# Laplace Transform of the input portion of this circuit

• Comp Sci
• Jason-Li
In summary, the conversation discusses solving for a circuit in phasor form using Laplace transforms. There is some confusion about the form of the transform and the use of conductance, susceptance, and admittance instead of impedance. However, the final answer is deemed correct.
Jason-Li
Homework Statement
a) Draw the Laplace form of the input portion of the circuit, as
represented in FIGURE 4(c).

b) Derive an expression for the Laplace transfer function as per figure 4(c).

I have attached the transfer function & figure 4 (c) in my attempt.
Relevant Equations
Standard look up tables for Laplace
So I have completed (a) as this (original on the left):

I have then went onto (b) and I have equated T(s)=Z(s) as follows:

and due to

hence
Does this look correct to you smarter people?

Thanks in advance! All replies are welcome :)

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Are you trying to solve for the circuit in phasor form?

The results look right to me if it's just phasor form. There's a step where things look a little confusing by my eye. This part below looks concerning to me although it appears less concerning later on the right side.

$${1 \over Z(s)} = {1 \over {sC_p}} \parallel {1 \over {sC_s}} \parallel R_L$$

When I read about Laplace transforms (including standard lookup tables among your relevant equations)... I'm actually expecting a transform from ##s## to ##t## or its converse. The one on the table I'd be looking at is in this form although you'll have to massage the equation (a little bit) on your own if you do require the additional steps.

$$e^{-at} \leftrightarrow {1 \over {s + a}}$$

Your transform is ## V_L (s)/(I_p(s). ## You should be using conductance G(s), susceptance B(s) and admittance Y(s) = G + jB(s) rather than impedance Z = R + jX(s). That is suggested by the all-parallel nature of the circuit. You need to think in these terms if the math is thereby greatly simplified. Similarly, use Norton's therem for the input section.

Anyway, your answer is correct. But think G, B and Y whenever a circuit has lots of loops and few independent nodes.

## 1. What is the Laplace Transform of the input portion of this circuit?

The Laplace Transform of the input portion of a circuit is a mathematical tool used to analyze the behavior of a circuit in the frequency domain. It converts a time-domain signal into a complex frequency-domain representation, which allows for easier analysis and understanding of the circuit's response to different inputs.

## 2. How is the Laplace Transform calculated for a circuit's input portion?

The Laplace Transform of a circuit's input portion is calculated by taking the integral of the input signal multiplied by the exponential function e^-st, where s is a complex variable representing frequency. This integral is evaluated from 0 to infinity.

## 3. What are the advantages of using the Laplace Transform for circuit analysis?

The Laplace Transform allows for the analysis of complex circuits with multiple inputs and outputs, which would be difficult to analyze using traditional methods. It also simplifies the analysis by converting differential equations into algebraic equations, making it easier to solve and understand the behavior of the circuit.

## 4. Can the Laplace Transform be used for all types of circuits?

Yes, the Laplace Transform can be used for all types of circuits, including linear and nonlinear circuits. However, it is most commonly used for linear time-invariant (LTI) circuits, as it provides the most accurate and efficient analysis in these cases.

## 5. How does the Laplace Transform of the input portion affect the overall behavior of a circuit?

The Laplace Transform of the input portion determines the frequency response of a circuit, which describes how the circuit responds to different frequencies of input signals. This allows for the prediction of the circuit's behavior and can help in designing circuits with specific frequency responses.

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