solution08

# solution08 - 1(a In"standard form the defining differential...

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1/. (a) In "standard" form the defining differential equation is IN OUT OUT OUT V CL V CL R R V CL R L CR V 1 ) / 1 ( / 2 1 2 1 = + + + + This is a second order system with CL b CL R R a C R L R CL R L CR a 1 ) / 1 ( 1 / 2 1 0 2 1 2 1 1 = + = + = + = Applying the formulae for ω N , ζ , and k gives 1 2 2 2 1 2 1 2 2 1 2 1 1 2 1 0 / 1 1 / ) / 1 ( / 1 ) / 1 ( 2 / 1 / 2 ) / 1 ( R R R R R CL R R CL b k CL R R C R L R a CL R R a N N N + = + = + = = + + = = + = = ω ζ The value of k reflects the fact that, in the steady state, the circuit acts as a voltage divider. (b) For the values given ω N , = 7.10669, ζ = 0.3589 and k = 0.99. Because ζ is less than 1, the system is underdamped. The response of an underdamped second order system to initial conditions is ) 1 sin( 1 ) 0 ( ) 0 ( ) 1 cos( ) 0 ( ) ( 2 2 2 t e y y t e y t y N t N N N t N N ζω + + = In this case ) 0 ( y is zero. Plugging in values and rearranging gives () ) 633 . 6 sin( ) 633 . 6 cos( 3845 . 0 24 ) ( 55 . 2 t t e t V t OUT + =

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2/. use logarithmic decrement ω N = 1.575 rads/sec ζ = 0.08 spring constant = mass * sqr( ω N ) = 4.9666 N/m damping co-efficient = ζ * 2 * sqrt(spring constant * mass) = 0.5043 N / (m/sec)
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## This note was uploaded on 07/16/2009 for the course SYSC 3600 taught by Professor John bryant during the Winter '08 term at Carleton CA.

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solution08 - 1(a In"standard form the defining differential...

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