Frequency Filtering Algorithm

Hello,
The requirements of a system to be designed are:

Transmitting frequency : 50 KHz
Received Echo Frequency to be detected : 45-55 KHz
Frequency Resolution : 90 hz
Sampling Rate : 250 KHz

We are transmitting 16 bursts of 50 KHz. 
So the total time period of transmission will be only 320micro
seconds.
Therefore within the same time duration we have to collect the samples
and detect the frequency of echo. The measurement duration is around
3ms.

Could anybody suggest frequency filtering algorithm for this.

My first intuition says that solution lies in 'multirate digital
signal processing'. But you are also welcomed to give suggestion
independent of my intuition.

Thanks and Regards,
Pawan Jadia.

pawan_jadia@yahoo.com (pawan jadia) wrote in message news:<698534c.0406082221.3a67fd47@posting.google.com>...
> Hello,
> The requirements of a system to be designed are:
> 
> Transmitting frequency : 50 KHz
> Received Echo Frequency to be detected : 45-55 KHz
> Frequency Resolution : 90 hz
> Sampling Rate : 250 KHz
> 
> We are transmitting 16 bursts of 50 KHz. 
> So the total time period of transmission will be only 320micro
> seconds.
> Therefore within the same time duration we have to collect the samples
> and detect the frequency of echo. The measurement duration is around
> 3ms.
> 
> Could anybody suggest frequency filtering algorithm for this.
> 
> My first intuition says that solution lies in 'multirate digital
> signal processing'. But you are also welcomed to give suggestion
> independent of my intuition.
> 
> Thanks and Regards,
> Pawan Jadia.

Pawan,

As an example, if you use selective-sampling you would synthesize 111
waveforms ( 10Khz/90 Hz) starting from 45 Khz and ending at 55 Khz in
90 Hz increments (45000, 45090, 45180, etc.) for a period each of
twice the lowest frequency (45 KHz). The A/D would oversample by
perhaps 6x55 KHz = 330 KHz (to get enough data points for resolution
of formulas - see http://www.signaldisplay.com/sampling.html) and the
total time to collect samples would be 2*111/45000 Hz = 4.9 msec. The
processor would then solve for magnitude components to locate the peak
frequency in the 45K - 55K band. With this you don't have to bandpass
or downsample (the bandpass will also introduce a delay).

Best regards,

Mike

pawan_jadia@yahoo.com (pawan jadia) wrote in message news:<698534c.0406082221.3a67fd47@posting.google.com>...
> Hello,
> The requirements of a system to be designed are:
> 
> Transmitting frequency : 50 KHz
> Received Echo Frequency to be detected : 45-55 KHz
> Frequency Resolution : 90 hz
> Sampling Rate : 250 KHz
> 
> We are transmitting 16 bursts of 50 KHz. 
> So the total time period of transmission will be only 320micro
> seconds.
> Therefore within the same time duration we have to collect the samples
> and detect the frequency of echo. The measurement duration is around
> 3ms.
> 
> Could anybody suggest frequency filtering algorithm for this.
> 
> My first intuition says that solution lies in 'multirate digital
> signal processing'. But you are also welcomed to give suggestion
> independent of my intuition.
> 
> Thanks and Regards,
> Pawan Jadia.

Pawan,

As an example, if you use selective-sampling you would synthesize 111
waveforms ( 10Khz/90 Hz) starting from 45 Khz and ending at 55 Khz in
90 Hz increments (45000, 45090, 45180, etc.) for a period each of
twice the lowest frequency (45 KHz). The A/D would oversample by
perhaps 6x55 KHz = 330 KHz (to get enough data points for resolution
of formulas - see http://www.signaldisplay.com/sampling.html) and the
total time to collect samples would be 2*111/45000 Hz = 4.9 msec. The
processor would then solve for magnitude components to locate the peak
frequency in the 45K - 55K band. With this you don't have to bandpass
or downsample (the bandpass will also introduce a delay).

Best regards,

Mike

"pawan jadia" <pawan_jadia@yahoo.com> wrote in message
news:698534c.0406082221.3a67fd47@posting.google.com...
> Hello,
> The requirements of a system to be designed are:
>
> Transmitting frequency : 50 KHz
> Received Echo Frequency to be detected : 45-55 KHz
> Frequency Resolution : 90 hz
> Sampling Rate : 250 KHz
>
> We are transmitting 16 bursts of 50 KHz.
> So the total time period of transmission will be only 320micro
> seconds.
> Therefore within the same time duration we have to collect the samples
> and detect the frequency of echo. The measurement duration is around
> 3ms.
>
> Could anybody suggest frequency filtering algorithm for this.
>
> My first intuition says that solution lies in 'multirate digital
> signal processing'. But you are also welcomed to give suggestion
> independent of my intuition.
>
> Thanks and Regards,
> Pawan Jadia.

