MATHEMATICS 267 April 2007 Final Exam Solutions
1. (a) An elastic string of length 4 with fixed ends has an initial shape u(x, 0) = f (x), where
f (x) =
�
0
1
0
if 0 ≤ x < 1
if 1 ≤ x ≤ 3
if 3 < x ≤ 4
It is released from rest at time t = 0. Assume that the displacement u(x, t) satisfies
0 ≤ x ≤ 4, t > 0.
uxx = utt ,
Find u(x, t).
(b) Sketch u(x, 0) and u(x, 1).
Solution. (a) From the formula sheet with L = 4, c = 1 and g(x) = 0
u(x, t) =
∞
�
�
k=1
�
kπt
kπx
Ak cos( kπt
4 ) + Bk sin( 4 ) sin( 4 )
with Bk = 0 and
Ak =
2
L
�
L
0
f (x) sin( kπx
L )
dx =
So
u(x, t) =
1
2
∞
�
k=1
�
3
1
2
kπ
�
1 4
kπx �3
sin( kπx
4 ) dx = − 2 kπ cos( 4 ) 1 =
�
2
kπ
�
3kπ
cos( kπ
4 ) − cos( 4 )
�
�
3kπ
kπt
kπx
cos( kπ
4 ) − cos( 4 ) cos( 4 ) sin( 4 )
(b) We are told that u(x, 0) = f (x), which has the graph on the left below. We are starting at t = 0
with a single bump of height 1 and width 2 centred at x = 2. As time increases from 0, this single bump
will divide itself into two half sized bumps, each of width 2, with one moving to the right with speed 1
and one moving to the left with speed 1. At t = 1 the right moving bump will run from 2 to 4 and the
left moving bump will run from 0 to 2. So, the graph of u(x, 1) is given on the right below.
2. Let g(x) = −x be defined for 0 ≤ x ≤ 1.
(a) Extend g(x) as a periodic function of period 1. Find the Fourier series for g(x) in complex form.
(b) Extend g(x) as an odd function of period 2. Find the Fourier series for g(x) in terms of sines and
cosines.
(c) One endto(xedit.
= 0) of a copper bar (α2 = 1) of length 1 is maintained at 0◦ C while the other is at
Double-click
◦
10 C. Initially the entire bar is at 0◦ C. Find the temperature u(x, t) in the bar if u(x, t) satisfies
0 ≤ x ≤ 1, t > 0
ut = uxx
u(0, t) = 0
t>0
u(1, t) = 10
t>0
Solution. (a) The given function has period 1 so that � =
g(x) =
∞
�
cn ei2πnx
with
n=−∞
1
1
2
and
cn =
�
1
0
(−x)e−i2πnx dx
Note that we are free to choose any full period for the domain of integration in the formula for cn . For
�1
n = 0, c0 = 0 (−x) dx = − 12 . For n �= 0,
cn = −
so
�
−i2πnx
x
−i2πn e
−i2πnx
1
(−i2πn)2 e
−
g(x) = − 12 −
�1
0
i
= − 2πn
∞
�
�
�
since e−i2πnx �
x=0,1
=1
i2πnx
i
2πn e
n=−∞
n�=0
(b) This function has period 2 and is also odd. So it has a Fourier sin series expansion with L = 1. The
Fourier coefficient
bk =
2
L
�
L
g(x) sin( kπx
L
0
�
= 2 x cos(kπx)
−
kπ
�
dx = 2
sin(kπx)
(kπ)2
�1
0
�
1
0
g(x) sin(kπx) dx = −2
= 2 cos(kπ)
= 2 (−1)
kπ
kπ
k
�
1
x sin(kπx) dx
0
The Fourier series expansion is
g(x) =
∞
�
bk sin(kπx) =
k=1
∞
�
2
(−1)k kπ
sin(kπx)
k=1
Double-click
to steady
edit. state temperature S(x) should increase linearly from 0 at x = 0 to 10 at x = 1. So
(c) The
S(x) = 10x = −10g(x)
Let the temperature u(x, t) = S(x) + w(x, t). Then S(x) obeys
0 = St = Sxx ,
S(0) = 0,
S(x) = −10g(x),
S(1) = 10,
0 ≤ x ≤ 1, t ≥ 0
Consequently u(x, t) obeys
ut = uxx ,
u(0, t) = 0,
u(1, t) = 10,
0 ≤ x ≤ 1, t ≥ 0
u(x, 0) = 0,
if and only if w(x, t) obeys
wt = wxx ,
w(0, t) = 0,
w(x, 0) = 0 − S(x) = 10g(x)
w(1, t) = 0,
From the formula sheet, with α = 1 and L = 1, the solution to the problem (∗) is
w(x, t) =
∞
�
Bk e−k
2
π2 t
sin(kπx)
k=1
with
Bk =
2
L
�
L
10g(x) sin(kπx) dx = 20
0
�
1
0
20
g(x) sin(kπx) dx = (−1)k kπ
from part (b). So the final answer is
u(x, t) = S(x) + w(x, t) = 10x +
∞
�
k=1
2
20 −k
(−1)k kπ
e
2
π2 t
sin(kπx)
(∗)
3. (a) By direct integration, find the Fourier Transform of the function
a(t) = rect
(b) Find the Fourier Transform of the function
b(t) =
�
�t�
cos(t)
0
2
cos(πt).
