MATH 267
Due: Feb 15, 2012, in the class
ASSIGNMENT # 5
You have FIVE problems to hand-in. Hand in written solutions for grading at the BEGINNING of the
lecture on the due date. Illegible, disorganized or partial solutions will receive no credit.
*Staple your HW. From now on, you will get F IV E marks OFF if you do not staple your HW!
Note that the instructor will NOT provide stapler.
1. (a) Let f (t) = |t| for −2 ≤ t ≤ 2. Let F (t) be a periodic extension of f with period 4.
i. Sketch the graph of F (t) (show at least three periods).
Solution We skip this.
ii. Write down the complex Fourier series of F (t).
Solution : The complex series for 2L-periodic function is
∞
X
F (t) =
π
ck eik L t
k=−∞
where
ck =
1
2L
Z
L
π
F (t)e−ik L t dt.
−L
Now, we see L = 2. So, (note F (t) = |t| for −2 < t < 2)
Z
π
1 2
ck =
|t|e−ik 2 t dt
4 −2
Z
Z 2
o
π
1n 0
t
−ik π
2
=
−te
dt +
te−ik 2 t dt
4
−2
0
Case: k = 0:
c0 =
1n
4
Z
0
Z
−tdt +
−2
2
o
tdt = 1
0
Case: k 6= 0:
Now, we use integration by parts to see that for constant α 6= 0,
Z B
Z B αt
teαt iB
e
αt
te dt =
dt
−
α A
α
A
A
eαt iB
teαt
− 2
=
α
α A
Thus, (notice that in the following computations it is OK to have k in the denominator since
k 6= 0),
Z
Z 2
o
π
1n 0
−ik π
t
2
dt +
te−ik 2 t dt
ck =
−te
4
0
−2
π
π
π
h te−ik π2 t
1 nh te−ik 2 t
e−ik 2 t i0
e−ik 2 t i2 o
=
−
+
+
−
4
−ik π2
(−ik π2 )2 −2
−ik π2
(−ik π2 )2 0
io
1 nh −2eikπ
1
eikπ i h 2e−ikπ
e−ikπ
1
=
π +
π 2 −
π 2 +
π −
π 2 +
π 2
4
−ik 2
(−ik 2 )
(−ik 2 )
−ik 2
(−ik 2 )
(−ik 2 )
nh
i
h
io
ikπ
−ikπ
1
1
e
e
1
=
+
π 2 −
π 2 + −
π 2 +
4
(−ik 2 )
(−ik 2 )
(−ik 2 )
(−ik π2 )2
n
o
2
=
1 − (−1)k (after cancelations using eikπ = e−ikπ = (−1)k and i2 = −1)
−k 2 π 2
(
− k24π2
k odd = ±1, ±3, ±5, · · · ,
=
0
k 6= 0 and even = ±2, ±4, ±6, · · · .
1
Therefore, the Fourier series is
X
F (t) = 1 +
−
odd k=±1,±3,±5,···
4
k2 π2
π
eik 2 t
iii. Write down the real (trigonometric) Fourier series of F (t).
Solution
Notice that in this case, the function F is even, so it should have the Fourier cosine series.
Indeed,
π
π π eik 2 t = cos k t + i sin k t .
2
2
So,
4 ik π t
e 2
k2 π2
odd k=±1,±3,±5,···
X
π 4 h
π i
=1+
− 2 2 cos k t + i sin k t
k π
2
2
odd k=±1,±3,±5,···
h
X
π 4
π π π i
=1+
− 2 2 cos k t + i sin k t + cos − k t + i sin − k t
k π
2
2
2
2
odd k=1,3,5,···
h
i
X
4
π =1+
− 2 2 2 cos k t
(here, used the even/ odd properties of cos, sin, resp.)
k π
2
odd k=1,3,5,···
X
π i
8 h
=1+
− 2 2 cos k t .
k π
2
1+
X
−
odd k=1,3,5,···
(b) Let g(t) = t for 0 < t < 2. Let G(t) be an odd periodic extension of g with period 4.
i. Sketch the graph of G(t) (show at least three periods). Solution We skip this.
ii. Write down the complex Fourier series of G(t).