"pawan jadia" <pawan_jadia@yahoo.com> wrote in message
news:698534c.0406082221.3a67fd47@posting.google.com...
> Hello,
> The requirements of a system to be designed are:
>
> Transmitting frequency : 50 KHz
> Received Echo Frequency to be detected : 45-55 KHz
> Frequency Resolution : 90 hz
> Sampling Rate : 250 KHz
>
> We are transmitting 16 bursts of 50 KHz.
> So the total time period of transmission will be only 320micro
> seconds.
> Therefore within the same time duration we have to collect the samples
> and detect the frequency of echo. The measurement duration is around
> 3ms.
>
> Could anybody suggest frequency filtering algorithm for this.
>
> My first intuition says that solution lies in 'multirate digital
> signal processing'. But you are also welcomed to give suggestion
> independent of my intuition.

Pawan,

No algorithm will make up for inadequate physics.

I interpret your description as follows:

You are transmitting 16 *cycles* of 50kHz for a total pulse length of
320usec.  A pulse of this length will yield a frequency resolution of
1/0.000032 = 3,125Hz.  This is not nearly the 90Hz you desire.

To get 90 Hz resolution, you need a pulse length of around 1/90 = 11msec.

Multirate really has nothing to do with it.

Fred

mharney@signaldisplay.com (Michael Harney) wrote in message news:<ce8a7803.0406091715.55877670@posting.google.com>...
> pawan_jadia@yahoo.com (pawan jadia) wrote in message news:<698534c.0406082221.3a67fd47@posting.google.com>...
> > Hello,
> > The requirements of a system to be designed are:
> > 
> > Transmitting frequency : 50 KHz
> > Received Echo Frequency to be detected : 45-55 KHz
> > Frequency Resolution : 90 hz
> > Sampling Rate : 250 KHz
> > 
> > We are transmitting 16 bursts of 50 KHz. 
> > So the total time period of transmission will be only 320micro
> > seconds.
> > Therefore within the same time duration we have to collect the samples
> > and detect the frequency of echo. The measurement duration is around
> > 3ms.
> > 
> > Could anybody suggest frequency filtering algorithm for this.
> > 
> > My first intuition says that solution lies in 'multirate digital
> > signal processing'. But you are also welcomed to give suggestion
> > independent of my intuition.
> > 
> > Thanks and Regards,
> > Pawan Jadia.
> 
> Pawan,
> 
> As an example, if you use selective-sampling you would synthesize 111
> waveforms ( 10Khz/90 Hz) starting from 45 Khz and ending at 55 Khz in
> 90 Hz increments (45000, 45090, 45180, etc.) for a period each of
> twice the lowest frequency (45 KHz). The A/D would oversample by
> perhaps 6x55 KHz = 330 KHz (to get enough data points for resolution
> of formulas - see http://www.signaldisplay.com/sampling.html) and the
> total time to collect samples would be 2*111/45000 Hz = 4.9 msec. The
> processor would then solve for magnitude components to locate the peak
> frequency in the 45K - 55K band. With this you don't have to bandpass
> or downsample (the bandpass will also introduce a delay).
> 
> Best regards,
> 
> Mike

----> Thanks Mike, I think this is what I wanted. Thanks a lot. But
total time to collect the sample will be 111/45000 = 2.5 ms as
described in the paper. am I right? and secondly bandpass filter is a
part of the block diagram that you suggested. I am trying to
understand the paper fully. I will get back to you very soon.

Thanks and Regards,
pawan jadia.

