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### Frequency Resolution of the DFT

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```Hello,

I am trying to determine the theoretical frequency resolution of the DFT. I
am sub-sampling an RF sign wave with a 8 bit ADC. I understand the
restrictions of the FFT's frequency resolution due to the nature of the
algorithm, but a DFT can compare the sampled sine wave to an infinite amount
different frequencies and take the best fit.

I think the answer to this question has something to do with the resolution
of the ADC (8 bit in my case), acquisition time, the amount of samples, and
noise in the system. I have yet to see a clear mathematical formula to
explain this. Please remember that I am sampling a filtered RF sine wave
that is AC coupled into the ADC, and I already know roughly the frequency of
the acquired sine wave. I have all the processing time in the world so a DFT
is ideal for this situation.

Thanks to all that respond.

```
 0
Reply Thomas 10/29/2003 6:46:39 PM

See related articles to this posting

```Thomas Magma wrote:

> Hello,
>
> I am trying to determine the theoretical frequency resolution of the DFT. I
> am sub-sampling an RF sign wave with a 8 bit ADC. I understand the
> restrictions of the FFT's frequency resolution due to the nature of the
> algorithm, but a DFT can compare the sampled sine wave to an infinite amount
> different frequencies and take the best fit.
>
> I think the answer to this question has something to do with the resolution
> of the ADC (8 bit in my case), acquisition time, the amount of samples, and
> noise in the system. I have yet to see a clear mathematical formula to
> explain this. Please remember that I am sampling a filtered RF sine wave
> that is AC coupled into the ADC, and I already know roughly the frequency of
> the acquired sine wave. I have all the processing time in the world so a DFT
> is ideal for this situation.
>
> Thanks to all that respond.

I can't connect with comparing the sampled signal with an infinite
number of frequencies. How is that conceivable with a finite amount of
computation?

Frequency is the reciprocal of period. The frequency resolution of a
Fourier transform is the reciprocal of the time encompassed by the
samples fed to it. (When sampling for one second, frequencies of one Hz
can be resolved. When sampling for a minute, 1/60th Hz can be resolved.
Assuming always, that the Nyquist criterion is met.

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

```
 0
Reply Jerry 10/29/2003 8:30:00 PM

```Jerry Avins wrote:

> Thomas Magma wrote:
>
>> Hello,
>>
>> I am trying to determine the theoretical frequency resolution of the
>> DFT. I
>> am sub-sampling an RF sign wave with a 8 bit ADC. I understand the
>> restrictions of the FFT's frequency resolution due to the nature of the
>> algorithm, but a DFT can compare the sampled sine wave to an infinite
>> amount
>> different frequencies and take the best fit.
>>
>> I think the answer to this question has something to do with the
>> resolution
>> of the ADC (8 bit in my case), acquisition time, the amount of
>> samples, and
>> noise in the system. I have yet to see a clear mathematical formula to
>> explain this. Please remember that I am sampling a filtered RF sine wave
>> that is AC coupled into the ADC, and I already know roughly the
>> frequency of
>> the acquired sine wave. I have all the processing time in the world so
>> a DFT
>> is ideal for this situation.
>>
>> Thanks to all that respond.
>
>
> I can't connect with comparing the sampled signal with an infinite
> number of frequencies. How is that conceivable with a finite amount of
> computation?
>
> Frequency is the reciprocal of period. The frequency resolution of a
> Fourier transform is the reciprocal of the time encompassed by the
> samples fed to it. (When sampling for one second, frequencies of one Hz
> can be resolved. When sampling for a minute, 1/60th Hz can be resolved.
> Assuming always, that the Nyquist criterion is met.
>
> Jerry

I neglected your question (surmise?) about converter resolution. With
infinite converter accuracy, a DFT's frequency resolution is

(sample frequency)/(number of samples).

Noise and finite converter accuracy and resolution keep that limit from
being reached. It takes good measurements to distinguish a 5/16-18 bolt
from am M8-1.4. Thin nuts will fit either, but thicker ones -- the
analog of longer sample times -- won't.

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

```
 0
Reply Jerry 10/29/2003 8:51:50 PM

```All an FFT does is multiple a known frequency (sine table) to an unknown
frequency (sampled data) and plot the real/imaginary vector magnitude of
each result. An FFT quantitively  increments the "known" frequency part
which limits the frequency resolution. A DFT is not restricted by this
quantitive increment and can increment it's known frequency on a much more
finite scale resulting in a higher frequency resolution.

Theoretically you should be able to achieve infinite frequency resolution
from only two samples of a sine wave given that you are sampling a perfect
sine wave with no noise and are using an ADC with 2^infinite resolution, and
also you know roughly the frequency so as to eliminate the images.

"Jerry Avins" <jya@ieee.org> wrote in message
news:bnp80a\$5vj\$1@bob.news.rcn.net...
> Thomas Magma wrote:
>
> > Hello,
> >
> > I am trying to determine the theoretical frequency resolution of the
DFT. I
> > am sub-sampling an RF sign wave with a 8 bit ADC. I understand the
> > restrictions of the FFT's frequency resolution due to the nature of the
> > algorithm, but a DFT can compare the sampled sine wave to an infinite
amount
> > different frequencies and take the best fit.
> >
> > I think the answer to this question has something to do with the
resolution
> > of the ADC (8 bit in my case), acquisition time, the amount of samples,
and
> > noise in the system. I have yet to see a clear mathematical formula to
> > explain this. Please remember that I am sampling a filtered RF sine wave
> > that is AC coupled into the ADC, and I already know roughly the
frequency of
> > the acquired sine wave. I have all the processing time in the world so a
DFT
> > is ideal for this situation.
> >
> > Thanks to all that respond.
>
> I can't connect with comparing the sampled signal with an infinite
> number of frequencies. How is that conceivable with a finite amount of
> computation?
>
> Frequency is the reciprocal of period. The frequency resolution of a
> Fourier transform is the reciprocal of the time encompassed by the
> samples fed to it. (When sampling for one second, frequencies of one Hz
> can be resolved. When sampling for a minute, 1/60th Hz can be resolved.
> Assuming always, that the Nyquist criterion is met.
>
> Jerry
> --
> Engineering is the art of making what you want from things you can get.
> �����������������������������������������������������������������������
>

```
 0
Reply Thomas 10/29/2003 9:18:56 PM

```Jerry Avins wrote:
>
> Frequency is the reciprocal of period. The frequency resolution of a
> Fourier transform is the reciprocal of the time encompassed by the
> samples fed to it. (When sampling for one second, frequencies of one Hz
> can be resolved. When sampling for a minute, 1/60th Hz can be resolved.

