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Laplace Transforms
& Functions exam
Unit impulse δ(t) - answer ✅✅ 1 A function that is 0 for all time, except at time 0, it has an infinite magnitude. Area underneath the curve = 1 Approximates signals for behaviors such as a hammer strike. Unit Step u(t) - answer ✅✅1/s A function that is 0 for all time up to t, except at time 0, it steps put to a value of 1. 1 - answer ✅✅1/s Area underneath the curve = 1 Unit Ramp At - answer ✅✅A/s² A straight line with a slope of 1 and y intercept of 0 Area underneath the curve =1/s² tⁿ ¹ ÷ (n - 1)! - answer⁻ ✅✅1/sⁿ
tⁿ - answer ✅✅n!/sⁿ ¹⁺ A parabola or cubic function e^(-at) - answer ✅✅1/(s+a) t * e^(-at) - answer ✅✅1/(s+a)² tⁿ ¹ ÷ (n - 1)! * e^(-at) - answer⁻ ✅✅1/(s+a)ⁿ tⁿ * e^(-at) - answer ✅✅n!/(s+a)ⁿ ¹⁺ sin (bt) - answer ✅✅b/(s²+b²) cos (bt) - answer ✅✅s/(s²+b²) sinh (bt) - answer ✅✅b/(s²-b²) cosh (bt) - answer ✅✅s/(s²-b²) 1/a [ 1 - e^(-at) ] - answer ✅✅1/s(s+a) 1/(b-a) [ e^(-at) - e^(-bt) ] - answer ✅✅1/(s+a)(s+b) 1/(b-a) [ be^(-bt) - ae^(-at) ] - answer ✅✅s/(s+a)(s+b)
Unit Step Function - answer ✅✅The function's value is defined as 0 up to a certain point (a), but then as a constant value of 1 thereafter f(t) = 1(t) Laplace of Unit Step Function - answer ✅✅1/s Ramp Function - answer ✅✅f(t) = At Laplace of Ramp Function - answer ✅✅A/s² Sinusoidal Function - answer ✅✅f(t) = A sin (bt) or f(t) = A cos (bt) Laplace of Sinusoidal Function - answer ✅✅Ab/(s² + b²) or As/(s² + b²) Pulse Function - answer ✅✅f(t) = A/t₀ l(t) - A/t₀ l(t-t₀) Laplace of Pulse Function - answer ✅✅f(t) =1/s * A/t₀ * (1-e^(-st₀)) Impulse Function - answer ✅✅f(t) = A/t₀ Laplace of Impulse Function - answer ✅✅A Linear Spring Element - answer ✅✅F = k(x₁-x₂) Torsional Spring Element - answer ✅✅T = k(θ₁-θ₂)
Linear Damping Element - answer ✅✅F = b(x°₁-x°₂) Where x° is the rate of change in displacement (velocity) Rotary Damping Element - answer ✅✅T = k(θ°₁-θ°₂) Where θ° is the rate of change in displacement (velocity) Rotary System Model - answer ✅✅ Spring Mass System Equation of Motion - answer ✅✅mx°° + kx = 0 (If From FBD Spring Mass System Characteristic Equation - answer ✅✅x = 0 or D² + k/m Spring Mass System Solution - answer ✅✅x = A sin{√(k/m)t} + B cos{√(k/m)t} Free Vibration with Viscous Damping Equation of Motion - answer ✅✅Spring Mass w/Damper System m°°x + bx° +kx = 0 From FBD = forcing function i.e. person on a bar...not allowing it to oscillate and return to initial equilibrium Free Vibration with Viscous Damping Characteristic Equation - answer ✅✅ms² + bs + k = 0 or = forcing function i.e. person on a bar...not allowing it to oscillate and return to initial equilibrium
Undamped Natural Frequency Equation - answer ✅✅wn = √(k/m) = sπƒ Damping Ratio - answer ✅✅ε = b/ 2√(k/m) Damping Frequency - answer ✅✅wd = wn √(1-ε²) Cycle of System - answer ✅✅T = 2π/wd Also known as the period Frequency - answer ✅✅ƒ = 1/T ƒ = √(k/m) / 2π Underdamped System - answer ✅✅Underdamped if: b²< 4mk OR 0 < ε < 1 There is damping, but not enough to immediately stop the oscillation Laplace of Underdamped System - answer ✅✅x(t) = e^(-εwnt) [ x(0) cos (wd t) + ε/√(1-ε²) sin (wd t) Critically Damped System - answer ✅✅Critically Damped if: b²= 4mk OR ε = 1 Damping forces are just enough to overcome the forces of the spring (will eventually damp out)
Overdamped System - answer ✅✅Overdamped if: b²> 4mk OR ε > 1 Damping impedes oscillatory motion Determining if System is Critically, Over or Underdamped from Equation of Motion - answer ✅✅mx°°+ 2(εw)x° + w^2x = 0 Where ε = Damping Ratio and w = Undamped Natural Frequency Transfer Function - answer ✅✅G(s) G(s) = x(s)/P(s) Take original equation of motion (mx°° + bx° + kx) = P Do Laplace transform of both sides using L[x(t)] = x(s) and L[P(t)] = P(s) Simplify and factor x(s) out Divide P(s) by x(s) to get x(s)/P(s) which = G(s)
