07 Lecture 10

8/8/2019 07 Lecture 10

http://slidepdf.com/reader/full/07-lecture-10 1/69

Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MITOpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].

JV: 6.777J/2.372J Spring 2007, Lecture 10 - 1

Lumped-Element System Dynamics

Joel Voldman*

Massachusetts Institute of Technology

*(with thanks to SDS)

8/8/2019 07 Lecture 10




Outline

> Our progress so far

> Formulating state equations

> Quasistatic analysis

> Large-signal analysis

> Small-signal analysis

> Addendum: Review of 2nd-order system dynamics

8/8/2019 07 Lecture 10




Our progress so far…

> Our goal has been to model multi-domain systems

> We first learned to create lumped models for each

domain

> Then we figured out how to move energy between

domains

> Now we want to see how the multi-domain systembehaves over time (or frequency)

8/8/2019 07 Lecture 10





> The Northeastern/ADI RF Switch

> We first lumped the mechanical domain

Images removed due to copyright restrictions.

Figure 11 on p. 342 in: Zavracky, P. M., N. E. McGruer, R. H. Morrison,and D. Potter. "Microswitches and Microrelays with a View Toward MicrowaveApplications." International Journal of RF and Microwave Comput-AidedEngineering 9, no. 4 (1999): 338-347.

Fixed plate

Fixed support

Dashpot b

g V

I

+

- z

Spring k

Mass m

Adapted from Figure 6.9 in Senturia, Stephen D. Microsystem Design . Boston, MA: Kluwer Academic Publishers, 2001, p.

138. ISBN: 9780792372462.

Silicon0.5

µm

1 µm

Pull-down

electrode

Cantilever

Anchor

Image by MIT OpenCourseWare.

Adapted from Rebeiz, Gabriel M. RF MEMS: Theory, Design, and

Technology . Hoboken, NJ: John Wiley, 2003. ISBN: 9780471201694.


8/8/2019 07 Lecture 10





> Then we introduced a two-port capacitor to convert

energy between domains• Capacitor because it stores potential energy

• Two ports because there are two ways to store energy

» Mechanical: Move plates (with charge on plates)

» Electrical: Add charge (with plates apart)

• The system is conservative: system energy only depends on

state variables

I

+

-

V C

1/kg

F m

b

+

-

.

AgQgQW

ε 2),(

2

=

8/8/2019 07 Lecture 10





> We first analyzed systemquasistatically

> Saw that there is VERY different

behavior depending on whether• Charge is controlled

stable behavior at all gaps

• Voltage is controlled pull-in at g=2/3g0

> Use of energy or co-energy

depends on what is controlled• Simplifies math

FdgVdQdW +=

A

Q

g

gQW

F Q

ε 2

),( 2

=∂

∂=

A

Qg

Q

gQW V

gε

=∂

∂=

),(

FdgQdV dW −=

*

2

2*

2g

AV

g

W F in

V

ε =

∂

∂=

in

g

V

g

A

V

W Q

ε =

∂

∂=

*

8/8/2019 07 Lecture 10




Today’s goal

> How to move from quasi-static to dynamic analysis

> Specific questions:

• How fast will RF switch close?

> General questions:

• How do we model the dynamics of non-linear systems?

• How are mechanical dynamics affected by electrical domain?

> What are the different ways to get from model to

answer?

8/8/2019 07 Lecture 10




Outline







8/8/2019 07 Lecture 10




Adding dynamics

> Add components to completethe system:

• Source resistor for the voltagesource

• Inertial mass, dashpot

> This is now our RF switch!

> System is nonlinear, so we

can’t use Laplace to gettransfer functions

> Instead, model with state

equations

Electrical domain Mechanical domain

V in

Resistor

R I

Dashpot b Spring k Mass m

z g

V

Fixed plate

Fixed support

+

+

-

-

C

F

W(Q,g)

++ R

V in V

--

I g z

m

1/k

b

+

-

Adapted from Figure 6.9 in Senturia, Stephen D. Microsystem Design . Boston,

MA: Kluwer Academic Publishers, 2001, p. 138. ISBN: 9780792372462.


