# Advanced Circuits, Laplace Transform, Find Initial Conditions

• Engineering
• Color_of_Cyan
In summary, the conversation discusses finding the solution for Vo(s) given relevant equations and the Laplace transform. The initial condition is v(0+), not v(0-), and the right-hand side can be simplified to L-1{Vi(s)N(s)}. Equating like powers of s to the given terms and solving for v(0+) and the necessary additional derivatives of v(0+) is the recommended approach, rather than using partial fraction expansions.
Color_of_Cyan
Homework Statement
For a circuit, applying Laplace transform, the output is related to input by:

V_o(S)=(N(s) V_i(s) - s^2+s-2)/D(s),

where D(S) = (s^3 + s^2 + 1).

Find ALL initial conditions (there is no need to know N(s) ).
Relevant Equations
Partial Fraction Expansion, Laplace Transforms:
d^n x(t) / dt^n <---> s^n X(S) - s^n - 1x(0-) – s^(n-2)x'(0-) – s^(n-3)]x''(0-) ….
dx(t)/dt <---> sX(s) – x(0-)

d^2x(t)/dt^2 <---> s^2X(s) – sx(0-) – x/(0-)
Vo(S) = [ N(s)Vi(s) + (- s2 + s - 2) ] / s3 + s2 + 1 ;

can ignore (-s^2 + s - 2).

From relevant equations:

Vo(S) = [N(s)*Vi(s)]/(s^3 + s^2 + 1); -> (d3Vo(t)/dt3) + (d2Vo(t)/dt2) + Vo(t) = N(t)(dvi)/dt

L[vi(t)] = t to s domain: [s3Vo(s) - s2Vo(0-) - SV'o(0-) - Vo''(0-)]Vo(s) + s2 - SVo - V'o(0)] Vo(s) + Vo(s) = N(Vi)(s)

= [s3 V(s) - s2 + s + 2] V0(s) + [s2 - s - 1]V0(s) + V0(s) = N(Vi(s))

I'm sort of lost simplifying this now. I know I'm supposed to set up for partial fraction expansion eventually but not sure how.

Color_of_Cyan said:
Vo(S) = [ N(s)Vi(s) + (- s2 + s - 2) ] / s3 + s2 + 1 ;

can ignore (-s^2 + s - 2).

From relevant equations:

Vo(S) = [N(s)*Vi(s)]/(s^3 + s^2 + 1); -> (d3Vo(t)/dt3) + (d2Vo(t)/dt2) + Vo(t) = N(t)(dvi)/dt
This is a good start but where did you get the right-hand side? It's just L-1{Vi(s)N(s)} so leave it at that. You will not be concerned with N(s) in any way anyway.

NOTE: You should use v(t) instead of V(t). Don't confuse the two. Also, always use lower-case s fot the Laplace variable, not S.

Also, the initial condition is v(0+), not v(0-) etc.

You have the right idea equating v(0+), v'(0+) etc. with the given terms not functions of N(s). But now I've kind of lost you. Don't use partial fraction expansions.

Equate like powers of s to your given terms & solve for v(0+) and the necessary additional derivatives of v(0+).

## 1. What is an advanced circuit?

An advanced circuit is a network of interconnected electronic components that perform a specific function, such as amplification, filtering, or signal processing. These circuits often contain complex configurations of resistors, capacitors, inductors, and transistors, and are used in a wide range of electronic devices.

## 2. What is the Laplace transform?

The Laplace transform is a mathematical tool used to simplify the analysis of complex systems, such as advanced circuits. It converts a function from the time domain to the frequency domain, making it easier to solve differential equations and determine the behavior of a system over time.

## 3. How is the Laplace transform used in advanced circuit analysis?

The Laplace transform is used to convert the differential equations that describe the behavior of an advanced circuit into algebraic equations. This makes it easier to solve for the voltage and current at different points in the circuit, and to determine the overall response of the circuit to different inputs.

## 4. What are initial conditions in advanced circuit analysis?

Initial conditions refer to the values of voltage and current in a circuit at a specific point in time. These values are typically used to determine the behavior of a circuit after a sudden change in input, such as turning on a switch or applying a pulse of electricity.

## 5. How do you find initial conditions in advanced circuit analysis?

To find initial conditions, you must first solve the differential equations that describe the circuit's behavior using the Laplace transform. Then, you can use the initial values of voltage and current at a specific point in time to determine the constants of integration and fully define the circuit's response to different inputs.

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