.1% frequency measure of a 6.6ms tone?

If I only have 6.6 millisecond worth of data can I measure the frequency of
a 7 kHz tone to within .1 percent ( 7 Hz)?

I think no because the FFT could only resolve 1/6.6e-3 ~= 150 Hz. He thinks
it can be done with a filter and an fm demodulator.

What do you think?

-Clark

cpope wrote:

> If I only have 6.6 millisecond worth of data can I measure the frequency of
> a 7 kHz tone to within .1 percent ( 7 Hz)?

You can measure the frequency to whatever accuracy. Time is irrelevant, 
all that matters is the SNR.

> I think no because the FFT could only resolve 1/6.6e-3 ~= 150 Hz. He thinks
> it can be done with a filter and an fm demodulator.
> What do you think?

You both are wrong :-)


Vladimir Vassilevsky
DSP and Mixed Signal Design Consultant
http://www.abvolt.com

On Oct 25, 4:10 pm, Vladimir Vassilevsky <antispam_bo...@hotmail.com>
wrote:
> cpope wrote:
> > If I only have 6.6 millisecond worth of data can I measure the frequency of
> > a 7 kHz tone to within .1 percent ( 7 Hz)?
>
> You can measure the frequency to whatever accuracy. Time is irrelevant,
> all that matters is the SNR.

Isn't the time required for a given frequency estimation
accuracy somewhat inversely proportional to the SNR?  Both
seem to be relevant (until you peg at the lower limit of
only 3 non-aliased points for a pure enough sinusoid).


IMHO. YMMV.
--
rhn A.T nicholson d.0.t C-o-M

cpope wrote:
> If I only have 6.6 millisecond worth of data can I measure the frequency of
> a 7 kHz tone to within .1 percent ( 7 Hz)?
> 
> I think no because the FFT could only resolve 1/6.6e-3 ~= 150 Hz. He thinks
> it can be done with a filter and an fm demodulator.
> 
> What do you think?
> 

Depends on how fine the data is sampled. It pretty much boils down to 
finding the zero crossings.

-- 
Regards, Joerg

http://www.analogconsultants.com/

Ron N. wrote:

>>>If I only have 6.6 millisecond worth of data can I measure the frequency of
>>>a 7 kHz tone to within .1 percent ( 7 Hz)?
>>
>>You can measure the frequency to whatever accuracy. Time is irrelevant,
>>all that matters is the SNR.
> 
> 
> Isn't the time required for a given frequency estimation
> accuracy somewhat inversely proportional to the SNR?  Both
> seem to be relevant (until you peg at the lower limit of
> only 3 non-aliased points for a pure enough sinusoid).

To be more exact, the SNR is defined as the total signal energy divided 
by the total noise energy (both measured as per given time). The 
accuracy is proportional to the log(SNR), and the coefficient of the 
proportionality is determined by the autocorrelation function of the 
waveform...


Vladimir Vassilevsky
DSP and Mixed Signal Design Consultant
http://www.abvolt.com

Joerg wrote:

> cpope wrote:
> 
>> If I only have 6.6 millisecond worth of data can I measure the 
>> frequency of
>> a 7 kHz tone to within .1 percent ( 7 Hz)?
>>
>> I think no because the FFT could only resolve 1/6.6e-3 ~= 150 Hz. He 
>> thinks
>> it can be done with a filter and an fm demodulator.
>>
>> What do you think?
>>
> 
> Depends on how fine the data is sampled. It pretty much boils down to 
> finding the zero crossings.

Sinewave = A cos (Wt + fi). You have three unknowns: A, W, fi. Now if 
you have any piece of a sinewave, you can determine those unknowns. The 
accuracy is determined only by the noise of the measurement. Of course, 
for the sampled waveform the aliasing conditions apply.

Vladimir Vassilevsky
DSP and Mixed Signal Design Consultant
http://www.abvolt.com

On Oct 25, 3:55 pm, "cpope" <cep...@nc.rr.com> wrote:
> If I only have 6.6 millisecond worth of data can I measure the frequency of
> a 7 kHz tone to within .1 percent ( 7 Hz)?
>
> I think no because the FFT could only resolve 1/6.6e-3 ~= 150 Hz.

An FFT doesn't resolve; it only transforms.  If you just
pick a peak bin, then that method resolves to only 150 Hz.

If you interpolate (quadratic, sinc, zero-pad) an FFT peak,
or use some other least squares frequency fit, and the S/N
ratio is high enough, then you might be able estimate
a frequency peak to within 1/20th of an FFT bin.


IMHO. YMMV.
--
rhn A.T nicholson d.0.t C-o-M

Ron N. wrote:
> On Oct 25, 4:10 pm, Vladimir Vassilevsky <antispam_bo...@hotmail.com>
> wrote:
>> cpope wrote:
>>> If I only have 6.6 millisecond worth of data can I measure the frequency of
>>> a 7 kHz tone to within .1 percent ( 7 Hz)?
>> You can measure the frequency to whatever accuracy. Time is irrelevant,
>> all that matters is the SNR.
> 
> Isn't the time required for a given frequency estimation
> accuracy somewhat inversely proportional to the SNR?  Both
> seem to be relevant (until you peg at the lower limit of
> only 3 non-aliased points for a pure enough sinusoid).

The statistics of the noise would be important too, wouldn't it?

Steve

cpope wrote:
> If I only have 6.6 millisecond worth of data can I measure the frequency of
> a 7 kHz tone to within .1 percent ( 7 Hz)?
> 
> I think no because the FFT could only resolve 1/6.6e-3 ~= 150 Hz. He thinks
> it can be done with a filter and an fm demodulator.
> 
> What do you think?
> 
> -Clark
> 
> 
What's the sample rate? I hope it's greater than 14kHz.

Allan

"Allan Wilson" <allanw@awwconcepts.co.nz> wrote in message
news:13i2nsbkja9ge87@news.supernews.com...
> cpope wrote:
> > If I only have 6.6 millisecond worth of data can I measure the frequency
of
> > a 7 kHz tone to within .1 percent ( 7 Hz)?
> >
> > I think no because the FFT could only resolve 1/6.6e-3 ~= 150 Hz. He
thinks
> > it can be done with a filter and an fm demodulator.
> >
> > What do you think?
> >
> >
> What's the sample rate? I hope it's greater than 14kHz.

Huh? The sample rate has only to be higher then 14Hz to accomodate for
+/-7Hz tolerance. We need 3 samples per 6 milliseconds thus ~500Hz is
enough.

Vladimir Vassilevsky
DSP and Mixed Signal Consultant
www.abvolt.com