8/8/2019 07 Lecture 10




State Equations

> Dynamic equations for general

system (linear or nonlinear) can beformulated by solving equivalentcircuit

> In general, there is one state variable

for each independent energy-storageelement (port)

> Good choices for state variables:

the charge on a capacitor(displacement) and the current in aninductor (momentum)

> For electrostatic transducer, need

three state variables

• Two for transducer (Q,g)

• One for mass ( dg/dt)

Goal:

⎟⎟ ⎠

⎞

⎜⎜⎝

⎛

=⎥⎥

⎥

⎦

⎤

⎢⎢

⎢

⎣

⎡

constantsor

of functions

gQ,g,gg

Q

dt

d

8/8/2019 07 Lecture 10




Formulating state equations

( )1

1

in

in

dQ I V V

dt R

dQ QgV

dt R Aε

= = −

⎛ ⎞= −⎜ ⎟

⎝ ⎠

> Start with Q

> We know that dQ/dt=I

> Find relation between I and

state variables and constants

KVL : 0

0

in R

in

V e V

V IR V

− − =

− − =

IRe R =

QgV Aε =

I

V in

R +-

+

-

V

C

W Q( ,g )

+ -e R

8/8/2019 07 Lecture 10





0

0

:KVL

=−−−

=−−−

zb zmkzF

eeeF bmk

> Now we’ll do

> We know that

k

m

b

e kz

e mz

e bz

=

=

=

g

gdt

gd

=

g zg zgg z −=−=⇒−= ,0

[ ]

⎥⎦

⎤⎢⎣

⎡+−−−=

+−−−=

=++−−

gbggk A

Q

mdt

gd

gbggk F m

g

gbgmggk F

)(2

1

)(1

0)(

0

2

0

0

ε

C

W (Q,g )

1/k zg

+

+

-

-

ek

eb

F e

m

m

b

+ +

- -

. .

F l i i

8/8/2019 07 Lecture 10





2

0

1

1( )

2

in

QgV Q R A

d g g

dt g Qk g g bg

m A

ε

ε

⎡ ⎤⎛ ⎞−⎢ ⎥⎜ ⎟⎡ ⎤ ⎝ ⎠⎢ ⎥⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎡ ⎤⎢ ⎥⎣ ⎦

− − − +⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦

> State equation for g is easy:

> Collect all three nonlinear state equations

> Now we are ready to simulate dynamics

gdt

dg=

O tli

8/8/2019 07 Lecture 10




Outline







Q i t ti l i

8/8/2019 07 Lecture 10




Quasistatic analysis

⎥⎥

⎥⎥⎥

⎦

⎤

⎢⎢

⎢⎢⎢

⎣

⎡

⎥⎦⎤⎢

⎣⎡ +−−−

⎟ ⎠

⎞⎜⎝

⎛ −

=⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

gbggk A

Qm

g A

QgV

R

g

gQ

dt

d in

)(2

1

1

0

2

ε

ε

State eqns

I

V in

R +

-

+

-V

C

W Q( ,g )

1/k zg

F m

b

+

-

. .

Equivalent circuit

Given static Vin, etc.

What is deflection,

charge, etc.?(Just now)

Followcausal path

(Wed)

Fixed-pointanalysis

State eq ations

8/8/2019 07 Lecture 10




State equations

),(),(

uxyuxx

g f

==

2

0

1

)

1 ( )2

in

Qg

V R A

f g

Q k g g bgm A

ε

ε

⎡ ⎤⎛ ⎞

−⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎢ ⎥=⎢ ⎥

⎡ ⎤⎢ ⎥− − − +⎢ ⎥⎢ ⎥

⎣ ⎦⎣ ⎦

(x,u

State variables

Inputs

Outputs

For transverse

electrostatic actuator:

[ ]inV =u

⎥⎥⎦

⎤

⎢⎢⎣

⎡=

ggQ

xState variables:

Inputs:

[ ] ),( uxy gg == Outputs:

8/8/2019 07 Lecture 10


Fixed points of the electrostatic actuator

8/8/2019 07 Lecture 10




Fixed points of the electrostatic actuator

> This analysis is analogous to what we did last time…

⎥⎦

⎤⎢⎣

⎡+−−−=

=⎟ ⎠

⎞

⎜⎝

⎛

−=

gbggk

A

Q

m

g A

Qg

V R in

)(

2

10

0

1

0

0

2

ε

ε

22

02( )

2 2

in

in

Qg

V A

V AQk g g

A g

ε

ε

ε

=

= = −

0 0.5 10

0.5

1

normalized voltage

n o r m a l i z e d g

a p

stable

2

2

02kg

AV gg inε

−=

Last time…

Operating point

Outline

8/8/2019 07 Lecture 10




Outline







Large signal analysis

8/8/2019 07 Lecture 10




Large-signal analysis

⎥⎥

⎥⎥⎥

⎦

⎤

⎢⎢

⎢⎢⎢

⎣

⎡

⎥⎦

⎤⎢⎣

⎡+−−−

⎟ ⎠

⎞⎜⎝

⎛ −

=⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

gbggk A

Qm

g A

QgV

R

ggQ

dt

d in

)(2

1

1

0

2

ε

ε

State eqns

I

V in

R +

-

+

-V

C

W Q( ,g )

1/k zg

F m

b

+

-

. .