No, that's the frequency *spacing* for the Fourier frequencies.  One
can resolve frequencies much closer than that if the S/N is high
enough.  The DFT evaluated at non-Fourier frequencies must be used,
and if the signal has harmonics it has to be massaged in a nontrivial
way.

For a single sinusoid with Gaussian noise, you'll find the formulas
you need by looking up parametric frequency estimation.  I think the
simplest way to get at them is from the Bayesian approach of Jaynes
and Bretthorst, though there are more conventional approaches.  A
quick reference for the needed formulas is Bretthorst's book at

http://bayes.wustl.edu/

He also shows what to do if you have harmonics (and I summarize
some of the algorithm in http://www.museweb.com/bha.pdf).  The
standard forumulas assume a single sinusoid and Gaussian noise
with known sigma.  For your case, presuming the only "noise" is
from the converter, I'd use the standard sqrt(12) formula to
get an equivalent sigma from your ADC resolution and plug that
into the formula.  It won't be exact, but it should be in the
ballpark of what you can hope to achieve.

-Tom

--

To respond by email, replace "somewhere" with "astro" in the
```
 0
Reply Tom 10/29/2003 9:25:34 PM

```Thomas Magma wrote:

> All an FFT does is multiple a known frequency (sine table) to an unknown
> frequency (sampled data) and plot the real/imaginary vector magnitude of
> each result. An FFT quantitively  increments the "known" frequency part
> which limits the frequency resolution. A DFT is not restricted by this
> quantitive increment and can increment it's known frequency on a much more
> finite scale resulting in a higher frequency resolution.

Where does that notion come from? It isn'r valid. "DFT" stands for
"discrete Fourier transform"; an operation on a series of numbers,
rather than a continuous function. FFTs -- fast Fourier transforms --
are a a group of algorithms that perform DFTs with good computational
efficiency. They don't differ from DFTs, they _are_ DFTs.

> Theoretically you should be able to achieve infinite frequency resolution
> from only two samples of a sine wave given that you are sampling a perfect
> sine wave with no noise and are using an ADC with 2^infinite resolution, and
> also you know roughly the frequency so as to eliminate the images.

If you know that you have a pure sinusoid, you don't need an FT. That's
a different (and not usually useful) subject.

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

```
 0
Reply Jerry 10/30/2003 4:18:00 AM

```Tom Loredo wrote:

> Jerry Avins wrote:
>
>>Frequency is the reciprocal of period. The frequency resolution of a
>>Fourier transform is the reciprocal of the time encompassed by the
>>samples fed to it. (When sampling for one second, frequencies of one Hz
>>can be resolved. When sampling for a minute, 1/60th Hz can be resolved.
>
>
> No, that's the frequency *spacing* for the Fourier frequencies.  One
> can resolve frequencies much closer than that if the S/N is high
> enough.  The DFT evaluated at non-Fourier frequencies must be used,
> and if the signal has harmonics it has to be massaged in a nontrivial
> way.

You can indeed resolve more detail, but with techniques that extend
Fourier transforms. The transform itself doesn't go there.

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

```
 0
Reply Jerry 10/30/2003 4:35:04 AM

```Jerry Avins wrote:
> If you know that you have a pure sinusoid, you don't need an FT. That's
> a different (and not usually useful) subject.

The case where the signal contains a *bounded* number of (possibly
decaying, i.e. complex-frequency) pure sinusoids is, actually, quite an
important and well-studied case.  (e.g. to extract eigenfrequencies and
damping rates from time series, either from experiment or from simulation.)

As you say, in that case, you don't want a Fourier transform per se,
since the FT assumes that you know basically nothing about your signal.
A beautiful and very readable paper on the subject is:

V. A. Mandelshtam and H. S. Taylor, "Harmonic inversion of time
signals," J. Chem. Phys., vol. 107, no. 17, p. 6756-6769 (Nov. 1
1997).  See also erratum, ibid, vol. 109, no. 10, p. 4128 (Sep. 8
1998).

(Note that directly doing a least-squares fit is usually a very poorly
conditioned approach.)

Cordially,
Steven G. Johnson

```
 0
Reply Steven 10/30/2003 7:41:38 AM

```Maybe I should be leaving Jean Fourier's name out of this to try to explain
myself. I'm talking about multiplying a band of frequencies (spectrum
analysis with a fixed span) with sampled data of a single sinusoidal
frequency and plotting the result and then determining frequency
resolution/accuracy.

So lets say I sampled a 10MHz sinusoidal signal which I know is accurate to
+/- 5 ppm. My reference to my ADC is known and accurate and I trying to
determine the actual frequency of the 10MHz. So I will want to analyze a
span of 100 Hz centered around the 10MHz which is equivalent to the +/-5ppm.
Without using a FFT, I can mathematical generate any frequency in that band
and multiply it with my captured data and then simply take the greatest
vector magnitude to determine the actual frequency of the 10MHz signal.

What I'm trying to determine is the limitations of this approach. If I
choose to break that 100 Hz into 100,1000, or 10000 discrete frequencies and
multiply it with my data, what is the advantage.? And where does the ADC bit
resolution, amount of samples and sample length fit in?

I hope that I am making myself clear enough as to what I am trying to
achieve.

"Jerry Avins" <jya@ieee.org> wrote in message
news:bnq3dq\$9tt\$1@bob.news.rcn.net...
> Thomas Magma wrote:
>
> > All an FFT does is multiple a known frequency (sine table) to an unknown
> > frequency (sampled data) and plot the real/imaginary vector magnitude of
> > each result. An FFT quantitively  increments the "known" frequency part
> > which limits the frequency resolution. A DFT is not restricted by this
> > quantitive increment and can increment it's known frequency on a much
more
> > finite scale resulting in a higher frequency resolution.
>
> Where does that notion come from? It isn'r valid. "DFT" stands for
> "discrete Fourier transform"; an operation on a series of numbers,
> rather than a continuous function. FFTs -- fast Fourier transforms --
> are a a group of algorithms that perform DFTs with good computational
> efficiency. They don't differ from DFTs, they _are_ DFTs.
>
> > Theoretically you should be able to achieve infinite frequency
resolution
> > from only two samples of a sine wave given that you are sampling a
perfect
> > sine wave with no noise and are using an ADC with 2^infinite resolution,
and
> > also you know roughly the frequency so as to eliminate the images.
>
> If you know that you have a pure sinusoid, you don't need an FT. That's
> a different (and not usually useful) subject.
>
> Jerry
> --
> Engineering is the art of making what you want from things you can get.
> �����������������������������������������������������������������������
>

```
 0
Reply Thomas 10/30/2003 8:02:02 PM

```Jerry Avins wrote:
>
> You can indeed resolve more detail, but with techniques that extend
> Fourier transforms. The transform itself doesn't go there.

