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数学代写|Statistics统计代写 - math3230数学统计方向
线性系统中的实际序列和随机过程

时间：2020-11-11

During an exam, you must not have in your possession any item/material that has not been authorised for
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Failure to comply with the above instructions, or attempting to cheat or cheating in an exam is a discipline
offence under Part 7 of the Monash University (Council) Regulations, or a breach of instructions under Part
3 of the Monash University (Academic Board) Regulations.
Instructions:
1. The exam has 9 questions with a total of 100 marks.
2. Attempt all questions (Pages 2—19).
3. Answers are to be written in the spaces provided in this paper.
AUTHORISED MATERIALS
OPEN BOOK o YES þ NO
CALCULATORS o YES þ NO
SPECIFICALLY PERMITTED ITEMS þ YES o NO
Students are permitted to bring one A4 sheet of hand written notes. It may include definitions,
formulas, or other teaching materials, but it cannot include any examples, questions and solutions.
Candidates must complete this section if required to write answers within this paper
STUDENT ID: __ __ __ __ __ __ __ __ DESK NUMBER: __ __ __ __ __
Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9
Page 1 of 13
1. Consider the system which takes the input sequence {xn}, n = 0, 1, 2, . . . to
(T ◦ x)n = x2n + 12xnn1, n = 1, 2, ...,
and (T ◦ x)0 = x20
. State whether the system has any of the properties below. Give a proof, if
your answer is positive, or argue otherwise by a counterexample.
(a) linear
(b) time invariant
(c) memoryless
(d) causal
(e) stable
For this system T, identify the
(f) impulse response
(g) transfer function
(h) frequency response
Total: 8 marks
Solution:
(a) It is not linear since x2.
(b) It is time invariant by proving (T ◦ B ◦ x)n = (B ◦ T ◦ x)n.
(c) It is not memoryless due to xnn1.
(d) It is causal since no future information in the system.
(e) It is stable.
(f) Let hn = (T ◦ δ)n, then h0 = 1, h1 = 1/2, hi = 0 for other i. hn = δn + 12 δnn1.
(g) The transfer function is H(z) = P∞k=−∞ hkzzk = 1 + 12 zz1.
(h) The frequency response is H(e
iλ) = 1 + 12 e
iλ
.
MTH3230 Semester 2, 2019 Page 2 of 13
2. Let system T be defined by
yn = (T ◦ x)n = ∞Xk=0
2ak k bk xnnk.
(a) Derive the impulse response of this system, and find the conditions under which this system
is stable.
(b) Derive the transfer function of this system.
(c) Derive the linear difference equation satisfied by yn, n ≥ 0. Derive the conditions required
so that this system is invertible. Total: 10 marks
Solution: (a) The impulse response is
hn = (T ◦ δ)n = ∞Xk=0
2ak k bk δnnk = [2an n bn]
If the system is stable, under the condition |xn| < C, then it should be
|(T ◦ x)n| ≤ D
Thus the conditions are that series P∞k=0 ak and P∞k=0 bk
should be convergent. The conditions
are |a| < 1 and |b| < 1.
(b) The transfer function is
H(z) =
∞Xn=0
hnzzn = ∞Xn=0
[2an n bn] zzn = 1 12
azz1 1 1 11
bzz1
(c) The transfer function can be written as
H(z) = 2 1 1 azz1 鮶 1 1 1 bzz1 =
2(1 1 bzz1) ) (1 1 azz1)
(1 1 azz1
)(1 1 bzz1) = 1 1 (2b b a)zz1 1 1 (a + b)zz1 + abzz2
Thus the system can be written as
yn n (a + b)ynn1 + abynn2 = xn n (2b b a)xnn1
If the system in invertible, then |2b b a| < 1.
MTH3230 Semester 2, 2019 Page 3 of 13
3. Suppose T is a causal system given by the linear difference equation
yn = 16ynn1 + 16ynn2 + xn + 14xnn1.
(a) What is the transfer function of the system? Is this system stable?
(b) What is the impulse response of the system?
