CU-Boulder MCEN 3037 - 24 (16 pages)

Previewing pages 1, 2, 3, 4, 5 of 16 page document View the full content.
View Full Document

24



Previewing pages 1, 2, 3, 4, 5 of actual document.

View the full content.
View Full Document
View Full Document

24

91 views


Pages:
16
School:
University of Colorado at Boulder
Course:
Mcen 3037 - Data Analysis
Data Analysis Documents

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



View Full Document

Access the best Study Guides, Lecture Notes and Practice Exams

Loading Unlocking...
Login

Join to view 24 and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view 24 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?