I think you need to be more precise with your terminology.  The DFT
is defined as a sum evaluated only at the Fourier frequencies.  But the
sum itself (which some authors call the discrete time Fourier transform)
can be evaluated at any frequencies between the Fourier frequencies.
There is even a fast algorithm to do it (using the fractional FFT).
So if by "the transform" you mean the DFT, you are right, and since
the OP phrased his question specifically in reference to the DFT,
I'll grant you that.  But if (as one might assume from your previous
sentence) you are referring to Fourier transforms in general, you can
evaluate them at any desired frequency, even for a finite, discretely
sampled time series.

In any case, the OP's question was about frequency resolution, and
the resolution of a frequency estimator calculated using Fourier methods
for a given data set has little to do with the Fourier spacing.  You
need to use at least that fine a spacing to *find* the peak in the
periodogram, but the actual peak location provides a frequency estimate
whose resolution depends on the S/N and can easily be much finer
than the Fourier spacing.

-Tom

--

To respond by email, replace "somewhere" with "astro" in the
```
 0
Reply Tom 10/30/2003 8:45:17 PM

```"Thomas Magma" <kholmes@beer.com> wrote in message
news:_6eob.226412\$pl3.60226@pd7tw3no...
> Maybe I should be leaving Jean Fourier's name out of this to try to
explain
> myself. I'm talking about multiplying a band of frequencies (spectrum
> analysis with a fixed span) with sampled data of a single sinusoidal
> frequency and plotting the result and then determining frequency
> resolution/accuracy.

This makes no sense to me...
Multiply a spectrum: "a band of frequencies (spectrum analysis with a fixed
span)"
with time samples "sampled data of a single sinusoidal frequency"

Fred

```
 0
Reply Fred 10/30/2003 8:49:33 PM

```Thanks Tom,

It  looks like we are on the same page. I should have been a little more
clearer in regards to my terminology. I am in fact (trying) to talk about
the 'frequencies between the Fourier frequencies'.

Have you seen any paper describing the limitations for frequency
resolution/accuracies that can be achieved by analyzing with sub Fourier
frequencies?

"Tom Loredo" <loredo@somewhere.cornell.edu> wrote in message
news:3FA1785D.EEB82A6B@somewhere.cornell.edu...
> Jerry Avins wrote:
> >
> > You can indeed resolve more detail, but with techniques that extend
> > Fourier transforms. The transform itself doesn't go there.
>
> I think you need to be more precise with your terminology.  The DFT
> is defined as a sum evaluated only at the Fourier frequencies.  But the
> sum itself (which some authors call the discrete time Fourier transform)
> can be evaluated at any frequencies between the Fourier frequencies.
> There is even a fast algorithm to do it (using the fractional FFT).
> So if by "the transform" you mean the DFT, you are right, and since
> the OP phrased his question specifically in reference to the DFT,
> I'll grant you that.  But if (as one might assume from your previous
> sentence) you are referring to Fourier transforms in general, you can
> evaluate them at any desired frequency, even for a finite, discretely
> sampled time series.
>
> In any case, the OP's question was about frequency resolution, and
> the resolution of a frequency estimator calculated using Fourier methods
> for a given data set has little to do with the Fourier spacing.  You
> need to use at least that fine a spacing to *find* the peak in the
> periodogram, but the actual peak location provides a frequency estimate
> whose resolution depends on the S/N and can easily be much finer
> than the Fourier spacing.
>
> -Tom
>
> --
>
> To respond by email, replace "somewhere" with "astro" in the

```
 0
Reply Thomas 10/30/2003 8:59:19 PM

```Jerry Avins wrote:
>
> Where does that notion come from? It isn'r valid. "DFT" stands for
> "discrete Fourier transform"; an operation on a series of numbers,
> rather than a continuous function. FFTs -- fast Fourier transforms --
> are a a group of algorithms that perform DFTs with good computational
> efficiency. They don't differ from DFTs, they _are_ DFTs.

Thomas, just to amplify Jerry's point here, "DFT" stands for a particular
discrete represetnation of the Fourier transform of a finite data set.
It is true that you can evaluate the transform at non-Fourier frequencies,
but the DFT itself is by definition the set of transform values at those
frequencies.  Those frequencies are important for both analytic and statistical
reasons.  Analytically, if you know the transform at just those frequencies,
you can calculate it any any other frequencies; that is, though the transform
is a continuous function of frequency, the function is determined by just
the N values at the Fourier frequencies.  Statistically, those N transform
values are statistically independent, which makes certain statistical
calculations a lot easier.

That said, even though the continuous transform (I think some authors
call it the discrete time fourier transform, DTFT) is determined by
its value at the Fourier frequencies, it is true that in some applications
one may want to know the values between them.  Frequency estimation is
one such application.

I think your use of "DFT" was a red herring and that it led the discussion
astray.  DTFT might have been better, though perhaps what you really
should have used is "periodogram."  This is basically the squared
magnitude of the transform, but in statistics it's often considered
as a continuous function of frequency.  The periodogram is all you
need to do good frequency estimation if there is only a single
sinusoid in the data.  For multiple sinusoids, you need the phase
information in the full DTFT, not just the modulus.

> If you know that you have a pure sinusoid, you don't need an FT. That's
> a different (and not usually useful) subject.

Well, that depends on the problem domain.  If it's just a pure sinusoid,
you may not "need" an FT, but you'll end up calculating it to get the
optimal answer (i.e., the function you'll calculate to do least squares
or max. likelihood frequency estimation is known to be proportional to the
periodogram).  If you allow a finite number of sinusoids, possibly >1,
then the problem arises in lots of places.  I'm an astronomer and we
encounter it a lot in astronomy.  It also arises in audio testing,
where the test signal has one or more sinusoids, and the output
has those sinusoids plus others as distortion products.  Many pieces
of hardware and software use DFTs in various ways to measure this stuff.

-Tom

--

To respond by email, replace "somewhere" with "astro" in the
```
 0
Reply Tom 10/30/2003 9:00:00 PM

```> I think your use of "DFT" was a red herring and that it led the discussion
> astray.

I agree. I retract the use of "DFT".

I am talking about the frequencies between the Fourier frequencies.

