Modeling Electrical Circuit Elements Using Laplace Transform

Study Case Modeling Electrical Circuit Elements Using Laplace Transform Michael Kurniawan (2018123456) University of Cincinnati, 2600 Clifton Ave, Cincinnati, OH 45221, United States kurniaml@mail.uc.edu Abstract Solving differential equations often encounters obstacles due to initial or boundary conditions. These obstacles are commonly encountered in the application of differential equations, including in electrical circuit analysis. One effective method for solving this problem is Laplace Transform. The Laplace transform, defined as ∞ 𝑒 −𝑠𝑡 𝑓(𝑡) 𝑑𝑡, ℒ{𝑓(𝑡)} = ∫ 0 can be used to find solutions to systems of differential equations with constant coefficients. The application of this method to electrical circuits includes: converting differential equations from the time domain (t) to the frequency domain (s), mapping initial conditions to auxiliary algebraic equations, solving with algebraic calculations, and taking the inverse Laplace transform to obtain specific solutions directly. Keywords: Laplace transform, differential equations, electrical circuits, initial conditions, frequency domain INTRODUCTION The Laplace transform is a method for solving differential equations by mapping initial value problems into algebraic equations that can be solved using algebraic techniques and Laplace transform tables. This method was first introduced by PierreSimon Laplace (1749–1827), a French mathematician and professor in Paris. The Laplace transform allows for the direct solution of problems that satisfy the initial conditions. An electrical circuit is a collection of electrical components that are interconnected to form at least one closed path. These circuits can be modeled using differential MICHAEL KURNIAWAN 1 Study Case equations, usually second order with constant coefficients, so that their solutions can be obtained using Laplace Transforms. The analysis is simple for circuits with only one element, but becomes complex for circuits with several different elements. Circuit components are divided into active and passive: • Active: Voltage and current sources that generate energy. • Passive: Do not generate energy, only absorb or store energy, such as resistors (absorb), inductors (store magnetic energy), and capacitors (store electrical energy). The difficulty of analysis increases because the characteristics of each element are different, even though mathematically v_R,v_L,v_C