if π < t < 3π
elsewhere
(c) Find the Inverse Fourier Transform of
�
c(ω) =
Solution. (a) Since
a(t) =
we have
�
a(ω) =
�
∞
rect
−∞
� 1
�t�
2
�
4
2 + 3iω − ω 2
cos(πt)
0
cos(πt)e−iωt dt
if −1 < t < 1
elsewhere
=
�
1
−1
�
1
2
�
�
eiπt + e−iπt e−iωt dt
ei(π−ω)t
e−i(π+ω)t
=
e
+e
dt
=
+
i(π − ω) −i(π + ω)
−1
�
�
1 2i sin(π − ω) 1 −2i sin(π + ω)
1
1
=
+
=
−
sin(ω)
2 i(π − ω)
2 −i(π + ω)
π−ω π+ω
2ω
= 2
sin(ω)
π − ω2
(b) Set
1
2
�
i(π−ω)t
d(t) = a
�
Then d(ω)
= π�
a(πω) and
�
−i(π+ω)t
�t�
π
=
�
cos(t)
0
1
2
(c)
so, by table lookup
�
c(ω) =
−1
if −1 < πt < 1 or −π < t < π
elsewhere
�
b(t) = d(t − 2π) =⇒ �b(t) = e−2πωi d(ω)
= e−2πωi π�
a(πω) = e−2πωi π
= e−2πωi
�1
2ω
sin(πω)
1 − ω2
2(πω)
sin(πω)
π 2 − (πω)2
4
4
4
4
=
=
−
2
2 + 3iω − ω
(2 + iω)(1 + iω)
1 + iω 2 + iω
�
�
c(t) = 4 e−t − e−2t u(t)
3
Double-click to edit.
4. In this problem you will analyze this circuit:
The input signal is a time-varying voltage x(t) and the output signal is the voltage y(t) measured across
the inductor. Low-frequency signals face little opposition to flow through the inductor, so they get
dissipated mostly by the resistor. High-frequency signals flow easily through the capacitor, so they
also get dissipated by the resistor. But signals of some intermediate frequency are opposed by both
reactive components, and produce large-amplitude outputs. The signals described above are related by
the constant coefficient differential equation
RLCy �� (t) + Ly � (t) + Ry(t) = Lx� (t).
(a) Let x
�(ω) and y�(ω) be the Fourier transforms of x(t) and y(t). Define
H(ω) =
y�(ω)
,
x
�(ω)
H(ω) = A(ω)eiφ(ω) .
A(ω) = |H(ω)|,
Find simple algebraic expressions for H(ω), A(ω) and tan(φ(ω)).
(b) Use calculus to find the value of ω > 0 at which A(ω) is maximized. This is the circuit’s resonant
frequency. Express your answer in terms of L, R, and C. [Hint: Maximize |A(ω)|2 . ]
Solution. (a) The Fourier Tranform of f � (t) is iω f�(ω). So taking Fourier transforms of the specified
ODE gives
RLC(iω)2 y�(ω) + L(iω)�
y (ω) + R�
y (ω) = L(iω)�
x(ω) =⇒ H(ω) =
Consequently
A(ω) = |H(ω)| =
|R −
|iωL|
ω 2 RLC
− iωL|
=�
y�(ω)
iωL
=
x
�(ω)
R − ω 2 RLC + iωL
L|ω|
2
R2 (1 − ω 2 LC) + ω 2 L2
Multiplying the top and bottom of H(ω) by the complex conjugate of the denominator gives
H(ω) =
iωL
R − ω 2 RLC − iωL
ω 2 L2 + iωRL(1 − ω 2 LC)
=
2
R − ω 2 RLC + iωL R − ω 2 RLC − iωL
R2 (1 − ω 2 LC) + ω 2 L2
and
tan φ(ω) =
R(1 − ω 2 LC)
ωL
(b) Notice that
A(ω)2 =
L2 ω 2
2
R2 (1 − ω 2 LC) + ω 2 L2
Let x = ω 2 and maximize over x > 0 the function
f (x) =
R2 (1
x
− LCx)2 + L2 x
The derivative
� 2
�
�
�
R (1−LCx)2 + L2 x − x − 2LCR2 (1 − LCx) + L2
R2 (1 − LCx)2 + 2LCxR2 (1 − LCx)
�
f (x) =
=
�
�2
�
�2
R2 (1 − LCx)2 + L2 x
R2 (1 − LCx)2 + L2 x
R2 (1 − LCx)(1 + LCx)
=�
�2
R2 (1 − LCx)2 + L2 x
4
Double-click to edit.
