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Lossy Inverting Integrator

  • Thread starter kushnee
  • Start date
  • #1
2
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Hey guys,

I've got this set of problems on RC/RL circuits and differentiators and integrators. I was able to solve all of them, but this one stumped me for some reason.

Here is the problem:

http://img31.imageshack.us/img31/4286/34370298.png [Broken]

Relevant methods:

1) multiplying both sides by e^(-t/RC) and differentiating?

2) integrating, multiplying by e^(-t/RC), and isolating Vout?

Thats the methods I used to solve the other problems, but I'm not even sure how to start on this one, so any hints would be appreciated.

Oh, also, the solution should look something like this (i think?):

Vout(t)=c1+(c2-c1)e^(-(t-t0)/RC) and c1,c2 are constant
 
Last edited by a moderator:

Answers and Replies

  • #2
berkeman
Mentor
57,306
7,292
Hey guys,

I've got this set of problems on RC/RL circuits and differentiators and integrators. I was able to solve all of them, but this one stumped me for some reason.

Here is the problem:

http://img31.imageshack.us/img31/4286/34370298.png [Broken]

Relevant methods:

1) multiplying both sides by e^(-t/RC) and differentiating?

2) integrating, multiplying by e^(-t/RC), and isolating Vout?

Thats the methods I used to solve the other problems, but I'm not even sure how to start on this one, so any hints would be appreciated.

Oh, also, the solution should look something like this (i think?):

Vout(t)=c1+(c2-c1)e^(-(t-t0)/RC) and c1,c2 are constant
It looks like you need to differentiate both sides to get a DiffEq that you can solve...
 
Last edited by a moderator:
  • #3
2
0
Thanks for the tip. So I've tried integrating it and this is what I got:

d/dt[Vout-Vout(t0)]=-[(A-Vout(infinity)+Vout(t0))/RC]

Is this even correct? I'm not sure what method to use to solve this DE and I'm doubting my differentiation skills at this point. I replaced V1 with A, since it's a constant i suppose, and I moved the derivative of Vout(t0) to the left side.

Thanks,
Adam.
 

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