Solution : The complex series for 2L-periodic function is
∞
X
G(t) =
π
ck eik L t
k=−∞
where
1
ck =
2L
L
Z
π
G(t)e−ik L t dt.
−L
Now, we see L = 2. So, (note G(t) = t for −2 < t < 2)
1
ck =
4
Z
2
−2
Case: k = 0:
c0 =
1
4
π
te−ik 2 t dt
Z
2
tdt = 0
−2
Case: k 6= 0:
Now, we use integration by parts to see that for constant α 6= 0,
Z
B
Z B αt
teαt iB
e
−
dt
α A
α
A
teαt
eαt iB
=
− 2
α
α A
teαt dt =
A
2
Thus, (notice that in the following computations it is OK to have k in the denominator since
k 6= 0),
Z
1 2 −ik π t
2 dt
te
ck =
4 −2
π
π
1 h te−ik 2 t
e−ik 2 t i2
=
−
4 −ik π2
(−ik π2 )2 −2
2eikπ
e−ikπ
eikπ o
1 n 2e−ikπ
+
−
+
=
4 −ik π2
(−ik π2 )2
−ik π2
(−ik π2 )2
=
(−1)k
(after cancelations using eikπ = e−ikπ = (−1)k and i2 = −1)
−ik π2
=
2(−1)k+1
ikπ
Therefore, the Fourier series is
∞
X
2(−1)k+1 ik π t
e 2
ikπ
G(t) =
k=−∞
iii. Write down the real (trigonometric) Fourier series of G(t).
Solution Notice that in this case, the function G is odd, so it should have the Fourier sine series.
Indeed,
π
π π eik 2 t = cos k t + i sin k t .
2
2
So,
∞
X
2(−1)k+1 ik π t
e 2
ikπ
G(t) =
k=−∞
∞
X
2(−1)k+1 h
π i
π =
cos k t + i sin k t
ikπ
2
2
=
=
=
k=−∞
∞
X
k=1
∞
X
k=1
∞
X
k=1
π π i 2(−1)−k+1 h
π π i
2(−1)k+1 h
cos k t + i sin k t +
cos − k t + i sin − k t
ikπ
2
2
−ikπ
2
2
2(−1)k+1 h
π i
2i sin k t (here, used the even/ odd properties of cos, sin, resp., and (−1)k = (−1)−k )
ikπ
2
π 4(−1)k+1 i
sin k t
ikπ
2
π
2. For L > 0, let ωk = k L
, for k = 0, ±1, ±2, ±3, ±4, · · · , and let ∆ω =
π
L.
(a) Show that the Fourier series of a 2L-periodic function f (t) can be written as
f (t) =
∞
1 X e
f (ωk )eiωk t ∆ω,
2π
k=−∞
where i =
√
−1 and
Z
L
f (t)e−iωk t dt.
fe(ωk ) =
−L
Solution : Note that the Fourier series of a 2L-periodic function f (t) is
f (t) =
∞
X
k=−∞
3
π
ck eik L t
where
First, using ωk =
π
kL
,
L
Z
1
ck =
2L
π
f (t)e−ik L t dt.
−L
we can write
∞
X
f (t) =
ck eiωk t
k=−∞
Now,
ck =
1
2L
Z
=
f (t)e−iωk t dt
−L
1 π
=
2π L
=
L
Z
1
∆ω
2π
Z
1 n
L
f (t)e−iωk t dt
−L
Z L
f (t)e−iωk t dt
(using the definition of ∆ω)
−L
L
o
f (t)e−iωk t dt ∆ω
2π
−L
1 e
=
f (ωk )∆ω
2π
(using the definition of fe(ωk ))
Back to the Fourier series, we get
f (t) =
∞
1 X e
f (ωk )eiωk t ∆ω
2π
k=−∞
as desired.
(b) Use the notation of (a). Assume L > 1. Let g(t) be the 2L-periodic function given by


1 if − 1/2 < t < 1/2,
g(t) = 0 if − L < t ≤ −1/2,


0 if 1/2 ≤ t < L.
