me375_sp11_hwk03_soln - ME 375 Spring 2011 Homework No 3...

ME 375 – Spring 2011
Homework No. 3
Due: Friday, February 4
Problem No. 1 (30%)
a) For
u t
( )
shown below, determine the Laplace transform L u t
( )
⎡
⎣
⎤
⎦
.
b) If Y s
( )
=
s + 6
5 s
2
+ 4 s
is the Laplace transform of y t
( )
:
i. what is
y t = 0
( )
?
ii. what is
y t → ∞
( )
?
c) If Y s
( )
=
s + 6
5 s
2
+ 4 s
is the Laplace transform of y t
( )
, what is the Laplace
transform of e
− 2 t
y t
( )
?
Part a): For the function
u t
( )
:
u t
( )
= 2 h t − 4
( )
+ 0.5 t − 4
( )
h t − 4
( )
where h t
( )
is the unit step function. Taking the Laplace transform of u t
( )
above gives:
L u t
( )
⎡
⎣
⎤
⎦
= 2 L h t − 4
( )
⎡
⎣
⎤
⎦
+ 0.5 L t − 4
( )
h t − 4
( )
⎡
⎣
⎤
⎦
= 2 e
− 4 s
L h t
( )
⎡
⎣
⎤
⎦
+ 0.5 e
− 4 s
L t
[ ]
=
2 e
− 4 s
s
+
0.5 e
− 4 s
s
2
=
2 s + 0.5
s
2
e
− 4 s
Part bi): y t = 0
( )
= lim
s →∞
sY s
( )
⎡
⎣
⎤
⎦
= lim
s →∞
s s + 6
( )
5 s
2
+ 4 s
⎡
⎣
⎢
⎤
⎦
⎥
= lim
s →∞
s + 6
5 s + 4
⎡
⎣
⎢
⎤
⎦
⎥
=
1
5
Part bii): y t → ∞
( )
= lim
s → 0
sY s
( )
⎡
⎣
⎤
⎦
= lim
s → 0
s s + 6
( )
5 s
2
+ 4 s
⎡
⎣
⎢
⎤
⎦
⎥
= lim
s → 0
s + 6
5 s + 4
⎡
⎣
⎢
⎤
⎦
⎥
=
6
4
= 1.5
Part c): L e
− 2 t
y t
( )
⎡
⎣
⎤
⎦
= Y s + 2
( )
=
s + 2
( )
+ 6
5 s + 2
( ) 2
+ 4 s + 2
( )
=
s + 8
5 s
2
+ 24 s + 28
2
2

Problem No. 2 (40%)
a) Determine the solution
x t
( )
for the following differential equation with the initial
conditions of
x 0
( )
= 2
:
10  x + 5 x = 20 h t
( )
, where
h t
( )
is the unit step function.
b) Determine the solution x t
( )
for the following differential equation with the initial
conditions of x 0
( )
= 0 and

x 0
( )
= 2 :  x + 12  x + 100 x = 0
c)