Additional exercises in Stationary Stochastic Processes

Transcription

1 Mathematical Statistics, Centre or Mathematical Sciences Lund University Additional exercises 8 * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * Additional exercises in Stationary Stochastic Processes X State, with motivation, which o the igures below that are realizations, covariance unctions and spectral densities. Combine the three dierent processes with their covariance unctions and spectral densities. A B C 4 6 D G 4 E.. H 4. 4 F.. 4 I X The igure below shows realizations o two dierent processes in discrete time with covariance unctions and spectral densities. Determine which o the igures that belong to the same process. A 6 B 4 C D E 8 F

2 X The igure below shows two realizations o two dierent processes in discrete time together with covariance unctions and spectral densities. Combine the igures which belong to the same process. X4 A real valued weakly stationary process X(t), t R has the spectral density { ( /6) 6 R X () = > 6 a) Determine the covariance unction o the process. b) The process X(t) is sampled and a new process Y t, t = ± d, ±d,... is computed where s = /d = 4. Determine the spectral density or the sampled process Y t. X A stationary process Y (t), t R has a spectral density that consists o two boxes : R Y () = or <.7 or.7 < or <. or. <.7 or.7 a) What smallest sampling requency s = /d is allowed i one should avoid aliasing? b) Determine the spectral density or the sampled process Z t, t =, ±d, ±d,... i d =..

3 X6 A weakly stationary process X(t), t R has the spectral density <., R X () = + <., otherwise. The process is sampled with s = /d =. Calculate the covariance unction and the spectral density or the sampled process Y t = X(t), t =, ±, ±,.... Also determine the variance or Ȳ = Y t + Y t + Y t+. X7 Let Y (t), t R be a stationary stochastic process in continuous time with the covariance unction r Y (τ) = + (πτ). The process is sampled at the time points t =, ±d, ±d.... Find the spectral density o the sampled process X t = Y (t), t =, ±d, ±d.... X8 The two pictures below show the expected values o the estimates o samples o an AR(4)- process rom two dierent methods, the modiied periodogram and the Welch method with 4 windows and % overlap. Both methods use Hanning windows. Combine the igures with the correct method and motivate your answer. Method Method E[ ˆRx()] E[ ˆRx()]

4 X9 A colleague gives you a data set, where a stationary stochastic process, X(t), is disturbed by noise that can be assumed to be zero-mean, white and stationary Gaussian. The data set should be analyzed and the stationary process X(t) should possibly be modelled. The data set are depicted in the igure below, spread out on the y-axis to be viewable. From these ten realizations it is diicult to say anything about X(t) and to get more inormation you compute the mean value o the ten realizations and also the individual periodograms (also spread out on the y-axis to be viewable). From these igures, could you judge what stationary stochastic process to use as a irst model o the data set? Ten realizations Periodogram t Mean value - t....4 X The ollowing data sequences are presented together with the periodogram in the lower igure. Can you judge rom which signal a) b) or c), the periodogram is computed? Or could the periodogram belong to more than one o the signals? Consider that all the presented sequences actually involves many requency components and that the periodogram does not speciy when in time dierent requencies appear a) Chirp signal -. t. b) Impulse signal t - - c) Noise realization - t Periodogram

5 X A real-valued weakly stationary process X(t), t R has the covariance unction r x (τ) = +τ and expected value E[X(t)] =. It is iltered through an ideal bandpass ilter with requency unction H() =, < <, and zero or all other values. Determine the expected value and the variance or the output signal rom the ilter. X A real valued stationary Gaussian process X(t), t R has the expected value and spectral density R X () = πe π. a) Determine the covariance unction. b) Determine the spectral density ater iltering through a linear ilter with requency unction H() = + i. c) A new process Y (t) = X(u)du is deined. Is the process Y (t) a Gaussian process? X From the stationary sequence X t, t =, ±, ±,..., a new stationary sequence Y t, t =, ±, ±,..., is deined as Y t = X t X t + X t + X t+. Determine the quote R Y ()/R X () between the spectral densities. X4 A weakly stationary process X(t), t R, has the spectral density { <, R X () = or all other values. a) Determine the covariance unction or X(t). b) The process X(t) is sampled with sample requency /d =. Determine and draw the spectral density ater the sampling? c) The process X(t) is sampled with /d = 4 which gives the discrete time sequence Z k, where k = ±, ±.... A new process Y k is created as Y k = ( ) k Z k. Determine and draw the spectral density or the new process Y k? (Guidance: Investigate what happens with the covariance unction or Y k or dierent values o k.)

