wombat said:
Fred said:
First of all this is not homework it's just that linear systems and
transient circuit analysis hasn't been in the job description for a
while, actually ever.
R1 A R2 B R3
+---/\/\/\/\----+-----/\/\/\/\-----+-----/\/\/\/\-------+
| | | |
x(t) | | C1 | C2 | y(t)
===== ----- ----- =====
=== ----- ----- ===
| | | |
| | | |
+---------------+------------------+--------------------+
GND
Anyway the circuit is shown above. Clearly in steady state it's just a
voltage divider of the difference of Vx and Vy. The problem is that Vx
and Vy vary with time (out of my control). I need to report VA and VB to
the user but it must be the steady state result. In other words I must
filter out the transient effects caused by x and y. Please note that I
can't modify the circuit in any way. I know all the values for caps and
I can also measure *all* voltages. I even know the nominal values for
the resistors. The point of all this is to 'see' if the resistors change
through the "fog" caused the time varying sources.
My idea was to somehow use the system response [h(t)] to work out the
steady state result for A and B. Perhaps divide VA(t) by h(t) ????
eg in the case of VA:
x(t) --->| |
| h(t) |---> VA(t)
y(t) --->| |
I guess the first thing is, am I on the right track?
Not even close.
Secondly I could do with some tips on calculating h(t) at A and B.
I really appreciate any help.
Your system is undetermined. The problem statement is to predict a new
steady state for Va and Vb as a function of R1,2,3. These resistors are
on the order of 2e15 ohms and the capacitors are on the order of 2e-12
for a time constant of 4e3, or thousands of seconds, and this holds for
relatively minor +/-10% change in R. Then Vx and Vy exhibit a drift
characteristic on the order of hundreds of seconds. You can get an idea
of what happens by thinking of C1 and C2 as DC sources, batteries, of
magnitude steady state Va and Vb. As the resistor fluctuate at a rate
nearly instantaneous relative to the circuit time constants, all voltages
remain unchanged, and charge will be circulated through the resistors to
maintain those node voltages constant. Looks like you have everything
wrong, attempting to measuring a circuit parameter that nature is forcing
to be constant, meaning you have to measure *current* to detect the
resistor changes, the voltage measurements will barely move by ppm and be
undiscernible from drift. And what does this have to do with your
original ill-posed resistor network that was another failed
identification problem? You're a starting to look like a big waste of
time.
My original post was probably a little premature and therefore misleading
with regard to the problem statement so I'll clarify.
"I need to know if the resistors change by more than 10% while having to
contend with the sources of x and y moving up and down."
Regarding the time constant, I have modelled the circuit. As an example:
When R2 decreases it's resistance by 10% (to 1800G) point B changes to
it's maximum voltage (however only 0.4% change) in under 8 secs.
Unfortunately the current through the resistors can't be measured so I
have to rely on voltage measurement. It sounds pretty extreme, 0.4%
accuracy is hard to come by but if I measure differentially across the
resistor it equates to ~5% change - definitely achievable.
The previous problem is related but my methodology changed when I realised
I couldn't do it that way. It wasn't solvable.
I'm still not sure what your after but if I interpret what you said above
correctly then all you want to do is determine the change in R?
This is quite easy if you can measure the sources and have some standard
value of R to compare to. All you have to do is measure the voltage across
R and the voltage sources and compare it to what would theoretically be
expected with the standard value that R is suppose to be.
i.e., You use the "theoretical" values of the resistanaces and capacitances
and the experimental voltages sources and then the experimental values for
all and then compare for differences.
If your circuit is fixed and you can solve it algebraically then all you
have to do is "plug in" the measured values and the theoretical values and
compare. If you can't solve the circuit algebraically then you will have to
do numerics on it to get the results.
You could do more advanced mathematics(stochastic DE's and such) but I think
for your problem it would be quite easy since you can measure the voltage of
the voltage source. (else it would be impossible for an arbitrary voltage
source because you can't tell if the extra voltage drop on the resistors are
coming from the change in resistance or from the voltage source)
for example,
R
+---/\/\/\/\----+
| |
If, say, you are trying to figure out the change in R then its quite easy.
Vx - I*R - Q/C = 0
==>
I = dQ/dt
so
Vx - dQ/dt*R - Q/C = 0
and Q(t) = exp(-t/RC)*(Q0 + int(Vx(s)*exp(s/RC)/R,s=0..t))
but VR(t) = dQ/dt*R =
so the voltage drop across the resistor is given by
VR(t) = Vx(t) - exp(-t/RC)/RC*int(Vx(s)*exp(s/RC),s=0..t) - VC0*exp(-t/RC)
But we can measure VR(t), Vx(t), and VC0 and hopefully we know the
theoretical value for C(else its more complicated and probably impossible to
measure) then we can calculate the value R for this(numerically).
i.e., say the theoretical value for R is 10ohms. Then we can measure and
plug into the equation above and test for different R's until we get a true
statement. This R then can be compared with 10 ohms to see how much it
varied to a "hidden" variable.
If, say, C = 1, VC0 = 0, and Vx(t) = t when we start measuring(So we will
have to sample the voltage sourc) and suppose that we measure the voltage
across the resistor at the end of 1 sec to be 1/2V
then
1/2 = 1 - exp(-1/R)*int(s*exp(s/R),s=0..1)
we get the equation
1/2 = 1 - R^2*exp(-1/R) - R^2 + R
We can solve this numerically to find out what R has to be to produce those
measurements: This equation has solutions at about 1.065 ohms.
So one would have a change in about 9 ohms.
Anyways, Thats just an example and the measurements are made up for the
purpose of demonstrating what you could do. It might not be the best way to
do this though depending on the circuit and such. If you could measure the
current then it would be hell of a lot easier as you would only need to
measure the voltage and current of the resistor you want to measure and then
compute its value using ohms law... and it would be correct regardless of
the circuit topology(for the most part).
The above method may not work well though since its possible to have
multiple solutions and there might be stability issues in trying to
implement it.
Ultimately it would probably be much easier to just measure the current
along with the voltage. You can also do the same to measure the capacitance
since V = Q/C and I = dQ/dt but you will need to sample the current enough
to build a history for the integration.
Ofcourse if this was some type of research then you would be using much more
complication mathematics and physics to get the answers but I doubt you want
to spend the next 10 years on that.
Hopefully I'm on the right track with what you want to do. If so then one
thing you have to realize is that there are probably thousands of factors
that can change the resistance and can complicate matters. Even the simple
RC circuit above is quite complicated in this aspect if both the resistance
and capacitance can change. If you don't know what is making them change and
cannot model that sufficiently then it can be very difficult to measure
there change by limiting yourself to measuring only one aspect of them(such
as there voltage). The reason is simple. What we measure as a change of
resistance my actually be from a change of capacitance and vice versa. So
you would have to be vary careful not to allow this type of situation to
occur in your method. Also, say, assuming the capacitance is constant when
it is not could contribute. So, in this problem is not only what you
measure but what you don't. This is not to say that there might be
simplifications and approximations that could make the problem much easier
to deal with. I'm just not completely clear on what you are trying to do.
Anyways,
Jon