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I was curious about the definition of the term detect samples went it pertains to PI detectors. I have read that most detectors need 5 samples, here is where I am confused. Is that 5 transmit sine way cycles, or 5 complete transmit pulses ? Thank you for any help. Best to all Jim
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good question James!
so funny book...
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That's usually samples per one Tx pulse. In PI you observe voltage on the coil after a flyback, which is generally an exponential decay. Sampling is done in a period from the very least 4us to over 200us, and there are different schemes to do that. Only the most complex ones take 5 or more samples. The most common scheme is with 2 samples.Comment
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I think the context of James' question refers to the number of samples/cycles required for the Integrator to produce a usable output signal.
Remember that the Integrator output is very small with one sample and increases as the number of samples increases.
Quote from a post by Carl:
The purpose of the integrator is to constructively add successive positive samples (resulting from a target) while
averaging out everything else. That is, random noise (which is both positive and negative) will average to zero. So the
integrator needs to have a charge rate that is fast enough (compared to the sample rate) to build up a signal, and a
discharge rate that will get rid of the signal fairly quickly one the target is gone, so you don't get an audio response
long past the target.
So let's make up some numbers. Sample rate is 1kHz, so sample period is 1ms. Sample width is 10us or 1%. Let's say a
weak target signal from the preamp is 10mV. A single sweep of the coil takes 1 second.
We get 1000 pulses per swing, so as we go over a target we might get 10-20 positive samples. Let's say exactly 10.
Without an integrator, this would result in 10 samples that are 10mV high with a 1% duty cycle.
This is a time-averaged signal of 100uV and probably will never trigger the audio.
Let's say we use a single-ended integrator shown below; R1=1k, R2=100k, C1=0.1uF. When the switch is closed vin (=10mV)
is converted to a 10uA current (i1) by R1 and charges the cap. dv/dt = i/C1 which gives 10uA/0.1uF = 100V/s, so in 10us the
cap charges up by a whopping 1mV. Wow, a whole millivolt? Yup.
When the switch opens the integrator is left with R2 discharging C1, at a time constant t = R2*C1 = 10ms. We have 990us
of discharge time (sample switch open) so the signal drops to e^(-990us/10ms) = 90% of whatever the sampled value is. So
when the switch is closed, the cap charges by 1mV, and when the switch is open the cap discharges by 10%.
We assumed a 10mV signal for 10 samples, so at the end of each of the 10 samples we have:
1: 1.00mV
2: 1.90mV
3: 2.71mV
4: 3.44mV
5: 4.10mV
6: 4.68mV
7: 5.22mV
8: 5.69mV
9: 6.12mV
10: 6.51mV
6.5mV doesn't sound like much, but it's 65 times better than the average 100uV without the integrator. And the integrator
is usually followed by a gain stage so if it has a gain of 100 (e.g., HH) then the final voltage applied to the audio is 650mV.
That's much better.
Finally, when there is no more target signal, the integrator cap will discharge in about 3t (95%) so a time constant of
10ms means the integrator output dies off in about 30ms.
- CarlComment
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Thank you Carl. Waltr, Kt315,an Davor. Thanks you folks really shined a light on a question I had about metal detectors, homemade and commercial. Thanks again have a great weekend, Best to all JimComment
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Jim
there is great article http://www.geotech1.com/pages/metdet...ead/HHv1p5.pdf
do not forget that. ufff.... and you read it in your native unlike me...
its so clear for you!Comment
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Thx kt315.Comment
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Tnx.KT315Comment
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