6 X A stationary Gaussian process X(t), t R, has the covariance unction r X (τ) = e τ /. a) Calculate the variances o X(t), X(t + ), and X (t). b) Calculate the covariance and cross-covariances, C[X(t), X(t+)], C[X(t), X (t)] and C[X (t), X(t+ )]. c) Compute the probability, P (X(t + ) > X(t) + X (t)). X6 A stationary Gaussian process X(t), t R, has the covariance unction r X (τ) = cos(πτ) + cos(4πτ) and expected value m X =. Motivate that the process is dierentiable and compute the probability, P (X(t + ) > +.X(t) + X (t)). X7 A stationary Gaussian process X(t), t R, has the covariance unction r X (τ) = e τ, and unknown expected value E[X(t)] = m. You will be able to estimate m with m X() + X() = α + ( α) X(t)dt. Determine the constant α so that the variance o the estimate will become as small as possible. X8 Let the input signal, X(t), t R, to a linear ilter be a zero-mean weakly stationary signal with covariance unction r X (τ) = +τ. The ilter is deined by the impulse response: h(t) = { t, otherwise. What is the cross-covariance, r X,Y (τ) = C[X(t), Y (t + τ)], between the input and the output signal? 6

7 X9 You know that X t, t =, ±, ±... is an AR()-process and that r X () =, r X () = and r X () =. Calculate r X ()? X Let X t, t =, ±, ±,... be a sequence o independent stochastic variables with E[X t ] = and V [X t ] =. A new process is deined as Y t = X t X t + X t. Calculate the covariance unction and spectral density o the process Y t, t =, ±, ±.... X From the white noise sequence, e t, t =, ±, ±,..., a new process is deined as which is known to have the covariance unction X t = e t + b e t + b e t b q e t q, r X (τ) = τ =, τ =, τ =, τ =, τ >. Determine the spectral density and the constants b... b q i V [e t ] =. X The igure shows one AR(p)-process and one MA(q)-process. Combine realizations, covariance unctions and spectral densities or the two process. Motivate. 7

8 X The igures below belong to processes o the type AR or MA. State, with motivation, which o the igures that are realizations, covariance unctions and spectral densities. Determine which realizations, covariance unctions and spectral densities that belong to the same process. Also determine o which type the processes are and decide which order that can be seen in the igures. Hint: No process should be assumed to have an order larger than. A B C D..4 G 4 4 E 4 H..4 4 F I X4 The igures below belong to processes o type AR. Determine which pole-zero-plots, covariance unctions and spectral densities that belong to the same process. Also, state which order each o the processes has. How does the order o the AR-process relate to the number o poles and to the number o peaks in the spectral density? r(τ). -. R() - 4 A - I τ B - II..4 4 τ C - III τ 8

9 X The igures below belong to processes o type AR or MA. Determine which pole-zero-plots, covariance unctions and spectral densities that belong to the same process and decide the type o each process. Also, state which order each o the processes has. A B C log R() r(τ).. τ α τ β τ γ....4 X6 The igure shows the poles and zeros o dierent ARMA(,)-processes and the corresponding periodogram estimates rom realizations. Combine the pole-zero plot with the corresponding periodogram. Motivate your answer. ˆRx() A - ˆRx() B - ˆRx() C

10 X7 The igure shows the poles and zeros o dierent ARMA-processes and the corresponding modiied periodogram estimates using a Hanning window rom realizations, both in db-scale and linear scale. Combine the pole-zero plot with the corresponding logarithmic periodogram. Motivate your answer. How useul are the linear scaled periodograms? A B C - log ( ˆRx()) - -4 log ( ˆRx()) - -4 log ( ˆRx()) α 6..4 β 6..4 γ ˆRx() 4 ˆRx() 4 ˆRx() X8 The two pictures below show the expected values in db-scale o the estimates o two dierent methods, the periodogram and the modiied periodogram using a Hanning window. The process is an ARMA-process o low order. Can you guess the order o the ARMA-process using the two expected values. Motivate your choice rom the properties o the two methods. Periodogram Modiied periodogram 4 4 log (E[ ˆRx()]) log (E[ ˆRx()])

