ME 530.343: Design and Analysis of Dynamic Systems Spring 2009
Lecture 10 - Introduction to the Laplace Transform
Week of February 23, 2009
Today’s Objectives
Reading: Chapter 19 • Introduction to the Laplace Transform
The Laplace Transform
Converts a function of a real variable (always t in this class) into a function of a complex variable, s. “Laplace transform of f ”: L[f (t)] = F (s) Time domain t → frequency domain s Defining equation: F (s) =
∞ −st dt 0 f (t)e ∞ 0−
(Another convention is F (s) =
f (t)e−st dt)
s is a complex quantity s = σ + iω (Note that some texts use j instead of i.) e−st = e−σt e−iωt
Representing functions in the frequency domain
Step Function
f (t) = A for t > 0 f (t) = 0 for t ≤ 0 recall that F (s) = L[A] =
∞ −st dt 0 Ae
f (t)e−st dt
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t=∞ L[A] = Ae |t=0 −s L[A] = 0 − A·1 −s L[A] = A s
−st
for a unit step function, U(t) L[U (t)] = 1 s
Exponential function
f (t) = e−at ∞ L[e−at ] = 0 e−at e−st dt ∞ L[e−at ] = 0 e−(s+a)t dt L[e−at ] = L[e−at ] =
e−(s+a)t t=∞ −(s+a) |t=0 1 s+a
Ramp function
f (t) = t for t > 0 ∞ L[t] = 0 te−st dt Use formula for integration by parts b b b a udv = uv|a − a vdu
t=∞ L[t] = te |t=0 − −s L[t] = 0 − 0 + s1 2 L[t] = s1 2
−st
1 −s
∞ −st dt 0 e
Trigonometric functions
f (t) = sin(ωt) ∞ L[sin(ωt)] = 0 sin(ωt)e−st dt Use integration by parts e−st t=∞ L[sin(ωt)] = (−s)2 +ω2 (−s sin(ωt) − ω cos(ωt))|t=0 ω L[sin(ωt)] = s2 +ω2 Can also show that L[cos(ωt)] =
s s2 +ω 2
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Dependent Variables and their Derivatives
Say we have a dependent variable, x L[x] = X(s) L[x] = sX(s) − x(0) ˙ L[¨] = s2 X(s) − sx(0) − x(0) x ˙ Where x(0), x(0) are initial conditions ˙ Also note that you can take the inverse Laplace: L−1 [X(s)] = x(t) Differentiation example: ∞ L[f˙] = 0 ( df )e−st dt dt integrate by parts ... ∞ t=∞ L[f˙] = e−st f (t)|t=0 − 0 f (t)(−se−st )dt L[f˙] = [0 − f (0)] + s f (t)e−st L[f˙] = −f (0) + sF (s) L[f˙] = sF (s) − f (0)
What is the point of all this?
To find solutions to ODEs, you take the Laplace Transform of your equation of motion, solve for the solution in frequency domain, then take the inverse Laplace to get the time domain solution.
Example:
Equation of motion: bx + kx = fa (t) ˙ Say we have a step input in the applied force: fa (t) = 0 for t < 0 fa (t) = A for t > 0 The step function is U (t) · A Take Laplace transform of the entire equation of motion L[bx + kx = fa (t)] ˙ L[bx] + L[kx] = L[fa (t)] ˙ b[sX(s) − x(0)] + kX(s) = A s Let the initial condition be zero, x(0) = 0, and we can solve for X(s) X(s) =
s(s+ k ) b
A b
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We usually write this in a form so that there are no coefficients to the s variables. Now, to find x(t), we perform the inverse Laplace, using the attached table (page 548 in Close, Frederick and Newell). From the table, you cannot find the exact form above, so we need to split it into two terms. Let A = α, k = β, so b b α X(s) = s(s+β) We want to split this into two terms: α m n s(s+β) = s + s+β Multiply out to find correct coefficients m and n: m(s+β)+n(s) = s(m+n)+mβ s(s+β) s(s+β) We now know that we need α = s(m + n) + mβ, so m + n = 0 and mβ = α Thus, m = α and n = − α β β In terms of our original variables, m =
A k
A k
and n = − A k
So the final form of our solution in frequency domain is: X(s) =
s
+
−A k s+ k b
Now we take the inverse Laplace of each term: k L−1 [X(s)] = A − A e− b t k k x(t) = ( A )(1 − e−( b )t ) k
k
Note that we have just solved an inhomogeneous ODE. Using the Laplace tables and following a few simple rules, we will never have to perform integrals in order to take the Laplace Transform (or the Inverse Laplace Transform) of a function.
Laplace Transform Properties
Assume 2 functions f(t) and g(t) F (s) = L[f (t)] G(s) = L[g(t)] • Multiplication by a constant L[af (t)] = 0 af (t)e−st dt ∞ L[af (t)] = a 0 f (t)e−st dt L[af (t)] = aF (s)
∞
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• Superposition L[f (t) + g(t)] = 0 [f (t) + g(t)]e−st dt ∞ L[f (t) + g(t)] = 0 f (t)e−st dt + g(t)e−st dt L[f (t) + g(t)] = F (s) + G(s) • Multiplication of Functions (in general) L[f (t) · g(t)] = 0 f (t)g(t)e−st dt L[f (t) · g(t)] =???? L[f (t) · g(t)] = F (s)G(s)!! • But there are some special cases: – multiplication by an exponential L[f (t)e−at ] = f (t)e−at e−st dt L[f (t)e−at ] = f (t)e−(s+a)t dt L[f (t)e−at ] = F (s + a)
d L[tf (t)] = − ds F (s) ∞ ∞
– multiplication by time
Some Important Properties:
• Differentiation – First derivative use integration by parts: L[f˙] =
∞ df −at −st e dt 0 dt e
– Higher order derivatives
udv = uv|∞ − 0 vdu 0 u = e−st du = −se−st dv = ( df )dt dt v = f (t) ∞ L[f˙] = e−st f (t)|∞ − 0 f (t)(−se−st )dt 0 ∞ L[f˙] = [0 − f (0)] + s 0 f (t)(−se−st )dt L[f˙] = sF (s) − f (0)
n
∞
• Integration
f L[ d n ] = sn F (s) − sn−1 f (0) − ... − f (n−1) (0) dt ¨ L[f ] = sL[f˙(t)] − f˙(0) ¨ L[f ] = s[sF (s) − f (0)] − f˙(0) ¨ L[f ] = s2 F (s) − sf (0) − f˙(0)
L[
t 0 f (λ)dλ]
= 1 F (s) s
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