Re: Envelope Detector using Hilbert Transform

"w106pjs" <w106pjs@xxxxxxxxx> wrote in
news:eIOdnTdh47fuVHrfRVn-ow@xxxxxxxxxxxx:

>>"w106pjs" <w106pjs@xxxxxxxxx> wrote in
>>news:9fGdnTz2ELZZLnrfRVn-tQ@xxxxxxxxxxxx:
>>
>>> All
>>>
>>> I have been kind going through previous threads in this group on
>>> similar concern and question that I have. Still I feel my feet is
>>> not on the ground yet...with this issue..
>>>
>>> Background:
>>> I have working with ultrasound signals (300 Khz) from a solid state
>>> sensor to a target at 3". The received signals from the sensor (300
>>> Khz) are digitized through a sampling card and imported into MATLAB,
>>> to preserve the time information as precisely as possible, sampling
>>> frequency is high Fs = 50Mhz.
>>> To generate a envelope on the 300Khz received echo, conventional
>>> magnitude of the analytic is used in MATLAB.
>>>
>>> Envelope = abs(hilbert(echo_signal));
>>> It works as expected. and then rest of the processing continues on
>>> the envelope signal.
>>>
>>> Question and Concern:
>>> 1. What should be an alternative approach to acheive the same. ?
>>> Say FIR hilbert transformer, All pass IIR design..
>>> With sampling frequency this high, my fear is FIR may generate real
>>> higher length filter.Correct me If I am wrong!!
>>>
>>> 2. Hilbert function in MATLAB works using the FFT and IFFT
>>> approach.What approach should be more efficient in terms of
>>> implementing it in real time on the DSP processor.
>>>
>>> thanks in advance.
>>> Paul
>>>
>>
>>I'm not sure why you need to oversample by so much, so I will only
>>comment on the hilbert implementation.
>>
>>The easiest way to do a hilbert transform is to use an odd length anti
>>symmetric FIR filter. I generally use a Parks McClelland (remez
>>exchange)
>
>>fit for the filter to make the passband ripple small.
>>
>>The fir filter is a bandpass filter. It doesn't need to be very long
>>if the signal of interest is not too close to DC. If you are sampling
>>at 50MHz, and the signal is a 300k, you are close to DC, so the filter
>>will
>
>>be fairly long. Sample at a few MHz, and the filter will be very easy
>>to
>
>>implement on a DSP.
>>
>>You get the real output from the center tap of the same delay line
>>that is used for the imaginary part.
>>
>>Rick Lyon's book: Understanding Digital Signal Processing ( 2nd
>>edition)
>
>>explains this approach. Marvin Frerking's book: Digital Signal
>>Processing
>
>>for Communication Systems is another excellent reference (but very
>>hard to find).
>>
>>The envelope is the SQRT (r^2 + i^2). This is easy to implemt on a
>>SHARC
>
>>or similiar floating point DSP.
>>
>>We have boards that can do this application.
>>
>>
>>--
>>Al Clark
>>Danville Signal Processing, Inc.
>>--------------------------------------------------------------------
>>Purveyors of Fine DSP Hardware and other Cool Stuff
>>Available at http://www.danvillesignal.com
>>
>
> Al Clark
>
> Thank you for your prompt reply.
> 1. The reason it is oversampled by a significant amount is to
> preserve the time resolution in the application. The DSP algorithm is
> being used to precisely measure the time of flight for the ultrasound
> signal through air which demands high resolution. Sampling at a low
> rate and interpolating for better resolution was experimented with
> very little sucess compared to oversampled routine.
>
> 2. Since 300K is not what signal of interest @ Fs=50 Mhz, the lenght
> would certainly be large, does it become really impractical to
> implement on the real hardware.
> Is IIR filters (All pass filters) would be of any help in this case.
> What band edges would you recommend for FIR BP filter.
>
> thanks again.
> Paul
>
>
> This message was sent using the Comp.DSP web interface on
> www.DSPRelated.com
>

I think I am going to get myself in trouble, but here goes:

Is the ultrasonic sensors transmitting at 300kHz and you are trying to
recover a reflection? It seems to me you will have an uncertainty due to
the wavelength of 300kHz since BT >= 1 (bandwidth time product). I don't
see how oversampling by approximately 167/1 is going to provide
additional resolution.

Assuming air:

sound travels at 344m/s

You have a spatial resolution of 344m/300k = 1.146mm

I think you achieve the same answer sampling at 2MHz as sampling at
50MHz.

This isn't really my area of expertise. Someone should jump in a clarify
all of this, please.



--
Al Clark
Danville Signal Processing, Inc.
--------------------------------------------------------------------
Purveyors of Fine DSP Hardware and other Cool Stuff
Available at http://www.danvillesignal.com
.

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