Equivalent circuit

Given a step inputV in (t )u (t )

What is g(t), Q(t), etc.?(Earlier)

SPICE

Integratestate eqns

Direct Integration

8/8/2019 07 Lecture 10




Direct Integration

> This is a brute force approach: integrate the state

equations

• Via MATLAB ® (ODExx)

• Via Simulink ®

> We show the SIMULINK ® version here

• Matlab ® version later

Electrostatic actuator in Simulink ®

8/8/2019 07 Lecture 10




Electrostatic actuator in Simulink

[ ]

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

⎥⎦

⎤⎢⎣

⎡+−−−

⎟ ⎠

⎞

⎜⎝

⎛

−

=

=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

gbggk A

Q

m

g

A

Qg

V R

V

g

g

Q

in

in

)(2

1

1

)

,

0

2

ε

ε

uf(x,

ux(u[1]^2)/(2*e*A)

Electrostatic force

k b Damping

Sum2

At-rest gap

V_in

1/R

1/R Charge

1/(e*A)

Position

Velocity

Inertia

-1/m

11

2

3

s

Q

g

gdot

Qg

*

1

g _ 0

_

_

Spring

+

+

+

+

+

1

s

1s

Adapted from Figure 7.8 in Senturia, Stephen D. Microsystem Design . Boston, MA: Kluwer

Academic Publishers, 2001, p. 174. ISBN: 9780792372462.


Electrostatic actuator with contact

8/8/2019 07 Lecture 10




<

>+

+

+

+

+

Electrostatic force

(u[1]^2)/(2*e*A)

Inertia

g_min

g > g_min?

ZeroDamping bk Spring

Sum2

At-rest gap

0

-1/m

Accel > 0 ?

g _ 0 1/(e*A)

1/R

1/R Charge

Qg

Q

+

+

_

_

12

3

s

*

1 1s

1s

1

V_in

g

Position

VelocitySwitch

gdot

Adapted from Figure 7.9 in Senturia, Stephen D. Microsystem Design . Boston, MA: Kluwer Academic Publishers, 2001,

p. 175. ISBN: 9780792372462.


Behavior through pull-in

8/8/2019 07 Lecture 10




Behavior through pull in

Time

3002502001501005000 10 0

1 10 8

2 10 8

3 10 8

4 10 8

5 10 8

6 10 8

7 10 8

Charge

Drive (scaled)

Time

Release

Pull-in

3002502001501005000.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

P o s i t i o n

Adapted from Figure 7.10 in Senturia, Stephen D. Microsystem Design . Boston, MA: Kluwer Academic Publishers, 2001,

p. 176. ISBN: 9780792372462.


Behavior through pull-in

8/8/2019 07 Lecture 10




Behavior through pull in

Time

Release

Pull-in

300250200150100500-1.5

-1.0

-0.5

0.0

0.5

1.0

V e l o c i t y

Time

Release

Drive

3002502001501005000.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

P o s i t i o n

Discharge

Pull-in

Adapted from Figure 7.11 in Senturia, Stephen D. Microsystem Design. Boston, MA: Kluwer Academic Publishers, 2001, p. 177. ISBN: 9780792372462.


Outline

8/8/2019 07 Lecture 10




Outline







Small-signal analysis

8/8/2019 07 Lecture 10




Small signal analysis

Given O.P.

What is g(t)

due to small

changes inVin(t)?

LinearizeState eqns

Equivalentcircuit

Transferfunctions

Frequencyresponse

Naturalsystem

dynamics

Linearizedstate eqns(Jacobians)

Linearizedcircuit

Linearize

Take LT

Form TFs

SSS

poles &zerosform TFs

E i g e n

f c

t n

A n a l y

s i s

I n t eg r a t e

Given O.P.

How fastcan I wiggle

tip?

Timeresponse


8/8/2019 07 Lecture 10




g y

Given O.P.

What is g(t)

due to small

changes inVin(t)?

LinearizeState eqns

Equivalentcircuit


I n t eg r a t e

Given O.P.


tip?

