[Caleb Fulton]

Originally Posted by MikeDubreuil

Exactly,
It's general knowledge that the Victors pulse 12V to the motors to control their speed. Except at the extremes when the FET's give the full 12V. It would be very difficult to measure this on a general purpose multimeter.

I am extremely skeptical of the voltages you are getting. A better test would be to see if the frequency is linear. I plan on checking that when I get back to the lab after the holidays.

I'll do the same, and try to remember to post my results as well.

In response to Andy's post:

I can't tell how familiar you are with circuit analysis in the time domain, but it seems like you have a pretty good grasp of it qualitatively.

Here's a "formal" way of going about the series inductor-resistor's response to an impulse.

The impulse can be modeled by the use of the unit step function:

Vin = Vm(u(t) - u(t-Tw)),

where Vm is the voltage magnitude of the pulse and Tw is the length of the pulse. The unit step function u(tau) is simply 1 for tau >= 0 and 0 otherwise.

The "physics" description of how the inductor responds to voltages and currents is good for qualitative analysis, but I often get confused with the signs corresponding to the use of the term EMF instead of voltage drops. Here's the way I visualize it (this should be fun in ascii):

 Current, i(t)
   |   |
   |   |
   |  \|/
   |   v
+  |
    ) <-this "comes out at you"
  (   <-this "goes into the background"
    )
  (
    )
-  |
   |

emf = -L * di(t)/dt

An increasing i(t) over some period of t will cause the magnetic flux cutting the coils to increase in the downward direction. This causes an emf to be generated in the opposite direction, which is in the upward direction for the orientation of the coils above. Here's where my memory gets fuzzy, but I *think* it's right: The emf going upward, having the same units as Voltage, causes a voltage to develop across the terminals as marked on the figure. THe reason for this orientation, as I think of it, is that the emf is the result of a circular electric field that causes positive charges to go up in the figure. Thus, if a "load" existed in parallel with the inductor, it would act like a "battery" with a "voltage" that caused the current to go in the direction of the emf.

Thus, we can write a nice voltage/current relationship for the inductor and eliminate the negative sign and the emf term:

v(t) = L * di(t)/dt,

where the voltage and current orientation are as in the figure above. This corresponds to the passive sign convention, commonly used in linear circuit analysis, and the sign convention it follows is also used for resistors and capacitors: the current enters the positive node of any two-terminal element.

So let's say that we have a voltage source, given by Vin = Vm(u(t) - u(t-Tw)), in series with our inductor, L, and reisistor, R.

current
 -->
 ______
|      |
|      R
+      |
Vin    L
-      |
|______|

Kirchoff's voltage law gives the following equation:

Vin(t) = R * i(t) + L * di(t)/dt

or:

di(t)/dt + R / L * i(t) = Vin(t) / L

Using the integrating factor method (you can look this up if you haven't seen it before):

Let the integrating factor be exp(R / L * t) (where exp(x) = e^x)

then:

exp(R / L * t) * (di(t)/dt + R / L * i(t)) = d/dt((exp(R / L * t) * i(t)) = exp(R / L * t) * Vin(t) / L

Changing the variable of integration to tau and integrating both sides from 0 to t (Vin is 0 for tau < 0):

int(d/dt(exp(R / L * tau) * i(tau)), tau, 0, t) = int(exp(R / L * tau) * Vin(tau) / L, tau, 0, t)

For 0 <= t < Tw, Vin(t) = Vm, so evaluating the integrals gives:

exp(R / L * t) * i(t) - i(0) = Vm / L * ( L / R) * exp(R / L * t) - Vm / L * ( L / R) = Vm / R * exp(R / L * t) - Vm / R

We can say, for this case, that i(0), the initial current through the inductor, is 0, so:

exp(R / L * t) * i(t) = Vm / R * exp(R / L * t) - Vm / R

i1(t) = Vm / R (1 - exp(-R / L * t))

I call this i1(t) because it is only valid for the first part of the Vin excitation to the system.

From here, one can define t' = t - Tw such that t' = 0 corresponds to t = Tw

We can now solve for i2(t'), which is valid for all t' > 0 (or t > Tw) using the same method. It will come down to this equation again:

di(t')/dt' + R / L * i(t') = Vin(t') / L = 0 for t' > 0

Separating variables:

di(t')/i(t') = -R / L * dt'
or
di(tau')/i(tau') = -R / L * dtau'

Integrating from 0 to t', since we have taken into account tau' < 0 in the previous section of the problem:

ln(|i(t')|) - ln(|i(0)|) = -R / L * t'

exp(ln(|i(t')|) - ln(|i(0)|)) = exp(-R / L * t')

i2(t') = i2(0) exp(-R / L * t') where i2(0) = i1(Tw), a constant

In terms of t, i(t) for t > Tw is:

i1(Tw) exp(-R / L * (t - Tw))

Thus, the current over all time is:

i(t) = {0, t < 0 
       {Vm / R (1 - exp(-R / L * t)), 0 <= t < Tw
       {i1(Tw) exp(-R / L * (t - Tw)), t >= Tw

So this is what the waveform for the current looks like when there is a resistor in series with an inductor as a response to a pulse.