Tom, I think you know what I am babbling about, so what the hell would it be
called. You say the DTFT uses Fourier frequencies and I'm not talking about
that. How about Discrete Frequency Transform?...oh wait...that could get
confusing:)

"Tom Loredo" <loredo@somewhere.cornell.edu> wrote in message
news:3FA17BD0.4F8C0617@somewhere.cornell.edu...
> Jerry Avins wrote:
> >
> > Where does that notion come from? It isn'r valid. "DFT" stands for
> > "discrete Fourier transform"; an operation on a series of numbers,
> > rather than a continuous function. FFTs -- fast Fourier transforms --
> > are a a group of algorithms that perform DFTs with good computational
> > efficiency. They don't differ from DFTs, they _are_ DFTs.
>
> Thomas, just to amplify Jerry's point here, "DFT" stands for a particular
> discrete represetnation of the Fourier transform of a finite data set.
> It is true that you can evaluate the transform at non-Fourier frequencies,
> but the DFT itself is by definition the set of transform values at those
> frequencies.  Those frequencies are important for both analytic and
statistical
> reasons.  Analytically, if you know the transform at just those
frequencies,
> you can calculate it any any other frequencies; that is, though the
transform
> is a continuous function of frequency, the function is determined by just
> the N values at the Fourier frequencies.  Statistically, those N transform
> values are statistically independent, which makes certain statistical
> calculations a lot easier.
>
> That said, even though the continuous transform (I think some authors
> call it the discrete time fourier transform, DTFT) is determined by
> its value at the Fourier frequencies, it is true that in some applications
> one may want to know the values between them.  Frequency estimation is
> one such application.
>
> I think your use of "DFT" was a red herring and that it led the discussion
> astray.  DTFT might have been better, though perhaps what you really
> should have used is "periodogram."  This is basically the squared
> magnitude of the transform, but in statistics it's often considered
> as a continuous function of frequency.  The periodogram is all you
> need to do good frequency estimation if there is only a single
> sinusoid in the data.  For multiple sinusoids, you need the phase
> information in the full DTFT, not just the modulus.
>
> > If you know that you have a pure sinusoid, you don't need an FT. That's
> > a different (and not usually useful) subject.
>
> Well, that depends on the problem domain.  If it's just a pure sinusoid,
> you may not "need" an FT, but you'll end up calculating it to get the
> optimal answer (i.e., the function you'll calculate to do least squares
> or max. likelihood frequency estimation is known to be proportional to the
> periodogram).  If you allow a finite number of sinusoids, possibly >1,
> then the problem arises in lots of places.  I'm an astronomer and we
> encounter it a lot in astronomy.  It also arises in audio testing,
> where the test signal has one or more sinusoids, and the output
> has those sinusoids plus others as distortion products.  Many pieces
> of hardware and software use DFTs in various ways to measure this stuff.
>
> -Tom
>
> --
>
> To respond by email, replace "somewhere" with "astro" in the

```
 0
Reply Thomas 10/30/2003 9:21:14 PM

```Tom Loredo wrote:
> Jerry Avins wrote:
>
>>Where does that notion come from? It isn'r valid. "DFT" stands for
>>"discrete Fourier transform"; an operation on a series of numbers,
>>rather than a continuous function. FFTs -- fast Fourier transforms --
>>are a a group of algorithms that perform DFTs with good computational
>>efficiency. They don't differ from DFTs, they _are_ DFTs.
>
>
> Thomas, just to amplify Jerry's point here, "DFT" stands for a particular
> discrete represetnation of the Fourier transform of a finite data set.
> It is true that you can evaluate the transform at non-Fourier frequencies,
> but the DFT itself is by definition the set of transform values at those
> frequencies.  Those frequencies are important for both analytic and statistical
> reasons.  Analytically, if you know the transform at just those frequencies,
> you can calculate it any any other frequencies; that is, though the transform
> is a continuous function of frequency, the function is determined by just
> the N values at the Fourier frequencies.  Statistically, those N transform
> values are statistically independent, which makes certain statistical
> calculations a lot easier.

Statistically they are only independent as N goes to infinity
and the input process is white or if the process just happens to be made
up of a set of N random amplitude sinusoids that happend to have
frequencies on the exact centers of the DFT bins and you use a boxcar
window.

Spectral leakage <==> statistically dependent bins

The other thing is that any analytic function is uniquely defined by its
complex poles or as the math people like to say the residues. There
really isn't anything special about bin center frequencies unless the N
samples correspond to a segment of signal with period N.

>
> That said, even though the continuous transform (I think some authors
> call it the discrete time fourier transform, DTFT) is determined by
> its value at the Fourier frequencies, it is true that in some applications
> one may want to know the values between them.  Frequency estimation is
> one such application.
>
> I think your use of "DFT" was a red herring and that it led the discussion
> astray.  DTFT might have been better, though perhaps what you really
> should have used is "periodogram."  This is basically the squared
> magnitude of the transform, but in statistics it's often considered
> as a continuous function of frequency.  The periodogram is all you
> need to do good frequency estimation if there is only a single
> sinusoid in the data.  For multiple sinusoids, you need the phase
> information in the full DTFT, not just the modulus.
>
>
>>If you know that you have a pure sinusoid, you don't need an FT. That's
>>a different (and not usually useful) subject.
>
>
> Well, that depends on the problem domain.  If it's just a pure sinusoid,
> you may not "need" an FT, but you'll end up calculating it to get the
> optimal answer (i.e., the function you'll calculate to do least squares
> or max. likelihood frequency estimation is known to be proportional to the
> periodogram).  If you allow a finite number of sinusoids, possibly >1,
> then the problem arises in lots of places.  I'm an astronomer and we
> encounter it a lot in astronomy.  It also arises in audio testing,
> where the test signal has one or more sinusoids, and the output
> has those sinusoids plus others as distortion products.  Many pieces
> of hardware and software use DFTs in various ways to measure this stuff.
>
> -Tom
>

```
 0
Reply Stan 10/30/2003 9:44:09 PM

```Here is a chunk of Java code to help explain what I am doing. The math
should be able to explain better than my words.

It analyzes the spectrum between 9.999950 MHz and 10.000050 MHz using 10000
different frequencies giving .01 Hz resolution. The sampled data is stored
in a large array call sampleddata[50000] and is a result of sub-sampling a
10.000014215 MHz sinusoidal wave at a rate 266 KHz. The algorithm analyzes
all 50000 points of data per frequency.

public void spectrum(){
double frequency;
double time;
int[] plot = new int[10000];
double sintotal;
double costotal;
double pi = 3.14159265359;

for(i = 0; i<10000; i++){
frequency = 9999950 + (i * .01);
sintotal = 0.0;
costotal = 0.0;
for(j = 0; j<50000; j++){
time = j * 1/266000;
sintotal = sintotal + (sampleddata[j] * Math.sin(2*pi*frequency*time));
costotal = costotal + (sampleddata[j] * Math.cos(2*pi*frequency*time));
}
plot[i] = Math.sqrt((sintotal*sintotal) + (costotal*costotal));
}

And that's it!!
Now I realize that the side lobes (tails) will be high because I don't
implement a window filter but the CW (carrier wave) should always have the
greatest magnitude anyways, and all I'm interested in is an accurate measure
of the CW frequency. I also didn't convert the spectrum to Log.

