CU-Boulder MCEN 3037 - 24 (16 pages)
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- Pages:
- 16
- School:
- University of Colorado at Boulder
- Course:
- Mcen 3037 - Data Analysis
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Unformatted text preview:
INPUT sensor transducer Signal A MEASUREMENT SYSTEM Signal B OUTPUT Signal C Zero order responds instantly to measurand no instrument is truly zero order but many can be approximated as zero order First order show capacitive type energy storage effects mechanical analogs to capacitors are springs and devices that store thermal energy thermometer is a first order system Second order have inertial effects of inductance or accelerated mass as well as capacitive effects common spring mass systems bathroom scale Second order systems include a characteristic called damping or energy loss Covers systems that possess inertia accelerometers loudspeakers pressure transducers d2y dy Here s the equation a2 2 a1 a0 y F t dt dt We ll see this more often in the standard form 1 n a0 a2 2 y n 2 n y y kF t a1 2 a0 a2 1 k a0 Let s examine the solution to this equation 1 2 y y y kF t 2 n n In the absence of a forcing function we have the homogeneous equation y 2 n y n2 y 0 This has the solution yh t Ce t If we plug this back in we get the characteristic equation 2 2 n 2n 0 This quadratic equation has two roots given by 1 2 n n 1 2 The nature of the solution depends on this 1 2 n n 1 2 yh t C1e 1t C2e 2t Example Under damped Case 0 1 1 2 n i n 1 2 1 2 n n 1 2 We can plug this in above y h t C1 e i 1 n n 2 t C2e i 1 n n 2 t We can rewrite this as yh t e nt B1 sin d t B2 cos d t We see that the homogeneous transient solution is a damped sine wave d n 1 2 Damped natural frequency In the absence of damping the response oscillates at n yh t e nt C1 C2 n t Critically Damped Case 1 Over damped Case 1 y h C1 e 2 1 t n C2e 2 1 t n 1 2 y n 2 n y y kF t Lets consider a step function input F t AH t We have for t 0 1 2 n y 2 n y y kA homogeneous particular The total solution is given by yT yh y p Here y p kA Let s take the case of a critically damped system as an example yh t e n t C1 C2 n t yT kA e nt C1 C2 nt Taking the initial conditions as y t 0 dy t 0 0 dt We get yT kA kA 1 nt e nt Critically
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Study Guide: Midterm 2
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