More about PI signal levels

[Anonymous]

To get ready for a digital PI project I have been doing some tests on my VLF test bed. It reads signals below 1 uV at the coil, but the difference is that it uses a narrower bandwidth amplifier. So there is less noise going into the A-D than there would be in a PI.
I increased the bandwidth of the amplifier to 6 kHz for a test, and the noise after integration increased by about a factor of 10 compared to what I normally see. This causes me to worry about the quality of the signal I am going to get if I just directly sample the output of the amplifier.
The first graph below shows the frequency response for a sampling rate of 4400 Hz with every other sample being subtracted and an amplifier bandwidth of 14 kHz. This is similar to a 2200 Hz pulse repetition frequency and differential sampling. The first peak in the response at 2200 Hz is the one that collects useful information. All the other peaks to the right pickup noise. These other peaks cannot be removed after sampling. They can be reduced by using a sharper roll off in the amplifier, or by taking more samples.
The most straight forward way to reduce those unwanted peaks is to increase the PRF. The second graph below shows the frequency response for a 7 kHz PRF and a 14 kHz amplifier. With fewer peaks to the right there is less opportunity to pick up noise.
In a sampled system there is normally a low pass filter before the sampling to cut off the peaks on the right. But that is a problem in a PI where there is a very large voltage spike just before the signal of interest.
Robert

 

[Anonymous]

Hi Robert,
Taking your last point first, you could gate the output of the amplifier so that all the transmitter transients are removed, just leaving the whole RX time up to the start of the next TX pulse. Maybe then you could use a low pass filter before sampling?
The Goldquest maximum prf is 10kHz, which pushes the first peak up to the 10kHz line on your graph and the succeeding noise sensitive peaks a bit further up. As you say, if we can roll off the amplifier a bit faster, this would help.
Is it possible to simulate how the responses are modified by the width of the sample and the time between the first and second sample. i.e. if I use a sample width of 5uS, or multiples of, then it greatly reduces interference from our 200kHz long wave transmitter, because the sample duration is equal to a full cycle, or cycles, of the interference. The sample spacing also can have a similar effect on lower frequency signals.
On one type of pulse generator that I use, altering the clock frequency alters everything else in proportion so you can nearly always find a quiet spot, whatever the interfering frequency.
Excluding rf interference before it ever reaches the front end amplifier would be nice. Sometimes you get interference from transmitters running way above the pass band of the amplifier, and this appears to be demodulation of the signal by the front end diodes or the non-linearity of the amplifier itself at those frequencies.
Efficient shielding of the coil and using ferrite sleeves on the RX input wires helps to some degree.
Eric.

 

[Anonymous]

Eric
One thing I neglected to mention about the previous graphs is that they include a 5 Hz low pass filter after the sampling. The first graph below is for a 2000 Hz pulse repetition frequency with no low pass filtering after the samples and no bandwidth limit before the sampling. There is a null at half the sampling frequency, and additional nulls at all odd harmonics of the first one.
The second graph shows the same thing with a 5 Hz low pass filter added after sampling. This filter cannot eliminate any peaks but it can reduce the response between the peaks making the system less sensitive to noise.
If the sampling period is P and you take a minus sample at a delay of D after the main sample then there will be nulls in the response at frequencies of 1/D and 1/(P-D) and all their harmonics and at 0. In this case P is 500 usec. The third graph is an example of taking the minus sample halfway between the main samples, so D is 250. In this case 1/D and 1/(P-D) are the same, 4000 Hz. So there are nulls at 4000 Hz and all its harmonics. These nulls fall right on half the peaks in the previous graph.
The fourth graph shows the response when D = 50 us and P-D is 450. So there are nulls at 20 kHz and at 2222 Hz and all their harmonics. The nulls at 20 kHz and its harmonics are easy to see in the graph. This is good for knocking out a narrow band source of interference. But notice the peaks at 4, 8, 12 kHz. that were eliminated from the previous graph but are back now. This is not good for random noise. If there is one dominant source of noise then it might be best to adjust D to knock out that source. But if there is no dominant source then I think D=P/2 probably eliminates the most noise.
I have not figured out an easy way yet to show the effects of the sample width.
I had been thinking about gating the amplifier, but I have not been able to figure out all the consequences yet.
Robert

 

[Anonymous]

Eric
Integrating a sample over a width of W causes a null in the response at a frequency of 1/W and all its harmonics. I can show those nulls in the frequency response the same way I did the nulls from subtracting a sample taken after a delay D.
But just adding together two simple samples taken W/2 apart causes the same null in the response as integrating over W. I cannot believe that 2 samples W/2 apart is equivalent to an infinite number of samples over W. So I think there is something else to show and I cannot figure out how to do it in a frequency response graph.
Robert