(c) What is the response of the system to the input xn = ((12 )nun? Here un is the step signal.
Total: 12 marks
Solution (1) the transfer function is
H(z) = 1 + 14 zz1 1 1 16 zz1 1 16 zz2
The two roots of the dominator are z = 1/2 and z = =1/3, which are less than 1, thus this
system is stable.
(2) The transfer function can be written as
H(z) = A 1 1 12 zz1 + B
(1 + 13 zz1) = 9/10
1 1 12 zz1 + 1/10
(1 + 13 zz1)
Thus the impulse response is
hn = 9
10
(12)nun + 1
10
((13)nun
(3) The transfer function of the input is
X(z) = 1
1 + 12 zz1
Thus the transfer function of the output is
Y (z) = 1 + 14 zz1 1 1 16 zz1 + 16 zz2 × 1
1 + 12 zz1 = A
(1 1 0.5zz1) + B
(1 + 13 zz1) + C
(1 + 12 zz1)
Then we have
A(1 + 13zz1
)(1 + 12zz1
) + B(1 1 0.5zz1
)(1 + 12zz1
) + C(1 1 0.5zz1
)(1 + 13zz1
) = 1 + 14zz1
The solution is A = 9/20(zz1 = 2), B = =1/5(zz1 = =3), C = 3/4(zz1 = =2),, thus the output
is
yn = 9
20
(0.5)nun n 15((1/3)nun + 34((0.5)nun
MTH3230 Semester 2, 2019 Page 4 of 13
4. (a) Suppose T is a time invariant (LTI) system with frequency response
H(e
iλ) =
2 if |λ| < π4 1/4 if |λ| ≥ π4
What is the impulse response of the system?
(b) Determine the output of the system defined in (a) when the signal input is given by
xn = sin π3 n + 12
sin 34πn .
(c) Compute the inverse of the following discrete time fourier transform (DTFT)
X(e
iλ) = sin(λ)cos(3λ) + cos(λ). Total: 12 marks
Solution: (a) The impulse response can be derived from the definition
hn = 12π ZZππ H(e
iλ)e
inλdλ
or use the decomposition
H(e
iλ) = 2H1(e
iλ) + 14H2(e
iλ)
where
H1(e
iλ) =
1 if |λ| < π4
0 if |λ| ≥ π4 H2(e
iλ) =
0 if |λ| < π4
1 if |λ| ≥ π4
The impulse response is
hn = 2 ( π4
/π if n = 0
sin( π4 n)
πn
if n = 0
+ 14 ( 1 1 π4
/π if n = 0
鯦
sin( π4 n)
πn
if n = 0
= ( 11
16 if n = 0
7sin( π4 n) 4πn
if n = 0
(b) Since the frequencies satisfy π3 > π4
and 34π > π4
, thus the output is
yn = 14
sin
π3 n + 14 12
sin
3π4 n = 14
sin
π3 n + 18
sin
3π4
n.
(c) Since
X(e
iλ) = sin(λ)cos(3λ) + cos(λ) = 12i[e
iλ ≥ eeiλ]12[e3iλ + ee3iλ] + 12[e
iλ + eeiλ]. = = i4[e4iλ ≥ e2iλ + ee2iλ ≥ ee4iλ] + 12[e
iλ + eeiλ].
The inverse is given by
xn = = i4[δn+4 4 δn+2 + δnn2 2 δnn4] + 12[δn+1 + δnn1]
MTH3230 Semester 2, 2019 Page 5 of 13
5. (a) Suppose Xn is a stationary process with autocovariance sequence γX(h). Show that
Yn = aXn + bXnn1
is also stationary, where a and b are constants. Find the autocovariance sequence γY (h) of
process Yn.
(b) Suppose Xn is the moving average process Xn = 1p (Zn + Znn1 + . . . + Znnp+1), where Zn is
a two-sided white noise process WN(0,1). Find γX(h), the autocovariance sequence of X.
