Chapter 4

Chapter 4 - EEL4657 Dr Haniph Latchman Chapter 4 Solutions...

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Unformatted text preview: EEL4657 - Dr. Haniph Latchman Chapter 4 Solutions 1. The second order system is given by: G ( s ) = ω 2 n s 2 + 2 ζs + ω 2 n The impulse response is determined by: G ( s ) = ω 2 n s 2 + sξω n s + ω 2 n = ω n p 1- ξ 2 · ω n p 1- ξ 2 ( s + ξω n ) 2 + ω 2 n (1- ξ 2 ) g ( t ) = ω n p 1- ξ 2 e- ξω n t sin( ω n p 1- ξ 2 t ) = ω n β e- ξω n t sin( ω n βt ) and set β = p 1- ξ 2 θ = tan- 1 β ξ 1 The step response is the integral of the impulse response y ( t ) = Z t ω n β e- ξω n τ sin( ω n βτ ) dτ =- 1 ξβ Z t sin( ω n βτ ) de- ξω n τ =- 1 ξβ e- ξω n τ sin( ω n βτ ) t + ω n ξ Z t e- ξω n τ cos( ω n βτ ) dτ =- 1 ξβ e- ξω n t sin( ω n βt )- 1 ξ 2 Z t cos( ω n βτ ) de- ξω n τ =- 1 ξβ e- ξω n t sin( ω n βt )- 1 ξ 2 e- ξω n τ cos( ω n βτ ) t- β 2 ξ 2 Z t ω n β e- ξω n τ sin( ω n βτ ) dτ =- 1 ξβ e- ξω n t sin( ω n βt )- 1 ξ 2 e- ξω n t cos( ω n βt ) + 1 ξ 2- β 2 ξ 2 y ( t ) = 1- 1 β e- ξω n t [ ξ sin( ω n βt ) + β cos( ω n βt )] Y STEP ( t ) = 1- 1 β e- ξω n t sin( ω n βt + θ ) 2. The impulse response of G ( s ) = ω 2 n s 2 + 2 ζs + ω 2 n 2 is given by G ( s ) = ω 2 n s 2 + sξω n s + ω 2 n = ω n p 1- ξ 2 · ω n p 1- ξ 2 ( s + ξω n ) 2 + ω...
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This note was uploaded on 02/01/2012 for the course EEL 4657 taught by Professor Latchman during the Fall '10 term at University of Florida.

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Chapter 4 - EEL4657 Dr Haniph Latchman Chapter 4 Solutions...

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