Midterm Solution

# x1 k 1 ku 3x2 k a 0 3 1 k 12 b k 1 c

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: u + ku y =E −2 E u − k 2 Ey + ku − 3y =E −1 u − E −1 (k 2 Ey ) + E −1 (ku − 3y ) =E −1 u − (k − 1)2 y + E −1 (ku − 3y ) x2 (k ) = y, x2 (k + 1) = u − (k − 1)2 x2 (k ) + x1 (k ) x1 (k + 1) = ku − 3x2 (k ), A= 0 −3 , 1 −(k − 1)2 B= k , 1 C= 0 1 , D= 0 , c) y + t2 y + 3y = u + tu = D2 y + t2 Dy + 3y = Du + tu ¨ ˙ ˙ y =D−2 Du − t2 Dy + tu − 3y =D−1 u − D−1 (t2 Dy ) + D−1 (tu − 3y ) D−1 (t2 Dy ) = t2 y − D−1 (2ty ) y = D−1 u − t2 y + D−1 (tu + 2ty − 3y ) x2 = y, A= x1 = tu + 2tx2 − 3x2 , ˙ 0 2t − 3 , 1 −t2 B= t , 1 x2 = u − t 2 x2 + x1 ˙ C= 0 1 , D= 0 , Problem 3 A ﬁxed linear system has the zero state impulse response h(t) = t exp(−2t)1(t). Find the zero state response of the system to the input u(t) = 1(t) − 1(t − 2) Solution: The zero state step response is given by 2 0 0 −∞ 1 τ exp(−2τ )dτ = − exp(−2τ )τ 2 1 = − t exp(−2t) + 2 1 = − t exp(−2t) + 2 a( t ) = τ exp(−2τ )dτ τ exp(−2τ )1(τ )dτ = 1(t) h(τ )dτ = −∞ t t t t a( t ) = − t t 0 − 0 1 − exp(−2τ )dτ 2 1 − exp(−2τ )τ 4 t 0 1 (1 − exp(−2t)) 4 1 t exp(−2t) + (1 − exp(−2t)) 1(t) 2 4 The response to the input u(t) = 1(t) − 1(t − 2) is given by a(t) − a(t − 2), i.e., 1 (t − 2) 1 t exp(−2(t−2))+ 1−exp(−2(t−2)) − exp(−2t)+ (1−exp(−2t)) 1(t)− − 2 4 2 4 Problem 4 Consider the following system x= ˙ 1 02 u x+ 1 1 −1 y= 1 0 x a) Find the transition...
View Full Document

## This document was uploaded on 03/23/2014 for the course EL 6253 at NYU Poly.

Ask a homework question - tutors are online