Section 4.6

```4.6 Laplace transform
• What else did he do?
∞
X ( s ) = Lx(t ) = ∫ e − st x(t )dt
0
converting differential
equation to algebraic ones
∞
∞
dx
∞
− st dx
dt = e − st x 0 + s ∫ e− st xdt
L
= ∫e
dt 0
dt
0
= sX ( s ) − x(0)
Pierre-Simon Laplace
1749 ( Beaumont-enAuge, Normandy) to
1827 ( Paris)
Generalized impedance
• Using the transforms of
d 2x
L 2 = s 2 X ( s ) − sx(0) − x (0)
second derivative
dt
• Write spring-mass(ms 2 + cs + k ) X ( s ) =
dashpot equation as
F ( s ) + (ms + c) x(0) + mx (0)
• Without initial
F (s)
= ms 2 + cs + k
Z (s) =
X (s)
conditions get
1
X (s)
generalized impedance
= 2
G(s) =
F ( s ) ms + cs + k
and transfer function
Laplace transform of delta and step
functions
• For delta function
∞
∆( s ) = ∫ e − stδ (t )dt = e − st
0
• Since X ( s ) = G ( s ) F ( s )
• Then
x(t ) = L-1G ( s ) F ( s )
Then
g (t ) = L G ( s )∆ ( s ) = L G ( s )
-1
• Similarly
-1
∞
∞
1
U ( s ) = ∫ e u (t )dt = ∫ e dt =
s
0
0
− st
− st
G(s)
s(t ) = L G ( s )U ( s ) = L
s
-1
t =0
=1
-1
Laplace transform tables
• Appendix B.8
• Boyce and Diprima, Elementary differential
equations and boundary value problems.
Example
• Use Laplace transform to find solution of
x + x = sin 2t
x(0) = 2
x (0) = 1
• Laplace transform of equation (see table for
sine)
s 2 X ( s ) − sx(0) − x (0) + X ( s ) = 2 /( s 2 + 4)
• Solution in s-plane
2 s + s + 8s + 6
X ( s) = 2
2
( s + 1)( s + 4)
3
2
Solution in time domain
• Partial fraction expansion
2s
5/3
2/3
X (s) = 2
+ 2
− 2
s +1 s +1 s + 4
• Back-transform using table
5
1
x(t ) = 2 cos t + sin t − sin 2t
3
3
• Checks?
Example 4.7 by table
• Find response of undamped single degree of
freedom system to the sawtooth pulse
F (t ) = F0 [tu (t ) − (t − T0 )u (t − T0 ) − u (t − T0 ) ]
1
Ltu (t ) = 2
s
(#3)
e − sT0
L (t − T0 )u (t − T0 ) = 2
(#13)
s
e − sT0
Lu (t − T ) =
(#12)
s
Example 4.7 - continued
⎡ T0 e − sT0 1
− sT0 ⎤
+ 2 1− e
⎢−
⎥
s
s
⎣
⎦
1
1
=
G(s) = 2
ms + k m( s 2 + ωn2 )
F
F (s) = 0
T0
• Solution in s-plane
(
)
X ( s ) = F ( s )G ( s )
• From Table B.8
1
1
L
= 2 (1 − cos ωnt )
2
2
s ( s + ωn ) ωn
-1
1
1
L 2 2
= 2 (ωnt − sin ωnt )
2
s ( s + ωn ) ωn
-1
Solution
• With some algebra
Checks?
4.7 General system response
• Applying Laplace transform to
mx + cx + kx = f (t )
x(0) = x0
x (0) = v0
• Get
X ( s) =
s + 2ζωn
F (s)
1
x
v
+
+
2
2
2
2 0
2
2 0
m( s + 2ζωn s + ωn ) s + 2ζωn s + ωn
s + 2ζωn s + ωn
• See development in textbook to get
1
x(t ) =
mωd
+
∫
t
0
F (τ )e −ζωn (t −τ ) sin ωd (t − τ )dτ
x0ωn
ωd
e
−ζωn t
cos(ωd t −ψ ) +
v0
ωd
e −ζωnt sin ωd t