Integration and RC Step response

[0tt0UK]

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

 

[Mark44]

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 = V_in - V_C, what will du be?

If the minus wasn't there the whole thing would change and the final equation would be different

Thanks in advance
Regards

 

[0tt0UK]

...u = V_in - V_C, what will du be?

du in function of Vc ?

du = -1 ? is that right?

 

[Mark44]

No, if u = V_in - V_C, then du = -dV_C.

 

[0tt0UK]

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

 

[Mark44]

Use and ordinary substitution with u = V_in - V_C.

What is du?
Make the substitution and do the integration. What do you get?

 

[0tt0UK]

- ln (Vin - Vc) + const
thanks very much Mark44