def
1
This has a critical point when x = x∗ = LC
. Since f � (x) > 0 for 0 < x < x∗ and f � (x) < 0 for x > x∗ ,
the point x∗ gives a global maximum for the function f . It follows that the maximizing frequency for
1
A(ω) obeys ω 2 = LC
. That is,
1
ω=√
LC
5. Consider the discrete time signal
x[n] = sin πn
2 cos(πn)
Is x[n] periodic? If so, find a period N .
Double-click to(a)
edit.
(b) Is the discrete Fourier transform x
�[k] of this signal periodic? If so, find a period for x
�[k].
(c) Find the discrete Fourier transform x
�[k] of this signal.
Solution. (a) Both sin t and cos t are periodic of period 2π. Consequently sin πn
2 is periodic of period
4 and cos(πn) is periodic of period 2. So x[n] obeys x[n + N ] = x[n] for all n if N = 4.
(b) x[n] and x
�[k] have the same fundamental period. So the fundamental period of x
�[k] is also N = 4.
(c) Comparing
� πn i
�
πn ��
1
2
− e− 2 i eπni + e−πni =
x[n] = sin πn
2 cos(πn) = 4i e
�
πn
3πn �
1
= 2i
−e 2 i+e 2 i
and
x[n] =
N
−1
�
k=0
gives
x
�[k]e
x
�[0] = x
�[2] = 0
6. Consider an LTI system for which
2πkn
N i
=
1
2i
3
�
k=0
�
e
πn
2 i
x
�[k]e
1
x
�[1] = − 2i
− e−
πn
2 i
�
eπni
(since e2nπi = 1)
πkn
2 i
x
�[3] =
1
2i
y[n] + 14 y[n − 1] − 18 y[n − 2] = x[n]
(a) Use X(z) and Y (z) to denote the z–transforms of x[n] and y[n], respectively. Express the z-transform
of y[n] + 41 y[n − 1] − 18 y[n − 2] in terms of Y (z).
Y (z)
(b) Find the system function H(z) = X(z)
for this system.
(c) Plot the poles and zeroes of H(z) and indicate the region of convergence, assuming that the system
is causal.
(d) Using z–transforms, determine y[n] if
x[n] =
� 1 �n
2
u[n]
Solution. (a) Write w[n] = y[n] + 14 y[n − 1] − 18 y[n − 2]. By definition, the z–transform of w[n] is
W (z) =
�
y[n]z −n +
n
=
�
n
1
4
�
n
y[n]z
−n
+
1 −1
4z
y[n − 1]z −n −
�
n
y[n − 1]z
= Y (z) + 14 z −1 Y (z) − 18 z −2 Y (z)
5
1
8
�
n
−(n−1)
y[n − 2]z −n
− 18 z −2
�
n
y[n − 2]z −(n−2)
(b) By part (a)
Y (z) + 14 z −1 Y (z) − 18 z −2 Y (z) = X(z)
=⇒ H(z) =
1
z2
Y (z)
1
�
��
�=�
��
�
=
=
1
1
1
1
1
X(z)
1 + 4 z −1 − 8 z −2
1 + 2 z −1 1 − 4 z −1
z + 2 z − 14
(c) The poles are at z = 14 , − 12 and the zeroes are at z = 0. The ROC is |z| > 12 , since we are told that
the system is causal.
(d) Since X(z) =
with
1
1 − 12 z −1
Y (z) = �
1
a
b
c
��
��
�=
+
+
1 + 12 z −1
1 − 14 z −1
1 − 12 z −1
1 + 12 z −1 1 − 14 z −1 1 − 12 z −1
�
�
1
1
�
�
a= �
= 3
=
1 −1
1 −1 ��
3
( 2 )(2)
1 − 4z
1 − 2z
1 −1
=−1
2z
�
�
1
1
1
�
��
�
b= �
=
=−
1 −1
1 −1 ��
(3)(−1)
3
1 + 2z
1 − 2z
1 −1
z =1
�4
�
1
1
�
��
�
c= �
=
=1
1 −1
1 −1 ��
(2)( 12 )
1 + 2z
1 − 4z
1 −1
1
��
2z
Hence
Y (z) =
1+
1
3
1 −1
2z
−
1−
1
3
1 −1
4z
+
1
1 − 12 z −1
6
=1
=⇒ y[n] =
� �
1
3
−
�
1 n
2
−
� �
1 1 n
3 4
+
� 1 �n �
2
u[n]