Freq_Domain_Analysis_FRF

Freq_Domain_Analysis_FRF - Frequency Domain Analysis...

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Burdisso - ME3514 1 Frequency Domain Analysis • Response of a linear system to harmonic inputs. X(t) = transient response (t) + steady-state response (t) This part of the response is harmonic with the same frequency as the input, ω sin : , oo mx bx kx P t and IC x x ω ++= ±± ± ±
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Burdisso - ME3514 2 Frequency Domain Analysis • Response of a linear system to harmonic inputs. X(t) = transient response (t) + steady-state response (t) sin : , oo mx bx kx P t and IC x x ω ++= ±± ± ± Transient part will die out with damping This part of the response is harmonic with the same frequency as the input, ω Frequency Domain Analysis is the computation of the steady-state response under harmonic inputs.
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Burdisso - ME3514 3 Frequency Domain Analysis • Example () 2 2 sin 2/ 1 s e c 1./ nn mx bx kx P t kN m rad s T mN s m ω π ωπ + += = =→ = = ±± ± steady-state response (t) transient response (t) + steady-state response (t) n T
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Burdisso - ME3514 4 Frequency Domain Analysis • Steady-state solution = particular solution where To find A and Φ in the time domain is difficult. sin ss ss ss mx bx kx P t ω + += ±± ± ( ) () ( )s in ( ) ss xt A t ωφ =+ Amplitude of harmonic response Relative phase of harmonic response w.r.t input
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Burdisso - ME3514 5 Frequency Domain Analysis • Frequency domain solution for A and Φ 1. Assume input is: 2. Particular solution is: 3. Replacing into EOM: 4. The amplitude and phase are: 2 ˆ () it ss P xt X e e km i b ω == + f tP e = ˆ ss X e = ˆˆ Magnitude of and Phase of ss ss A xx φ =
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Burdisso - ME3514 6 Frequency Domain Analysis The amplitude and phase are: () 2 1 2 2 2 2 1 2 2 2 2 ˆ ˆ tan tan it ss i ss P xt X e e km i b Pe b x t Xe e with kmb Pb A
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This note was uploaded on 04/18/2008 for the course ME 3514 taught by Professor J.renno during the Fall '08 term at Virginia Tech.

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Freq_Domain_Analysis_FRF - Frequency Domain Analysis...

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