Midterm Solution

# 3z10 z10 5 122 311 therefore f exp 12 t t2

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Unformatted text preview: (−2)k−k0 1 + 2(−2)k−k0 b) h(k, k0 ) = h(k − k0 − 1) = Cφ(k − k0 − 1)B 1(k − k0 − 1) + Dδ (k − k0 ) h(k − k0 ) = 1 4 − (−2)k−k0 −1 1(k − k0 − 1) 3 Problem 6 Consider the following system 1 0 2t u x+ t t −t x= ˙ y= 1 0 x a) Find the transition matrix φ(t, t0 ) b) Find the impulse response h(t, t0 ) ˆ Solution: In this case, A(t) = tA, with 02 ˆ , A= 1 −1 p(λ) = (λ − 1)(λ + 2) ˆ φ(t, t0 ) = exp β (t, t0 )A , t β (t, t0 ) = t0 φ(t, t0 ) = exp 1 τ dτ = (t2 − t2 ) 0 2 12 ˆ ˆ (t − t2 )A = f (A) = f (1)Z10 + f (−2)Z20 0 2 Using the method of trial functions, f1 (λ) = λ − 1 yields ˆ A − I = −3Z20 , Z20 = 1 1 −2 3 −1 2 and f2 (λ) = λ + 2 gives ˆ A + 2I = 3Z10 , Z10 = 5 122 311 Therefore, f (λ) = exp 12 (t − t2 )λ , 0 2 ˆ f (A) = exp 12 ˆ (t − t2 )A 0 2 and φ(t, t0 ) = exp 12 1 1 ˆ (t − t2 )A = exp (t2 − t2 )(1) Z10 + exp (t2 − t2 )(−2) Z20 0 0 0 2 2 2 1 2 exp (t2 − t2 ) + exp − (t2 − t2 ) 0 0 1 2 = 1 3 exp (t2 − t2 ) − exp − (t2 − t2 ) 0 0 2 12 (t − t2 ) − 2 exp − (t2 − t2 ) 0 0 2 12 2 2 2 1 exp (t − t0 ) + 2 exp − (t − t0 ) 2 2 exp b) h(t, t0 ) = C (t)φ(t, t0 )B (t0 )1(t − t0 ) + D(t)δ (t − t0 ) h(t, t0 ) = 1 1 2 exp (t2 − t2 ) + exp − (t2 − t2 ) 0 0 3 2 + 2 exp 12 (...
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## This document was uploaded on 03/23/2014 for the course EL 6253 at NYU Poly.

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