11 X9 A stationary sequence S t, t =, ±, ±... is disturbed by noise N t, t =, ±, ±.... One wants to improve the measured signal Y t = S t + N t by iltering Y t in a ilter with requency unction H() = a + a e iπ, giving the output Z t. Determine a and a so that the squared error, E [ (Z t S t ) ] becomes as small as possible. The processes S t and N t are assumed to be independent with expected value and covariance unctions r S (τ) =. τ and r N (τ) = (.) τ. X A weakly stationary process S t, t =, ±, ±... has expected value zero and covariance unction, τ = r S (τ) = τ = or other values. The process is disturbed by colored noise N t, t =, ±, ±... with expected value zero and covariance unction, τ = r N (τ) = τ = or other values. The process and the disturbance are independent. a) Determine the requency unction H(), / < / and the impulse response h(t), t =, ±, ±..., or the optimal Wiener ilter that minimizes E[(Y t S t ) ] when Y t is the output signal rom the ilter with X t = S t + N t as input signal. b) In communication systems it is quite usual with echo, i.e.,you have a signal according to X t = S t +.S t + N t, where.s t is the damped and delayed original process realization. Determine the coeicients a, b or the non-causal impulse response b t = h(t) = a t = or other values. or an optimal ilter that minimizes E[(Y t S t ) ], when Y t is the output signal and X t is the input signal to the optimal ilter.

12 X A signal in discrete time is deined as s t = t =, 4 t =, t =, t =, t = 4. The measurements are disturbed by Gaussian white noise, N t, t =, ±, ±,... with expected value E[N t ] = and variance V [N t ] =. The received signal is Y t = s t + N t or Y t = N t. Determine the decision time, the decision threshold and the impulse response o the causal matched ilter so the error probabilities becomes equal and as small as possible. α = P (detect signal i no signal sent) β = P (detect no signal i signal sent) X a) A matched ilter or detection o the signal t =, s t = t =, 4 t =, should be designed. The signal is disturbed by Gaussian zero mean white noise N t, t =, ±, ±,... with variance V [N t ] = 4. The received signal is Y t = s t + N t or Y t = N t. Determine the coeicients o the impulse response h t, t =,,, the decision threshold and the decision time so the error probabilities becomes equal and as small as possible. α = P (detect signal i no signal sent) β = P (detect no signal i signal sent) b) Now assume that the noise has the covariance unction, 4 τ =, r N (τ) = τ =, τ. Determine the coeicients h t, t =,, or optimal detection o the signal s t. Also, determine the equal error probabilities or this optimal ilter and compare with the error probabilities you would have received i you used the non-optimal ilter you designed in a).

13 Solutions X Figures A,C and G are realizations: they can not be covariance unctions as the largest value are not at zero and they can not be spectral densities as they are negative. Figures B, E and I are covariance unctions as they are negative and the maximum is at zero, and accordingly D, F and H must be spectral densities. The scales o the igures are dierent and to combine the right igures, the periods o realizations and covariance unctions need to be determined and matched to the corresponding requency. It might be easier to think o periods in seconds (s) and requency in Hertz, (Hz=/s). We ind the same period o. rom igures C and E which corresponds to the requency, i.e., D. The period is ound rom igures G and I which correponds to requency.8 in F. Finally the period o / in igures A and B corresponds to the strongest requency at in H. X Figures D and E are realizations: they can not be covariance unctions as the largest value are not at zero and they can not be spectral densities as they are negative. Figures A and C are covariance unctions as they are negative and the maximum is at zero, and accordingly B and F must be spectral densities (also seen rom the scale o the x-axis, to.). B has a strong peak or a low requency,.8 (period ) and a smaller one at. (period ). The corresponding covariance unction should have a period o and a smaller ast variation (jittering) as well, which its with C and the realization D, (note that the scales o the x-axes are dierent). The spectral density F corresponds to a ast variation as there is a strong peak at.4 (period.), which its with A and E. X Figures C and F are spectral densities (non-negative and scale to.) and B and D are covariance unctions (maximum at zero). Then A and E are realizations. C has a strong peak at., where F has one at a lower value.. B has a higher requency than D. This is also the case or E compared to A. Thereore, A-D-F belong to one process and E-B-C to the other process. Note the tricky time-scaling o the igures o the realizations compared to the igures o the covariance unctions, which might lead to the conclusion that A and B belong together, which is wrong. X4 a) The covariance unction is and r X () = 6. b) The spectral density is r X (τ) = 6π τ sin (6πτ) τ, R Y () = { ( /6) 8 / 8. X a) The maximum requency is =.7. To avoid aliasing the sampling interval d should ulill d = s, dvs s /, or d /. b) With s =, we get, R Z () = R Y ( + k s ) = R Y ( + k) k= k= = R Y () + R Y ( ) + R Y ( + ), giving <., R Z () =., <.

