This is a simplified equivalent circuit of the toroid. The loop on the left represents the test loop or beam, and the loop on the right represents the toroid inductor and the terminating resistance of the cable. To determine the current coming out of the toroid, I(t) :

Next, we take an approximation of a ping, and run it through the left hand loop (the beam is assumed to be a current source)

A good approximation would be:

for t = 0 to , then zero from then on. This happens to be easily integrated, and easily reproduced by a signal generator...

After differentiating this, plugging it into the above equation, and working through about two pages of algebra (reproduction of which is left as an exercise to the student), you get the following equation for the current coming out of the toroid:

for t = 0 to . After which time:

where happens to be equation (1) evaluated at .

An important thing to notice is that the signal coming out of the toroid is linear with respect to , the the amplitude of the input pulse.

Below is a plot of both the input and output signals, for toroid A. In this case:

The dotted curve is the input pulse, , and has been scaled down by a factor of 50 to fit in the same plot as the output pulse (solid line). Notice that at this frequency (corresponding to a 4 mS wide ping) there is a fairly large undershoot in the signal coming out of the toroid. This will cause the integrator signal to decay.