Unformatted text preview: 1 EE113 Homework 2 Solutions Problem 2.2 Consider the complex numbers √
1
3
z1 = + j
,
4
4 √
3
1
z2 =
−j
4
4 ∗
∗
2
3
∗
Find the polar representations of the numbers: z1 , z2 , z1 z2 , z1 /z2 , z1 z2 , and z1 z2 . Solution: The polar representations for these numbers are
∗
z1 = z2 =
∗
z1 z2 =
2
z1
=
z2
3
z1 z2 =
∗
z 1 z 2 = October 5, 2011 1 j 5π
e3
2
1 j 11π
e6
2
1 j 3π
e2
4
1 j 5π
e6
2
1 j 11π
e6
16
1 jπ
e6
4 DRAFT 2 Problem 2.6 Plot in polar coordinates the terms of the sequence x(n) = √
2
2 +j √
2
2 2n u(n). What is the energy and average power of this sequence?
Solution: The sequence can be rewritten as
√
2
x(n) =
(1 + j )
2
= 2n u(n) n (1 + j )2
2 u(n) = j n u(n)
See the plot below. 90 1.5 120 60
1
30 150
0.5 180 0 210 330 240 300
270 The energy of this sequence is thus
E= ∞
n=0 x(n)2 = ∞
n=0 1→∞ and the power is
N N 1
1
x(n)2 = lim
1=1
N →∞ N + 1
N →∞ N + 1
n=0
n=0 P = lim October 5, 2011 DRAFT 3 Problem 2.8 Let
x(n) = 1
4 n−2 ·e j ( π n− π )
6
3 √ 3
1
−j
2
2 · and denote its polar representation by x(n) = ρ(n)ejθ(n) , where both ρ(n) and θ(n) are functions of n.
(a) Determine ρ(n) and θ(n).
(b) Determine the even and odd parts of ρ(n).
(c) Determine the even and odd parts of θ(n).
Solution:
(a) ρ(n) and θ(n) are
n−2 1
4
ππ
π
π
π
θ(n) = n − − = n −
6
3
6
6
2 ρ(n) = x(n) = (b) By (2.25) and (2.26) on page 32 of the textbook, the even and odd parts of ρ(n) are
ρe (n) = ρ(n) + ρ(−n)
1
=
2
2 1
4 n−2 ρo (n) = ρ(n) − ρ(−n)
1
=
2
2 1
4 n−2 + 1
4 −n−2 − 1
4 −n−2 (c) By (2.25) and (2.26) on page 32 of the textbook, the even and odd parts of θ(n) are
θ(n) + θ(−n)
π
=−
2
2
π
θ(n) − θ(−n)
θo (n) =
=n
2
6 θe (n) = October 5, 2011 DRAFT 4 Problem 2.10 Express the complex exponential sequence
1
1
+j
2
2 x(n) = 2 · √
3
1
−j
2
2 2n in polar form and plot its terms at the time instants n = −1, 0, 1.
Solution: The polar form of the sequence is
x(n) = 1 j (− π n+ π )
3
2
e
2 See the following ﬁgure. 90
120 0.5
0.4 60 0.3
150 30 0.2
0.1 180 0 210 330 240 300
270 October 5, 2011 DRAFT 5 Problem 2.16 Consider the sequence
n−1 1
2 x(n) = 1 + u(n − 2) (a) Find its energy and average power.
(b) Let y (n) = x(−2n + 3). For what values of n is y (n) zero?
(c) Find the energy of x(2n).
Solution:
(a) Its energy and average power are
E= ∞ 2 n=−∞ x(n) = 1
P = lim
N →∞ 2N + 1 ∞ n−1 2 1
2 1+ n=2 →∞ N n=−N x(n)2 N N 1
2 1
1+2
N →∞ N − 1
n=2 = lim N n−1 2 1
2 1
1+
= lim
N →∞ N − 1
n=2 n−1 2n−2 N 1
2 2
N →∞ N − 1
n=2 = 1 + lim 1
2 + n−1 1
2 −2n+2 1
N →∞ N − 1
n=2 + lim 1
2 2n−2 =1
(b) By y (n) = x(−2n + 3), we get
y (n) = 1 + u(−2n + 1) One can see that
y (n) = 0, ∀ n≥1 (c) The energy of x(2n) is
E1 = October 5, 2011 ∞
n=−∞ 2 x(2n) = ∞
n=1 1+ 1
2 2n−1 2 →∞ DRAFT 6 Problem 2.17 Give, if possible, examples of nonzero sequences x(n) such that:
(a) x(n) and x(n − 2) are identical.
