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MATH 267 Problem Set 10 Solutions
1. Let f [n] and g[n] be two discrete–time signals, each of period 4, determined by f [0] = f [1] = f [2] = f [3] = 1
and g[0] = g[2] = 1, g[1] = g[3] = 0.
3
�
(a) Find h[n] = (f∗g)[n] =
f [n − m]g[m] directly from the definition. Then use your result to find ĥ.
m=0
(b) Find fˆ and ĝ and compare the componentwise product fˆĝ with the value of ĥ found in part (a).
�
Solution. (a) By definition h[n] = (f ∗ g)[n] = 0≤m≤3 f [n − m]g[m]. In this case f [n] = 1 for all n,
�
since f is to have period 4, so that h[n] = 0≤m≤3 g[m] = 2, for all n. By Problem 1 of Problem Set 9,



ĥ[0]
1
 ĥ[1]  1  1


 ĥ[2]  = 4  1
1
ĥ[3]
1
−i
−1
i
1
−1
1
−1



1
h[0]
1
i   h[1]  1  1

= 
−1
h[2]
4 1
−i
h[3]
1
1
−i
−1
i
1
−1
1
−1
 
 
1
2
2
i 2
0
  =  
−1
2
0
−i
2
0
(b) By Problem 2, vectors a and e, of Problem Set 9,
ˆ   
f [0]
1
 fˆ[1]   0 


 fˆ[2]  =  0 
0
fˆ[3]


ĝ[0]
 ĝ[1] 

=
ĝ[2]
ĝ[3]
 
1
0

1
2 1
0
The componentwise product is
ˆ
 
 1
f [0]ĝ[0]
1 × 12
2
 fˆ[1]ĝ[1]   0 × 0   0 

=
=
  =
 fˆ[3]ĝ[2] 
0 × 12
0
0×0
0
fˆ[3]ĝ[3]


ĥ[0]


1  ĥ[1] 
4 
ĥ[2] 
ĥ[3]
by the convolution part of Problem 4 of Problem Set 9. This gives the same �
h[k] as above.
2. Let a[n] and b[n] be two discrete–time signals, each of period 4, determined by
a[0] = 10 a[1] = −2 − 2i
b[0] = 27
b[1] = −7 − 16i
a[2] = −2
b[2] = −5
a[3] = −2 + 2i
b[3] = −7 + 16i
Find a discrete–time signal x[n], again of period 4, such that a ∗ x = b.
Solution. By the convolution property of period discrete–time sequences
a ∗ x = b ⇐⇒ N â[k]x̂[k] = b̂[k] ⇐⇒ x̂[k] =
b̂[k]
N â[k]
By the first part of Problem 1 of Problem Set 9,


â[0]
 â[1] 

=
â[2]
â[3]


b̂[0]
 b̂[1] 


 b̂[2]  =
b̂[3]

1
1 1

4 1
1

1
1 1

4 1
1
1
−i
−1
i
1
−i
−1
i
1
−1
1
−1
1
−1
1
−1


1
a[0]
i   a[1] 

=
−1
a[2]
−i
a[3]


1
b[0]
i   b[1] 

=
−1
b[2]
−i
b[3]
1

1
1 1

4 1
1

1
1 1

4 1
1
1
−i
−1
i
1
−i
−1
i
1
−1
1
−1
1
−1
1
−1


 
1
10
1
i   −2 − 2i 
2

 = 
−1
−2
3
−i
−2 + 2i
4

  
1
27
2
i   −7 − 16i   0 

= 
−1
−5
9
−i
−7 + 16i
16
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So
x̂ =
�
2
0
9
16
4×1 , 4×2 , 4×3 , 4×4
By the second part of Problem 1 of Problem Set 9,