Timeresponse

Linearization about a fixed point

8/8/2019 07 Lecture 10




p

> This is EXTREMELY common in MEMS literature

> This is also done in many other fields, with different

names

• Small-signal analysis• Incremental analysis

• Etc.

Linearization About an Operating Point

8/8/2019 07 Lecture 10




p g

> Using Taylor’s theorem, a system can

be linearized about any fixed point

> We can do this in one dimension or

many

xdx

df X f x X f

X

δ δ

0

)()( 00 +≈+

),( uxx f =

Operating point

)()()()(

t t t t

uUuxXx

0

0

δ δ

+= +=

( ) ( )uUxXxX 000 δ δ δ ++=+ , f dt

d

Multi-dimensional Taylor

Cancel

( ) ( ) )()(,

0000 ,,

t u

f t

x

f f

dt

d

U X j

i

U X j

i uxUXx

X 000 δ δ δ

⎟⎟ ⎠

⎞

⎜⎜⎝

⎛

∂

∂+⎟⎟

⎠

⎞

⎜⎜⎝

⎛

∂

∂+=+

( ) )()()(

0000 ,,

t uu

f t x

x

f

dt

t xd i

U X j

ii

U X j

ii δ δ δ

⎟⎟

⎠

⎞⎜⎜

⎝

⎛ ∂∂+⎟

⎟

⎠

⎞⎜⎜

⎝

⎛ ∂∂=

J1J2

x

f(x)

X 0

f X =Y ( )0 0

f X + x( )0 δ

df/dx

δ x

~

Linearization About an Operating Point

8/8/2019 07 Lecture 10




p g

> The resulting set of

equations are linear, andhave dynamics described

by the Jacobians of f (x,u)

evaluted at the fixed point.

> These describe how much

a small change in one state

variable affects itself or

another state variable

> The O.P. must be evaluated

to use the Jacobian

> Example – linearization of

the voltage-controlled

electrostatic actuator

( )inV R

g

g

Q

mb

mk

AmQ

A R

Q

A R

g

g

g

Q

dt

d δ

δ

δ

δ

ε

ε ε

δ

δ

δ

⎟

⎟⎟⎟⎟

⎠

⎞

⎜

⎜⎜⎜⎜

⎝

⎛

+⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎟⎟⎟⎟⎟⎟⎟⎟⎟

⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎜

⎝

⎛

−−−

−−

=⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

0

0

1

100

0

ˆ

0

00

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

=

..

3

..

3

..

3

..

2

..

2

..

2

..

1

..

1

..

1

POPOPO

POPOPO

POPOPO

g

f

g

f

Q

f

g

f

g

f

Q

f

g

f

g

f

Q

f

1J

u(t)JxJtx 21 δ δ δ +=)(

J1J2

State Equations for Linear Systems

8/8/2019 07 Lecture 10




> Normally expressed with:

• x: a vector of state variables

• u: a vector of inputs

• y: a vector of outputs

• Four matrices, A,B,C, D

> For us, Jacobian matrices take the place of A and B

> C and D depend on what outputs are desired• Often C is identity and D is zero

> Can use to simulate time responses to arbitrarySMALL inputs

• Remember, this is only valid for small deviations from O.P.

= +x Ax Bu

= +y Cx Du

Direct Integration in Time

8/8/2019 07 Lecture 10




> Can integrate via Simulink ®

model (as before) orMATLAB ®

> First define system in

MATLAB ®

• using ss(J1,J2,C,D) or

alternate method

> Can use MATLAB ®

commands step, initial,

impulse etc.

> Response of electrostatic

actuator to impulse of voltage

• Parameters from text (pg167)

0

500

1000

-4

-2

0

2

D

i s p l a c e m e n t

0 5 10 15 20 25-6-4-2024

V e l o c i t y

Impulse Response

Time (sec)

C h a r g e


8/8/2019 07 Lecture 10




Given O.P.

What is g(t)

due to smallchanges inVin(t)?

LinearizeState eqns

Equivalentcircuit

Transferfunctions Frequencyresponse


Take LT

Form TFs

SSS

poles &zeros

I n t eg r a t e

Given O.P.


tip?

