During an exam, you must not have in your possession any item/material that has not been authorised for your exam. This includes books, notes, paper, electronic device/s, mobile phone, smart watch/device, calculator, pencil case, or writing on any part of your body. Any authorised items are listed below. Items/materials on your desk, chair, in your clothing or otherwise on your person will be deemed to be in your possession. No examination materials are to be removed from the room. This includes retaining, copying, memorising or noting down content of exam material for personal use or to share with any other person by any means following your exam. 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