Using analog computer to solve 2nd-order diff eq

[jim hardy]

One of the problem now is that for practical reason, I use C=0.1uF and R=100k for the integrators, so Vout=100*integrate(Vin),

i'd have made R=10meg to keep it real time...

i remember doing time scaling but not quite exactly how.

have you tried on your simulator placing 100::1 attenuators at inputs of both integrators?
see section 9 at pdf page 35 of 51 here:
https://courses.engr.illinois.edu/ece486/labs/lab1/analog_computer_manual.pdf

 

[dilloncyh]

i'd have made R=10meg to keep it real time...

i remember doing time scaling but not quite exactly how.

have you tried on your simulator placing 100::1 attenuators at inputs of both integrators?
see section 9 at pdf page 35 of 51 here:
https://courses.engr.illinois.edu/ece486/labs/lab1/analog_computer_manual.pdf

I tried other values of resistance and capacitance for the integrator, but 100k ohm and 0.1uF seems to work best. If I use any other larger resistor (~1M ohm), the output signal is just to small to be seen on the oscilloscope). Since y'=-100*integrate(y''), and y=-100*integrate(y')=10000*integrate(y''), my first thought is that the diff eq becomes y''+100by'+10000cy=-f(t), but it doesn't seem to be true when I check the results.
For the lecture notes u mention, I can't really understand what it means and what it has to do with 'time' scaling, as now it is the output (y-axis of a plot) of each integrator that is changed.

PS: for the circuit diagram, just ignore Ud (Vsignal and Vin are shorted together). It predicts the actual Vout of my circuit in the lab quite well, and based on output of Vout and Vout2 using different values of R2 and R3, I believe that it actually works. Only problem is I don't know what equation is it actually solving and I need to work backward to get the coefficients b and c (which I have no idea how to do).

 

[dilloncyh]

Problem solved already. I guess my project's done. thanks for all the replies and help

 

[jim hardy]

Problem solved already.

Just now tuned in. What'd you do ?

 

[dilloncyh]

let 1/CR = k
the equation should be y''+k*b*y'+k^2*c*y=-k^2*f(t)
at least the ltspice and the actual circuit give the solution to this equation

 

[jim hardy]

Great !So it's still solving real time?

 

[dilloncyh]

Great !So it's still solving real time?

I'm still not sure what 'real time' means, but if u throw any diff eq of the form y''+by'+cy = f(t) to me, I can solve it using my circuit provided that the values of coefficients are not too extreme