Midterm Solution

Expt t0 2 exp2t t0 1 t t0 3 expt t exp2t

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Unformatted text preview: matrix φ(t, t0 ) b) Find the impulse response h(t, t0 ) Solution: φ(t, t0 ) = exp(A(t − t0 )) = f (A) = f (1)Z10 + f (−2)Z20 A= 02 , 1 −1 p(λ) = (λ − 1)(λ + 2) Using the method of trial functions, f1 (λ) = λ − 1 yields A − I = −3Z20 , Z20 = 1 1 −2 3 −1 2 and f2 (λ) = λ + 2 gives A + 2I = 3Z10 , Z10 = 3 122 311 1(t−2) Thus, φ(t, t0 ) = 1 1 22 1 −2 + exp(−2(t − t0 )) exp(t − t0 ) 11 −1 2 3 3 2 exp(t − t0 ) + exp(−2(t − t0 )) 2 exp(t − t0 ) − 2 exp(−2(t − t0 )) 1 φ(t, t0 ) = 3 exp(t − t ) − exp(−2(t − t )) exp(t − t ) + 2 exp(−2(t − t )) 0 0 0 0 b) h(t, t0 ) = Cφ(t, t0 )B 1(t − t0 ) + Dδ (t − t0 ), therefore h(t, t0 ) = h(t − t0 ) = 1 4 exp(t − t0 ) − exp(−2(t − t0 )) 1(t − t0 ) 3 The same result could be obtained with φ(t) = exp(At) = ρ0 I + ρ1 A, 1 1 ρ0 = (2 exp(t) + exp(−2t)), ρ1 = (exp(t) − exp(−2t)) 3 3 Problem 5 Consider the following system x(k + 1) = 1 02 u( k ) x( k ) + 1 1 −1 y ( k ) = 1 0 x( k ) a) Find the discrete transition matrix φ(k, k0 ) b) Find the delta response h(k, k0 ) Solution: φ(t, t0 ) = Ak−k0 = f (A) = f (1)Z10 + f (−2)Z20 A= 02 , 1 −1 p(λ) = (λ − 1)(λ + 2) Using the method of trial functions, f1 (λ) = λ − 1 yields A − I = −3Z20 , Z20 = 1 1 −2 3 −1 2 and f2 (λ) = λ + 2 gives A + 2I = 3Z10 , Z10 = 4 122 311 Thus, φ(k, k0 ) = Ak−k0 = (1)k−k0 Z10 + (−2)k−k0 Z20 k − k0 2 − 2(−2)k−k0 1 2 + (−2) φ(k, k0 ) = 3 1...
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