(b) x(n)x(−n + 1) = 0 for n = 0, 1, 2.
(c) x(n) + x∗ (n) = cos[ π (n − 1)].
3
Solution:
(a) For example,
x(n) = 1
(b) For example,
x(n) = u(n − 1)
(c) For example,
x(n) = October 5, 2011 π
1
cos (n − 1)
2
3 DRAFT 7 Problem 2.19 The DC level of a sequence x(n) is deﬁned as the average value
x
¯ 1
·
2N + 1 lim N →∞ N x(n)
n=−N Let x(n) be an odd sequence. Show that its DC level is zero.
Proof: Be deﬁnition, the DC level of an odd sequence is
x
¯ lim N →∞ = lim N →∞ = lim N →∞ = lim N →∞ = lim N →∞ 1
·
2N + 1
1
·
2N + 1
1
·
2N + 1
1
·
2N + 1 N x(n)
n=−N
−1 N x(n) + x(0) + x(n)
n=1 n=−N
1
n′ =N N x(−n′ ) + x(0) + x(n)
n=1 (let n′ = −n) N x(0) + (x(n) + x(−n)) n=1 x(0)
2N + 1 =0
where we also notice that for odd sequences
x(0) = −x(0) = 0 October 5, 2011 DRAFT 8 Problem 2.21 Determine the DC level of the following sequences:
(a) x(n) = ( 1 )n−2 u(n − 3).
2 (b) x(n) = ( 1 )n sin( π n).
3
7
Solution: (a) The DC level of this sequence is
x
¯ lim N →∞ 1
·
2N + 1 N N x(n) = lim n=−N N →∞ 1
·
N − 2 n=3 1
2 n−2 =0 (b) The DC level of this sequence is
x
¯ October 5, 2011 lim N →∞ 1
·
2N + 1 N x(n)
n=−N = lim N →∞ 1
·
2N + 1 N n=−N 1
3 n sin π
n
7 =0 DRAFT 9 Problem 2.23 Plot the sequence x(n) = δ (n + 1) + ( 1 )n u(n − 3). Plot also its even and odd components.
2
Solution: The even and odd parts of x(n) are
x(n) + x(−n)
2
x(n) − x(−n)
xd =
2
xe = See the following ﬁgure. 1 x(n), xe (n), xo (n) x(n)
xe (n)
xo (n)
0.5 0 −0.5
−20 October 5, 2011 −15 −10 −5 0 n 5 10 15 20 DRAFT 10 Problem 2.38 Given the sequence x(n) = ( 1 )n u(n), plot the sequences x(3n) and x(n/3). Find the energies of
4
the latter sequences as well. How do the even and odd components of x(3n) and x(n/3) relate to those of x(n)?
Solution: See the following ﬁgure. 1 x(n)
x(3 n)
x(n/ 3) 0.9 x(n), x(3 n), x(n/ 3) 0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0 5 10 15 n 20 The energy of x(3n) is
E1 = ∞
n=−∞ 2 x(3n) = ∞
n=0 1
4 3n 2 = ∞
n=0 12n 1
2 = 1
1 − 21
12 and the energy of x(n/3) is
E2 = ∞ x n=−∞ n
3 2 = ∞
n′ =−∞ By (2.25) and (2.26), the even and odd parts of y (n) x(n′ )2 = ∞
n=0 x(3n) and z (n) 1
4 n2 = 1
1 − 21
4 x(n/3) are x(3n) + x(−3n)
y (n) + y (−n)
=
= xe (3n)
2
2
y (n) − y (−n)
x(3n) − x(−3n)
yo (n) =
=
= xo (3n)
2
2
ye (n) = and
z (n) + z (−n)
x(n/3) + x(−n/3)
=
= xe (n/3)
2
2
z (n) − z (−n)
x(n/3) − x(−n/3)
zo (n) =
=
= xo (n/3)
2
2
ze (n) = October 5, 2011 DRAFT ...
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 Fall '08
 Walker
 Energy, lim, Complex number, Fibonacci number, odd parts

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