My question is (and was), is this valid? Am I really getting .01 Hz
resolution. If so, what are the limits to this approach? And is there a
formula that governs this discrete frequency analysis approach?

"Tom Loredo" <loredo@somewhere.cornell.edu> wrote in message
news:3FA1785D.EEB82A6B@somewhere.cornell.edu...
> Jerry Avins wrote:
> >
> > You can indeed resolve more detail, but with techniques that extend
> > Fourier transforms. The transform itself doesn't go there.
>
> I think you need to be more precise with your terminology.  The DFT
> is defined as a sum evaluated only at the Fourier frequencies.  But the
> sum itself (which some authors call the discrete time Fourier transform)
> can be evaluated at any frequencies between the Fourier frequencies.
> There is even a fast algorithm to do it (using the fractional FFT).
> So if by "the transform" you mean the DFT, you are right, and since
> the OP phrased his question specifically in reference to the DFT,
> I'll grant you that.  But if (as one might assume from your previous
> sentence) you are referring to Fourier transforms in general, you can
> evaluate them at any desired frequency, even for a finite, discretely
> sampled time series.
>
> In any case, the OP's question was about frequency resolution, and
> the resolution of a frequency estimator calculated using Fourier methods
> for a given data set has little to do with the Fourier spacing.  You
> need to use at least that fine a spacing to *find* the peak in the
> periodogram, but the actual peak location provides a frequency estimate
> whose resolution depends on the S/N and can easily be much finer
> than the Fourier spacing.
>
> -Tom
>
> --
>
> To respond by email, replace "somewhere" with "astro" in the

```
 0
Reply Thomas 10/31/2003 1:24:39 AM

```"Tom Loredo" <loredo@somewhere.cornell.edu> wrote in message
news:3FA0304E.6A636416@somewhere.cornell.edu...
> Jerry Avins wrote:
> >
> > Frequency is the reciprocal of period. The frequency resolution of a
> > Fourier transform is the reciprocal of the time encompassed by the
> > samples fed to it. (When sampling for one second, frequencies of one Hz
> > can be resolved. When sampling for a minute, 1/60th Hz can be resolved.
>
> No, that's the frequency *spacing* for the Fourier frequencies.  One
> can resolve frequencies much closer than that if the S/N is high
> enough.  The DFT evaluated at non-Fourier frequencies must be used,
> and if the signal has harmonics it has to be massaged in a nontrivial
> way.
>
> For a single sinusoid with Gaussian noise, you'll find the formulas
> you need by looking up parametric frequency estimation.  I think the
> simplest way to get at them is from the Bayesian approach of Jaynes
> and Bretthorst, though there are more conventional approaches.  A
> quick reference for the needed formulas is Bretthorst's book at
>

I measure frequency in similar apps by applying the samples to an FM
demodulator / instantaneous frequency detector (typically after converting
to complex baseband) and averaging the output. It is a very simple method.

> http://bayes.wustl.edu/
>
> He also shows what to do if you have harmonics (and I summarize
> some of the algorithm in http://www.museweb.com/bha.pdf).  The
> standard forumulas assume a single sinusoid and Gaussian noise
> with known sigma.  For your case, presuming the only "noise" is
> from the converter, I'd use the standard sqrt(12) formula to
> get an equivalent sigma from your ADC resolution and plug that
> into the formula.  It won't be exact, but it should be in the
> ballpark of what you can hope to achieve.
>
> -Tom
>
> --
>
> To respond by email, replace "somewhere" with "astro" in the

```
 0
Reply John 10/31/2003 1:38:48 AM

```Thomas Magma wrote:

...

> What I'm trying to determine is the limitations of this approach. If I
> choose to break that 100 Hz into 100,1000, or 10000 discrete frequencies and
> multiply it with my data, what is the advantage.? And where does the ADC bit
> resolution, amount of samples and sample length fit in?

...

As is claimed in much of the spam I've been getting lately, size
matters. No finite set of samples can represent a pure sine wave. There
will be end effects that introduce vibrato. It's no good deciding to
include an integer number of cycles in the analyzed data; you don't know
how many samples that includes unless you already know the frequency,
and there's no reason to believe that the period of the waveform is an
integer multiple of the sampling interval. This is not to say that good
judgment can't reduce the end effects, but they are there to stay.

The problem you are trying to solve is analogous to finding that nut
which is the best fit on a threaded rod. If the diametral fit is very
close or the nut very long, its pitch must be very accurate. As the
clearance increases or the length decreases, the pitch mismatch can
increase. Read pitch as spatial frequency and length as number of
samples. Clearance depends on precision, degraded both by noise and the
limits of measurement.

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

```
 0
Reply Jerry 10/31/2003 6:13:21 AM

```Jerry Avins <jya@ieee.org> writes:

> The problem you are trying to solve is analogous to finding that nut
> which is the best fit on a threaded rod. If the diametral fit is very
> close or the nut very long, its pitch must be very accurate. As the
> clearance increases or the length decreases, the pitch mismatch can
> increase. Read pitch as spatial frequency and length as number of
> samples. Clearance depends on precision, degraded both by noise and the
> limits of measurement.

Great analogy, Jerry!

Ciao,

Peter K.

--
Peter J. Kootsookos

"I will ignore all ideas for new works [..], the invention of which
has reached its limits and for whose improvement I see no further
hope."

- Julius Frontinus, c. AD 84
```
 0
Reply p 10/31/2003 6:19:52 AM

```"Thomas Magma" <kholmes@beer.com> schrieb
> > I think your use of "DFT" was a red herring and that it led
> > the discussion astray.
>
> I agree. I retract the use of "DFT".
>
> I am talking about the frequencies between the Fourier
> frequencies.
>
> Tom, I think you know what I am babbling about, so what the hell
> would it be called. You say the DTFT uses Fourier frequencies
> and I'm not talking about that. How about Discrete Frequency
> Transform?...oh wait...that could get confusing:)
>

You are looking for the digital equivalent of a frequency indicator,
such as is often found on transportable power generators: lamella of
different lengths resonate more or less with the frequency of the
generated current. I don't remember how the mechnical devices are
called, for your device I'd suggest "digital resonator" :-).

You could run your data through a Goertzel Algorithm, varying the
frequency and see where the amplitude of your resonator is highest.