pawan_jadia@yahoo.com (pawan jadia) wrote in message news:<698534c.0406100137.2a94562c@posting.google.com>...
> mharney@signaldisplay.com (Michael Harney) wrote in message news:<ce8a7803.0406091715.55877670@posting.google.com>...
> > pawan_jadia@yahoo.com (pawan jadia) wrote in message news:<698534c.0406082221.3a67fd47@posting.google.com>...
> > > Hello,
> > > The requirements of a system to be designed are:
> > > 
> > > Transmitting frequency : 50 KHz
> > > Received Echo Frequency to be detected : 45-55 KHz
> > > Frequency Resolution : 90 hz
> > > Sampling Rate : 250 KHz
> > > 
> > > We are transmitting 16 bursts of 50 KHz. 
> > > So the total time period of transmission will be only 320micro
> > > seconds.
> > > Therefore within the same time duration we have to collect the samples
> > > and detect the frequency of echo. The measurement duration is around
> > > 3ms.
> > > 
> > > Could anybody suggest frequency filtering algorithm for this.
> > > 
> > > My first intuition says that solution lies in 'multirate digital
> > > signal processing'. But you are also welcomed to give suggestion
> > > independent of my intuition.
> > > 
> > > Thanks and Regards,
> > > Pawan Jadia.
> > 
> > Pawan,
> > 
> > As an example, if you use selective-sampling you would synthesize 111
> > waveforms ( 10Khz/90 Hz) starting from 45 Khz and ending at 55 Khz in
> > 90 Hz increments (45000, 45090, 45180, etc.) for a period each of
> > twice the lowest frequency (45 KHz). The A/D would oversample by
> > perhaps 6x55 KHz = 330 KHz (to get enough data points for resolution
> > of formulas - see http://www.signaldisplay.com/sampling.html) and the
> > total time to collect samples would be 2*111/45000 Hz = 4.9 msec. The
> > processor would then solve for magnitude components to locate the peak
> > frequency in the 45K - 55K band. With this you don't have to bandpass
> > or downsample (the bandpass will also introduce a delay).
> > 
> > Best regards,
> > 
> > Mike
> 
> ----> Thanks Mike, I think this is what I wanted. Thanks a lot. But
> total time to collect the sample will be 111/45000 = 2.5 ms as
> described in the paper. am I right? and secondly bandpass filter is a
> part of the block diagram that you suggested. I am trying to
> understand the paper fully. I will get back to you very soon.
> 
> Thanks and Regards,
> pawan jadia.

Pawan,

Yes - theoretically you can use only one period of each sinewave (one
period is all that is required to satisfy the Fourier integral so that
when you multiply two sines you get the sum and difference products).
Two periods just makes it a little cleaner (one period gives you 2.5
msec and two periods each give you 5 msec). Yes you are right - the
algorithm does use bandpass filtering on the front-end and there is
about 100 usec delay (1/10Khz). This delay is hopefully OK but it is
necessary to bandpass so the sum and difference products are limited
to the low and high frequencies you are seeking components for. If you
assume zero or constant phase, the matrix of equations that results
for magnitude and phase (set equal to the A/D values) is easy to solve
for by just inverting the matrix to find the magnitude components. Of
course, the phase is usually not constant so the matrix will be
quadratic equations which it is still easy to find a solution for (I
have done this on MathCad and there are algorithms for solving for a
system of quadratics that can be ported to code). You may need to
increase the A/D sampling rate to generate more points as the more
components you are solving for (in this case - 111 bins) the same
number of equations you need. This is the trade off - high A/D
oversampling for narrow resolution). Accurate mixers are also
important - I recommend a straight multiplier like the MPY634 from
Burr-Brown (now TI). I am working on developing a program that allows
you to import data and it does all the math - of course it won't be in
real time for most systems but it will make it easier to do data
analysis after the data is captured. Let me know if you have more
questions.

Best regards,

Mike

> > 
> > ----> Thanks Mike, I think this is what I wanted. Thanks a lot. But
> > total time to collect the sample will be 111/45000 = 2.5 ms as
> > described in the paper. am I right? and secondly bandpass filter is a
> > part of the block diagram that you suggested. I am trying to
> > understand the paper fully. I will get back to you very soon.
> > 
> > Thanks and Regards,
> > pawan jadia.
> 
Pawan,

Yes - theoretically you can use only one period of each sinewave (one
period is all that is required to satisfy the Fourier integral so that
when you multiply two sines you get the sum and difference products).
Two periods just makes it a little cleaner (one period gives you 2.5
msec and two periods each give you 5 msec). Yes you are right - the
algorithm does use bandpass filtering on the front-end and there is
about 100 usec delay (1/10Khz). This delay is hopefully OK but it is
necessary to bandpass so the sum and difference products are limited
to the low and high frequencies you are seeking components for. If you
assume zero or constant phase, the matrix of equations that results
for magnitude and phase (set equal to the A/D values) is easy to solve
for by just inverting the matrix to find the magnitude components. Of
course, the phase is usually not constant so the matrix will be
quadratic equations which it is still easy to find a solution for (I
have done this on MathCad and there are algorithms for solving for a
system of quadratics that can be ported to code). You may need to
increase the A/D sampling rate to generate more points as the more
components you are solving for (in this case - 111 bins) the same
number of equations you need. This is the trade off - high A/D
oversampling for narrow resolution). Accurate mixers are also
important - I recommend a straight multiplier like the MPY634 from
Burr-Brown (now TI). I am working on developing a program that allows
you to import data and it does all the math - of course it won't be in
real time for most systems but it will make it easier to do data
analysis after the data is captured. Let me know if you have more
questions.