indicate the voltage across the resistor, inductor, and capacitor. SOLVING DIFFERENTIAL EQUATIONS WITH LAPLACE TRANSFORMS Laplace transforms are very useful for solving linear differential equations with constant coefficients. In this study, the focus is on second-order differential equations. General steps: 1. Take the Laplace transform of the given differential equation, including the initial conditions. 2. The result is an algebraic equation in the form Y(s). 3. If necessary, use partial fractions to solve the equation. 4. Take the inverse Laplace transform to obtain the solution in the time domain y(t). Given a second-order linear differential equation: ∞ 𝑒 −𝑠𝑡 𝑓(𝑡) 𝑑𝑡, ℒ{𝑓(𝑡)} = ∫ 0 Steps: 1. Laplace transform: ℒ{𝑦(𝑡)} = 𝑌(𝑠), ℒ{𝑟(𝑡)} = 𝑅(𝑠) 𝑎[𝑠 2 𝑌(𝑠) − 𝑠𝑦(0) − 𝑦 ′ (0)] + 𝑏[𝑠𝑌(𝑠) − 𝑦(0)] + 𝑐𝑌(𝑠) = 𝑅(𝑠) 2. Arrange the algebraic equation for Y(s): 𝑌(𝑠) = (𝑎𝑠 + 𝑏)𝑦(0) + 𝑎𝑦 ′ (0) + 𝑅(𝑠) 𝑎𝑠 2 + 𝑏𝑠 + 𝑐 3. If necessary, perform partial fractions. 4. Take the inverse Laplace transform to obtain y(t). MICHAEL KURNIAWAN 2 Study Case SAMPLE PROBLEMS AND SOLUTIONS Example 1: 2𝑦 ′′ + 4𝑦 ′ − 3𝑦 = 0, 𝑦(0) = 4, 𝑦 ′ (0) = 9 Laplace Transform: 2[𝑠 2 𝑌(𝑠) − 4𝑠 − 9] + 4[𝑠𝑌(𝑠) − 4] − 3𝑌(𝑠) = 0 (2𝑠 2 + 4𝑠 − 3)𝑌(𝑠) = 8𝑠 + 38 𝑌(𝑠) = 8𝑠 + 38 (2𝑠 − 1)(𝑠 + 3) Using partial fractions: 𝑌(𝑠) = 12 2 − 2𝑠 − 1 𝑠 + 3 Inverse Laplace transform: 𝑦(𝑡) = 6𝑒 0.5𝑡 − 2𝑒 −3𝑡 Example 2: 𝑦 ′′ − 4𝑦 ′ − 4𝑦 = 8𝑡, 𝑦(0) = 0, 𝑦 ′ (0) = 12 Laplace transform: (𝑠 2 − 4𝑠 − 4)𝑌(𝑠) − 12 = 𝑌(𝑠) = 8 𝑠2 8 12 + 𝑠 2 (𝑠 − 2)2 (𝑠 − 2)2 After inverse transformation: 𝑦(𝑡) = 14𝑡𝑒 2𝑡 − 2𝑒 2𝑡 + 2𝑡 + 2 Example 3: 𝑦 ′′ − 4𝑦 ′ + 8𝑦 = 8𝑒 2𝑡 cos⁡ 2𝑡 + 6𝑒 2𝑡 sin⁡ 2𝑡, 𝑦(0) = 4, 𝑦 ′ (0) = 10 Laplace transform: (𝑠 2 − 4𝑠 + 8)𝑌(𝑠) − 4𝑠 + 6 = 8(𝑠 − 2) 12 4𝑠 − 8 + 2 + + 2 2 2 (𝑠 − 2) + 4 [(𝑠 − 2) + 4] (𝑠 − 2)2 + 4 With the inverse Laplace transform: 3 7 𝑦(𝑡) = 2𝑡𝑒 2𝑡 sin⁡ 2𝑡 − 2 𝑡𝑒 2𝑡 cos⁡ 2𝑡 + 4𝑒 2𝑡 cos⁡ 2𝑡 + 2 𝑒 2𝑡 sin⁡ 2𝑡 ELECTRICAL CIRCUIT ELEMENTS IN THE S-DOMAIN In order for an electrical circuit to be transformed into the s-domain via Laplace transformation, we must express the circuit elements in s-domain form. According to John Bird (2007:640), the transformation of each element is as follows: MICHAEL KURNIAWAN 3 Study Case Resistor (R) In the time domain, resistors follow Ohm's law: 𝑣𝑅 (𝑡) = 𝑅𝑖(𝑡) After applying the Laplace transform: 𝑣𝑅 (𝑠) = 𝑅𝐼(𝑠) Capacitor (C) A capacitor in the time domain can be written as: 𝑖(𝑡) = 𝐶 𝑑𝑣𝐶 (𝑡) 1 atau𝑣𝐶 (𝑡) = ∫ 𝑖(𝑡)𝑑𝑡 𝑑𝑡 𝐶 The Laplace transform yields: 𝑣𝐶 (𝑠) = 1 𝐼(𝑠) 