HTH
Martin

```
 0
Reply Martin 10/31/2003 10:53:50 AM

```
Fred Marshall wrote:
>
> "Thomas Magma" <kholmes@beer.com> wrote in message
> news:_6eob.226412\$pl3.60226@pd7tw3no...
> > Maybe I should be leaving Jean Fourier's name out of this to try to
> explain
> > myself. I'm talking about multiplying a band of frequencies (spectrum
> > analysis with a fixed span) with sampled data of a single sinusoidal
> > frequency and plotting the result and then determining frequency
> > resolution/accuracy.
>
> This makes no sense to me...
> Multiply a spectrum: "a band of frequencies (spectrum analysis with a fixed
> span)"
> with time samples "sampled data of a single sinusoidal frequency"

I think he is saying that given F(x) is the DFT of f(x) then his analysis
of the time samples  amounts to F[x[t]*exp(iw0t] where w0 is the estimated
frequency and the results are relative to that frequency. But it still
comes down to how many samples and how fine the sampling rate is.

-jim

-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
http://www.newsfeeds.com - The #1 Newsgroup Service in the World!
-----==  Over 100,000 Newsgroups - 19 Different Servers! =-----
```
 0
Reply jim 10/31/2003 1:19:55 PM

```"Peter J. Kootsookos" <p.kootsookos@remove.ieee.org> wrote in message
news:s68k76m2ak7.fsf@mango.itee.uq.edu.au...
> Jerry Avins <jya@ieee.org> writes:
>
> > The problem you are trying to solve is analogous to finding that nut
> > which is the best fit on a threaded rod. If the diametral fit is very
> > close or the nut very long, its pitch must be very accurate. As the
> > clearance increases or the length decreases, the pitch mismatch can
> > increase. Read pitch as spatial frequency and length as number of
> > samples. Clearance depends on precision, degraded both by noise and the
> > limits of measurement.
>
> Great analogy, Jerry!
>

Yeah, like Peter said!

Fred

```
 0
Reply Fred 10/31/2003 5:51:34 PM

```"Thomas Magma" <kholmes@beer.com> wrote in message
news:ehfob.227393\$pl3.65894@pd7tw3no...
> > I think your use of "DFT" was a red herring and that it led the
discussion
> > astray.
>
>...............
..so what the hell would it be called. You say the DTFT uses Fourier
frequencies and I'm not talking about
> that. How about Discrete Frequency Transform?...oh wait...that could get
> confusing:)

It is the good old classical continuous Fourier Transform.
This is computed over a continuous time function from -inf to +inf and is
normally computed to yield a continuous frequency function from inf to +inf.
The forward transform is a continuous integral.
The inverse transform is a continuous integral.

+inf
F(w)=int[f(t)*e^(-jwt) dt]; and you pick the values of w for which to
compute it, most often -inf to +inf
-inf

In contrast to Tom Loredo's comment, I believe the term "Discrete Time
Fourier Transform" (DTFT) is reserved for situations where the temporal
information is at discrete times and the Fourier Transform is continuous.
Thus the emphasis on "discrete time".
The forward transform is a discrete sum (taken at distinct times) over an
infinity of frequencies.
The inverse transform is a continuous integral taken from inf to +inf.

Starting with the Fourier Transform

+inf
F(w)=int[f(t)*e^(-jwt) dt]; and you pick the values of w for which to
compute it, most often -inf to +inf
-inf                    and then realize that f(t)=0 except at times
"Ti",
the points where f(t) is nonzero; i.e. the
samples
there is no necessity for the Ti to be spaced
regularly here
this simplifies to:

F(w)=sum[f(Ti)*e^(-jwTi)];  and you pick the values of w for which to
compute it, most often -inf to +inf
i                     i=1,N samples
Now the sum can be taken over a finite temporal epoch as long as the samples
cover a finite epoch.

Then, the "Discrete Fourier Transform" (DFT) is with discrete time samples
*and* discrete frequency samples.
The forward transform is a discrete sum (taken at distinct times) over a
finite temporal epoch computed at a discrete set of frequencies.
The inverse transform is a discrete sum (taken at distinct frequencies) over
a finite frequency period at a discrete set of times.

F(Wk)=sum[f(Ti)*e^(-jWiTi)]; k=1,M; i=1:N and generally M=N but I don't have
time to say why today
.....it gets into uniform sampling / shifting to periodic representations of
f(t) and F(w)......

In your case, you can use the Fourier Transform and you can choose whichever
frequencies may be of interest if you like.  (Computing over all frequencies
is but one choice but is necessary if you ever want to do an inverse
transform).
You can envision how this works by using the time samples that you have with
an infinite number of zeros inserted between samples and zeros at the
beginning and end of the sequence of samples from -inf to +inf.
Then you could use the DFT/FFT to compute something close couldn't you?  At
least it would be an interesting experiment.  Just add 255 zeros between
samples giving around 256*N samples and zero pad the ends with 10*256*N
zeros each.  Then compute the entire DFT/FFT or, similarly, compute the DFT
at frequencies of interest.

What is important about the situation you have and doing this?

- The sample inteval in time should be related to the highest frequency of
the underlying data that was sampled (Nyquist).  However, because the time
samples cover a finite temporal epoch, the Fourier Transform as described
above can have infinite extent.  The continuous transform itself is as if
the sample rate is infinite but with lots of zeros in the data.

- The temporal epoch determines the actual frequency resolution (or the
ability to resolve adjacent frequency components) that you obtain in the
Fourier Transform.  So, even though it's continuous, it can only change
value at a limited rate ... one way to think about it anyway.  If the
temporal epoch is of length L seconds, then real resolution of 1/L Hz is
about what you have.  However, if there is a single sinusoidal input then
you should expect to see the peak in frequency at the right place - there
may be pathological cases but I'm not aware of any in the simple situation
described here.

The idea of computing only certain frequency values lends itself to
envisioning how one might construct algorithms for finding a local peak,
etc.....  You might look at the Goertzel algorithm as an example.

I think this unifies the discussion if you can wade through it all - and
assuming I didn't make a dumb mistake here and there.

Fred

```
 0
Reply Fred 10/31/2003 6:49:40 PM

```"Tom Loredo" <loredo@somewhere.cornell.edu> wrote in message
news:3FA1785D.EEB82A6B@somewhere.cornell.edu...
> Jerry Avins wrote:
> >
> > You can indeed resolve more detail, but with techniques that extend
> > Fourier transforms. The transform itself doesn't go there.

> I think you need to be more precise with your terminology.  The DFT
> is defined as a sum evaluated only at the Fourier frequencies.  But the
> sum itself (which some authors call the discrete time Fourier transform)
> can be evaluated at any frequencies between the Fourier frequencies.
> There is even a fast algorithm to do it (using the fractional FFT).
> So if by "the transform" you mean the DFT, you are right, and since
> the OP phrased his question specifically in reference to the DFT,
> I'll grant you that.  But if (as one might assume from your previous
> sentence) you are referring to Fourier transforms in general, you can
> evaluate them at any desired frequency, even for a finite, discretely
> sampled time series.