14 X6 The spectral density is R Y () = k= R X ( + k s ) = + =, / < < /. The spectral density o the sampled process is constant, the values Y t are thereore uncorrelated and V [Ȳ ] = /. X7 We have r Y (τ) = and R +(πτ) Y () = π π e π /π = e. The spectral density o the sampled process is R X () s = = k= ( e + R Y ( + k s ) = e sk + k= ( = e + e = ( e + e s e s k= k= e + e s ( )) e + e, e +sk e sk+ ) ) e s e s s < s X8 Method is the modiied periodogram and method is the Welch method. With n =, the Hanning window mainlobe in the modiied periodogram is 4/n =.4. The Hanning window mainlobe o the Welch method will be much wider as the Welch method with 4 windows and % overlap will use Hanning windows o length n = 4 with corresponding mainline width o 4/n =.. Thereore the more peaked spectrum must belong to the modiied periodogram and the more smoothed one to the Welch method. X9 In the ten periodograms, peaks are viewable in almost all cases at the requency o =.. This indicates that a sinusoid, or a periodic signal, is present in the realization. The periodic signal is diicult to see in the dierent realizations, but also in the mean value. We can then suspect that the phase is dierent in the realizations. The mean value average level seems to dier rom zero, rather it is close to.. This indicates that there is at least some o the realizations that have a mean value diering rom zero. This is conirmed rom some o the periodograms which show strong peaks at =. It is then reasonable to include a stochastic level in the model, alternatively make sure that each realization o the data set has a mean value o zero. An appropriate model could be, X(t) = A + A cos(π.t + φ), t =... 99, where A and A are probably independent stochastic variables and φ is uniorm in the interval [, π]. 4

15 X The igure shows the periodogram corresponding to all three o the proposed signals! When computing the Fourier transorms o the sequences, the dierences will show up in the argument, the phase. These dierences are then hidden in the squared absolute value, the periodogram. Please veriy using the ollowing Matlab code, where x is the chirp signal, y is the impulse signal and z is a noise realization. >> n=; >> x=cos(*pi*.*([:n-] ).^); >> y=real(it(abs(t(x)).*exp(j**pi*[[:-:-n/] ;[n/-:-:] ]/n*))); >> b=rand(n/,)-.; >> z=real(it(abs(t(x)).*exp(j**pi*[b;;-b(n/:-:)]))); The ollowing expression with x replaced or y and z produces the periodogram. >> plot([:n-]/n,/n*abs(t(x,n)).^) X The spectral density is The variance is and the expected value is V [Y (t)] = R X () = πe π. πe π d = e π e 4π, m Y = H()m X = =. X a) The transorm table with α = gives r X (τ) = /( + τ ). b) R y () = + i πe π = ( + )πe π. c) A linear combination o Gaussian processes becomes a Gaussian process. X The quote is given rom R Y ()/R X () = H() as The sequence can be simpliied into R Y () = H() R X (). Y t = X t+ + X t X t, and rom deinition the output rom a linear ilter is given as Y t = u= h(u)x t u = u= h(u)x t u, where the impulse response is identiied as h( ) = /, h() = /, and h() = /. The requency unction is H() = u h(u)e iπu = eiπ + e iπ = ( cos(π)). resulting in R Y () R X () = H() = 4 9 ( cos(π)).

