## Integration and RC Step response

1. The problem statement, all variables and given/known data

Hi everyone,

I'm new here. This is my first post. I found physicsforums when researching on google a solution for a doubt I've had when trying to solve rc step response. Basically, the RC analyse showed at http://freespace.virgin.net/ljmayes....ory/Rcstep.htm solves step by step the mathematical behave of RC circuits.

I understand the whole anylise but the line with the follow part (integrating):

2. Relevant equations

$$\int$$dt = RC$$\int$$$$\frac{1}{Vin - Vc}$$dVc

t = - CR ln(Vin - Vc) + const

 PhysOrg.com science news on PhysOrg.com >> Hong Kong launches first electric taxis>> Morocco to harness the wind in energy hunt>> Galaxy's Ring of Fire

Mentor
 Quote by 0tt0UK 1. The problem statement, all variables and given/known data Hi everyone, I'm new here. This is my first post. I found physicsforums when researching on google a solution for a doubt I've had when trying to solve rc step response. Basically, the RC analyse showed at http://freespace.virgin.net/ljmayes....ory/Rcstep.htm solves step by step the mathematical behave of RC circuits. I understand the whole anylise but the line with the follow part (integrating): 2. Relevant equations $$\int$$dt = RC$$\int$$$$\frac{1}{Vin - Vc}$$dVc t = - CR ln(Vin - Vc) + const 3. The attempt at a solution I don't understand the rason for the "-" just befor CR. Where did it come from? I though $$\int$$$$\frac{1}{x}$$dx = ln x + const
Yes, that's correct (or at least close. The right side should be ln |x| + const.

Similarly,
$$\int\frac{du}{u}~=~ln|u| + C$$

Using an ordinary substitution in your problem, u = Vin - VC, what will du be?
 Quote by 0tt0UK If the minus wasn't there the whole thing would change and the final equation would be different Thanks in advance Regards

 Quote by Mark44 ...u = Vin - VC, what will du be?
du in function of Vc ?

du = -1 ? is that right?

Mentor

## Integration and RC Step response

No, if u = Vin - VC, then du = -dVC.

 right!? so how would you solve $$\int$$$$\frac{1}{Vin - Vc}$$dVc ?? Would you differentiate 1/Vin-Vc in function of Vc and then integrate the result? sorry I still don't understand the "-" on the right side of the eq. thanks
 Mentor Use and ordinary substitution with u = Vin - VC. What is du? Make the substitution and do the integration. What do you get?
 - ln (Vin - Vc) + const thanks very much Mark44

 Tags integration, integrator, low pass filter, step response