Laplace Transform Techniques in Circuit Analysis

Laplace Transforms of Important Functions

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Basic Laplace Transforms

  1. Unit Step Function: L[u(t)]=1sL[u(t)] = \frac{1}{s}

    • The unit step function is often used to model systems that switch on at a specific time.
  2. Constant Function: L(t)=1s2L(t) = \frac{1}{s^2}

    • Represents a constant rate or acceleration in a system.
  3. Polynomial Functions: L(t2)=2s3L(t^2) = \frac{2}{s^3}

    • General form: L(tn)=n!sn+1L(t^n) = \frac{n!}{s^{n+1}}
    • Useful for analyzing free response of systems and solving differential equations involving polynomials.
  4. Exponential Decay: L(eat)=1s+aL(e^{-at}) = \frac{1}{s+a}

    • Models processes like radioactive decay and cooling.
  5. Exponential Growth: L(eat)=1saL(e^{at}) = \frac{1}{s-a}

    • Used in contexts like population growth or compounding interest.
  6. Exponential Times Polynomial: L(teat)=1(s+a)2L(t e^{-at}) = \frac{1}{(s+a)^2}

    • Useful in more complex system responses involving time-delay effects.

Trigonometric Functions

  1. Cosine Function: L(cosωt)=ss2+ω2L(\cos \omega t) = \frac{s}{s^2 + \omega^2}

    • Describes oscillatory systems, such as electrical circuits or mechanical vibrations.
  2. Sine Function: L(sinωt)=ωs2+ω2L(\sin \omega t) = \frac{\omega}{s^2 + \omega^2}

    • Similar applications to cosine, often used with initial conditions.
  3. Damped Sine Wave: L(eatsinωt)=ω(s+a)2+ω2L(e^{-at} \sin \omega t) = \frac{\omega}{(s+a)^2 + \omega^2}

    • Models systems where oscillation amplitude decreases over time, like shock absorbers.
  4. Damped Cosine Wave: L(eatcosωt)=s+a(s+a)2+ω2L(e^{-at} \cos \omega t) = \frac{s+a}{(s+a)^2 + \omega^2}

    • Applies to damped oscillatory systems.

Derivatives and Integrals

  • Derivatives:

    • If L[f(t)]=F(s)L[f(t)] = F(s), then: L[ddtf(t)]=sF(s)f(0)L\left[\frac{d}{dt}f(t)\right] = sF(s) - f(0)
    • The derivative formula links the time domain to the s-domain by introducing a multiplicative factor ss and accounting for initial conditions.
  • Integrals:

    • If L[f(t)]=F(s)L[f(t)] = F(s), then: L[f(t)dt]=F(s)sL\left[\int f(t) dt \right]= \frac{F(s)}{s}
    • Integration in the Laplace domain corresponds to division by ss, useful for solving differential equations with integral terms.

These Laplace transform relationships are essential tools for engineers and scientists to analyze and solve linear time-invariant systems, especially in control systems, signal processing, and differential equations. They simplify the handling of complex systems by converting differential equations into algebraic ones.

Extended readings:

byjus.com
Laplace Transform (Definition, Formula, Properties and Examples)
lpsa.swarthmore.edu
Laplace Transform Properties - Linear Physical Systems Analysis
math.libretexts.org
9.7: The Laplace Transform - Mathematics LibreTexts

Transformed Circuits

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Resistor

  • Equation: V(t)=i(t)RV(t) = i(t)R

    • A resistor follows Ohm's Law, where the voltage across the resistor is directly proportional to the current through it, multiplied by its resistance RR.
  • Laplace Transform: V(s)=I(s)RV(s) = I(s) \cdot R

    • In the Laplace domain, the transformation simplifies the analysis of linear time-invariant systems. The impedance of a resistor remains RR.
  • Circuit Representation:

    • Original: A simple resistor with input voltage V(t)V(t) and current i(t)i(t).
    • Transformed: Resistor with impedance RR and input current I(s)I(s).

