Time-Frequency Resolution, Uncertainty Principle

Hi

I am little bit confused about the issue of time-frequency resolution
and the principle of uncertainty. The latter states that product
DT*Dw, where DT, Dw denotes respectively the time and frequency
resolution is bounded. If the time resolution *increases* then the
frequency resolution *decreases*

However i am confused in the case where we interpolate a signal, then
its time resolution *increases*. Since we get larger number of samples,
the frequency resolution *increases* as well. The fourier transform will
contain larger number of samples spanning the interval [-pi,pi].
Therefore the frequency resolution has increased even that the time
resolution has increased as well which goes against the uncertainty
principle.


I can't see what's wrong with my analysis? any help .










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"Hristo Stevic" <hristostev@yahoo.com> wrote in message
news:d32b134353ea65fb7da3e1a13028a28c.52609@mygate.mailgate.org...
> Hi
>
> I am little bit confused about the issue of time-frequency resolution
> and the principle of uncertainty. The latter states that product
> DT*Dw, where DT, Dw denotes respectively the time and frequency
> resolution is bounded. If the time resolution *increases* then the
> frequency resolution *decreases*
>
> However i am confused in the case where we interpolate a signal, then
> its time resolution *increases*. Since we get larger number of samples,
> the frequency resolution *increases* as well. The fourier transform will
> contain larger number of samples spanning the interval [-pi,pi].
> Therefore the frequency resolution has increased even that the time
> resolution has increased as well which goes against the uncertainty
> principle.
>
>
> I can't see what's wrong with my analysis? any help .

Hristo Stevic,

With interpolation, the number of points increases but the resolution
doesn't.  This means that a given resolution will remain.
Example:
- two sinusoids are 1Hz apart
- they are sampled for .2 seconds giving around 5Hz resolution.
Spectral analysis of the .2 second record won't resolve the two sinusoids
because the resolution is 5Hz.
If the frequency domain is interpolated, you still won't be able to resolve
the two sinusoids because there's still only 0.2 seconds of input data.  The
sinusoids will still be smushed together (that's the technical term for it).

Same thing if we interpolate in time.  We add zeros to the spectrum - is at
least one way.  There's still only the data that we originally had in
frequency.

That's a quickie answer at least.

Fred

Hristo Stevic wrote:

> Hi
>
> I am little bit confused about the issue of time-frequency resolution
> and the principle of uncertainty. The latter states that product
> DT*Dw, where DT, Dw denotes respectively the time and frequency
> resolution is bounded. If the time resolution *increases* then the
> frequency resolution *decreases*
>
> However i am confused in the case where we interpolate a signal, then
> its time resolution *increases*.

Interpolation doesn't really add new samples.  Interpolation lets you find
peaks
between the grid points.

You really need to be careful about "uncertainty principles" in classical
macroscopic systems.

In most sorts of audio to RF signals you can measure both the amplitude and
phase simultaneously at the output of a DFT bin fairly easily. The
precision of the measurement is bounded by the signal to noise ratio.  In
some cases, at a high enough SNR, you can resolve differences that the
"uncertainty principle" would tell you that you can't.

I'll give you a real easy example,  we have 2 signals   A sin(wt)  and
(-A) sin(wt) .

If are constrained to look at the magnitude squared of a DFT bin centered
at radian frequency w, you can't tell the difference between the two, but
these 2 signals are the basic examples in binary communications.  They have
exactly the same frequency and the ability to distinguish between them is a
function of local clock synchronized to the transmitter's clock.

In a quantum mechanical system, the measurement has to be real.  There is a
significant difference between a classical and quantum mechanical
measurement.





> Since we get larger number of samples,
> the frequency resolution *increases* as well. The fourier transform will
> contain larger number of samples spanning the interval [-pi,pi].
> Therefore the frequency resolution has increased even that the time
> resolution has increased as well which goes against the uncertainty
> principle.
>
> I can't see what's wrong with my analysis? any help .
>
> --
> Posted via Mailgate.ORG Server - http://www.Mailgate.ORG

Thanks for both of you, however i am still confused....