This has been discussed before, with a variety of disagreements.  I don't
expect that to change.

In the beginning there was the Fourier series, where periodic functions were
represented as sums of harmonically related (integer multiples of a
fundamental) sines and cosines.   f(t) is periodic and continuous, F(w) is
not (necessarily) periodic, but is discrete (evaluated at only specific
frequencies).   The frequency points in radians/second, are spaced 2 pi/
period, though there may be an infinite number of them.

As the period is increased, lim T --> infinity, F(w)  becomes continuous,
the Fourier transform.  I remember this one well, as it took me a while to
learn it.  I had understood Fourier series for a long time before learning
Fourier transforms in my college physics class.   I didn't understand so
quickly that it was the limit as T --> infinity (possibly my TA's accent
slowed it down, or that class was at 8:00 AM.)    When I finally figured out
what was happening it seemed so obvious.

Now, from the symmetry of the transform one could also have a function f(t)
that was not periodic but discrete (sampled at specific points in time), and
expect a transform that, F(w) that is periodic and continuous.  For F(w) to
be continuous there must be an infinite number of sample points.

For discrete periodic functions f(t), one gets a discrete periodic
transform.  The periodicity is exactly that expected from Nyquist sampling
and aliasing.   Given a finite number of discrete sample points, as for
example the data on an audio CD, if one wants to apply Fourier analysis to
such data, the easiest and most obvious is to consider the signal as
periodic.  (All car CD players that I know of seem to believe this, too.  I
only have two samples, but both start over once they get to the end.)

Note that for the discrete periodic case, if N sample points go in, N
independent frequency points come out.  The transform can be represented as
an N by N matrix multiplied by N sample points generating N frequency
values.

If one makes assumptions about the points not sampled,  such as that they
are all zero for example, one can evaluate F(w) between the points it would
otherwise be defined at.

(replace (w) with (2 pi f) above if you like cycles/second instead of
radians/second.  You get a lot more 2 pi's to keep track of.)

> In any case, the OP's question was about frequency resolution, and
> the resolution of a frequency estimator calculated using Fourier methods
> for a given data set has little to do with the Fourier spacing.  You
> need to use at least that fine a spacing to *find* the peak in the
> periodogram, but the actual peak location provides a frequency estimate
> whose resolution depends on the S/N and can easily be much finer
> than the Fourier spacing.

I am not sure what you mean by Fourier spacing.   As Jerry said earlier, the
frequency resolution is (sample frequency)/(number of samples).    The
sample frequency, f=1/T, so the frequency resolution is 1/( T * number of
samples), which is the total sampling time.

As was also mentioned previously, if you know the signal contains a single
sinusoid, you can get the frequency, phase, and amplitude if you have three
exact sample values.  Since for a real signal there is always finite noise
you never get exact sample values.   In that case, more sample points can be
used to reduce the effect of the noise, in a least squares sense.   Given a
set of sample points, one could apply a non-linear least squares fit between
A*sin(w*t+phi), with A, w, and phi as fit parameters, and determine the best
fit.  Least squares should be usable if the system is overdetermined.
Fourier analysis assumes that the system is the sum of a set of sinusoids,
and is, in general, never overdetermined.  This sounds more like the claim
in the OP, with an infinite number of frequencies, sin(w) is a continuous
function of w, for example.  The fit can only be done as accurately as you
can compute w, phi, and sin(w*t+phi), though usually somewhat less
accurately.  Double precision floating point should be plenty good enough
for this case, though.

Note, though, that for the original problem, a 10MHz signal with +/- 5ppm,
the bandwidth is only 100Hz for a 200Hz nyquist frequency.  N points taken
at 200Hz determines the frequency, as Jerry said, to 200Hz/N.  (This assumes
certain properties of the A/D conversion, which may not be true.)   The
accuracy of the time base is important.  Sampling jitter should average out,
decreasing as sqrt(N).   It doesn't take many points before you are likely
better than the time base.  1000000 points gives you 200uHz (microHertz)
resolution.   (Also, if the frequency is outside +/-5ppm, you will be off by
a multiple of 100Hz due to aliasing.)

I do hope this isn't a homework problem.  It doesn't sound like one, though.
If it is be sure to reference the newsgroup.

-- glen

```
 0
Reply Glen 10/31/2003 8:22:48 PM

```Stan Pawlukiewicz wrote:
>
> Statistically they are only independent as N goes to infinity
> and the input process is white or if the process just happens to be made
> up of a set of N random amplitude sinusoids that happend to have
> frequencies on the exact centers of the DFT bins and you use a boxcar
> window.
>
> Spectral leakage <==> statistically dependent bins
>
> The other thing is that any analytic function is uniquely defined by its
> complex poles or as the math people like to say the residues. There
> really isn't anything special about bin center frequencies unless the N
> samples correspond to a segment of signal with period N.

Yes, all absolutely true.  I was just trying to give the jist of some of the
reasons for focusing on the Fourier frequencies.  Thanks for taking the
time to fill in the details.

-Tom
```
 0
Reply Tom 10/31/2003 9:13:12 PM

```Hi Thomas-

> Tom, I think you know what I am babbling about, so what the hell would it be
> called. You say the DTFT uses Fourier frequencies and I'm not talking about
> that. How about Discrete Frequency Transform?...oh wait...that could get
> confusing:)

IIRC, I have not seen uniform terminology regarding DTFT, but it is
used to refer to the *continuous frequency* transform of a
discretely sampled time series.  Where I think there is some slop in the
definition is whether the time series is allowed to be finite.

If I understand you correctly, one way I might have worded the question
is:  "What kind of resolution can I expect from using the peak of the
periodogram to estimate a frequency?"  This is well known (in the case
where the signal is well approximated by a single sinusoid and the noise
is Gaussian and uncorrelated) and I think I already gave you some references
in an earlier post.  Section 2.4 of Bretthorst's book is a nice discussion
of this, with explicit formulas for the frequency resolution as a function
of the length of the time series and the S/N level:

http://bayes.wustl.edu/

(the link to the book is in the "G. Larry Bretthorst" paragraph).  Larry
uses the Bayesian approach to get the equations, and I personally think
it's the clearest way to handle such problems, but that's controversial
in some fields.  In any case, for this particular problem the Bayesian
result is the same as more conventional "frequentist" results for the
frequency resolution (look at the literature for "parametric frequency
estimation").  You don't have Gaussian noise (roughly speaking you have
a box distribution for your "noise" due to the ADC bit depth), but if
you use the std. deviation corresponding to the range spanned by 1 bit
(from the usual sqrt(12) rule), you should get the right ballpark.  If
you want to see Larry's approach in action on an audio problem, with
resolution much finer than the Fourier spacing, see my AES paper from
a couple years ago, http://www.museweb.com/bha.pdf.  No nice formulas
for the freq. res'n, though.