Timeresponse

Solve via Laplace transform

8/8/2019 07 Lecture 10




> Use Laplace Transforms

to solve in frequencydomain

• Transform DE to algebraic

equations

• Use unilateral Laplace toallow for non-zero IC’s ( ) )()0()(

)()()0()(

)()()0()(

sss

ssss

ssss

LaplaceUnilateral

BUxXAI

BUAXxIX

BUAXxX

BuAxx

+=−

+=−

+=−

⇓

+=

Transfer Functions

8/8/2019 07 Lecture 10




> Transfer functions H(s) are useful for obtaining

compact expression of input-output relation• What is the tip displacement as a function of voltage

> Most easily obtained from equivalent circuit

> But can also be obtained from linearized state eqns

• Depends on A, B, C (or J1, J2, C) matrices

• Can do this for fun analytically (see attachment at

end)• Matlab can automatically convert from s.s to t.f.

formulations

> For our actuator, we would get three transfer

functions

( )

( )

( )( ) ( )

( )

( )

s

s

ss s

s

s

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥

= ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

gH

g

Q

in

in

in

V

V

V

Sinusoidal Steady State

8/8/2019 07 Lecture 10




> When a LTI system is

driven with a sinusoid, the

steady-state response is a

sinusoid at the same

frequency

> The amplitude of the

response is |H(jω)|

> The phase of the response

relative to the drive is the

angle of H(jω)

> A plot of log magnitude vs

log frequency and anglevs log frequency is called

a Bode plot

)cos()( 0 θ ω += t Y t ysss

)()()( ω ω ω jU j H jY =

{ }{ })(Re

)(Imtan

)( 00

ω

ω θ

ω

j H

j H

U j H Y

=

=

)(cos)( 0 t U t u ω =

Bode plot of electrostatic actuator

8/8/2019 07 Lecture 10




> Use Matlab ® command

bode with previouslydefined system sys

> Evaluate only one of TFs

> This tells us how quicklywe can wiggle tip!

• At a certain OP!

( )( )

( )

ss

s=

gH

inV -140

-120

-100

-80

-60

-40

-20

0

20

40

M a g n i t u d e ( d B )

10-2

10-1

100

101

102

103

-90

-45

0

45

90

135

180

P

h a s e ( d e g )

Bode Diagram

Frequency (rad/sec)


8/8/2019 07 Lecture 10




Given O.P.

What is g(t)


LinearizeState eqns

Equivalentcircuit


Naturalsystem

dynamics


Take LT

Form TFs

SSS

I n t eg r a t e

poles &zeros

Given O.P.


tip?

Timeresponse

Poles and Zeros

8/8/2019 07 Lecture 10




> For our models, system function is a ratio of polynomials in s

> Roots of denominator are called poles• They describe the natural (unforced) response of the system

> Roots of the numerator are called zeros

• They describe particular frequencies that fail to excite any output

> System functions with the same poles and zeros have the

same dynamics

> MATLAB solution for poles is VERY long

0

23 2 0

2 2

0 0 0

( )( )

( ) 1 1 1 1

Q

s ARmss Qb b k k

s s s RC m RC m m RC m A Rm

ε

ε

−

= =⎛ ⎞ ⎛ ⎞ ⎛ ⎞

+ + + + + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

in

gH

V

0

0ˆ

AC g

ε =where

Pole-zero diagram

8/8/2019 07 Lecture 10




-9 -8 -7 -6 -5 -4 -3 -2 -1 0-1

-0.5

0

0.5

1Pole-Zero Map

I m a g i n a r y A x i s

> Displays information

about dynamics ofsystem function

• Matlab command

pzmap

> Useful for examining

dynamics, stability,

etc.

polezero

-9 -8 -7 -6 -5 -4 -3 -2 -1 0-1

-0.5

0

0.5

1Pole-Zero Map

I m

a g i n a r y A x i s

Real axis

Real axis

gap

tip velocity ωd


8/8/2019 07 Lecture 10




Given O.P.

What is g(t)


LinearizeState eqns

Equivalentcircuit


Naturalsystem

dynamics


Take LT

Form TFs

SSS

poles &zeros

I n t eg r a t e

E i g e n

f c t n

A n a l y s

i s

Given O.P.


tip?

