Laplace transforms in circuit analysis, finding missing values

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The discussion focuses on using Laplace transforms in circuit analysis to find missing values for resistance (R) and capacitance (C). The participant is trying to determine these values from a given time-domain response equation, specifically 6 + 12e^-2t. It is suggested that instead of using Laplace transforms, one can directly analyze the steady-state voltage to find R and C. The approach involves recognizing that at steady state, a capacitor behaves as an open circuit, allowing for a voltage divider calculation to determine R. Ultimately, the emphasis is on simplifying the process by focusing on steady-state conditions rather than complex transformations.
Cocoleia
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Homework Statement


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The Attempt at a Solution


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At this point, usually I would replace the values and sometimes separate into partial fractions, and then take the inverse Laplace transformation. So I know that the inverse Laplace needs to give me 6+12e^-2t.

I am given the answers in my notes, R=6ohm and C=0.25F

Is my logic correct, that's to say finding the transfer function and then going about taking an inverse Laplace and kind of "matching" my equation with the unknowns to the equation given to me ?
Thanks
 
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Cocoleia said:
Is my logic correct, that's to say finding the transfer function and then going about taking an inverse Laplace and kind of "matching" my equation with the unknowns to the equation given to me ?
I suppose you could go that route. It may be more effort than is warranted though.

Since you're given the time domain response equation you should be able to determine the steady-state (final) voltage. That should let you determine the value of R quite easily. Can you think of what other information the equation gives you that would lead to a value for C?
 
gneill said:
I suppose you could go that route. It may be more effort than is warranted though.

Since you're given the time domain response equation you should be able to determine the steady-state (final) voltage. That should let you determine the value of R quite easily. Can you think of what other information the equation gives you that would lead to a value for C?
Is it that I take the Laplace of the given equation:
V(s) = 6/s + 12/(s-2)
and that's how I can say R=6?
 
No Laplace required! Just take the given time domain expression and determine the steady state (final) voltage. Then you should be able to set the value of R in the circuit accordingly. So what is the final voltage?

Hints:
At steady state, what does a capacitor "look like" to the circuit?

What type of circuit remains? (It's a common enough configuration)
 
gneill said:
No Laplace required! Just take the given time domain expression and determine the steady state (final) voltage. Then you should be able to set the value of R in the circuit accordingly. So what is the final voltage?

Hints:
At steady state, what does a capacitor "look like" to the circuit?

What type of circuit remains? (It's a common enough configuration)
Do I open circuit the capacitor and then do a voltage divider and since 6 is the "real part" of the given vo, when I solve for R in my voltage divider it equals 6. Solving for R will give 6.
 
Cocoleia said:
Do I open circuit the capacitor and then do a voltage divider and since 6 is the "real part" of the given vo, when I solve for R in my voltage divider it equals 6. Solving for R will give 6.
You should be confident in how to find the steady state conditions of a RC or LC circuit. Yes, you open circuit capacitors and short circuit inductors.

I don't understand what you mean by "real part" of the given vo. The given time domain expression is purely real valued for all time. What's the procedure for finding the steady state value of a time-domain expression? (what does "steady state" imply?)
 
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