Integration and RC Step response

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Homework Help Overview

The discussion revolves around the mathematical analysis of the RC step response in circuits, specifically focusing on the integration involved in deriving the relationship between time and voltage across the capacitor.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to understand the integration process related to the equation involving the RC circuit. Questions are raised about the origin of the negative sign in the equation and the implications of variable substitution in the integration process.

Discussion Status

The discussion is ongoing, with participants exploring the integration steps and questioning the assumptions made in the derivation. Some guidance has been provided regarding variable substitution, but no consensus has been reached on the understanding of the negative sign.

Contextual Notes

Participants are navigating through the mathematical intricacies of the integration process, with some expressing confusion over specific steps and the implications of the substitution used in the integration.

0tt0UK
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Integration

Homework Statement



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.mal/circuittheory/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):

Homework Equations



[tex]\int[/tex]dt = RC[tex]\int[/tex][tex]\frac{1}{Vin - Vc}[/tex]dVc

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


Thanks in advance
 
Last edited by a moderator:
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0tt0UK said:

Homework Statement



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.mal/circuittheory/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):

Homework Equations



[tex]\int[/tex]dt = RC[tex]\int[/tex][tex]\frac{1}{Vin - Vc}[/tex]dVc

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

The Attempt at a Solution



I don't understand the rason for the "-" just befor CR. Where did it come from?

I though [tex]\int[/tex][tex]\frac{1}{x}[/tex]dx = ln x + const
Yes, that's correct (or at least close. The right side should be ln |x| + const.

Similarly,
[tex]\int\frac{du}{u}~=~ln|u| + C[/tex]

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

Thanks in advance
Regards
 
Last edited by a moderator:
Mark44 said:
...u = Vin - VC, what will du be?

du in function of Vc ?

du = -1 ? is that right?
 
No, if u = Vin - VC, then du = -dVC.
 
right!? so how would you solve

[tex] \int[/tex][tex]\frac{1}{Vin - Vc}[/tex]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
 
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
 
Last edited:

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