Πανεπιστήµιο Κρήτης - Τµήµα Επιστήµης Υπολογιστών ΗΥ-570: Στατιστική Επεξεργασία Σήµατος 2015 ιδάσκων : Α. Μουχτάρης εύτερη Σειρά Ασκήσεων Λύσεις Ασκηση 1. 1. Consder the gven expresson for R 1/2 : R 1/2 = q q H R 1/2 R 1/2 = q q H j=1 j q q H R 1/2 R 1/2 = j=1 j q q H q q H If λ 1, λ 2,..., λ M correspond to dstnct egenvalues of the correlaton matrx R, then the egenvectors are orthogonal and q H q j = 1 f = j and 0 f j. So: R 1/2 R 1/2 = λ q q H = R showng that the ntended summaton expresson for R 1/2 when squared does produce R. 2. To fnd the square root of matrx R we proceed as follows: (a) Compute the egenvalues and egenvectors of matrx R (b) Take the square root of each egenvalue of matrx R (c) Compute the square root of matrx R usng the expresson of the prevous part,.e., R 1/2 = q q H Ασκηση 2. 1. Let U 1 and U 2 be untary matrces that s U H = U 1. Then, (U 1 U 2 ) H = U H 2 U H 1 = U 1 2 U 1 1 = (U 1 U 2 ) 1,
ΗΥ-570: Στατιστική Επεξεργασία Σήµατος εύτερη Σειρά Ασκήσεων Λύσεις 2015 2 showng that U 1 U 2 s untary. 2. Let V = U 1 wth U a untary matrx. Snce V H = (U 1 ) H = (U H ) H = U, we have V V H = U 1 U = I, showng that V s untary. Ασκηση 3. From the factorzaton that s gven A n k n I = Q n L n L n = Q 1 n (A n k n I) L n = Q H n (A n k n I) (1) snce Q n s untary. From the defnton of matrx A n+1 : A n+1 = L n Q n + k n I (1) = A n+1 = Q H n (A n k n I)Q n + k n I A n+1 = Q H n A n Q n k n Q H n IQ n + k n I A n+1 = Q H n A n Q n Ασκηση 4. 1. From the gven expresson for v we have v H v 1 = 1 and v H v j = 0, f j, thus we compute c (n) as c (n) = v H u(n) for 0 M 1. 2. Consder E[c (n)c j(n)] = E[v H u(n)(v H j u(n)) ] = E[v H u(n)u(n) H v j ] = v H E[u(n)u(n) H ]v j = v H Rv j 0, snce R s postve semdefnte. We have assumed that u(n) s zero mean. Thus the Fourer coeffcents are correlated 3. From the above we see that E[ c (n) 2 ] = v H Rv, or the power n the th Fourer mode.
ΗΥ-570: Στατιστική Επεξεργασία Σήµατος εύτερη Σειρά Ασκήσεων Λύσεις 2015 3 Ασκηση 5. Recall that the condton number of a matrx A s gven by χ(a) = A A 1. If we use the spectral norm for the defnton of the norm, then one way to express ths norm s A 2 s = max Ax 2 x 2. Consder the matrx A but multpled by a untary matrx U or the matrx UA. We wll show that A s = UA s and A 1 s = (UA) 1 s from whch we can conclude that χ(a) and χ(ua) are the same. To begn note that UAx 2 = (UAx) H (UAx) = x H A H U H UAx = x H A H Ax = Ax 2. Thus UA 2 s = max UAx 2 x 2 = max Ax 2 x 2 = A 2 s Lets now consder (UA) 1 2 s. We have (UA) 1 2 s = max A 1 U 1 x 2 x 2 = max A 1 U H x 2 x 2 To smplfy the above note that U H x 2 = (U H x) H (U H x) = x H UU H x = x H x = x 2, thus the norm of x s not changed by applyng a untary transformaton to t. We can use ths to replace the denomnator n the expresson for (UA) 1 2 s above and get (UA) 1 2 s = max A 1 U H x 2 U H x 2 = max A 1 u 2 u 2 = A 1 2 s, by replacng the maxmzaton over x wth a maxmzaton over u U H x snce U H s an nvertble transform. Combnng these two results gves the desred result.
ΗΥ-570: Στατιστική Επεξεργασία Σήµατος εύτερη Σειρά Ασκήσεων Λύσεις 2015 4 Ασκηση 6. If we take the functon g to be g(x) = ln x then the left-hand sde of the gven expresson s: lm M ( 1 M ) ln(λ ) = lm M ln ( M ) 1/M λ. Takng exponentals of ths expresson and the rght-hand sde (the ntegral of g (S (ω))) gves lm M ( M ) 1/M λ ( 1 π ) = exp ln(s(ω)) dω. 2π π Snce the determnant of R s equal to the product of the egenvalues we have shown the desred expresson. Ασκηση 7. 1. We are told to consder Rw = p and R + δr)(w + δw) = p. Expandng the left-hand sde of the second equaton gves Rw + Rδw + δr(w + δw) = p Usng Rw = p n the above gves Rδw + δr(w + δw) = 0 Solvng for δw we get δw = R 1 δr(w + δw). Takng the vector norms on both sdes we get δw R 1 δr ( w + δw ) = R 1 ( ) δr ( w + δw ) = χ(r) δr w +χ(r) δr δw,
ΗΥ-570: Στατιστική Επεξεργασία Σήµατος εύτερη Σειρά Ασκήσεων Λύσεις 2015 5 usng the defnton of the condton number χ(r) R 1. When we solve for δw n the above we get ( 1 χ(r) δr ) ( ) δr δw χ(r) w. It can be shown that f our matrx perturbaton δr s small enough so that the new matrx R + δr s stll nvertble then δr 1. In that case left-hand sde has a leadng coeffcent that s postve χ(r) and we can dvde by t to get δw w χ(r) 1 χ(r) δr δr If we assume that χ(r) δr 0 then we get δw w χ(r) δr whch s the desred expresson. 2. We are told to consder Rw = p and R(w + δw) = p + δp. Expandng the left-hand sde of the second equaton and usng Rw = p gves Rδp = δp, or solvng for δw we get δw = R 1 δp. (2) Takng vector norms on both sdes we get δw R 1 δp = χ(r) δp, usng the defnton of the condton number χ(r) R 1. 1 w so w and we get p Snce Rw = p we have p
ΗΥ-570: Στατιστική Επεξεργασία Σήµατος εύτερη Σειρά Ασκήσεων Λύσεις 2015 6 δw χ(r) δp p w, or δw w χ(r) δp p, whch s the desred expresson. Ασκηση 8. When we apply low-rank modelng the output vector at the recever s y ndrect = c (n)q + u (n)q. Then snce u(n) = c (n)q we have the error gven by [ ɛ ndrect = E y ndrect (n) u(n) 2] = E =p+1 c (n)q + 2 u (n)q = E 2 d (n)q, where we have defned d (n) as u (n) d (n) = c (n) 1 p p + 1 M Thus evaluatng ɛ ndrect we have ɛ ndrect = E d (n)d j (n)q H q j. j=1 Unless = j the term n the above summaton vanshes due to the orthogonalty of q and q j and we get ɛ ndrect = E [d (n)d (n)] = E [u (n)u (n)] + E [c (n)c (n)] = =p+1 σ 2 + λ = λ + pσ 2, =p+1 =p+1 as we were to show.