Dynamics. Today: Linear System Dynamics. EEL5225: Principles of MEMS Transducers (Fall 2003) Instructor: Dr. Hui-Kai Xie

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1 Today: Instructor: Dr. Hui-Kai Xie Dynamics Linear System Dynamics Direct integration System functions Superposition Convolution Sinusoidal steady state Reading: Senturia, pp. 7-8, Chapter 7, pp Lecture 3 by H.K. Xie 0/7/003

2 Linear Dynamics First power of state variables Dynamics Powerful analytical techniques available Laplace transform, Fourier transform, convolution, superposition, eigenfunction analysis, etc. Nonlinear Dynamics Nonlinear dependence on state variables Difficult to solve analytically Approimate methods to evaluate behavior near operating point (linearization)

3 Linear System Dynamics Eample: Mass-spring-damper system: Solved previously using LEM i e = 0 = F + e + e + e or i k m m dv F = M + bv + k v dt or dt F = Ma+ bv+ k Define state variables = = (position) (velocity) Ref. Senturia, Microsystem Design, p

4 Linear System Dynamics Eample: Mass-spring-damper system: Obtain state equations F = ma + bv + k F = m +b +k F = m + b + k = k b = + F m m m 4

5 = k b = + F m m m Linear System Dynamics 0 0 k b = + m m m F y [ 0] 0 0 = + F 5

6 Linear System Dynamics Compact Form = A+ Bu where =state variable vector y = C+ Du Eample u=system input vector y=output vector = F =,, F y = u = k b = + F m m m 0 A = k b m m 0 B = m C = [ 0 ] 0 D = 0 6

7 Direct Integration Time Response Via Numerical Integration Use numerical solver (Matlab, Mathcad, Mathematica, etc.) Simulink: state-space Eample: Step response (position, velocity) m=, k=, b=0.5 (MKS units) 7

8 Laplace Transform Definition (single-sided) 0 System Functions st L{ f()} t = f() t e dt F( s) where s= α +j ω. Apply to state equations L{} = L{ A+ Bu} sx() s (0) = AX() s + BU() s Rearranging, ( ) ( ) X() s = si A (0) + BU() s = A+ Bu y = C+ Du Initial state 8

9 System Functions ( ) ( ) X() s = si A (0) + BU() s Split system response into two parts () Zero-input response ( natural response) Let U(s)=0. ( ) Xzir () s = si A (0) () Zero-state response Let (0)=0. ( ) Xzsr () s = si A BU() s 9

10 System Functions Laplace Transform of System Output L( y) = L( C+ Du) Y() s = C[ X () s + X ()] s + DU() s Three parts zir zsr () Zero-input response () Zero-state response (3) Feedthrough from input Y () s = C[ X ()] s zsr zsr [ ] Y () s = C U() s zsr ( ) H() s = Yzsr () s U() s = C si A B H():system s function or transfer function 0

11 System function System Functions Number of rows equal to number of state variables Number of columns equal to number of inputs ( ) H ( s) = C si A B General form H ij ( s) = Num Den(s) Roots of Den(s) are natural frequencies or poles of system Roots of Num ij (s) are zeros of system ij Correspond to input time-dependence that yield no output

12 System Functions: Eample Mass-spring-damper system 0 0 k b u = + m m m 0 y = [ ] u where [ ] selects and as outputs. + 0 C = Taking the Laplace transform, ms + bs + k H() s = C( si A) B= s ms + bs + k X ( s) F ( s) X ( s) F ( s)

13 Roots of System Function System Functions H(s) to represent in terms of roots (poles and zeros) m s s s s H() s = = s s ( )( ) ms + bs + k ( )( ) ms + bs + k m s s s s Velocity has zero at s=0 Position and velocity response has two poles given by s, b b k = ± m m m 3

14 System Functions Physics of poles and zeros Undamped resonant frequency k ω 0 = m Damping constant b α = m Poles in terms of ω 0 and α Viscous damping coefficient b b α ξ = = = mk mω ω 0 0 s b b k, 0 = ± = α ± α ω = α ± jωd m m m Damped resonant frequency: ω = ω α = ω ξ d 0 0 4

15 General Solutions System Functions If ω > α ξ >, the response is underdamped. 0 or αt t () = Ae cosω t+ Ae sinω t αt d d If ω0 = α or ξ =, the response is critically damped. t () = Ate + Ae α t α t If ω0 < α or ξ <, the response is overdamped. t () = Ate + Ae st s t 5

16 System Functions Eample m=, k=, b=0.5 k b ω0 = = rad/s; α = = 0.5 and m m s, 0, ( ) ( ) = α ± α ω = 0.5 ± 0.5 s = 0.5 ± j = α ± jω where ω is the damped resonant frequency. d d Underdamped Step Response 6

17 System Functions Pole-zero diagram Underdamped Step Response Given m=, k=, b=0.5, 0 ( ) s = α ± α ω = 0.5 ± j = σ ± jω d 7

18 System Functions Quality Factor ω0 mω 0 Q= α = Q = b ξ Eample: Q= (moderately underdamped) Three cases Underdamped Critically damped Overdamped Q > 0.5 ω mω α = b 0 0 Q= where α= ω0 and Q= Q < 0.5 8

19 Transfer Function Position: Frequency response H( ω) = / + / * H( ω) = H( ω) H ( ω) = ( ω/ ωn) + ξ( ω/ ωn) H( ω) = tan ( ω ω ) j ξ( ω ω ) n ( ω ωn ) ( ) / ξ ω/ ω For position, 40dB/Dec at high frequency! n n Homework: How fast does the velocity amplitude decay at high frequency? Ref. Senturia, Microsystem Design, p

20 Fourier Transform: Y ( ω) = H( jω) U( ω) zsr Eample: Assume ut () = U costω t 0 0 Sinusoidal Steady State [ ] U( ω) = U π δ( ω+ ω ) + δ( ω ω ) Y zsr H( s) = or H( jω) = + sτ + jωτ [ ( + ) + ( )] U0π δω ω0 δω ω0 ( ω) = + jωτ e e yzsr () t = + = U jω0t jω0t U cosω t ωτsinωt jωτ 0 jωτ 0 + ( ωτ 0 ) 0