Compute ge(ωk ) =
Solution
RL
−L
g(t)e−iωk t dt, for all integer k.
Z
ge(ωk ) =
L
g(t)e−iωk t dt
−L
Z 1/2
=
e−iωk t dt.
−1/2
Now, two cases:
Case: ωk = kπ/L = 0, i.e. k = 0:
Z
ge(ωk ) =
4
1/2
dt = 1.
−1/2
Case: ωk = kπ/L 6= 0, i.e. k 6= 0:
Z 1/2
ge(ωk ) =
e−iωk t dt
−1/2
=
=
=
=
=
h e−iωk t i1/2
−iωk −1/2
e−iωk /2
eiωk /2
−
−iωk
−iωk
1 iωk /2
e
− e−iωk /2
(rearranged)
iωk
1
ωk (notice that 2i sin θ = eiθ − e−iθ )
2i sin
iωk
2
2
ωk
sin( ).
ωk
2
Thus,
ge(ωk ) =
(
1
2
ωk
if ωk = 0
otherwise.
sin( ω2k )
(Notice that the last function is exactly sinc(ωk /2). )
(c) For the function g(t) in (b), do the following.
i. Let L = 1. Sketch the graph of ge(ωk ) for −6π < ωk < 6π. For the values ωk , use the ω-axis (i.e.
the horizontal axis with ω variable)
ii. Let L = 2. Sketch the graph of ge(ωk ) for −6π < ωk < 6π. Use the ω-axis for the values ωk .
iii. Let L = 4. Sketch the graph of ge(ωk ) for −6π < ωk < 6π. Use the ω-axis for the values ωk .
Solution We skip this part.
3. Sketch the graph of the following functions and calculate their Fourier transforms. (For a given function
R∞
f (t) its Fourier transform is given by fb(ω) = −∞ f (t)e−iωt dt.)
(
1 if 0 < t < 1,
(a) f (t) =
0 otherwise.
Solution : Sketch is skipped.
∞
Z
f (t)e−iωt dt
fb(ω) =
−∞
Z 1
e−iωt dt
=
0
Case : ω = 0
Z
fb(0) =
1
dt
0
=1
Case: ω 6= 0
Z
fb(ω) =
1
e−iωt dt
0
=
h e−iωt i1
−iω 0
1
=
e−iω − 1
−iω
5
Thus,
fb(ω) =
(
1
1
−iω
e
−iω
−1
if ω = 0
otherwise.
Remark: Notice that the function e−iω − 1 = e−iω/2 e−iω/2 − eiω/2 = −2ie−iω/2 sin(ω/2). Using
this, in fact, we can write fˆ(ω) = e−iω/2 sinc (ω/2). This observation is of course related to the
time-shifting property of the Fourier transform, since f (t) = rect (t − 1/2).


if 0 < t ≤ 1,
1
(b) g(t) = −2 if 1 < t ≤ 2,


0
otherwise.
Solution : Sketch is skipped.
Z
gb(ω) =
∞
g(t)e−iωt dt
−∞
Z 1
e−iωt dt − 2
=
2
Z
0
e−iωt dt
1
Case : ω = 0
Z
gb(0) =
1
Z
dt − 2
0
2
dt
1
= 1 − 2 = −1
Case: ω 6= 0
Z
gb(ω) =
1
e−iωt dt − 2
0
Z
2
e−iωt dt
1
h e−iωt i2
−2
=
−iω 0
−iω 1
1
1
−iω
e
−1 −2
e−i2ω − e−iω
=
−iω
−iω
h
i
1 −iω
= − (e
− 1) 1 − 2e−iω
iω
h e−iωt i1
Thus,
gb(ω) =
(
−1
1
− iω
(e−iω
i if ω = 0
h
−iω
otherwise.
− 1) 1 − 2e
Remark: Notice that the function e−iω − 1 = e−iω/2 e−iω/2 − eiω/2 = −2ie−iω/2 sin(ω/2). Using
this, in fact, we can write ĝ(ω) = (1 − 2e−iω )e−iω/2 sinc ω.
