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Boise State University Department of Electrical and Computer Engineering ECE 360 – System Modeling and Control
The Laplace Transform II
Reading Assignment: Read Secs. 2.4-2.5.
Lecture Objectives:
- To derive the Laplace transforms of the derivative and integral of a time function.
- To apply the partial-fraction technique in the solution of linear differential equations.
Property 5: Derivative of a Time Function
L {f (t)} =
∫ (^) ∞
0
f (t)e−st^ dt = F (s)
L
{ f ′(t)
} = L
{ (^) df dt
} = sF (s) − f (0−)
Proof:
L
{ f ′(t)
} = L
{ (^) df dt
}
∫ (^) ∞ 0 −
df dt
e−st^ dt
=
∫ (^) ∞
0 −
e−st^ df
= [f (t)e−st]∞ 0 − −
∫ (^) ∞
0 −
f (t) d(e−st)
= [f (∞)e−s∞^ − f (0−)e−s^0 −] + s
∫ (^) ∞ 0 −
f (t)e−st^ dt = sF (s) − f (0−)
Higher Derivatives of a Time Function:
L
{ f ′′(t)
} = L
{ (^) d dt
[ f ′(t)
]} = sL
{ f ′(t)
} − f ′(0−) = s[sF (s) − f (0−)] − f ′(0−) = s^2 F (s) − sf (0−) − f ′(0−) L
{ f ′′′(t)
} = L
{ (^) d dt
[ f ′′(t)
]} = sL
{ f ′′(t)
} − f ′′(0−)
= s[s^2 F (s) − sf (0−) − f ′(0−)] − f ′′(0−) = s^3 F (s) − s^2 f (0−) − sf ′(0−) − f ′′(0−)
Property 6: Integral of a Time Function
L {f (t)} =
∫ (^) ∞
0
f (t)e−st^ dt = F (s)
L
{∫ (^) ∞
0
f (t)
}
F (s) s
Proof:
F (s) = L {f (t)} = L
{ (^) d dt
∫ (^) t
0
f (t) dt
} = sL
{∫ (^) t
0
f (t) dt
} −
∫ (^0)
0
f (t) dt
Examples:
0 1 2 t 0 1 2 t 0 1 2 t
1
2 2
1
2
u(t) tu(t) 0.5t 2 u(t)
1
f (t) = u(t) f (t) = tu(t) f (t) = 1 2
t^2 u(t)
F (s) = 1 s
F (s) = 1 /s s
s^2
F (s) = 1 /s
2 s
s^3
Solution of Linear Differential Equations:
Example 2: Solve
dx dt
Solution: Let X(s) = L {x(t)}.
sX(s) − x(0) + 2X(s) =
s + 2 (s + 2)X(s) = 5 +
s + 2 =^
5 s + 11 s + 2 X(s) =
5 s + 11 (s + 2)^2 =^
A
s + 2 +^
B
(s + 2)^2 =^
A(s + 2) + B (s + 2)^2 X(s) = 5 s^ + 11 (s + 2)^2
= As^ + (2A^ +^ B) (s + 2)^2
{ A = 5 2 A + B = 11 =⇒
{ A = 5 B = 1
X(s) = 5 s + 2
(s + 2)^2 x(t) = 5 e−^2 t^ + te−^2 t, t ≥ 0 x(0) = 5 + 0 = 5
Alternate Method for Partial-Fraction Expansion:
X(s) = 5 s^ + 11 (s + 2)^2
= A
s + 2
+ B
(s + 2)^2 B =
[ (s + 2)^2 X(s)
]
[^ s=−^2 =^ [5s^ + 11]s=−^2 =^1 (s + 2)^2 X(s)
] s=0 =^ [A(s^ + 2) +^ B]s= [5s + 11]s=0 = [A(s + 2) + B]s= 11 = 2 A + 1 =⇒ A = 5 X(s) = 5 s^ + 6 (s + 2)^2
s + 2
(s + 2)^2
Solution of Linear Differential Equations:
Example 3: Solve
d^2 x dt^2 + 4^
dx dt + 13x^ =^6 e
−t, x(0) = 0 , x′(0) = 0
Solution: Let X(s) = L {x(t)}.
s^2 X(s) − sx(0) − x′(0) + 4[sX(s) − x(0)] + 13X(s) = 6 s + 1 (s^2 + 4s + 13)X(s) = 6 s + 1
X(s) = 6 (s + 1)(s^2 + 4s + 13)
= A
s + 1
= A(s
(^2) + 4s + 13) + (Bs + C)(s + 1) (s + 1)[(s + 2)^2 + 3^2 ]
=
(A + B)s^2 + (4A + B + C)s + (13A + C) (s + 1)[(s + 2)^2 + 3^2 ]
A + B = 0
4 A + B + C = 0
13 A + C = 6
A = 0. 6
B = − 0. 6
C = − 1. 8
X(s) = 0.^6 s + 1
− 0.^6 s^ + 1.^8 [(s + 2)^2 + 3^2 ] X(s) = 0.^6 s + 1
− 0 .6(s^ + 2) + 0.^2 ×^3 [(s + 2)^2 + 3^2 ] x(t) = 0. 6 e−t^ − 0. 6 e−^2 t^ cos 3t − 0. 2 e−^2 t^ sin 3t, t ≥ 0 x(0) = 0. 6 − 0 .6 + 0 = 0