16 X4 a) The covariance unction is given by b) The sample requency causes aliasing, 8 { 6 τ = r X (τ) = ( cos(πτ)) τ (πτ) R X () c) The sampled process Z k, k =, ±, ±... has the same spectral density as X(t) with the normalized requency scale ν = / s = d. The covariance unction r Z (k) = r X (τ), with k = τ/d. The covariance unction r Y (k) = ( ) k r Z (k) = e iπk r Z (k) = e iπk/ r Z (k) = e iπτ/(d) r X (τ). The transorm table gives that g(τ)e iπ τ has the Fourier transorm G( ). With = /(d) = we ind R Y () = R X ( ), according to the igure below. 8 R X () R Y () 6 4 6

17 X The derivatives o r X (τ) = e τ / are r X (τ) = τe τ / and r X (τ) = (τ )e τ /. a) The variances are V [X(t)] = V [X(t + )] = r X () =, V [X (t)] = r X () =. b) The covariance and cross-covariance unctions are C[X(t), X(t + )] = r X () = e, C[X(t), X (t)] = r X() =, Always the case, see remark 6. p. 49 C[X (t), X(t + )] = r X() = e. c) Rewrite P (X(t) X(t + ) + X (t) < ). For the new Gaussian process Y (t) = X(t) X(t + ) + X (t) we get P (Y (t) < ) = Φ( m Y σ Y ). Usually, to ind σ Y the variance V [Y (t)] is needed including variances covariance and crosscovariances as calculated above. However, as the expected value is E[Y (t)] = E[X(t) X(t + ) + X (t)] = m X m X + =, then without calculating V [Y (t)]! P (Y (t) < ) = Φ( σ Y ) =., X6 The covariance unction is (indeinite many times) dierentiable and thereore the process is dierentiable in quadratic mean. We have C[X(t + ), X(t)] = r X () =, C[X(t + ), X (t)] = r X( ) =, C[X(t), X (t)] =, V [X (t)] = r X() = 4π, E[X(t + ).X(t) X (t)] = =, V [X(t+).X(t) X (t)] = V [X(t+)]+.V [X(t)]+4V [X (t)] C[X(t+), X(t)]+ = 96π +.7 P (X(t + ) > + X(t) + X (t)) = P (X(t + ) X(t) X (t) ) = = Φ( ) = Φ(.49) =.688 =.. 96π +.7 7

18 X7 The variance is given as V [m ] = V [α X() + X() ] + V [ α X() + X() X(t)dt] + C[α, α = α ( α) (V [X()] + V [X()] + C[X(), X()]) α( α) ( = α 4 ( e ) + = α 4 ( e ) + α + ( α) 9 4 Derivation or the optimal α, gives C[X(), X(t)]dt + ( α) t ( α) ( α( α) +. V α e 9 C[X(), X(t)]dt), e (t s) dsdt + α( α) ) + α( α) e, = α ( α) 9 + α = α = 6. X8 We start with the deinition o the output signal o a linear ilter: X(t)dt], C[X(t), X(s)]dtds, e t dt, Y (t) = = h(u)x(t u)du, X(t u)du, where X(t) is the input signal. We would like to compute the cross-covariance between the input and the output (in that order), which is, by its deinition: r X,Y (τ) = C[X(t), Y (t + τ)] = C = [ X(t), C [X(t), X(t + τ u)] du = [ = + (τ u) du = arctan(τ u) = arctan(τ) arctan(τ ). ] X(t + τ u)du, r X (τ u)du, ], 8

