Note#20 Summary of 2nd order systems

# Note#20 Summary of 2nd order systems - 10 2 s s or 10 100 =...

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Note #20. Pole locations for Second Order Systems ECSE-2410 Signals & Systems - Fall 2008 The general transfer (system) function, ( ) s H , for a second order system is () 2 2 2 2 n n n s s s H ωζ ω + + = , where the two parameters are ζ Æ damping factor n Æ natural frequency. Factoring the denominator results in roots (poles), 2 1 ζω ± = n n j s . These poles are plotted in a polar coordinate system: Note #20 p.1 Keeping n constant and varying from 0 to 1 > , the pole locations of () H and the resulting time responses change as shown below. Circles of constant radius m e θ = cos 2 1 n n m e n n Lines of constant Impulse Response m e damped) ( 0 un = damped) ( 1 0 under < < (2) damped) ( 1 critically = damped) ( 1 over > (slow decay)

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Example. Construct the pole-zero plot of () 100 10 100 2 + + = s s s H . Find ζ , n ω and classify the damping. Finally, determine the impulse response, ( ) t h , of this system. Equating like powers of in the denominators of s ( ) s H above and of the general second-order form, () 2 2 2 2 n n n s s s H ωζ + + = , we get = 2 2 2 n n s s + + 100
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Unformatted text preview: 10 2 + + s s or 10 100 = = n 10 2 = n ζω Æ 5 . 10 2 10 = ⋅ = . Consequently, the system is underdamped. Note #20 p.2 The poles of the system are 3 5 5 4 3 10 ) 10 )( 5 . ( 1 , 2 2 1 j j s s n n ± − = ± − = − ± − = . The impulse response of this system is simply the inverse Laplace transform of ( ) 100 10 100 2 + + = s s s H . In this case, we use the transform pair, ( ) ( ) ( t u t e s t o 2 2 sin ωα ) α − ↔ + + . Thus completing squares, ( ) ( ) ( ) s s s s H 3 5 5 3 5 3 20 100 10 100 2 2 2 ↔ + + ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = + + = ( ) ( ) ( ) t u t e t h t 3 5 sin 3 20 5 − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = o 60 5 . cos cos 1 1 = = = − − ζθ 3 5 1 2 = − n o 60 m ℑ e ℜ 10 = n 5 − = − n...
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Note#20 Summary of 2nd order systems - 10 2 s s or 10 100 =...

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