Timeresponse

Eigenfunction Analysis

8/8/2019 07 Lecture 10




> For an LTI system, we can find the eigenvalues and

eigenvectors of the J1 (or A) matrix describing the internaldynamics

( )xJ

x1δ

δ =

dt

d

u(t)JxJtx 21 δ δ δ +=)(

Axx

=dt

d

10)( K eK t x xdt

dx t +=⇒= λ λ t et

λ Kx =)(

Axx =λ

For scalar 1st-order system:

Our linear (or linearized)homogeneous systems look like:

If we try solution:

Plug into DE:

•This is an eigenvalueequation

•If we find λ we can findnatural frequencies ofsystem


8/8/2019 07 Lecture 10




> These λ are the same as the poles si of the system

> Can solve analytically

• Find λ from det(A-λI)=0

> Or numerically eig(sys)

-8.9904

-0.2627 + 0.8455i

-0.2627 - 0.8455i

Linearized system poles

8/8/2019 07 Lecture 10




σ

> We can use either λi or si to

determine natural frequenciesof system

> As we increase applied

voltage• Stable damped resonant

frequency decreases

> Plotting poles as systemchanges is a root-locus plot

Increasing voltage

jω

Spring softening> Plot damped resonant frequency versus applied voltage

8/8/2019 07 Lecture 10




> Plot damped resonant frequency versus applied voltage

> Resonant frequency is changing because net spring constant

k changes with frequency

> This is an electrically tuned mechanical resonator

2

3'

AV k k

g

ε = −

This is called

spring softening Voltage

D a m p e d r

e s o n a n t f r e q u e n c

y

0.060.050.040.030.020.0100.30

0.40

0.50

0.60

0.70

0.80

0.90

1.00

Adapted from Figure 7.5 in Senturia, Stephen D. Microsystem Design .

Boston, MA: Kluwer Academic Publishers, 2001, p. 169. ISBN: 9780792372462.



8/8/2019 07 Lecture 10




Given O.P.

What is g(t)

due to smallchanges in

Vin(t)?

LinearizeState eqns

Equivalentcircuit

Naturalsystem

dynamics


Linearizedcircuit

Linearize

Take LT

Form TFs

poles &zeros

I n t eg r a t e

E i g e n

f c t n

A n a l y s

i s

Given O.P.


tip?

Timeresponse

form TFs


SSS

Linearized Transducers

8/8/2019 07 Lecture 10




> Can we directly linearize

our equivalent circuit?YES!

> This is perhaps the most

common analysis in the

literature

> First, choose what is load

and what is transducer

• Here we include spring

with transducer

I

V (t)in

+

-

V +

-

C

W

1/k U

F m

b

+

-

F out

+

-

R

Source Transducer Load

Find OP

Linearize

δ I

δV (t)in

+

-

δV +

-

δU

m

b

δF out

-

R

Source Load

LinearizedTransducer

+

Linearized Transducer Model

> First find O P

8/8/2019 07 Lecture 10




> First, find O.P.

> Next, generate matrix to

relate incremental port

variables to each other• Start from energy and force

relations

• Linearize (take partials…)

> Recast in terms of port variables

> Define intermediate variables

> Final expression

0 0

0

ˆ

out

g QV Q A A

F Q gk

A

δ δ ε ε

δ δ

ε

⎡ ⎤⎢ ⎥⎡ ⎤ ⎡ ⎤= ⎢ ⎥⎢ ⎥ ⎢ ⎥

⎣ ⎦⎣ ⎦ ⎢ ⎥⎢ ⎥⎣ ⎦

I Q s

g U s

δ δ

δ δ

⎡ ⎤⎡ ⎤ ⎢ ⎥=⎢ ⎥ ⎢ ⎥⎣ ⎦

⎣ ⎦

0 0 0ˆ, ,V g Q

2

0( )2

out

QgV

AQ

F k g g A

ε

ε

=

= − −

00 0

0 0

,ˆ

Q AC V

g C

ε = =

0

0 0

0

0

1

ˆ

ˆout

V

sC sgV I F V U k

sg s

δ δ δ δ

⎡ ⎤⎢ ⎥

⎡ ⎤ ⎡ ⎤⎢ ⎥=⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎣ ⎦⎣ ⎦⎢ ⎥⎣ ⎦


> Now we want to convert this relation into a circuit

8/8/2019 07 Lecture 10




> Now we want to convert this relation into a circuit> Many circuit topologies are consistent with this matrix relation

> THIS IS NOT UNIQUE!

-1/k'

1/k

1/k*

+++

Co

_ _ _

+

_ {

F'

1:Γ i' u'

υ υ'

ui

F

1/k

+

_

+

_

1:Γ/κ 2

υ

ui

F

Co

∗

+

_

+

_

1:Γ

1 _ Γ

Γ 2/κ ∗

υ

ui

F

Co

+

_

+

_

υ

ui

F

Co

Γ

Co

k _ Γ/Co

1

Image by MIT OpenCourseWare.Adapted from Figure 5 on p. 163 in Tilmans, Harrie A. C. "Equivalent Circuit Representations of ElectromechanicalTransducers: I. Lumped-parameter Systems." Journal of Micromechanics and Microengineering 6, no. 1 (1996): 157-176.


