MATH 3150: PDE FOR ENGINEERS
MIDTERM TEST #1
1.
Calculate the integrals:
(a)
Z
eax cos x dx
where a is a constant.
(b)
Z
eax cos bx dx
where a, b are constants, with b 6= 0.
(c) Check your answer to this last question by differentiating the result.
Solution:
(a)
Z
eax cos x dx =
(b)
Z
eax cos bx dx =
eax
(a cos x + sin x)
a2 + 1
eax
(a cos bx + b sin bx)
a2 + b2
Date: October 30, 2000.
1
2
MATH 3150: PDE FOR ENGINEERS
MIDTERM TEST #1
1
0.9
0.8
0.7
–1 –0.8 –0.6 –0.4 –0.2
0
0.2
0.4
x
0.6
0.8
1
Figure 1. The periodic function equal to exp(−|x|) for −1/2 ≤
x ≤ 1/2
2.
(a) Use the result from the previous question to find the real Fourier amplitudes of the function
f (x) = ce−c|x|
(b)
(c)
(d)
(e)
− T /2 ≤ x ≤ T /2
with period T , where c is a positive constant.
Find the energy in each amplitude.
How much energy is in the function f (x)?
Taking the limit as c → ∞, what happens to the energy in am for m c?
Draw the function f (x) with c = 1 and T = 1. Include at least 2 full
periods in your graph.
Solution:
(a) We have bm = 0 because the function is even.
2
a0 =
1 − e−cT /2
T
4
1
am =
1 + (−1)m+1 e−cT /2
2
T 1 + 2πm
cT
(b)
2
4
1 − e−cT /2
T
!2
2
8
1
1 + (−1)m+1 e−cT /2
=
2
T 1 + 2πm
T a20 =
T 2
a
2 m
cT
(c)
2
Z
T /2
kf k =
f (x)2 dx = c 1 − e−cT
−T /2
(d) The energy of a0 goes to 4/T while the energy of am goes to 8/T .
(e) See figure 1.
MATH 3150: PDE FOR ENGINEERS
3.
MIDTERM TEST #1
3
Calculate the value of
e4001πi/2
Solution:
e4001πi/2 = e4000πi/2 eπi/2
1000
= e2πi
i
=i
4.
Suppose that f (x) has period 2π, and average value 0.
(a) Show that the energy in df /dx satisfies
2
df ≥ kf k2 .
dx (Hint: you will need to write out f (x) and df /dx as complex Fourier
series, and use Parseval’s theorem to write the energy of f (x) and of
df /dx as infinite sums of energy from each amplitude.) [This is Wirtinger’s
inequality; Wirtinger was your instructor’s thesis adviser’s thesis adviser’s
thesis adviser’s thesis adviser’s thesis adviser.]
(b) Which functions f (x) satisfy
2
df = kf k2 ?
dx Solution:
(a) Let ck be the complex amplitudes of f (x). The amplitudes of df /dx are
2
ikck so that the energy in this amplitude is 2πk 2 |ck | . The total energy
of df /dx is
2
∞
X
df 2
= 2π
k 2 |ck |
dx k=−∞
while f (x) has energy
2
kf k = 2π
∞
X
2
|ck |
k=−∞
Because f (x) has average value 0, we know c0 = 0. Therefore, in all of the
terms above, we have k 2 ≥ 1, giving the energy in each terms of df /dx a
factor k 2 larger than occurs in the sum for f (x).
(b) Only the c−1 and c1 terms of f (x) can be nonzero, because only they get
hit by factors of k 2 ≤ 1 in the infinite sum. Therefore
f (x) = c−1 e−ix + c1 eix
4
MATH 3150: PDE FOR ENGINEERS
5.
MIDTERM TEST #1
You may assume that
Z 1
2
(1 − x2 ) cos(αx) dx = 3 (sin α − α cos α)
α
0
Let f (x) = 1 − x2 for −1 ≤ x ≤ 1 and let f (x) be periodic with period 2.
(a) Draw the graph of f (x) over 2 periods.
(b) Calculate the real amplitudes for f (x).
(c) Calculate the energy of f (x).
(d) Find the energy of f (x) stored in the amplitudes am and bm in terms of
m.
(e) Show that more than 90% of the energy is contained in the terms
a0 , a1 , b1 , a2 , b2 . Use the approximation π 4 ∼ 97.
Solution:
(a) See the picture in figure 2 on the next page.
(b) The function is even, so bm = 0.
Z
1 T /2
a0 =
f (x) dx
T −T /2
Z 1
=
(1 − x2 ) dx
0
am
= 2/3
Z
2 T /2
2πmx
=
f (x) cos
dx
T −T /2
T
Z 1
=2
(1 − x2 ) cos (πmx) dx
0
(−1)m+1 4
=
π 2 m2
(c)
Z
2
1
f (x)2 dx
kf k = 2
0
Z
=2
1
(1 − x2 )2 dx
0
=
16
15
(d)
2
2
8
=
3
9
16
= 4 4
π m
T a20 = 2
T 2
a
2 m
(e)
T a20 + T2 a21 + a22
8/9 + 16/π 4 + 1/π 4
=
R T /2
16/15
f (x)2 dx
−T /2
15
=
· 8/9 + 17/π 4
16
15
4645
∼
· (8/9 + 17/97) =
> 90%
16
4656
MATH 3150: PDE FOR ENGINEERS
MIDTERM TEST #1
1
0.8
0.6
0.4
0.2
–2
–1
1
x
2
Figure 2. The periodic function equal to 1 − x2 for −1 ≤ x ≤ 1
5