(
e−2t if 0 < t ≤ 1,
(c) h(t) =
0
otherwise.
Solution
6
∞
Z
h(t)e−iωt dt
b
h(ω) =
−∞
Z 1
e−2t e−iωt dt
=
0
1
Z
e−(2+iω)t dt
=
0
h e−(2+iω)t i1
=
−(2 + iω) 0
i
1 h
− e−(2+iω) + 1
=
2 + iω
4. For real variable t, consider a complex-valued function z(t). For example, z(t) = e−t+2i . We say “z(t) → 0
as t → ∞” if the modulus (in other words, the magnitude) |z(t)| → 0 as t → ∞. Also, recall that by
R∞
RT
R∞
RT
definition −∞ f (t)dt = limT →∞ −T f (t)dt. Similarly, 0 f (t)dt = limT →∞ 0 f (t)dt.
√
2
(a) Let z(t) = e−t+it . (Here i = −1.) Show that z(t) → 0 as t → ∞. (Hint: first, compute |z(t)|.)
2
2
2
Solution |z(t)| = |e−t+it | = |e−t ||eit | = e−t since |eit | = 1. Thus, |z(t)| = e−t → 0 as t → ∞.
This shows that z(t) → 0 as t → ∞.
(b) Let
(
e−at
f (t) =
0
if t > 0,
otherwise.
where the constant a > 0. Compute the Fourier transform fb(ω), i.e.
RT
f (t)e−iωt dt and let T → ∞.)
−T
Solution
Z ∞
Z ∞
Z ∞
−iωt
−at −iωt
b
f (ω) =
f (t)e
dt =
e e
dt =
e−(a+iω)t dt
−∞
0
Z
= lim
T →∞
R∞
−∞
f (t)e−iωt dt. (Compute
0
T
e−(a+iω)t dt
0
h e−(a+iω)t iT
T →∞ −(a + iω) 0
h e−(a+iω)T
i
1
= lim
−
T →∞ −(a + iω)
−(a + iω)
1
=
((since |e−(a+iω)T | = e−aT → 0, thus, e−(a+iω)T → 0 as T → ∞)
a + iω
= lim
fb(ω) =
1
a + iω
(c) Let g(t) = e−a|t| where the constant a > 0. Compute the Fourier transform gb(ω), i.e.
RT
(Compute −T g(t)e−iωt dt and let T → ∞.)
7
R∞
−∞
g(t)e−iωt dt.
Solution
Z
gb(ω) =
∞
T →∞
−∞
= lim
T →∞
= lim
nZ
T →∞
= lim
0
eat e−iωt dt +
−T
nZ 0
e(a−iω)t dt +
a − iω
g(t)e−iωt dt
−T
Z T
0
Z T
−T
o
e−at e−iωt dt
(|t| = −t for t < 0, and |t| = t for t > 0)
o
e−(a+iω)t dt
0
nh e(a−iω)t i0
T →∞
T
Z
g(t)e−iωt dt = lim
−T
h e−(a+iω)t iT o
+
−(a + iω) 0
i
1
e−(a−iω)T
e−(a+iω)T
1
−
−
+
T →∞ (a − iω)
(a − iω)
(a + iω)
(a + iω)
1
1
=
+
((since |e−(a±iω)T | = e−aT → 0, thus, e−(a±iω)T → 0 as T → ∞)
a − iω a + iω
= lim
h
Thus,
gb(ω) =
1
1
+
a − iω a + iω
5. Find the Fourier transform of the following functions:
(
sin(2t) if − 1 < t < 1,
(a) f (t) =
(Hint: express sin(2t) in terms of complex exponentials.)
0
otherwise.
Solution : Note that sin(2t) =
1
2it
2i (e
Z
− e−2it ). Thus,
∞
fb(ω) =
f (t)e−iωt dt
−∞
1
Z
1 2it
(e − e−2it )e−iωt dt
2i
−1
Z
Z 1
o
1 n 1 2it −iωt
=
e e
dt −
e−2it e−iωt dt
2i
−1
−1
Z 1
Z 1
n
o
1
=
ei(2−ω)t dt −
e−i(2+ω)t dt
2i
−1
−1
=
There are three cases (due to the possibility of 2 − ω = 0 or 2 + ω = 0, which may give zero
denominators).