 
x[0]
1
 x[1]   1

=
x[2]
1
x[3]
1
1
i
−1
−i
�

 
1
x̂[0]
1
−i   x̂[1]   1

=
−1
x̂[2]
1
i
x̂[3]
1
1
−1
1
−1
�1
�
3
2 , 0, 4 , 1
=
1
i
−1
−i
1
−1
1
−1
 1 
 9 
1
2
4
−i   0 
 − 14 − i 
 3  =  1 
−1
4
4
i
1
− 14 + i
3. Compute, using the defining sum, the discrete–time Fourier transform of each of the following functions.
Also determine the fundamental period of the Fourier transform.
� �n−1
� �|n−1|
(a) 12
u[n − 1]
(b) 12
(c) δ[n + 2] − δ[n − 2]
Here
�
�
1 if n = 0
1 if n ≥ 0
u[n] =
δ[n] =
0 if n �= 0
0 if n < 0
Solution. (a)
∞
�
x̂(ω) =
x[n]e−iωn =
n=−∞
2
n=−∞
−iω
e
1 − 12 e−iω
=
∞
�
� 1 �n−1
u[n − 1]e−iωn =
∞
�
� 1 �n−1
2
e−iωn = e−iω
n=1
∞
�
�
n=1
�
1 −iω n−1
2e
period = 2π
(b)
∞
�
x̂(ω) =
x[n]e−iωn =
n=−∞
= e−iω
∞
�
�1
2e
m=0
=
e
(c)
x̂(ω) =
2
e−iωn =
n=−∞
m=n−1
�=−n
−iω
∞
�
� 1 �|n−1|
(1 −
(1 −
�
−iω m
+
1
2
∞
�
�
∞
�
�
n=−∞
=
2
e−iωn +
�
iω �
5
4
=
−iω
e
+
1 − 12 e−iω
1
3 −iω
4e
1 −iω
− 12 eiω
2e
−
0
�
� 1 �1−n
2
e−iωn
n=−∞
n=1
1
2e
�=0
1 iω
1
1 −iω
)
2 e ) + 2 (1 − 2 e
1 −iω
1 iω
)(1 − 2 e )
2e
∞
�
� 1 �n−1
=
1
2
− 12 eiω
3 −iω
4e
5
4
period = 2π
− cos ω
�
δ[n + 2] + δ[n − 2] e−iωn = ei2ω − e−i2ω = 2i sin(2ω)
period = π
4. Compute, using the defining integral, the discrete–time inverse Fourier transform of
Solution. For n �= 0
x[n] =
1
2π
�
π
x̂(ω)e
−π
1
= − πn
+
iωn
1 −iπn
πn e
dω =
+
1
2π
1 iπn
πn e
�
x̂(ω) =
�
(−2i)e
iωn
2i
−2i
if 0 < ω ≤ π
if −π < ω ≤ 0
0
−π
1
πn
−
=
2
πn
�
dω +
1
2π
�
�
(−1)n − 1
2
π
0
(2i)eiωn dω = − πi
�
�
1 iωn 0
in e
−π
+
�
�
i 1 iωn π
π in e
0
For n = 0
x[0] =
1
2π
Double-click to edit.
�
�
π
x̂(ω) dω =
−π
1
2π
�
0
(−2i) dω +
−π
1
2π
π
(2i) dω = 0
0
5. Find the discrete–time signals x[n] whose Fourier transforms are the following. Do not compute the
�π
1
integral x[n] = 2π
x̂(ω)eiωn dω directly.
−π
(a) x̂(ω) = cos2 ω + sin2 (3ω)
(b) x̂(ω) =
Solution. (a)
e−iω − 15
1 − 15 e−iω
x̂(ω) = cos2 ω + sin2 (3ω) =
1
4
�
(c) x̂(ω) =
eiω + e−iω
= − 14 e−6iω + 14 e−2iω + 1 + 14 e2iω −
�∞
Matching coefficients with x̂(ω) = n=−∞ x[n]e−iωn gives
x[n] =
 1

 −4

1
�2
1
4
1 6iω
e
4
−
�
ei3ω − e−3iω
1 − 13 e−iω
1 − 14 e−iω − 18 e−2iω
�2
if n = ±6
if n = ±2
4


1


0
if n = 0
otherwise
(b) We know that the DFT of an u[n] is 1−ae1 −iω (assuming that |a| < 1) and that the DFT of δn0 [n] =
δ[n − n0 ] is e−iωn0 . So we use partial fractions to try and write x̂(ω) as a sum of constants times
1
−iωn0
’s.
1−ae−iω ’s and constants times e
x̂(ω) =
−5(1 − 15 e−iω ) +
1 − 15 e−iω
24
5
= −5 +
1−
24
5
1 −iω
5e
=⇒ x[n] = −5δ[n] +
� �
24 1 n
u[n]
5 5
(c) Factoring the denominator,
x̂(ω) = �
�
�
�
�
A 1 + 14 e−iω + B 1 − 12 e−iω
1 − 13 e−iω
A
B
��
�
�
��
�
=
+
=
1 − 12 e−iω
1 + 14 e−iω
1 − 12 e−iω 1 + 14 e−iω
1 − 12 e−iω 1 + 14 e−iω
if A + B = 1 and
A
4
−
B
2
= − 13 . So A =
2
9
x[n] =
and B =
� 2 � 1 �n
9 2
7
9
+
and
7
9
�
−
� �
1 n
u[n]
4
6. Let x̂(ω) denote the Fourier transform of the discrete–time signal x[n] depicted in the figure below.
Perform the following calculations without explicitly evaluating x̂(ω).
(a) Evaluate x̂(0).
�π
(b) Evaluate −π x̂(ω) dω.
(c) Evaluate x̂(π).
�π
(d) Evaluate −π |x̂(ω)|2 dω.
dx̂
(e) Evaluate dω
(0).
2
1
-3
7
-2 -1 0 1 2 3 4 5 6
−1
3
8
n
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�
Solution. (a) x̂(0) = n x[n] = −1 + 1 + 2 + 1 + 1 + 2 + 1 − 1 = 6 .
�π
(b) −π x̂(ω) dω = 2πx[0] = 4π .
�
(c) x̂(π) = n (−1)n x[n] = 1 + 0 − 1 + 2 − 1 + 0 − 1 + 2 − 1 + 0 + 1 = 2 .
�
�
�π
�
(d) −π |x̂(ω)|2 dω = 2π n |x[n]|2 = 2π 1 + 1 + 4 + 1 + 1 + 4 + 1 + 1 = 28π .
(e)
dx̂
dω
is the DFT of −inx[n]. Hence
dx̂
dω (0)
= −i
�
nx[n]
n
�
�
= −i (−3) × (−1) + (−1) × 1 + 0 × 2 + 1 × 1 + 3 × 1 + 4 × 2 + 5 × 1 + 7 × (−1)
�
�
= −i
3
−
1 + 0 + 1 + 3 + 8 + 5 − 7
= −12i
4