19 X9 The Yule-Walker equations give + a = σ + a + a = a + a = with solution a = /8 and a = /8. From the Yule-Walker equations r X () = a r X () a r X () = /8 /8 = /8. X The covariance unction is given rom r Y (τ) = C[Y t, Y t+τ ] = C[X t X t + X t, X t+τ X t+τ + X t+τ ], where the variance r X () = and r X (τ) = or all other values o τ. We get, τ =, 8, τ = ±, r Y (τ) =, τ = ±,, or all other values. The spectral density is given rom the Fourier transorm o the covariance unction, R Y () = τ= r Y (τ)e iπτ, = e iπ 8e iπ + 8e iπ + e iπ, = 6 cos(π) + 4 cos(4π). X The spectral density is given rom the known covariance unction as R X () = τ r X (τ)e iπτ = + e iπ + e iπ. The spectral density output o the ilter with impulse response coeicients, b, b,..., b q or the white noise spectral density input R e () = σ = is given rom R X () = H() R e () = q b u e iπu. u= Reormulating into q q R X () = b u b v e iπ(u v) u= v= and comparing with R X () = + e iπ + e iπ yields R X () = b + b + b b e iπ + b b e iπ, where only the possible non-zero coeicients are retained. The solution becomes b = b = and all other b k =. 9

20 X Only B and D can be spectral densities, they are positive and the scale range between and.. Then only C and E can be covariance unctions and A and F are realizations. D has one low and one high strong requency, which corresponds with the covariance unction in C and the realization in A, and is most possibly an AR(4)-process. The covariance unction in E is probably an MA()-process, with corresponding spectral density B and realization F. X The spectral density is positive which coincides with D,F,H. The covariance unction is largest at zero, i.e., B,G,I. Accordingly, A,C,E are realizations. The covariance unction I belongs to a MA()-process as it is zero when τ > where B and G belong to AR-processes. The highest requencies are ound in E connected to F. Somewhat lower requencies are ound in A and D where C must be connected with H with the lowest requency content. The covariance unction (the MA()-process) its with the spectral density F where the other spectral densities have peaks (AR-processes). Realization A its with the covariance unction B, where the realization C has the same period as G. D is probably an AR(4)-process as we see two requency peaks and H is possibly an AR()-process as we see two requency peaks and a peak at the requency zero. In summary: MA()-process E,F,I, AR(4)-process D,A,B, AR()-process H,C,G. X4 The pole-zero plots and the spectral densities can be combined as A-II, B-III and C-I. The strongest peak in III has a high requency which connects to the covariance unction in. The spectral density in I is more damped than the one in II although they have the same requency, which corresponds to the more damped covariance unction in compared to the one in. Thereore I and belong together and II and. The number o poles ound in the upper hal, with angle to π including the real axis, correspond to the number o peaks in the spectral density, i.e., A-II- is AR(4), B-III- is AR(), C-I- is AR(). X The ollowing combinations belong together: A--α which is an MA()-process, as there are two zeros and the covariance unction is zero or τ > and the spectral density shows a clear zero (low value) at the requency corresponding to the angle π i A. B--γ is an AR(4)-process as there is 4 poles and C--β is an AR()-process. The covariance unction in has a lower requency than the one in which correspond to the notes in B and C. Similarly, the spectral density in γ has the two peaks at a lower requency than the peaks in the spectral density in β. X6 Periodogram belongs to a high requency process as it has the main power at >., which could be B, with the pole at a high angle. Periodogram is the most low-requency process, which then must be C, which has the pole at a low angle. Periodogram has most power at =. which corresponds to A. X7 The logarithmic scaled spectrum estimates indicate where the zeros are located, which are where the dierences are ound. We then combine periodogram with C, with A, with B, according to the location o the zeros. The linear called periodograms do not give any inormation on the zeros location and as the power rom the poles are ound around =. in all cases, the igures do not give enough inormation. It is always advisable to visualize in both scales!

21 X8 The periodogram shows two peaks (poles) close to =. where the modiied periodogram shows a broader peak but also one zero at =.. As we know that the periodogram has a more narrow mainlobe than the modiied periodogram and thereore can resolve peaks that are close, the assumption o two peaks is reasonable. The modiied periodogram, however, has lower sidelobes, which makes it possible to also resolve zeros, and the zero seen at =. is probably one zero. We ind the mirrored picture on the negative side o the requency axis, and thereore in total 4 poles and zeros, i.e., an ARMA(4,)-model. Periodogram Modiied periodogram 4 4 log (E[ ˆRx()]) log (E[ ˆRx()]) X9 The impulse response coeicients, h() = a and h() = a are identiied rom the given requency unction. The output rom the ilter is and the squared error becomes Z t = a Y t + a Y t, E [ (a Y t + a Y t S t ) ] = E [ (a S t + a S k S t + a N t + a N t ) ]. As all processes are zero-mean we rewrite and expand the expression as Dierentiation yields a V [S t ] + a V [S t ] + a V [N t ] + a V [N t ] + V [S t ]+ a a C[S t, S t ] + a a C[N t, N t ] a vv [S t ] a C[S t, S t ] = a + a + a a = ɛ. ɛ a = 4a = a = ɛ a = 4a = a = 4

