MATH 267
Due: March 28, 2012, in the class
ASSIGNMENT # 9
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. You will get F IV E marks OFF if you do not staple your HW! Note that
the instructor will NOT provide stapler.
Note:
In the following, we do not distinguish between “length= N discrete-time signals” and “N -periodic discretetime signals”.
1. [Discrte Fourier transform for periodic signals]
(a) Find the discrete Fourier transform of the following periodic signals with period N .
i. x[n] = cos(2πn). N = 4
ii. y[n] = cos(πn/3) + sin(πn/2). N = 12
(Hint: Express sin and cos using complex exponentials and try to use ‘orthogonality’ to compute the
summation. )
(b) Suppose x[n] is a periodic discrete-time signal with period = N . Let x
b[k] be its discrete Fourier
PN −1
2
transform. Assume x[0] = N and k=0 |b
x[k]| = N . Find x[n] and x
b[k] for all 0 ≤ n, k ≤ N − 1.
(Hint: Use Parseval’s relation. This is similar to one of the class examples.)
(c) Let a[n] be a periodic signal with period N = 16 with


1 0 ≤ n ≤ 8,
a[n] = 0 9 ≤ n ≤ 12,


1 13 ≤ n ≤ 15.
Compute the discrete Fourier transform b
a[k]. (This is similar to one of the class examples.)
Solution
(a):
(i) Note cos(2πn) = 1. Thus, x = [1, 1, 1, 1]. And
x
b[k] =
3
1X
x[n]e−i2πkn/4
4 n=0
3
1 X −i2πkn/4
e
4 n=0
(
1 for k = 0 ,
=
0 for k = 1, 2, 3.
=
Thus, x
b = [1, 0, 0, 0].
1
(ii) cos(πn/3) + sin(πn/2) = 12 (eiπn/3 + e−iπn/3 ) +
yb[k] =
1
iπn/2
2i (e
− e−iπn/2 ). Thus,
11
i
1 X h 1 iπn/3
1
(e
+ e−iπn/3 ) + (eiπn/2 − e−iπn/2 ) e−i2πkn/12
12 n=0 2
2i
=
11
11
1 X iπn/3
1 X iπn/2
(e
(e
+ e−iπn/3 )e−i2πkn/12 +
− e−iπn/2 )e−i2πkn/12
24 n=0
24i n=0
=
11
11
1 X iπn/3 −i2πkn/12
1 X −iπn/3 −i2πkn/12
e
e
+
e
e
24 n=0
24 n=0
+
=
11
11
1 X −iπn/2 −i2πkn/12
1 X iπn/2 −i2πkn/12
e
e
−
e
e
24i n=0
24i n=0
11
11
1 X i2π2n/12 −i2πkn/12
1 X −i2π2n/12 −i2πkn/12
e
e
+
e
e
24 n=0
24 n=0
+
11
11
1 X i2π3n/12 −i2πkn/12
1 X −i2π3n/12 −i2πkn/12
e
e
−
e
e
24i n=0
24i n=0
11
11
1 X i2π(2−k)n/12
1 X i2π(−2−k)n/12
e
+
e
=
24 n=0
24 n=0
+
11
11
1 X i2π(−3−k)n/12
1 X i2π(3−k)n/12
e
−
e
24i n=0
24i n=0

12


24 + 0 + 0 − 0


12


0
 + 24 + 0 − 0
12
= 0 + 0 + 24i
−0


12


0 + 0 + 0 − 24i

0
k=2,
k = 10 ,
k = 3,
k = 9,
k = 0, · · · , 11 but k 6= 2, 3, 9, 10.
(using orthogonality)
Therefore,
yb[k] =

