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Unformatted text preview: Solutions to HW 2 Exercise 3.9 (a) X t = 0.3 X t 1 + Z t (1  0.3B ) X t = Z t . B >1 stationary. X t = Z t /(1  0.3B ) = (1 + 0.3B + 0.32 B 2 + ) Z t = Z t + 0.3Zt 1 + 0.32 Z t 2 + = (0.3)i Z t i
i =0 (b) X t = Z t  1.3Z t 1 + 0.4 Z t  2 X t = (1  1.3B + 0.4 B 2 ) Z t . B = 1.25, 2 > 1 invertible. (1  0.5 B ) X t = (1  1.3B + 0.4 B 2 ) Z t ,
stationary and invertible. (c) X t = 0.5 X t 1 + Z t  1.3Z t 1 + 0.4 Z t  2 Problem 1 Stationary: E [ X t ] = cos(ct ) E [ Z1 ] + sin(ct ) E [ Z 2 ] = 0 ,
2 (h) = E [ X t X t + h ] = cos(ct ) cos(c(t + h)) E[ Z12 ] + sin(ct )sin(c(t + h)) E[ Z 2 ] + + {cos(ct )sin(c(t + h)) + sin(ct )cos(c(t + h))}E[ Z1Z 2 ] = = 2{cos(ct ) cos(c(t + h)) + sin(ct )sin(c(t + h))} = 2 cos(ch). Problem 2 a) b) ( B ) = 1 + .2 B  .48 B 2 = 0, B1 = 1.25, B2 = 1.6667 ( B ) = 1 + 1.9 B + .88B 2 = 0, B1 = 1.25, B2 = .909 2 causal. noncausal; invertible. ( B) = 1 + .2 B + .7 B = 0, B1,2 = 0.143 1.187i,  B1,2 = 1.195 c) ( B) = 1 + .6 B = 0, B = 1.667 causal; ( B) = 1 + 1.2 B = 0, B = 0.833 noninvertible. ( B) = 1 + 1.8 B + .81B 2 = 0, B1,2 = 1.11 ( B) = 1 + 1.6 B = 0, B1 = .625 ( B) = 1  .4 B + .04 B = 0, B1,2 = 5
2 d) e) causal. noncausal; invertible. ...
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This note was uploaded on 09/23/2009 for the course MATH 200 taught by Professor Gur during the Spring '09 term at Accreditation Commission for Acupuncture and Oriental Medicine.
 Spring '09
 gUR

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