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

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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.
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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 (

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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) =

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