I was speaking about the fourier transform of a discrete signal. 
My point is that if we upsample the signal, we ll get more samples at
the frequency
. Thus higher frequency resolution. you can see that the time resolution
has not changed also. However, the uncertainty principle claims that
these two types of resolution are inversely proportional :-(

so what i am missing...

you can see from the DFT expression, the frequency resolution is defined
by 2pi/N, where N is the number of input signal sample
Thus if N increases, the time resolution increases,also did the
frequency resolution


with decimation, time resolution decreases and also the frequnecy
resolution
decreases [the number of samples in both domains diminish]

with interpolation, time resolution increases and also the frequnecy
resolution
increases [the number of samples in both domains increase]


these two cases contradict the uncertainty principle according to my
analysis





-- 
Posted via Mailgate.ORG Server - http://www.Mailgate.ORG

"Hristo Stevic" <hristostev@yahoo.com> wrote in message news:<ed0641c77d39163f7c4d348e1db9e921.52609@mygate.mailgate.org>...
> Thanks for both of you, however i am still confused....
> 
> I was speaking about the fourier transform of a discrete signal. 
> My point is that if we upsample the signal, we ll get more samples at
> the frequency
> . Thus higher frequency resolution.

You have to distinguish between the "true smples" that were actually 
measured or otherwise used to produce your original time series, and 
the "artifical points" you introduce to the sequence by interpolating 
your original data sequence.

> you can see that the time resolution
> has not changed also. However, the uncertainty principle claims that
> these two types of resolution are inversely proportional :-(

"Resolution" refers to the original data sequence. Assume you sample 
some time series at, say, 1 Hz at times 0s, 1s, 2s, 3s,...  and that 
the resulting discrete sequence looks nice and stationary. Assume then, 
that if you had sampled it at 4 Hz, at times 0s, 0.25s, 0.5s,0.75s,...
you had detected a single peak sample at, say 0.5s. There are no way 
an interpolation scheme can reproduce that from your 1 Hz data.

One can of course say that "that's not valid, if the peak was at 1s 
the 1 Hz data would show it!". Well, the uncertainty principle states
what it takes to be guaranteed to find a feature. 

> so what i am missing...
> 
> you can see from the DFT expression, the frequency resolution is defined
> by 2pi/N, where N is the number of input signal sample
> Thus if N increases, the time resolution increases,also did the
> frequency resolution

No. There is a trick called "zero padding" that often is interpreted as 
"increasing the resolution of the spectrum" but in reality only 
interpolates it. 

To see the difference, try the following exercise:

Generate a data sequence that consists of two sinusoidals at f1=0.14*fs
and 0.20*fs (fs is sampling frequency). Let both sinusoidals have unit 
amplitude. Make the "original" sequence length N=8 samples. Next, 
zero pad (i.e. aappend 1016 zeroes at the end) this sequence to a total 
length of 1024 samples. Compute the DFT of both the original and the 
zero padded sequence and plot them in the same plot. Make sure both 
spectra use the frequency range [0,fs>. 

Repeat the above but with N=32 original samples. Compare the two plots, 
particularly in the [0.14*fs,0.20*fs] range. What you see is due to 
the uncertainty principle.

> 
> with decimation, time resolution decreases and also the frequnecy
> resolution
> decreases [the number of samples in both domains diminish]
> 
> with interpolation, time resolution increases and also the frequnecy
> resolution
> increases [the number of samples in both domains increase]
> 
> 
> these two cases contradict the uncertainty principle according to my
> analysis

Try the exercise above, and you will see that the uncertainty principle 
only relates to actually measured data points and is not affected 
by artificially introduced samples.

