# Laplace transforms in circuit analysis, finding missing values

• Cocoleia
In summary: 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?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.
Cocoleia

## The Attempt at a Solution

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

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?)

## 1. What is a Laplace transform and how is it used in circuit analysis?

A Laplace transform is a mathematical tool used to convert a time-domain function into its equivalent frequency-domain representation. In circuit analysis, it is used to simplify the analysis of complex circuits by converting differential equations into algebraic equations.

## 2. How do I find the missing values in a circuit using Laplace transforms?

To find missing values in a circuit using Laplace transforms, you will need to first write the circuit equations in the s-domain using the appropriate circuit elements and their corresponding transfer functions. Then, solve for the unknown variables and use inverse Laplace transform to convert the results back to the time-domain.

## 3. What are the advantages of using Laplace transforms in circuit analysis?

One of the main advantages of using Laplace transforms in circuit analysis is that it simplifies the process of solving complex circuits. It also allows for the use of algebraic equations instead of differential equations, making it easier to analyze and solve circuits. Additionally, Laplace transforms can also be used to analyze circuits with varying input signals, making it a versatile tool for circuit analysis.

## 4. Are there any limitations to using Laplace transforms in circuit analysis?

While Laplace transforms are a powerful tool for circuit analysis, they do have some limitations. One limitation is that it can only be used for linear circuits, meaning that circuits with non-linear elements cannot be analyzed using Laplace transforms. Additionally, the use of Laplace transforms can become complicated for circuits with multiple input sources.

## 5. How can I improve my understanding of Laplace transforms in circuit analysis?

To improve your understanding of Laplace transforms in circuit analysis, it is important to have a strong understanding of basic circuit theory and mathematical concepts such as complex numbers and differential equations. Additionally, practicing with different circuit problems that involve Laplace transforms can help improve your skills and understanding. Seeking additional resources, such as textbooks or online tutorials, can also be helpful in improving your understanding of Laplace transforms in circuit analysis.

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