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Posted: January 29, 2011 09:37 pm
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Hello everyone

Problem: An RC circuit initially at rest has a step input signal.The response c(t) across C is c(t)=1-e^-at
If now theres is an initial voltage of 2v on the capacitor then find out the c(t) for the same input step voltage.

My attempt at solution:

C(s)=1/s - 1/(s+a)
U(s)= 1/s

Comparing it with transfer function of RC circuit, Vout(s)/U(s)=1/(1+RCS) so here a=1/RC.

Now the new Vout(s)=c(s) + 2v
hence c(s)=Vout(s) - 2v

The answer given is 1 + e^-at. I just cannot come to that answer using the above method. Please can anyone guide me. Also can someone verify whether what i did was right.


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Posted: January 30, 2011 01:46 am
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You omitted the partial fractions step. Your final step before the inverse Laplace is factoring the equation from a polynomial fraction to a series of low-order fractions. Specifically, you'll have a s(s+a) denominator which must be seperated to obtain the 1/s and 1/(s+a) terms. Performing this process, you should find the initial condition affects the coefficients, thus giving the expected result (the voltage should fall asymptotically to the input level).


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Posted: December 15, 2011 01:14 pm
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I agree with the previous post in that you did not really show how you got C(s) so it's hard to say what you are doing is correct or not.

There are several ways to approach this problem mostly because it only involves one storage component, the capacitor.

One standard way would be to write a first order differential equation and use Laplace equivalents for the derivatives (simply Laplace transforms of the derivatives).

If you dont want to write a differential equation (after all it's only one capacitor) then you can use superposition calling the input source source 1 and the initial cap voltage source 2.

Most simply, since the initial cap voltage can be modeled as a simple battery in series with the cap lower lead (that connects to ground) we can shift the ground to the center tap of the cap and this new battery source. We can then solve the standard equation because now we only have one source which is the battery voltage and input source voltage (added or subtracted as needed):
The only difference now is that since we made the source Vs the combined original input source minus the initial voltage, we have to add that initial voltage back to the equation to get the right answer:
and since Vs was the input source minus the initial cap voltage we get:
That solution didnt require much work at all.

This post has been edited by MrAl on December 15, 2011 01:18 pm

Sorry i stepped on your T.O.E.
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