Rune

"Fred Marshall" <fmarshallx@remove_the_x.acm.org> wrote in message
news:hCoMa.2112$Jk5.1132401@feed2.centurytel.net...
>
> "Hristo Stevic" <hristostev@yahoo.com> wrote in message
> news:d32b134353ea65fb7da3e1a13028a28c.52609@mygate.mailgate.org...
> > Hi
> >
> > I am little bit confused about the issue of time-frequency resolution
> > and the principle of uncertainty. The latter states that product
> > DT*Dw, where DT, Dw denotes respectively the time and frequency
> > resolution is bounded. If the time resolution *increases* then the
> > frequency resolution *decreases*
> >
> > However i am confused in the case where we interpolate a signal, then
> > its time resolution *increases*. Since we get larger number of samples,
> > the frequency resolution *increases* as well. The fourier transform will
> > contain larger number of samples spanning the interval [-pi,pi].
> > Therefore the frequency resolution has increased even that the time
> > resolution has increased as well which goes against the uncertainty
> > principle.
> >
> >
> > I can't see what's wrong with my analysis? any help .
>
> Hristo Stevic,
>
> With interpolation, the number of points increases but the resolution
> doesn't.  This means that a given resolution will remain.
> Example:
> - two sinusoids are 1Hz apart
> - they are sampled for .2 seconds giving around 5Hz resolution.
> Spectral analysis of the .2 second record won't resolve the two sinusoids
> because the resolution is 5Hz.
> If the frequency domain is interpolated, you still won't be able to
resolve
> the two sinusoids because there's still only 0.2 seconds of input data.
The
> sinusoids will still be smushed together (that's the technical term for
it).
>
> Same thing if we interpolate in time.  We add zeros to the spectrum - is
at
> least one way.  There's still only the data that we originally had in
> frequency.
>
> That's a quickie answer at least.

One that isn't completely true.  Consider that you might have 0.2s samples
of such sinusoids.  Maybe sampled at 1MHz with a 30 bit A/D converter, and
that there is no (measurable) noise in the sinusoids.   I believe, then,
that you will have plenty enough information to separate the sinusoids.   (I
overdid the numbers slightly to emphasize the point.)

It is only when you can't measure the signal itself that the uncertainty
principle comes in.

There was a story about someone designing a complicated RADAR system,
believing in the uncertainty principle decided that it couldn't be built.
It was something like using a radar to measure the speed and distance of an
airplane.  You are, of course, limited in the measurement of the airplane.
It is that, and not the measurement of the phase and amplitude of the
reflected electromagnetic wave, that is limiting.

OK, consider this one, maybe applicable to DSP.  The amplitude of the
vibration of your eardrum for the softest sounds you can hear is much less
than the diameter of an atom.  With the uncertainty prinicple you wouldn't
be able to measure such an atom.  But the vibration averaged over the many
atoms of your eardrum is measurable.

In the RADAR case, it is not one photon that is being measured, and would be
limited by the uncertainty principle, but an average over many photons, and
the fact that you can directly measure the electromagentic field.

-- glen

"Fred Marshall" <fmarshallx@remove_the_x.acm.org> wrote in message
news:hCoMa.2112$Jk5.1132401@feed2.centurytel.net...

> Example:
> - two sinusoids are 1Hz apart
> - they are sampled for .2 seconds giving around 5Hz resolution.
> Spectral analysis of the .2 second record won't resolve the two sinusoids
> because the resolution is 5Hz.
> If the frequency domain is interpolated, you still won't be able to
resolve
> the two sinusoids because there's still only 0.2 seconds of input data.
The
> sinusoids will still be smushed together (that's the technical term for
it).

To continue my previous post, yes, spectral analysis won't.  If, for example
you put the signal through a balanced mixer with an appropriate LO, you
would have a hard time distinguishing them.

But if you directly measure the sinusoids amplitude as a function of time,
and you know that only two sinusoids are possible, (or a linear combination)
you can separate them.