Best regards,

Mike

"Fred Marshall" <fmarshallx@remove_the_x.acm.org> wrote in message news:<4fSdnWnhXpIre1rdRVn-jw@centurytel.net>...
> "pawan jadia" <pawan_jadia@yahoo.com> wrote in message
> news:698534c.0406082221.3a67fd47@posting.google.com...
> > Hello,
> > The requirements of a system to be designed are:
> >
> > Transmitting frequency : 50 KHz
> > Received Echo Frequency to be detected : 45-55 KHz
> > Frequency Resolution : 90 hz
> > Sampling Rate : 250 KHz
> >
> > We are transmitting 16 bursts of 50 KHz.
> > So the total time period of transmission will be only 320micro
> > seconds.
> > Therefore within the same time duration we have to collect the samples
> > and detect the frequency of echo. The measurement duration is around
> > 3ms.
> >
> > Could anybody suggest frequency filtering algorithm for this.
> >
> > My first intuition says that solution lies in 'multirate digital
> > signal processing'. But you are also welcomed to give suggestion
> > independent of my intuition.
> 
> Pawan,
> 
> No algorithm will make up for inadequate physics.
> 
> I interpret your description as follows:
> 
> You are transmitting 16 *cycles* of 50kHz for a total pulse length of
> 320usec.  A pulse of this length will yield a frequency resolution of
> 1/0.000032 = 3,125Hz.  This is not nearly the 90Hz you desire.
> 
> To get 90 Hz resolution, you need a pulse length of around 1/90 = 11msec.
> 
> Multirate really has nothing to do with it.
> 
> Fred

----->Fred, I am agree with you. I am new in the project so I am just
getting the requirements. Actually i don't need 90 hz resolution at
all. Now the problem is my transmission duration is 320 micro seconds.
so receive duration will also be 320 usec. with 250 khz sample rate
within this 320 usec i can collect only 80 samples. now how to detect
the frequency with these 80 samples. Here the receiving frequency may
change, if the object is moving, due to doppler effect. So the new
problem is unknown frequency detection (estimation) of chirp signal.

Thanks and regards,
pawan jadia.

"Fred Marshall" <fmarshallx@remove_the_x.acm.org> wrote in message news:<4fSdnWnhXpIre1rdRVn-jw@centurytel.net>...
> "pawan jadia" <pawan_jadia@yahoo.com> wrote in message
> news:698534c.0406082221.3a67fd47@posting.google.com...
> > Hello,
> > The requirements of a system to be designed are:
> >
> > Transmitting frequency : 50 KHz
> > Received Echo Frequency to be detected : 45-55 KHz
> > Frequency Resolution : 90 hz
> > Sampling Rate : 250 KHz
> >
> > We are transmitting 16 bursts of 50 KHz.
> > So the total time period of transmission will be only 320micro
> > seconds.
> > Therefore within the same time duration we have to collect the samples
> > and detect the frequency of echo. The measurement duration is around
> > 3ms.
> >
> > Could anybody suggest frequency filtering algorithm for this.
> >
> > My first intuition says that solution lies in 'multirate digital
> > signal processing'. But you are also welcomed to give suggestion
> > independent of my intuition.
> 
> Pawan,
> 
> No algorithm will make up for inadequate physics.
> 
> I interpret your description as follows:
> 
> You are transmitting 16 *cycles* of 50kHz for a total pulse length of
> 320usec.  A pulse of this length will yield a frequency resolution of
> 1/0.000032 = 3,125Hz.  This is not nearly the 90Hz you desire.
> 
> To get 90 Hz resolution, you need a pulse length of around 1/90 = 11msec.
> 
> Multirate really has nothing to do with it.
> 
> Fred

----->Fred, I am agree with you. I am new in the project so I am just
getting the requirements. Actually i don't need 90 hz resolution at
all. Now the problem is my transmission duration is 320 micro seconds.
so receive duration will also be 320 usec. with 250 khz sample rate
within this 320 usec i can collect only 80 samples. now how to detect
the frequency with these 80 samples. Here the receiving frequency may
change, if the object is moving, due to doppler effect. So the new
problem is unknown frequency detection (estimation) of chirp signal.

Thanks and regards,
pawan jadia.