As far as actually *calculating* the periodogram/DTFT/power spectrum
between Fourier frequencies, if you can use an FFT to narrow down
your region of interest sufficiently, simply doing a slow transform
on a fine grid in that region may be as fast as you can do.  You can
zero-pad the data and do the FFT, but you'll be wasting effort if
you have been able to narrow down the range of interest to one or
a few Fourier bins.  In some cases you may find the fractional FFT
will speed up evaluation on a fine grid in part of the frequency
space; Bailey and Swarztrauber explain exactly how to do it here:

http://citeseer.nj.nec.com/bailey95fractional.html

If you follow up on the fFFT, make sure you look up Bailey's web site,
where he has a page fixing mistakes in a couple of the key equations
of that famous fFFT paper.

Peace,
Tom

--

To respond by email, replace "somewhere" with "astro" in the
```
 0
Reply Tom 10/31/2003 9:33:49 PM

```Hi Tom,
Thanks for yours responses. I am looking into the references that you gave
me.

I posted some Java code yesterday under this subject which I really think
helps explain what I am looking for. Do you read code? If so, could you look
at what I posted and let me know what you think. Remember I am only trying
to recover an accurate frequency measure of a CW which should be the peak
value in the spectrum plot which my algorithm produces.

Thanks in advance Tom or anyone else that will read my code.

"Tom Loredo" <loredo@somewhere.cornell.edu> wrote in message
news:3FA2D53D.6DD6EBFC@somewhere.cornell.edu...
>
> Hi Thomas-
>
> > Tom, I think you know what I am babbling about, so what the hell would
it be
> > called. You say the DTFT uses Fourier frequencies and I'm not talking
> > that. How about Discrete Frequency Transform?...oh wait...that could get
> > confusing:)
>
> IIRC, I have not seen uniform terminology regarding DTFT, but it is
> used to refer to the *continuous frequency* transform of a
> discretely sampled time series.  Where I think there is some slop in the
> definition is whether the time series is allowed to be finite.
>
> If I understand you correctly, one way I might have worded the question
> is:  "What kind of resolution can I expect from using the peak of the
> periodogram to estimate a frequency?"  This is well known (in the case
> where the signal is well approximated by a single sinusoid and the noise
> is Gaussian and uncorrelated) and I think I already gave you some
references
> in an earlier post.  Section 2.4 of Bretthorst's book is a nice discussion
> of this, with explicit formulas for the frequency resolution as a function
> of the length of the time series and the S/N level:
>
> http://bayes.wustl.edu/
>
> (the link to the book is in the "G. Larry Bretthorst" paragraph).  Larry
> uses the Bayesian approach to get the equations, and I personally think
> it's the clearest way to handle such problems, but that's controversial
> in some fields.  In any case, for this particular problem the Bayesian
> result is the same as more conventional "frequentist" results for the
> frequency resolution (look at the literature for "parametric frequency
> estimation").  You don't have Gaussian noise (roughly speaking you have
> a box distribution for your "noise" due to the ADC bit depth), but if
> you use the std. deviation corresponding to the range spanned by 1 bit
> (from the usual sqrt(12) rule), you should get the right ballpark.  If
> you want to see Larry's approach in action on an audio problem, with
> resolution much finer than the Fourier spacing, see my AES paper from
> a couple years ago, http://www.museweb.com/bha.pdf.  No nice formulas
> for the freq. res'n, though.
>
> As far as actually *calculating* the periodogram/DTFT/power spectrum
> between Fourier frequencies, if you can use an FFT to narrow down
> your region of interest sufficiently, simply doing a slow transform
> on a fine grid in that region may be as fast as you can do.  You can
> zero-pad the data and do the FFT, but you'll be wasting effort if
> you have been able to narrow down the range of interest to one or
> a few Fourier bins.  In some cases you may find the fractional FFT
> will speed up evaluation on a fine grid in part of the frequency
> space; Bailey and Swarztrauber explain exactly how to do it here:
>
> http://citeseer.nj.nec.com/bailey95fractional.html
>
> If you follow up on the fFFT, make sure you look up Bailey's web site,
> where he has a page fixing mistakes in a couple of the key equations
> of that famous fFFT paper.
>
> Peace,
> Tom
>
> --
>
> To respond by email, replace "somewhere" with "astro" in the

```
 0
Reply Thomas 10/31/2003 10:09:02 PM

```Fred Marshall wrote:

> "Peter J. Kootsookos" <p.kootsookos@remove.ieee.org> wrote in message
> news:s68k76m2ak7.fsf@mango.itee.uq.edu.au...
>
>>Jerry Avins <jya@ieee.org> writes:
>>
>>
>>>The problem you are trying to solve is analogous to finding that nut
>>>which is the best fit on a threaded rod. If the diametral fit is very
>>>close or the nut very long, its pitch must be very accurate. As the
>>>clearance increases or the length decreases, the pitch mismatch can
>>>increase. Read pitch as spatial frequency and length as number of
>>>samples. Clearance depends on precision, degraded both by noise and the
>>>limits of measurement.
>>
>>Great analogy, Jerry!
>>
>
>
> Yeah, like Peter said!
>
> Fred

Umm. Err. Gee, thanks. I suppose I earn my keep once in a while.

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

```
 0
Reply Jerry 11/1/2003 12:58:40 AM

```On Fri, 31 Oct 2003 09:51:34 -0800, "Fred Marshall"
<fmarshallx@remove_the_x.acm.org> wrote:

>
>"Peter J. Kootsookos" <p.kootsookos@remove.ieee.org> wrote in message
>news:s68k76m2ak7.fsf@mango.itee.uq.edu.au...
>> Jerry Avins <jya@ieee.org> writes:
>>
>> > The problem you are trying to solve is analogous to finding that nut
>> > which is the best fit on a threaded rod. If the diametral fit is very
>> > close or the nut very long, its pitch must be very accurate. As the
>> > clearance increases or the length decreases, the pitch mismatch can
>> > increase. Read pitch as spatial frequency and length as number of
>> > samples. Clearance depends on precision, degraded both by noise and the
>> > limits of measurement.
>>
>> Great analogy, Jerry!
>>
>
>Yeah, like Peter said!
>
>Fred

I'll third, that's pretty clever.  I'll be using that one myself...
;)

Eric Jacobsen
Minister of Algorithms, Intel Corp.
My opinions may not be Intel's opinions.
http://www.ericjacobsen.org
```
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