.001 Hz) whereas for the case using the DFT it's only
. on the order of the inverse of the DFT size.
In the DFT case its proportional to the inverse of the window (and
thus data) size.
.
. Whether this means you've lost information with the DFT I'm
. not entirely sure. At least superficially you don't have
. the same information.
.
. Steve
As has already been pointed out by others in this thread, for the
discrete case, by zero extending the data to twice the (windowed) data
size and performing this larger DFT, the result is exactly the same as
the discrete version of the PSD-via-autocorrelation. You have exactly
the same information. There's nothing superficial about it.
The trade-offs in bin width and sidelobe rejection available by
windowing are available in both approaches as well.
Dale B. Dalrymple
Reply by Jerry Avins●March 2, 20092009-03-02
HardySpicer wrote:
> On Mar 2, 6:05 pm, "m26k9" <maduranga.liyan...@gmail.com> wrote:
>> 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.
>
> You can say the same about any discrete process. Everything sampled is
> an approximation to analogue.
So you reject the sampling theorem? :-)
Jerry
--
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������
Reply by Steve Pope●March 2, 20092009-03-02
HardySpicer <gyansorova@gmail.com> wrote:
>On Mar 2, 6:49�pm, spop...@speedymail.org (Steve Pope) wrote:
>> 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. �
>Surely the cc is also an approximation? It is found from samples and
>must be?
I think for the PSD calculation, your frequency resolution/accuracy
is on the order of the inverse of the original sample length (in this
case, 0.001 Hz) whereas for the case using the DFT it's only
on the order of the inverse of the DFT size.
Whether this means you've lost information with the DFT I'm
not entirely sure. At least superficially you don't have
the same information.
Steve
Reply by HardySpicer●March 2, 20092009-03-02
On Mar 2, 6:49�pm, spop...@speedymail.org (Steve Pope) wrote:
> m26k9 <maduranga.liyan...@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
Surely the cc is also an approximation? It is found from samples and
must be?
Reply by HardySpicer●March 2, 20092009-03-02
On Mar 2, 6:05�pm, "m26k9" <maduranga.liyan...@gmail.com> wrote:
> 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.
You can say the same about any discrete process. Everything sampled is
an approximation to analogue.
Hardy