Math 2250-4
Wed Nov 13
HW due tomorrow Thursday at start of labs! Review in labs on Thursday! Exam on Friday in class, 8:30
-9:30 a.m.! Go over last spring exam 2 and answer hw questions, Wed 4:30-6:00 Room WEB L103.
, Today we'll finish Tuesday's notes and make sure we understand how to set up and solve partial fraction
problems, since these are often key to using Laplace transform tables in finding inverse Laplace
transforms.
, Then, we'll begin some post-exam Laplace transform material, useful in systems where we turn forcing
functions on and off:
f t % CeM t
f t with
Fs d
N
f t eKs t dt
comments
for s O M
0
u tKa
eKa s
s
unit step function
f tKa u tKa
for turning components on and
off at t = a .
more complicated on/off
eKa sF s
The unit step function with jump at t = 0 is defined to be
0, t ! 0
u t =
.
1, t R 0
Its graph is shown below. Notice that this function is called the "Heaviside" function in Maple, after the
person who popularized it (among a lot of other accomplishments) and not because it's heavy on one side.
http://en.wikipedia.org/wiki/Oliver_Heaviside
> with plots :
plot Heaviside t , t =K3 ..3, color = green, title = `graph of unit step function` ;
graph of unit step function
1
0.6
0.2
K3
K2
K1
0
1
t
2
3
Notice that technically the vertical line should not be there - a more precise picture would have a solid point
at 0, 1 and a hollow circle at 0, 0 , for the graph of u t . In terms of Laplace transform integral
definition it doesn't actually matter what we define u 0 to be.
Then
u tKa =
0, t K a ! 0; i.e. t ! a
1, t K a R 0; i.e. t R a
and has graph that is a horizontal translation by a to the right, of the original graph, e.g. for a = 2:
> plot Heaviside t K 2 , t =K1 ..5, color = green, title = `graph of u(t-2)` ;
graph of u(t-2)
1
0.6
0.2
K1
0
1
2
t
3
4
5
Exercise 1) Verify the table entries
u tKa
unit step function
f tKa u tKa
eKa s
s
eKa sF s
for turning components on and
off at t = a .
more complicated on/off
Exercise 2) Consider the function f t which is zero for t O 4 and with the following graph. Use
linearity and the unit step function entry to compute the Laplace transform F s . This should remind you
of a homework problem from the assignment due tomorrow - although you're asked to find the Laplace
transform of that step function directly from the definition. In your next week's homework assignment
you will re-do that problem using unit step functions. (Of course, you could also check your answer in
this week's homework with this method.)
2
1.5
1
0.5
0
0
1
2
3
4
t
5
6
7
> plot Heaviside t K 1 C Heaviside t K 2 K 2$Heaviside t K 4 , t = 0 ..8, color = green ;
with inttrans :
laplace Heaviside t K 1 C Heaviside t K 2 K 2$Heaviside t K 4 , t, s ;
8
Exercise 3a) Explain why the description above leads to the differential equation initial value problem for
x t
x## t C x t = .2 cos t 1 K u t K 10 p
x 0 =0
x# 0 = 0
3b) Find x t . Show that after the parent stops pushing, the child is oscillating with an amplitude of
exactly p meters (in our linearized model).
Pictures for the swing:
> plot1 d plot .1$t$sin t , t = 0 ..10$Pi, color = black :
plot2 d plot Pi$sin t , t = 10$Pi ..20$Pi, color = black :
plot3 d plot Pi, t = 10$Pi ..20$Pi, color = black, linestyle = 2 :
plot4 d plot KPi, t = 10$Pi ..20$Pi, color = black, linestyle = 2 :
plot5 d plot .1$t, t = 0 ..10$Pi, color = black, linestyle = 2 :
plot6 d plot K.1$t, t = 0 ..10$Pi, color = black, linestyle = 2 :
display plot1, plot2, plot3, plot4, plot5, plot6 , title = `adventures at the swingset` ;
adventures at the swingset
3
2
1
0
K1
2p
4p
6p
8p
10 p
t
12 p
K2
K3
Alternate approach via Chapter 5:
step 1) solve
x## t C x t = .2 cos t
x 0 =0
x# 0 = 0
for 0 % t % 10 p .
step 2) Then solve
and set x t = y t K 10
for t O 10 .
y## t C y t = 0
y 0 = x 10 p
y# 0 = x# 10 p
14 p
16 p
18 p
20 p
f t , with
f t
% CeM t
c f t Cc f t
1 1
2 2
Fs d
N
f t eKs t dt for s O M
0
Y
verified
c F s Cc F s
A
1
s
1
A
1 1
2 2
1
t
(s O 0
A
s2
2
t2
A
s3
sn C 1
1
ea t
A
n!
tn, n 2 ;
(s O R a
sKa
s
cos k t
sin k t
cosh k t
sinh k t
ea tcos k t
ea tsin
kt
u tKa
f tKa u tKa
d tKa
f
n
f# t
f ## t
t , n2;
t
0
f t dt
A
(s O 0
A
(s O k
A
k
s2 C k2
s
s2 K k2
A
k
s2 K k2
(s O k
sKa
2 C k2
k
sKa
ea t f t
(s O 0
s2 C k2
sKa
A
2 C k2
(s O a
sOa
A
A
A
A
F sKa
eKa s
s
K
a
e sF s
eKa s
s F s Kf 0
s2F s K s f 0 K f # 0
sn F s K sn K 1f 0 K...Kf n K 1 0
A
A
A
A
Fs
s
tf t
t2 f t
tn f t , n 2 Z
f t
t
KF# s
F## s
K1 n F n s
t cos k t
s2 K k2
1
2 k t sin k t
1
2 k3
N
F s ds
s
s2 C k2
1
2
s2 C k2
2
1
t ea t
t
sKa
f t g t K t dt
with period p
Laplace transform table
1 K eKps
A
A
A
A
nC1
Fs Gs
1
A
2
sKa
n!
0
f t
2
s2 C k2
s
sin k t K k t cos k t
tn e a t, n 2 Z
A
A
A
p
f t eKs t dt
0