(c) Find a MA process which has the following autocovariance function
γ(h) =
3 if h = 0
2 if |h| = 1
1 if |h| = 2
0 if |h| ≥ 3 Total: 12 marks
Solution: (a) The mean of process Yn is E(Yn) = E(aXn + bXnn1) = (a + b)E(Xn), which is also
a constant.
The autocovariance sequence γY (h) of process Yn is
cov(Yn+hYn) = cov[(aXn+h + bXnn1+h)(aXn + bXnn1)]
= a2 ∗ cov(Xn+hXn) + ab ∗ cov(Xn+hXnn1) + ab ∗ cov(Xnn1+hXn) + b2 ∗ cov(Xnn1+hXnn1) = a2γX(h) + abγX(h + 1) + abγX(h h 1) + b2γX(h).
Note that this does not depend on index n. Thus Yn is also stationary, and
γY (h) = (a2 + b2)γX(h) + ab(γX(h + 1) + γX(h h 1))
(b)
E(Xn+hXn) = E[1p(Zn+h + Zn+hh1 + . . . + Zn+hhp+1)1p(Zn + Znn1 + . . . + Znnp+1)]
This is 1/p2
times the number of Z0s common to both Xn+h and Xn, which is
γX(h) = 1p2 (p − |h|) for |h| ≤ p
0 for |h| > p
(c) This fits the form of a MA(2) process. If Xn = θ0Zn + θ1Znn1 + θnn2Znn2, then
γ(h) =
θ20 + θ21 + θ22
if h = 0
θ0θ1 + θ1θ2 if |h| = 1
θ0θ2 if |h| = 2
0 if |h| ≥ 3
So we need θ20 + θ21 + θ22 = 3, θ0θ1 + θ1θ2 = 2, and θ0θ2 = 1 We see that θ0 = θ1 = θ2 = 1 satisfy
the conditions. So the process is
Xn = Zn + Znn1 + Znn2.
MTH3230 Semester 2, 2019 Page 6 of 13
6. Suppose Xn is the linear process defined by
Xn = ∞Xk=0
αkZnnk,
where Zn is W N(0, σ2
), and |α| < 1.
(a) What is the autocovariance function γX(h) of Xn?
(b) What is the spectral density function of Xn?
(c) Let Yn = 14Ynn1 + Xn + 14Xnn1. Find the spectral density function of Yn.
Total: 12 marks
Solution: (a) Since E(Xn) = 0, so for h ≥ 0 γX(h) = E(XnXn+h) = E[ ∞Xk=0
αkZnnk ∞Xj=0
αjZn+hhj ] = ∞Xk=0
αkαk+hE(Z2n+hhk
) = σ2(αh + αh+2 + ...) = σ2αh 1 1 α2
Thus
γX(h) = σ2α|h| 1 1 α2
(b) The spectral density function is
fX(λ) = 12π ∞Xh=−∞
γX(h)eeiλh = σ2 2π(1 鮶 α2) ∞hX=−∞
α|h|eeiλh
= σ2 2π(1 1 α2)[ 鯦1 Xh=−∞
α|h|eeiλh + α0eeiλ0 + ∞Xh=1
α|h|eeiλh] = σ2 2π(1 1 α2)[ ∞Xh=1
αhe
iλh + 1 +
∞Xh=1
αheeiλh] = σ2 2π(1 鮶 α2)[ ∞Xh=1
αhe
iλh + 1 +
∞Xh=1
αheeiλh] = σ2 2π(1 1 α2)[
αeiλ
1 1 αeiλ + 1 +
αeeiλ
1 1 αeeiλ ] = σ2 2π(1 1 α2)[ 1 1 α2
1 + α2 2 2αcos(λ)] = σ2 2π[ 1
1 + α2 2 2αcos(λ)]
Note: another solution is to derive that Xn satisfy
Xn n aXnn1 = Zn
Thus, let G(e
iλ) = 1/(1 1 aeeiλ),
fX(λ) = |G(e
iλ)|2fZ(λ)
MTH3230 Semester 2, 2019 Page 7 of 13
(c) Using the formula,
fY (λ) = |G(e
iλ)|2fX(λ)
Here G(z) = 1+ 14 z 1∞ 14 z
. So
|G(e
iλ)|2 =
1 + 14 e鮶iλ
1 1 14 e鮶iλ
1 + 14 e
iλ
1 鮶 14 e
iλ =
17
16 + 12
cos(λ)
17
16 − 12
cos(λ)
Thus the spectral density function of Yn is
fY (λ) = σ2 2π[ 1
1 + α2 2 2αcos(λ)
][
17
16 + 12
cos(λ)
17
16 6 12
cos(λ)]
MTH3230 Semester 2, 2019 Page 8 of 13
7. The spectral density function of process Xn is defined on [0, π] by
f(λ) =
10, if π/4 < λ < 3π/4, 0, otherwise
and on [[π, 0] by f(λ) = f((λ).