1


2


1


2
1
2i


1

−

 2i

0
k=2,
k = 10,
k = 3,
k = 9,
k = 0, · · · , 11 but k 6= 2, 3, 9, 10.
In other words,
1 1
1 1
yb = [0, 0, , , 0, 0, 0, 0, 0, − , , 0]
2 2i
2i 2
(b): Recall the Parseval’s relation:
N −1
N
−1
X
1 X
|x[n]|2 =
|b
x[k]|2
N n=0
k=0
Notice that all the entries in the
summation are all nonnegative. By the given condition, the right
PN
−1
2
2
hand side is N . Thus, we see
n=0 |x[n]| = N . Now, since x[0] = N , the other entries in the
last sum should all vanish, i.e. x[1] = x[2] = · · · = x[N − 1] = 0. Thus, x = [1, 0, 0, · · · , 0]. Now,
PN −1
x
b[k] = N1 n=0 x[n]e−2πikn/N = 1 + 0 + 0 + · · · + 0 = 1. Thus, x
b = [1, 1, 1, · · · , 1].
(c) By time-shift we see that a[n] = b[n + 3] where b[n] is the 16-periodic signal with
(
1 0 ≤ n ≤ 11,
b[n] =
0 12 ≤ n ≤ 15,
2
2π
Therefore, b
a[k] = ei 16 ×3kbb[k] and
15
X
bb[k] = 1
b[n]e−i2πkn/16
16 n=0
=
11
1 X −i2πkn/16
e
16 n=0
=
11
1 X −i2πk/16 n
[e
]
16 n=0
=
1
1 − e−i 16 ×12
×
16
1 − e−i2πk/16
2πk
Therefore,
2πk
1 − e−i 16 ×12
1
b
a[k] = e
×
16
1 − e−i2πk/16
3πk
1 − e−i 2
×3k 1
i 3π
8
=e
×
16 1 − e−iπk/8
i 2π
16 ×3k
2. [NOT TO HAND IN] (This problem is similar to Example 1 in the online notes “Discrete-Time Fourier
Series and Transforms ”.)
Consider the “discrete square wave” function x[n] with one period given by
1, −N1 ≤ n ≤ N1
x[n] =
0, otherwise
for some positive integer N1 < N , where N is the fundamental period of x[n].
(a) Show that
(
x
b[k] =
2N1 +1
,
N
1 sin[2πk(N1 +1/2)/N ]
,
N
sin(πkN )
k = 0, ±N, ±2N, . . .
otherwise
(b) Use part (a) and the time shift property to compute the discrete Fourier series coefficients of the
function y[n] with one period given by:

0≤n≤8
 1,
0, 9 ≤ n ≤ 12
y[n] =

1, 13 ≤ n ≤ 15
3. [Periodic convolution]
Consider the folloing signals with period N = 4:
a = [1, 0, 1, −1],
b = [2, i, 1 + i, 3]
(e.g. a[0] = 1, a[3] = −1, b[2] = 1 + i, etc. )
(a) Calculate the periodic convolution a ∗ b by directly calculating the convolution sum.
(b) Calculate the Fourier coefficients b
a[k] and bb[k]. Use this to compute the Fourier coefficients ad
∗ b[k]
for a ∗ b by using the convolution property of the Fourier transform.
(c) Find a signal x[n] of period N = 4, such that (a ∗ x)[n] = b[n].
(Hint: you may want to use the convolution property of the Fourier transform/inversion. Remember
how we handle the circuit problem. This is similar.)
3
Solution
(a) :
(a ∗ b)[n] =
3
X
a[m]b[n − m]
m=0
Thus, (noting b[−1] = b[4 − 1] = b[3]; b[−2] = b[4 − 2] = b[2]; b[−3] = b[4 − 3] = b[1]),
(a ∗ b)[0] =
3
X
a[m]b[0 − m] = 1 × 2 + 0 × 3 + 1 × (1 + i) + (−1) × i = 3
m=0
(a ∗ b)[1] =
3
X
a[m]b[1 − m] = 1 × i + 0 × 2 + 1 × 3 + (−1) × (1 + i) = 2
m=0
(a ∗ b)[2] =
3
X
a[m]b[2 − m] = 1 × (1 + i) + 0 × i + 1 × 2 + (−1) × 3 = i
m=0
(a ∗ b)[3] =
3
X
a[m]b[3 − m] = 1 × 3 + 0 × (1 + i) + 1 × i + (−1) × 2 = 1 + i
m=0
So, a ∗ b = [3, 2, i, 1 + i].
(b): Note that e−i2π/4 = eiπ/2 = −i. Thus, e−i2π/4kn = (−i)kn . So,
3
1
1X
a[n]e−i2π/4kn = (1 × 1 + 0 × (−i)k + 1 × (−i)2k + (−1) × (−i)3k )
4 n=0
4
b
a[k] =
1
(1 + (−1)k − ik )
4
=
Thus, b
a = [1/4, −i/4, 3/4, i/4].
On the other hand,
3
X
1
bb[k] = 1
b[n]e−i2π/4kn = (2 × 1 + i × (−i)k + (1 + i) × (−i)2k + 3 × (−i)3k )
4 n=0
4
=
1
(2 + (−1)k ik+1 + (1 + i)(−1)k + 3ik )
4
Thus, bb = [(6 + 2i)/4, (2 + 2i)/4, 0, −i].
Use convolution property for N = 4, to see
ad
∗ b[k] = 4b
a[k]bb[k]
for k = 0, · · · , 3. So,
ad
∗ b = [(6 + 2i)/4, (−2i + 2)/4, 0, 1]
(c): Take Fourier transform:
a[
∗ x[k] = bb[k]
By convolution property (for N = 4),
a[
∗ x[k] = 4b
a[k]b
x[k]
4
Thus, we see
4b
a[k]b
x[k] = bb[k]
So,
x
b[1] = (2i − 2)/4,
x
b[0] = (6 + 2i)/4,
x
b[2] = 0,
x
b[3] = −1
i.e.
x
b = [(3 + i)/2, (−1 + i)/2, 0, −1].
To find x[n], apply the inverse discrete Fourier transform:
x[n] =
3
X
x
b[k]ei2πkn/4
k=0
−1 + i k
3+i
+(
)i + 0 + (−1)i3k
2
2
3 + i −ik + ik+1
=
+
+ 0 + (−1)k+1 ik
2
2
=
Thus,
x = [i, 1 + i, 3, 2]
4. [Convolution of non-periodic signals] (This problem is related to the topic and examples we covered in the
class on Wed. March 21.)
Recall for integers n ∈ Z,
(
1 if n ≥ 0 ,
u[n] =
0 otherwise.
(
δ[n] =
1 if n = 0,
0 otherwise.
(
δn0 [n] =
1 if n = n0 ,
0 otherwise.
Recall the class example (u ∗ u)[n] = (n + 1)u[n].
100 times
z
}|
{
(a) Find (δ2 ∗ δ2 ∗ · · · ∗ δ2 )[n].
(b) Let f [n] = u[n − 2]. g[n] = u[n + 3].
i. Find (f ∗ u)[n].
ii. Find (f ∗ g)[n].
(c) Let
(
1
h[n] =
0
|n| ≤ 3 ,
otherwise.
Find (h ∗ u)[n]
i. first, by computing the convolution sum directly;
ii. second, by using the algebraic properties of the convolution and using (u ∗ u)[n] = (n + 1)u[n].
Solution
(a): Recall δa [n] = δ[n − a] for a ∈ Z. δa ∗ δb = δa+b . By associativity, δa ∗ δb ∗ δc = δa+b+c , and so on.
100 times
z
}|
{
Thus, (δ2 ∗ δ2 ∗ · · · ∗ δ2 )[n] = δ2×100 [n] = δ200 [n].
5
(b); Note that we can write f [n] = (δ2 ∗ u)[n] and g[n] = (δ−3 ∗ u)[n]. Thus,
(f ∗ u)[n] = (δ2 ∗ u ∗ u)[n] = (u ∗ u)[n − 2] = (n − 2 + 1)u[n − 2] = (n − 1)u[n − 2]
(f ∗ g)[n] = (δ2 ∗ u ∗ δ−3 ∗ u)[n] = (δ2 ∗ δ−3 ∗ u ∗ u)[n] = (δ−1 ∗ u ∗ u)[n] = (n + 1 + 1)u[n + 1] = (n + 2)u[n + 1].
(c):
(i):
(h ∗ u)[n] =
=
∞
X
h[m]u[n − m] =
m=−∞
n
X
n
X
h[m]u[n − m]
(require n − m ≥ 0)
m=−∞
h[m]
m=−∞


