A
Anonymous
Guest
One way to remove some of the ground signal is by subtracting off an old value that was saved before the target signal arrived. This removes the DC component of the ground signal. The purpose of this example is to show that it is possible to remove even more of the ground signal.
For a target signal I am using the function (sin t)^2. For a ground signal I am using 1+.06sin t/2. The DC part of the ground signal is 1 and the AC part is 6% of the DC. The frequency of the ground signal is about 25% of the frequency of the target signal.
Filter 1 just subtracts the DC part of the ground signal.
Filter 2 is shown in a partial diagram below. There are 3 sample and holds that are clocked on each pulse. So they hold data from the last 3 pulses. The resistor values determine the filter coefficients. The filter function is out(t) = -in(t) +2*in(t-1) - in(t-2).
There are 3 graphs. The left graph shows the ground signal and a target signal that has one half the size of the ground signal. This is just to show the shapes of the two signals. In the other graphs the target signal will be .01 times the ground signal size.
The middle graph shows the .01 target signal and the ground signal after being filtered by filter 1.
The right graph shows the .01 target signal and the ground signal after each of them have been filtered by filter 2. Notice that the ground signal in this graph is much smaller than in the previous graph. This filter alters the shape of the target signal so both the X and R signals would have to go through identical filters before calculating an ID.
Filter 2 is very sensitive to noise, so this may not be a practical filter. It was just the simplest filter I could think of that would significantly reduce the ground signal.
Robert
For a target signal I am using the function (sin t)^2. For a ground signal I am using 1+.06sin t/2. The DC part of the ground signal is 1 and the AC part is 6% of the DC. The frequency of the ground signal is about 25% of the frequency of the target signal.
Filter 1 just subtracts the DC part of the ground signal.
Filter 2 is shown in a partial diagram below. There are 3 sample and holds that are clocked on each pulse. So they hold data from the last 3 pulses. The resistor values determine the filter coefficients. The filter function is out(t) = -in(t) +2*in(t-1) - in(t-2).
There are 3 graphs. The left graph shows the ground signal and a target signal that has one half the size of the ground signal. This is just to show the shapes of the two signals. In the other graphs the target signal will be .01 times the ground signal size.
The middle graph shows the .01 target signal and the ground signal after being filtered by filter 1.
The right graph shows the .01 target signal and the ground signal after each of them have been filtered by filter 2. Notice that the ground signal in this graph is much smaller than in the previous graph. This filter alters the shape of the target signal so both the X and R signals would have to go through identical filters before calculating an ID.
Filter 2 is very sensitive to noise, so this may not be a practical filter. It was just the simplest filter I could think of that would significantly reduce the ground signal.
Robert