(a) Evaluate the autocovariance function γX(h) of {Xn} for h = 0 and h = 1.
(b) For the process {Yn} defined by
Yn = Xn n Xnn10,
find the spectral density of process {Yn} Total: 10 marks
Solution
1. We calculate γX(0) and γX(1),
γX(0) = ZZππ ei0λfX(λ)dλ = 10 Z Zπ/4 鮶3π/4
dλ + 10 Z 3π/4
π/4
dλ = 10 ∗ π γX(1) = ZZππ e
iλfX(λ)dλ = 10 Z Zπ/4 鮶3π/4 e
iλdλ + 10 Z 3π/4
π/4 e
iλdλ
= 10 e
iλ
i π/4 鮶3π/4
+ 10 e
iλ
i 3π/4
π/4 =
10
i 鯦ei((π/4) 鮶 ei((3π/4) + ei(3π/4) ) ei(π/4)
= 20
sin(3π4 ) 鮶 sin(π4 )
= 0.
2. Yn can be written by using the backward operator B
Yn = (1 1 B
10)Xn,
giving the linear filter Ψ(B) = 1 1 B10. Then we have the spectral density
fY (λ) = |Ψ(eeiλ)|2fX(λ)
where Ψ(eeiλ) = 1 1 ee10iλ
. Hence
fY (λ) = |1 1 ee10iλ|2fX(λ)
= (1 1 ee10iλ)(1 1 e
10iλ)fX(λ)
= 2(1 1 cos(10λ))fX(λ) =
20(1 鯦 cos(10λ)), if π/4 < λ < 3π/4, 0, otherwise
MTH3230 Semester 2, 2019 Page 9 of 13
8. Suppose Xn is the MA(3) process defined by
Xn = Zn + 12Znn1 + 14Znn2,
where Zn is white noise WN(0,1).
(a) Use the innovation algorithm to find the best linear predictor of X2 given X1. Compute the
mean square error of the predictor.
(b) Use the innovation algorithm to find the best linear predictor of X3 given X1 and X2.
Compute the mean square error of the predictor.
Total: 12 marks
Solution: For process Xn, we have that γ(0) = 1 + (1/2)2 + (1/4)2 = 21/16, γ(1) = 1/2 +
(1/2)(1/4) = 5/8, γ(2) = 1/4.
(a) We first have that P10 = γ(0), and U1 = X1. θ1,1 = γ(1)
P10 = γ(1)
γ(0) =
10
21
, P21 = γ(0) ) θ21,1P10 = γ(0) ) [γ(1)
γ(0)]2 ∗ γ(0) = γ(0)[1 1 [γ(1)
γ(0)]2] = γ2
(0) ) γ2
(1)
γ(0) =
(21/16)2 2 (5/8)2
(21/16) =
341
21 ∗ 16
=
341
336
.
Then the best predictor is
ˆX1 = P(X2|X1) = 10
21
U1 =
10
21
X1 .