Case : ω = 2
Z
Z 1
o
1n 1
fb(2) =
dt −
e−i4t dt
2i
−1
−1
n
o
−i4t 1
e
1
=
2−
−1
2i
−i4
1
1
=
2−
− e−i4 + ei4 ]
2i
i4
1
1
=
2 − sin 4
2i
2
1
sin 4
= (1 −
)
i
4
8
Case : ω = −2
nZ
fb(−2) = 2i
1
ei4t dt −
−1
=
=
=
=
Z
1
o
dt
−1
o
1 n ei4t 1
−
2
2i
i4 −1
1 1 i4
e − e−i4 ] − 2
2i i4
1 1
sin 4 − 2
2i 2
1 sin 4
(
− 1)
i 4
Case: ω 6= 2, −2
Z
Z 1
o
1 n 1 i(2−ω)t
fb(ω) =
e
dt −
e−i(2+ω)t dt
2i
−1
−1
e−i(2+ω)t i1 o
1 nh ei(2−ω)t i1
+
=
2i i(2 − ω) −1 i(2 + ω) −1
1 nh ei(2−ω)
e−i(2−ω) i h e−i(2+ω)
ei(2+ω) io
=
−
−
+
2i i(2 − ω)
i(2 − ω)
i(2 + ω)
i(2 + ω)
o
2
1n 2
sin(2 − ω) −
sin(2 + ω)
=
2i 2 − ω
2+ω
Thus,

1
sin 4


 i (1 − 4 )
1 sin 4
fb(ω) = i (n 4 − 1)


2
1
sin(2 − ω) −
2i
2−ω
if ω = 2
if ω = −2
2
2+ω
o
sin(2 + ω)
otherwise.
Remark: Note that in fact, using the definition of sinc ω we can rewrite
1
sinc (2 − ω) − sinc (2 + ω)
fb(ω) =
i
(b) g(t) = e−5|t| cos(2t). (Hint: express cos(2t) in terms of complex exponentials.)
9
Solution : Note that cos(2t) = 21 (e2it + e−2it ). Thus,
gb(ω)
Z ∞
g(t)e−iωt dt
=
−∞
Z ∞
1 2it
=
(e + e−2it )e−5|t| e−iωt dt
2
−∞
Z ∞
Z 0
1 2it
1 2it
−2it 5t −iωt
(e + e
)e e
dt +
(e + e−2it )e−5t e−iωt dt
=
2
2
0
−∞
Z ∞
Z 0
1 2it
1 2it
(e + e−2it )e(5−iω)t dt +
(e + e−2it )e−(5+iω)t dt
=
2
0
−∞ 2
Z
1 0
(e 5−i(ω−2) t + e 5−i(ω+2) t dt
=
2 −∞
Z
1 ∞ − 5+i(ω+2) t
+
(e
+ e− 5+i(ω−2) t dt
2 0
1 h e 5−i(ω−2) t
e 5−i(ω+2) t i0
=
+
2 (5 − i(ω − 2)) (5 − i(ω + 2)) −∞
e− 5+i(ω−2) t i∞
1 h e− 5+i(ω+2) t
+
+
2 −(5 + i(ω + 2)) −(5 + i(ω − 2)) 0
(|t| = −t for t < 0, and |t| = t for t > 0)
Therefore,
gb(ω) =
i
1h
1
1
+
2 (5 − i(ω − 2)) (5 − i(ω + 2))
i
1h
1
1
+
+
2 (5 + i(ω + 2)) (5 + i(ω − 2))
(Remark: Here in the last three steps, strictly speaking, we have to use the similar procedure of
taking limit T → ∞ as in Problem 4 (b) and (c). )
By making common denominators, we can simplify the expression as
gb(ω) =
5
5
+
2
25 + (ω + 2)
25 + (ω − 2)2
*Staple your HW. From now on, you will get F IV E marks OFF if you do not staple your HW!
Note that the instructor will NOT provide stapler.
10