> This is the one used in the text

8/8/2019 07 Lecture 10




> This is the one used in the text

> Uses a transformer

• Transforms port variables• Doesn’t store energy

> What we want to do now is

identify ZEB, ZMS and ϕ, andfigure out what they mean…

⎟⎟ ⎠

⎞⎜⎜⎝

⎛

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ −=⎟⎟

⎠

⎞⎜⎜⎝

⎛

1

1

2

2

10

0

f

e

f

e

ϕ

ϕ

2

1

EB EB

out EB MO

EB MS MO

MO

V Z Z I

F Z Z U

Z Z Z Z

δ ϕ δ δ ϕ δ

ϕ

⎡ ⎤ ⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦ ⎣ ⎦

⎛ ⎞

= −⎜ ⎟⎝ ⎠


⎡ ⎤ ⎡ ⎤ ⎡ ⎤

8/8/2019 07 Lecture 10




⎥

⎦

⎤⎢

⎣

⎡

⎥⎥

⎥⎥

⎦

⎤

⎢⎢

⎢⎢

⎣

⎡

=⎥

⎦

⎤⎢

⎣

⎡

U

I

s

k

gs

V

gs

V

sC

F

V

δ

δ

δ

δ

0

0

0

0

0

ˆ

ˆ

1

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

⎟⎟

⎠

⎞⎜⎜

⎝

⎛ −=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎟⎟

⎠

⎞⎜⎜

⎝

⎛ −=

0

2

0

0

2

0

00

2

0

0

ˆ1

1

ˆ1

1

ˆ1

g Ak

Q

s

k

k C g

Q

s

k

s

k

sC

g

Q

s

k Z MS

ε 0

2

0

ˆ'

g A

Qk k

ε −=

2

1

EB EB

out EB MO

EB MS MO

MO

V Z Z I

F Z Z U

Z Z Z

Z

δ ϕ δ

δ ϕ δ

ϕ

⎡ ⎤ ⎡ ⎤ ⎡ ⎤=

⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦ ⎣ ⎦

⎛ ⎞= −⎜ ⎟

⎝ ⎠

0 0

0ˆ

C V

gϕ =


8/8/2019 07 Lecture 10




> C0

represents the

capacitance of the

structure seen from the

electrical port

> It is simply the capacitance

at the gap given by the

operating point

> As Vin increases, C0 will

increase until the structure

pulls in

> This is a tunable capacitor

0

0

ˆC

g

ε =


> k’ represents the effective

8/8/2019 07 Lecture 10




> k represents the effective

spring

> A combination of the

mechanical spring k and

the electrical spring

> This is an electrically

tunable spring!• Spring softening shows up

in k ’

> As Vin increases, k’ willdecrease from k (at Vin=0)

to 0 (at Vin=Vpi)

0

2

0

ˆ'

g A

Qk k

ε −=


> represents the

8/8/2019 07 Lecture 10




> ϕ represents the

electromechanicalcoupling

>Represents how much the

capacitance changes with

gap

> A measure of sensitivity

0 0 0

0 0ˆ ˆ

C V Q

g gϕ = =

0 0

. . . .

0 20

0 0

0

ˆ

ˆ

O P O P

C AV V

g g g

AV

g

C V

g

ε ϕ

ε

∂ ∂= − = −

∂ ∂

=

=

Transfer Functions

8/8/2019 07 Lecture 10




> Can use linearized

circuit to construct H(s)

using complex

impedances

> Usually helpful to“eliminate” transformer

> Transformer changes

impedances

1:ϕ

Z 1

Z 2

122

Z Z

ϕ =

δV i n +-

ϕ2 /k’ δU

m/ ϕ2

b/ ϕ2

R

C0

δV in +-

1/k’

δU

m

b

R

C0

1:ϕ

Can now get any transferfunction using standardcircuit analysis

Linearized Transducer Models

> Now we can understand Nguyen’s filter!

8/8/2019 07 Lecture 10




g y

Image removed due to copyright restrictions.

Figure 12 on p. 62 in: Nguyen, C. T.-C. "Micromechanical

Filters for Miniaturized Low-power Communications."

Proceedings of SPIE Int Soc Opt Eng 3673 (July 1999): 55-66.

Image removed due to copyright restrictions.

Figure 9 on p. 17 in Nguyen, C. T.-C. "Vibrating RF MEMS

Overview: Applications to Wireless Communications."