Inductor

  • Equation: V(t)=Ldi(t)dtV(t) = L \frac{di(t)}{dt}

    • The voltage across an inductor is proportional to the rate of change of current through it, scaled by inductance LL.
  • Laplace Transform:

    • V(s)=L[sI(s)i(0)]V(s) = L[sI(s) - i(0)]
    • I(s)=1LsV(s)+i(0)sI(s) = \frac{1}{Ls} V(s) + \frac{i(0)}{s}
    • These transformations account for the initial current i(0)i(0), showing how the inductance's stored energy affects current and voltage over time.
  • Circuit Representation:

    • Original: An inductor with initial current i(0)i(0).
    • Transformed: Inductor LsLs in series with a voltage source Li(0)L \cdot i(0).

Capacitor

  • Equation: i(t)=CdV(t)dti(t) = C \frac{dV(t)}{dt}

    • The current through a capacitor is proportional to the rate of change of voltage across it, determined by the capacitance CC.
  • Laplace Transform:

    • I(s)=C[sV(s)V(0)]I(s) = C[sV(s) - V(0)]
    • V(s)=1CsI(s)+V(0)sV(s) = \frac{1}{Cs} I(s) + \frac{V(0)}{s}
    • These equations highlight how a capacitor initially charged to V(0)V(0) affects system behavior in the Laplace domain.
  • Circuit Representation:

    • Original: A capacitor with initial voltage V(0)V(0).
    • Transformed: Capacitor 1Cs\frac{1}{Cs} in parallel with a voltage source V(0)s\frac{V(0)}{s}.

Insights:

  • Laplace Transformation: A mathematical tool used to convert differential equations into algebraic equations, making complex systems easier to handle, especially for control systems and signal processing.

  • Impedance in s-domain: For analysis in the Laplace domain, resistors, inductors, and capacitors are treated with their respective impedances RR, sLsL, and 1sC\frac{1}{sC}.

  • Initial Conditions: When transforming to the ss-domain, initial conditions are crucial as they represent the stored energy in capacitors and inductors, influencing system response.

Extended readings:

web.stanford.edu
[PDF] Lecture 7 Circuit analysis via Laplace transform
www.youtube.com
Laplace Domain Circuit Analysis - YouTube
maecourses.ucsd.edu
[PDF] s-Domain Circuit Analysis

LAPLACE CIRCUIT ANALYSIS NOTES

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Circuit Diagram

Description

  • Components:
    • Current Source: V0/sV_0/s.
    • Capacitor: 1Cs\frac{1}{Cs}.
    • Voltage Source: VsV_s.
  • Current: I(s)I(s).

Insights

  • Laplace Transform: The use of V(s)V(s) and I(s)I(s) indicates the application of Laplace transforms to circuit analysis, a powerful method to handle circuits in the s-domain.
  • Complex Frequency: The variable ss represents complex frequency, s=σ+jωs = \sigma + j\omega, where σ\sigma is the growth rate and jωj\omega is the oscillation.

Additional Concepts

Capacitive Reactance

  • Reactance in S-Domain: Xc(s)=1CsX_c(s) = \frac{1}{Cs}
    • Represents how a capacitor behaves as an impedance in the s-domain, inversely proportional to both the capacitance CC and ss.

Initial Conditions

  • Initial Charge or Current:
    • The term V0s\frac{V_0}{s} suggests an initial voltage, which is transformed from the time domain to the s-domain, accommodating initial conditions directly in Laplace analysis.

Analysis Strategy

  • Use Kirchhoff’s laws in the s-domain to set up equations involving voltage and current.
  • Solve algebraically for desired quantities, such as node voltages or branch currents, using known transforms.
  • Convert back to the time domain using inverse Laplace transform, if necessary.

Example Application

  1. Find I(s)I(s) in terms of known quantities:

    • Write the equation using KVL or KCL in the ss domain.
    • Example: Vs=I(s)1Cs+V0sV_s = I(s) \cdot \frac{1}{Cs} + \frac{V_0}{s}
    • Solve for I(s)I(s).
  2. Time-Domain Analysis:

    • Use inverse Laplace transform to revert known quantities back to the time domain.
    • Apply initial conditions where applicable to determine specific solutions in physical time.

General Tips

  • Make sure to account for all initial conditions due to their impact on the s-domain equations.
  • Familiarize yourself with standard Laplace transforms and their inverses, as they simplify solving differential equations.

Extended readings:

www.geeksforgeeks.org
Laplace Transform in Circuit Analysis - GeeksforGeeks
library.fiveable.me
Circuit analysis using Laplace transforms - Fiveable
www.dummies.com
Analyze a First-Order RL Circuit Using Laplace Methods - Dummies