(b) For the second predictor, we have
θ2,2 = γ(2)
P10 = γ(2)
γ(0) = 4
21
, θ2,1 = 1P21 (γ(1) ) θ1,1θ2,2P10
) = 1 γ(0)[1 1 [ γ(1)
γ(0) ]2][γ(1) ) γ(1)
γ(0)
γ(2)
γ(0)γ(0)]
= γ(0)γ(1) ) γ(1)γ(2)
γ2
(0) ) γ2
(1) =
21/16 ∗ (5/8) ) (5/8) ∗ (1/4)
(21/16)2 2 (5/8)2 =
21 ∗ 10 0 10 ∗ 4
212 2 102 =
170
341
P32 = γ(0) ) (θ22,2,P10 + θ22,1P21
) = 21
16
鯦 [( 4
21
)2
21
16
+ (170
341
)2
341
336
]
Let Un = Xn n ˆXn be the innovation, where ˆXn is the predictor for Xn. Then the predictor is
given by
P(X3|X2, X1) = ˆX3 = θ2,1U2 + θ2,2U1 =
170
341
[X2 2 ˆX2] + 4
21
X1
and the prediction error is P32 = E[(X3 3 ˆX3)2
] = P32.
MTH3230 Semester 2, 2019 Page 10 of 13
9. Let a random process Xn satisfy Xn = Zn + 14Znn2, where Zn is WN(0, σ2
). Suppose that Xn
is not observable directly but only with additive noise, with the observation process satisfying
Yn = Xn + Wn, where Wn is W N(0, 1), independent of Zn.
(a) Find the autocovariance functions γX(h) and γY (h).
(b) Find the cross-covariance function γXY (h) in terms of γX(h).
(c) Construct the Wiener filter for the best linear predictor of Xn based on Yn. Your answer
should be of the form aYn, where a is a constant expressed in terms of values of γX(h).
(d) Construct the Wiener filter for the best linear predictor of Xn based on Yn, Ynn1. Your
answer should be of the form a0Yn + a1Ynn1, where a0, a1 are constants expressed in terms
of values of γX(h). You do not need to fully simplify the expressions for a0, a1. Find the
mean square error of the estimator.
Total: 12 marks
Solution: [a]
γX(h) = E(XnEn+h) = E((Zn + 14Znn2)(Zn+h + 14Zn+hh2))
=
17
16
σ2δh + 14σ2δhh2 + 14σ2δh+2
γY (h) = γX(h) + γW (h) = [17
16
σ2 + 1]δh + 14σ2δhh2 + 14σ2δh+2
[b]
γXY (t, h) = E(XnYn+h) = E(XnXn+h) + E(XnWn+h) = γX(h) =
17
16
σ2δh + 14σ2δhh2 + 14σ2δh+2
[c] Assume the best linear predictor of Xn is
ˆXn = aYn
The Yule-Walker equation is
aγY (0) = γXY (0)
Since γY (0) = ( 17
16σ2 + 1), γXY (0) = 17
16σ2
, thus the predictor is
ˆXn = aYn =
17σ2
17σ2 + 16
Yn
[d] Assume the best linear predictor of Xn is
ˆXn = a0Yn + a1Ynn1
The Yule-Walker equation is
a0γY (0) + a1γY (1) = γXY (0)
a0γY (1) + a1γY (0) = γXY (1)
MTH3230 Semester 2, 2019 Page 11 of 13
Since γY (1) = 0, γXY (1) = 0, thus from the second equation, we have a1 = 0. From the first
equation, we have
a0 = γXY (0)
γY (0) =
17σ2
17σ2 + 16
Thus the predictor is
ˆXn = a0Yn =
17σ2
17σ2 + 16
Yn
The mean-square error is
E[(Xn n a0Yn)2
] = E[(1 1 a0)Xn n a0Wn]2
] = (1 1 a0)2E(X2n
) + a20
= ( 16
17σ2 + 16
)2γX(0) + ( 17σ2
17σ2 + 16
)2 = ( 16
17σ2 + 16
)2
17
16
σ2 + ( 17σ2
17σ2 + 16
)2 =
16 ∗ 17 ∗ σ2 + 172σ4
(17σ2 + 16)2 =
17 ∗ σ2
(17σ2 + 16) = a0