for n < −3,
0
= n + 4 for −3 ≤ n ≤ 3,


7
for n > 3.
(ii) Note that h[n] = u[n + 3] − u[n − 4] = (δ−3 ∗ u)[n] − (δ4 ∗ u)[n]. Therefore,
(h ∗ u)[n] = (δ−3 ∗ u ∗ u)[n] − (δ4 ∗ u ∗ u)[n] = (n + 3 + 1)u[n + 3] − (n − 4 + 1)u[n − 4]
= (n + 4)u[n + 3] − (n − 3)u[n − 4]
Notice that


0
(n + 4)u[n + 3] − (n − 3)u[n − 4] = n + 4


(n + 4) − (n − 3) = 7
for n < −3,
for −3 ≤ n ≤ 3,
for n > 3.
Thus, the answers in (i) and (ii) coincide.
5. [Discrete-time Fourier transform for non-periodic signals] (We will cover this topic on Monday, March 26.)
(a) x[n] = δ2 [n] + δ−2 [n]
n
(b) y[n] = 51 u[n − 1]
|n+1|
(c) z[n] = 15
Solution
(a)
x
b(ω) = δb2 (ω) + δb−2 (ω) = e−2iω + e+2iω = 2 cos(2ω)
Here in the first equality, we used the linearity and in the second equality we used the time-shift property.
(Of course, in this simple case, we can just apply the definition of Fourier transform and the Delta
function.)
n
n−1
(b): y[n] = 51 u[n − 1] = 15 15
u[n − 1]. Thus, using time-shift (using F to denote the discrete-time
Fourier transform),
1
1 n−1
1
1 n
F[
u[n − 1]](ω) = e−iω F[
u[n]](ω)
5
5
5
5
1
1
= e−iω
1 −iω
5
1 − 5e
yb(ω) =
(c) Notice that since 1 = u[n] + u[−n − 1], one can write any x[n] as x[n] = x[n]u[n] + x[n]u[−n − 1]
6
z[n] =
1 |n+1|
5
=
1 |n+1|
u[n]
5
+
1 |n+1|
u[−n
5
z[n] =
− 1] which then can be written as
1 n+1
1 −(n+1)
u[n] +
u[−(n + 1)]
5
5
Thus,
zb(ω) =
i
h 1
i
1 h 1 n
−(n+1)
F
u[n] (ω) + F
u[−(n + 1)] (ω)
5
5
5
Note
i
h 1
1
n
u[n] (ω) =
5
1 − 51 e−iω
h 1
i
h 1
i
−(n+1)
−n
F
u[−(n + 1)] (ω) = e+iω F
u[−n] (ω)
5
5
i
h 1
n
u[n] (−ω)
= e+iω F
5
1
= eiω
1 − 15 eiω
F
(time-shift)
(time-reversal)
Therefore,
zb(ω) =
1
1
1
×
+ eiω
5 1 − 15 e−iω
1 − 15 eiω
Remark: The time-reversal property is in the online notes page 12 in the table, and it can also be proved
very easily:
F[x[−n]](ω) =
∞
X
∞
X
x[−n]e−inω =
n=−∞
x[m]e−im(−ω)
change of index m = −n
m=−∞
= F[x[n]](−ω)
6. [Inverse discrete-time Fourier transform for non-periodic signals] (We will cover this topic on Monday,
March 26.)
Recall the discrete-time Fourier transforms of δn0 [n] and an u[n] (for |a| < 1) are e−iωn0 and
respectively.
1
1−ae−iω ,
Use these to find discrete-time signals x[n], y[n], z[n], whose Fourier transforms are given below. (Here,
each answer should be a signal defined on the set of integers: n ∈ Z.)
(a) x
b(ω) = cos2 ω + cos ω sin ω. (Hint: Can we express this as combination of complex exponentials?)
(b)
yb(ω) = 1 +
ei2ω
1 + 31 e−iω
(Hint; you may want to use time-shift property: see the table in page 12 in the online note ”
Discrete-time Fourier series and Fourier Transforms”.)
(c)
zb(ω) =
1
.
(1 + 12 e−iω )(1 + 13 e−iω )
(Hint: use partial fractions.)
Solution
7
(a): Note that
eiω + e−iω eiω − e−iω
2
2
2i
1 2iω
1
= (e + 2 + e−2iω ) + (e2iω − e−2iω )
4
4i
cos2 ω + cos ω sin ω =
eiω + e−iω 2
+
Thus, denoting the inverse discrete-time Fourier transform by F −1 , we see
1
1 −1 2iω
(F [e ][n] + F −1 [2][n] + F −1 [e−2iω ][n]) + (F −1 [e2iω ][n] − F −1 [e−2iω ][n])
4
4i
1
1
= (δ[n + 2] + 2δ[n] + δ[n − 2]) + (δ[n + 2] − δ[n − 2])
4
4i
1
1
for n = 2 ,

