Name________________________
Student I.D.___________________
Math 2280-001
Quiz 12 SOLUTIONS
April 24, 2015
You may choose to do #1 (Laplace transform), or #2 (Fourier series). Indicate clearly which problem
you'd like graded.
1) Find a formula for solutions to the forced oscillation problem below, as convolution integrals involving
the forcing function f and the "weight function". Hint: when you take the Laplace transform of the
differential equation and solve for X s if will be of the form X s = F s G s .
x## t C 2 x# t C 10 x = f t
x 0 =0
x# 0 = 0
(10 points)
A solution to the IVP makes both sides of the DE equal, so the Laplace transforms will be too:
s2 X s K 0 K 0 C 2 sX s K 0 C 10 X s = F s
X s s2 C 2 s C 10 = F s
1
X s =F s 2
=F s G s .
s C 2 s C 10
Since
1
G s =
,
sC1 2C9
1
g t = eKt sin 3 t
3
and by the convolution table entry
t
t
f t g t K t dt =
x t = f)g t =
0
0
t
( = g)f t =
0
f t eK t K t
1
sin 3 t K t
3
1
eKt sin 3 t f t K t dt )
3
dt
2) Find the Fourier series for the square wave of period P = 2 L = 4 and amplitude 3. In other words, for
the period 4 extension of
K3 K2 ! t ! 0
f t =
.
3 0!t!2
You may do this directly from the definition
N
N
a0
p
p
fw
C
an cos n
t C
bn sin n t
2
L
L
n=1
n=1
with
>
>
L
1
a0 =
L
an d
f, cos n
bn d
f, sin n
p
t
L
=
p
t
L
=
g t dt
KL
L
1
L
f t cos n
p
t dt, n 2 ;
L
f t sin n
p
t dt, n 2 ;
L
KL
L
1
L
KL
or by rescaling the 2 pKperiodic unit square wave function satisfying
square t =
K1 Kp ! t ! 0
0!t!p
1
which has Fourier series
square t =
4
p
1
sin n t
n
>
n odd
(10 points)
rescaling solution: We claim
p
t .
2
f t = 3 square
Check:
3 square
=
p
t =
2
3
0!
K3 K2 ! t ! 0
p
t!p
2
=f t .
3 0!t!p
Thus, using the formula for square t above,
f t = 3 square
p
t!0
2
K3 Kp !
p
12
1
np
t =
sin
t .
2
2
p n odd n
>
via computing Fourier coefficients: Since f t is odd its cosine and constant coefficients are zero
(because they are obtained by integrating an even*odd=odd function over a symmetric interval). So we
only need to compute the bn :
bn d
p
f, sin n t
L
1
=
L
L
f t sin n
KL
p
t dt .
L
Because an odd function times and odd function is even, we can compute these integrals by taking twice
the integral
over the left half of the interval:
1
bn = 2
2
2
3 sin n
0
6
p
bn =K
cos n t
np
2
Since cos np = K1
n
p
t dt
2
2
6
=K
cos np K 1
np
0
,
bn =
12
np
n odd
0
n even
so
f t =
12
1
np
sin
t .
2
p n odd n
>