Chapter SDOF Vibration Control.1 Transfer Function mx ɺɺ( t) + cxɺ ( t) + kx( t) = F( t) Defines the transfer function as output over input X ( s) 1 = G( s) = (1.39) F( s) ms + cs + k s is a complex number: s = α ± jβ = ζω ± jω n d 1
Complex s-plane of the Poles ω n ζω n jω n ω d Poles: the roots of the denominator of G(s): σ ms cs k + + = s = ζω ± jω 1 ζ 1, n 0 n
Block Diagram Representation X(s) U(s) = 1 ms + cs + k U(s) 1 ms + cs + k X(s) Input Plant or Structure Output SDOF General System 3
Frequency Response Function (FRF) 4
Bode Plots: Magnitude and Phase Fig 1.15, phase vs frequency for various damping ratios Fig 1.16, magnitude vs frequency for various damping rations 5
. Measurement and Testing Damping must be measured dynamically. Mass and stiffness can be measured in static experiments. Free decay allows the exponent to be measured, and hence the damping ratio. 6
Damping Measured by Time Response Definition of log decrement: δ = ln x( t) x( t + T ) d δ = ln e sin( ω t + φ) ζωnt d ζωn ( t+ Td ) e dt dtd sin( ω + ω + φ) ζωntd = ln( e ) = ζω T (1.46) n d δ = ζω n T d = πζ (1.47) 1-ζ δ ζ = (1.48) 4π + δ 7
Frequency Domain Estimates of Mass, Damping and Stiffness log G( jω) ω 0 1 1 ω ζω 1 = log log 1 log (1.49) + = k ωn ωn k 8
Damping Measured by Frequency Response Function (FRF) At the half-power points ζ 1 ω ω 1 = ωd 9
.3 system Stability Stability is defined for the solution of free response case: Stable: x( t) < M, t > 0 Asymptotically Stable: lim x( t) = 0 t Unstable: if it is not stable or asymptotically stable 10
Stability x t sin. 0.1. t t y t e.. x t z t e 0.1 t r t z t. x t Stable 1 1 Asymptotically Stable x t 0 5 10 y t 0 5 10 3 1 t 1 Fig 1.19 t Fig 1.0 4 z t r t 0 5 10 1 0 5 10 4 t t Divergent instability Flutter instability 11
.4 Vibration Control Design refers to choosing m, c and k in order to produce a more desirable response (or ζ and ω n ). Often this involves working with the spring constants of various materials. If vibration suppression is the goal, then situations arise when adjusting m, c and k still does not produce the desired response, then control methods are used. 1
Shape the Response by m, c, and k Design a system so that its response as a desired settling time, overshoot and time to peak. Fig 1.10 OS == e t t p s = ω n 3. = ω ζ n ζπ 1 ζ π 1 ζ fixes ζ fixes ω n is determined 13
Vibration Control If m, c and k cannot be adjusted to suit design goals, then control is a possibility Control consists of adding hardware to affect the response in some way Passive control: a fixed device or permanent change in physical parameters (such as constrained layer damping) Active control: use of some external adjustable or active (electronic) device, called an actuator, to provide a means of shaping or controlling the response 14
Active Control Open Loop Control: the control force applied to the system is independent of any measurement. Closed Loop Control: the control force depends in some way on a measurement of the system (requires a sensor). 15
Open Loop Control U(s) K G(s) X(s) Choose K such that X(s) has a desired shape, called constant gain (K) control. X ( s) U ( s) = KG( s) = ms K + cs + k mx ɺɺ( t) + cxɺ ( t) + kx( t) = Ku( t) 16
Closed Loop Control F(s) + - K G(s) structure X(s) H(s) Control law X(s) F(s) = KG(s) 1+ KG(s)H(s) ( ) 17
State Feedback State feedback: X(s) F(s) = X(s) F(s) = KG(s) 1+ KG(s)H(s) ( ) H(s) = g 1 s + g K ms + ( Kg 1 + c)s + Kg + k 1 ( ) mx ɺɺ( t) + ( Kg + c) xɺ ( t) + ( Kg + k) x( t) = KF( t) Now we have two more parameters to adjust in order to meet the desired performance. 18
Feedforward Control disturbance system filter Error signal Mostly used in acoustics and high frequency applications. 19
An Inverted Pendulum (Example) kl θ ( ) θ ( ) ( ) ml ɺɺ t + mg t = F t Need to find bounded F(t) Choose F( t) = aθ ( t) b ɺ θ ( t), then kl ( ) + ( ) + + ( ) = 0 ml ɺɺ θ t bɺ θ t mgl a θ t AS and BIOB if b > 0, and kl mgl + a > 0 AS: asymptotic stable BIOB: bounded input and bounded output 0
.5Vibration of Nonlinear Systems ɺɺ x( t) + f ( x, xɺ ) = 0 k = 1000, b = 10 k 1000 b 10 g( x) = g x k. x f x k. x b. x 3 h x k. x b. x 3 3 f ( x) = k x kx bx h( x) = kx + bx f x 3 5000 3750 500 150 Hardening Spring Softening Spring g x h x 4 0 4 150 Linear Spring 500 3750 5000 x 1
State Space Formulation x 1 and x are called state variables 0 ) ( by Define the equilibrium point ) ( ), ( 1 1 = = = = e e F x F x x f x x x x x x ɺ ɺ ɺ
A Nonlinear System of Soft Spring Compute the equilibrium for: ɺɺx x β x 3 + = 0 First order form: xɺ = x 1 xɺ = x ( β x 1) 1 1 3
Equilibrium points x = 0 x (β 1 x 1 1) = 0 = 0 1 x 0, e β, 0 1 β 0 4
Multiple Equilibrium Each equilibrium has a potentially different solution with a different stability behavior. One equilibrium may be stable another not. Depending on the initial conditions, the solution may favor the response of one equilibrium versus another so that the concept of stability must be associated with the equilibrium. 5