-- glen

"Hristo Stevic" <hristostev@yahoo.com> wrote in message news:<d32b134353ea65fb7da3e1a13028a28c.52609@mygate.mailgate.org>...
> Hi
> 
> I am little bit confused about the issue of time-frequency resolution
> and the principle of uncertainty. The latter states that product
> DT*Dw, where DT, Dw denotes respectively the time and frequency
> resolution is bounded. 

I cannot find any such statement in "Modern Physics" by Serway et al.
Can you please provide a very precise and specific reference? Serway
talks of uncertainty in energy and time, and in momentum and position
(which I believe is the standard form of Heisenberg's uncertainty
principle), but not between time and frequency. 

> If the time resolution *increases* then the
> frequency resolution *decreases*

I submit that the frequency of a perfect noiseless complex sinusoid can 
be measured perfectly in two samples: measure the phase at sample 1 and 
the phase at sample 2 and then use 2*pi*f = (phase 2 - phase 1)/T, where
the phases are in radians and T is the sample period in seconds. 

> However i am confused in the case where we interpolate a signal, then
> its time resolution *increases*. Since we get larger number of samples,
> the frequency resolution *increases* as well. The fourier transform will
> contain larger number of samples spanning the interval [-pi,pi].
> Therefore the frequency resolution has increased even that the time
> resolution has increased as well which goes against the uncertainty
> principle.
> 
> 
> I can't see what's wrong with my analysis? any help .

The bin width of the DFT decreases as the number of samples
increases. This is due to the process involved in the DFT and
not some inherent limitation in frequency measurement, as the
gedanken I suggested above shows. I think your basic problem
(and many others have made the same mistake) is thinking that
the uncertainty that crops up in physics has some equivalent
in signal processing. I assert that it does not.

"Hristo Stevic" <hristostev@yahoo.com> wrote in message news:<d32b134353ea65fb7da3e1a13028a28c.52609@mygate.mailgate.org>...
> Hi
> 
> I am little bit confused about the issue of time-frequency resolution
> and the principle of uncertainty. The latter states that product
> DT*Dw, where DT, Dw denotes respectively the time and frequency
> resolution is bounded. 

I cannot find any such statement in "Modern Physics" by Serway et al.
Can you please provide a very precise and specific reference? Serway
talks of uncertainty in energy and time, and in momentum and position
(which I believe is the standard form of Heisenberg's uncertainty
principle), but not between time and frequency. 

> If the time resolution *increases* then the
> frequency resolution *decreases*

I submit that the frequency of a perfect noiseless complex sinusoid can 
be measured perfectly in two samples: measure the phase at sample 1 and 
the phase at sample 2 and then use 2*pi*f = (phase 2 - phase 1)/T, where
the phases are in radians and T is the sample period in seconds. 

> However i am confused in the case where we interpolate a signal, then
> its time resolution *increases*. Since we get larger number of samples,
> the frequency resolution *increases* as well. The fourier transform will
> contain larger number of samples spanning the interval [-pi,pi].
> Therefore the frequency resolution has increased even that the time
> resolution has increased as well which goes against the uncertainty
> principle.
> 
> 
> I can't see what's wrong with my analysis? any help .

The bin width of the DFT decreases as the number of samples
increases. This is due to the process involved in the DFT and
not some inherent limitation in frequency measurement, as the
gedanken I suggested above shows. I think your basic problem
(and many others have made the same mistake) is thinking that
the uncertainty that crops up in physics has some equivalent
in signal processing. I assert that it does not.

Randy Yates wrote:
> 
  ...
> 
> I submit that the frequency of a perfect noiseless complex sinusoid can
> be measured perfectly in two samples: measure the phase at sample 1 and
> the phase at sample 2 and then use 2*pi*f = (phase 2 - phase 1)/T, where
> the phases are in radians and T is the sample period in seconds.
> 
  ...

Isn't the amplitude also needed? A*sin(Bt) has two unknowns. Perhaps you
mean two complex samples?

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