4 − 4i


 1 + 1 for n = −2 ,
= 14 4i

for n = 0 ,


2
0
otherwise.
x[n] = F −1 [b
x(ω)][n] =
(b)
h i
h
y[n] = F −1 1 [n] + F −1
ei2ω i
[n]
1 + 31 e−iω
h i2ω i
n+2
i2ω
u[n + 2]](ω), therefore, F −1 1+e1 e−iω [n] =
Note that by the time-shift property, 4 1+e1 e−iω = F[ − 13
3
3
n+2
u[n + 2]. Thurs,
− 31
1 n+2
y[n] = δ[n] + −
u[n + 2]
3
(c): Use partial fractions to see,
1
(1 + 12 e−iω )(1 + 13 e−iω )
A
B
=
+
(1 + 12 e−iω ) (1 + 13 e−iω )
zb(ω) =
where
1=A+
A −iω
B
e
+ B + e−iω
3
2
Thus, A + B = 1 and A/3 + B/2 = 0, and A = 3, B = −2.
Thus,
z[n] = 3F −1
h
1
i
1 −iω
2e
[n] − 2F −1
h
1
1 −iω
3e
1+
1+
1 n
1 n
=3 −
u[n] + 2 −
u[n]
2
3
1 n
1 n = 3 −
−2 −
u[n]
2
3
(
n
n
3 − 12 − 2 − 13
for n ≥ 0 ,
=
0
otherwise.
Second method: Use the convolution property: for x
b(ω) =
z[n] = (x ∗ y)[n].
8
1
,
1+ 12 e−iω
i
[n]
yb(ω) =
1
,
1+ 13 e−iω
we see
Notice that x[n] = −
1 n
u[n],
2
y[n] = −
1 n
u[n].
3
Therefore,
z[n] =
=
∞
X
m=−∞
∞
X
m=−∞
x[m]y[n − m]
−
1 n−m
1 m
u[m] −
u[n − m]
2
3
∞
1 n X 3 m
u[m]u[n − m]
(require m ≥ 0 and n − m ≥ 0 )
3 m=−∞ 2
(
n Pn
3 m
− 13
for n ≥ 0 ,
m=0 2
=
0
otherwise.

n+1
3

1 n 1− 2
for n ≥ 0 ,
−
3
3
1−
=
2

0
otherwise.
(
1 n
1 n
for n ≥ 0 ,
−2 − 3 + 3 − 2
=
0
otherwise.
= −
Check that the two methods gave the same answer.
*Staple your HW. You will get F IV E marks OFF if you do not staple your HW! Note that
the instructor will NOT provide stapler.
9