22 X a) The Wiener ilter is given by H() = R s () R S () + R N (), where and We get R S () = τ R N () = τ r S (τ)e iπτ = cos(π), r N (τ)e iπτ = + cos(π). H() =.. cos(π) =..(e iπ + e iπ ). <., and the impulse response h() =., h(±) =. and zero or all other t. b) For the given impulse response min a,b E[(aX t + bx t+ S t ) ], min a,b E[(aS t +.as t + an t + bs t+ +.bs t + bn t+ S t ) ], min a,b ((a +.b) +.a + b )r S () + (a + b )r N ()+ +(.a(a +.b) + b(a +.b))r S () + abr N ()). We get min(a, b) = 7 a + 7 b + ab a +. Dierentiation with respect to a, b yield a b = 7a + b =, = 7b + a =, resulting in b = 8/87.96 and a = 84/ The impulse response is h() =.449, h( ) =.96 and zero or all other t. X The causal impulse response is given as h() = c, h() = c, h() = c, h() = 4c, h(4) = c, where the constant is assumed to be c = and the decision time is T = 4. The expected value or received signal when the signal is sent becomes E[Y T signal sent] = E[ 4 h(u)(s T u + N T u )] = ( ) =. u= With equal error probabilities the threshold is k = E[Y T signal sent]/ = 7.. The variance or o the received signal is V [Y T ] = V [N 4 + N + N + 4N + N ] = =. The equal error probabilities become α = P (Y T > 7. no signal sent) = Φ(7./ ) = Φ(.78) =.44 4 β = P (Y T < 7. signal sent) = Φ((7. )/ ) = Φ(( 7.)/ ) =.44 4

23 X a) The coeicients h(t) are determined by h(t) = s T t. I T = we get the causal ilter 4 t =, h(t) = s t = t =, t =. For the case when signal is sent we get E[Y ] = E[ h(u)(s u + N u )] = u= u= h(u) = =, u= V [Y ] = V [ h(u)(s u + N u )] = V [ h(u)n u ] = = u= h(u) V [N u ] = ( )4 = 84. u= We get Y N(, 84) i no signal is sent and Y N(, 84) i signal is sent. The error probabilities become equal with the decision threshold k = =. and β = α = P (Y >. no signal sent) = Φ(. ).. b) The optimal coeicients are received rom the solution o s t = u= h(u)r N(t u) or t =,,, (c = ). 4 = 4h() + h(), = h() + 4h() + h(), = h() + 4h(), with solution h() =.9, h() =. and h() =.. E[Y ] = E[ h(u)(s u + N u )] = h()s + h()s + h()s = 4.4, u= V [Y ] = V [ h(u)(s u + N u )] = V [ h(u)n u ] = u= u= u= v= = 4(h() + h() + h() ) + h()h() + h()h() = 4.4. The error probabilities with the threshold k = E(Y ) =. become h(u)h(v)r N (u v) = β = α = P (Y >. no signal sent) = Φ(. 4.4 ).47. I we instead had used the ilter rom a) (which not is optimal any longer as the noise is not white) we receive or signal sent E[Y ] = E[ h(u)(s u + N u )] = h()s + h()s + h()s =, u= V [Y ] = V [ h(u)(s u + N u )] = V [ h(u)n u ] = u= u= u= v= = 4(h() + h() + h() ) + h()h() + h()h() = 4. The error probabilities with the threshold k = E(Y ) =. become β = α = P (Y >. no signal sent) = Φ(. 4 )., which is not much, but still larger than the optimal value.47. h(u)h(v)r N (u v) =