𝑠𝐶 Inductor (L) The inductor equation in the time domain: 𝑣𝐿 (𝑡) = 𝐿 𝑑𝑖(𝑡) 𝑑𝑡 After the Laplace transform: 𝑣𝐿 (𝑠) = 𝐿[𝑠𝐼(𝑠) − 𝑖(0)] APPLICATION OF LAPLACE TRANSFORM TO ELECTRICAL CIRCUITS Given an electrical circuit, the procedure for finding its solution through Laplace transform is (John Bird, 2007:642): 1. Apply the appropriate laws to form the differential equation of the circuit (e.g., Kirchhoff's laws and Ohm's law). 2. Take the Laplace transform of the equation. 3. Insert the given initial values or boundary conditions to form the auxiliary Laplace equation. 4. Use the inverse Laplace transform to obtain the solution in the time domain. MICHAEL KURNIAWAN 4 Study Case EXAMPLE PROBLEMS AND SOLUTIONS Example 4: Series L-R-C Circuit Using Kirchhoff's law: 𝑣(𝑡) = 𝑣𝐿 + 𝑣𝑅 + 𝑣𝐶 = 𝐿 𝑑𝑖 1 + 𝑅𝑖(𝑡) + ∫ 𝑖(𝑡)𝑑𝑡 𝑑𝑡 𝐶 Laplace transform: 𝐼(𝑠) = 𝑣 1 𝑠 2 𝐿 + 𝑠𝑅 + 𝐶 = 𝑣/𝐿 𝑅 1 𝑠 2 + 𝐿 𝑠 + 𝐿𝐶 By solving through perfect squares and Laplace inverse: 𝑖(𝑡) = 𝑣/𝐿 √ 1 − (𝑅/𝐿)2 𝐿𝐶 𝑒 −(𝑅/𝐿)𝑡 sin⁡(√ 1 − (𝑅/𝐿)2 𝑡) 𝐿𝐶 Example 5: RL circuit with a switch Differential equation: 0,1 𝑑𝑖 1 + 5𝑖 + ∫ 𝑖𝑑𝑡 − 2 = 0 𝑑𝑡 20 × 10−6 Laplace transform and solution: 𝑖(𝑡) = 0,0283𝑒 −25𝑡 sin⁡(706,7𝑡) A MICHAEL KURNIAWAN 5 Study Case Example 6: Two-loop circuit Loop equations I and II with Laplace transform: 𝐼1 (𝑠) = 6,667 𝐼2 (𝑠) = 𝑠 + 250 ⇒ 𝑖1 (𝑡) = 10 − 3,334𝑒 −166,667𝑡 𝑠(𝑠 + 166,667) 6,667 ⇒ 𝑖2 (𝑡) = 6,667𝑒 −166,667𝑡 𝑠 + 166,667 Example 7: Two-loop circuit with voltage output 𝐼1 (𝑠) = 5 1 1 ⇒ 𝑖1 (𝑡) = 5𝑒 −0,625𝑡 dan𝐼2 (𝑠) = − ⇒ 𝑖2 (𝑡) 𝑠 + 0,625 𝑠 𝑠 + 0,625 = 1 − 𝑒 −0,625𝑡 Example 8: Circuit with capacitor and inductor, initial conditions given Laplace transform yields: 𝑉𝑜𝑢𝑡 (𝑠) = MICHAEL KURNIAWAN 2𝑠(𝑠 + 3) 𝐴 𝐵𝑠 + 𝐶 = + 𝑠 3 + 8𝑠 2 + 10𝑠 + 4 𝑠 + 6,57 𝑠 2 + 1,43𝑠 + 0,61 6 Study Case With Laplace inverse: 𝑣𝑜𝑢𝑡 (𝑡) = 1,36𝑒 −6,57𝑡 + 0,64𝑒 −0,715𝑡 cos 0,316𝑡 − 1,84𝑒 −0,715𝑡 sin 0,316𝑡 CONCLUSION 1. Laplace transformation simplifies the solution of second-order linear differential equations with constant coefficients, so that the particular solution from the initial conditions can be obtained directly. 2. Steps for application: • Form the time-domain differential equation of the circuit (Ohm's/Kirchhoff's law). • Transform to the s-domain. • Insert initial values or boundary conditions. • Solve the algebraic equations, including partial fractions. • Use the inverse Laplace transform to obtain the current or voltage solution. MICHAEL KURNIAWAN 7 Study Case REFERENCE MICHAEL KURNIAWAN 8