Proceedings of SPIE Int Soc Opt Eng 5715 (January 2005): 11-25.

Small-signal analysis summary

8/8/2019 07 Lecture 10




Given O.P.

What is g(t)

due to smallchanges in

Vin(t)?

LinearizeState eqns

Equivalentcircuit

Transferfunctions

Frequencyresponse

Naturalsystem

dynamics


Linearizedcircuit

Linearize

Take LT

Form TFs

SSS

poles &zerosform TFs

E i g e n

f c t n

A n a l y s

i s

I n t eg r a t e

Given O.P.

How fast

can I wiggletip?

Timeresponse

Conclusions

> We can now analyze and design both quasistatic and

8/8/2019 07 Lecture 10




> We can now analyze and design both quasistatic and

dynamic behavior of our multi-domain MEMS

> We have much more powerful tools to analyze linear

systems than nonlinear systems

> But most systems we encounter are nonlinear

> Linearization permits the study of small-signal inputs> Next up: special topics in structures, heat transfer,

fluids

Review: analysis of a 2nd-order linear system

> Spring-mass-dashpot

8/8/2019 07 Lecture 10




( )⎥⎥⎦

⎤

⎢⎢⎣

⎡

−−=⎥⎦

⎤

⎢⎣

⎡

xbkxF m

x

x

x

dt

d

1

State

eqns

b

m

1/ k x.

F +

-

+

++

- -

-

eb

em

ek

= +x Ax Bu

= +y Cx Du

Direct Integration in Time

> Example: Spring-mass-dashpot step response

8/8/2019 07 Lecture 10




• k=m= 1;b =0.5;

>> A=[0 1;-1 -0.5]; B=[0;1];>> C=[1 0;0 1]; D=[0;0];>> sys=ss(A,B,C,D);>> step(sys) 0

0.5

1

1.5

P o s i t i o n

0 5 10 15 20 25-0.5

0

0.5

1

V e l o c i t y

Step Response

Time (sec)

A m p l i t u d e

Transfer Functions

> Can get TFs from A,B,C matrices

8/8/2019 07 Lecture 10




( ) ( )[ ] )()()0()(

)()()(

11sssss

sss

DUBUAIxAICY

DUCXY

+−+−=

+=−−

( )

( )

1

1

( ) ( )

( ) ( ) ( )

( )

s s s

s s s

s s

−

−

⎡ ⎤= −⎣ ⎦

=

⎡ ⎤= −⎣ ⎦

Y C I A B U

Y H U

H C I A B

Assume transient has died out (XZIR=0)

No feed-through (D=0)

Transfer Functions> Let’s do analytically & via MATLAB

⎥⎤

⎢⎡

⎥⎤

⎢⎡ s 1)(X

⎤⎡

8/8/2019 07 Lecture 10




⎥⎥⎥

⎥

⎦⎢⎢⎢

⎢

⎣ ++

++=

⎥⎥⎥

⎥

⎦⎢⎢⎢

⎢

⎣

=

k sbms

s

k sbms

s

s

s

s

2

2

)(

)(

)(

)(

F

X

F

H

( )

mk mbssmk mbss

sm

k m

bss

m

bs

m

k

s

mb

mk s

s

s

++=++=Δ

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

−

+

Δ=−

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

+

−=

⎥⎥⎦

⎤

⎢⎢⎣

⎡

−−−⎥⎦

⎤

⎢⎣

⎡

=−

−

2

1

)(

11

1

10

0

0

AI

AI

( )

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

Δ=

⎥⎥⎦

⎤

⎢⎢⎣

⎡

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

−

+

Δ⎥⎦

⎤⎢⎣

⎡=−

−

ms

m

msm

k m

bss

11

1

011

10

011BAIC

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

++

++=

15.0

15.0

1

)(

2

2

ss

s

sssH

>> [n,d]=ss2tf(A,B,C,D)

n =0 -0.0000 1.00000 1.0000 -0.0000

d =1.0000 0.5000 1.0000

s2 s1 s0

Transfer Functions> Can also construct H(s) directly

i l i d ds 1)(X

8/8/2019 07 Lecture 10




using complex impedances and

circuit model

xs xe

xe

xs xe

eeeF

m

b

k

bmk

m==

=

==

=−−−

m

b

k

k

0

kbm

kmb

++=

++=

==

ss

s

ss

s

s

s

s

2

2

1)(

1)(

)(

)(

Z

H

F

X

kbm ++

==

=

ss

s

s

s

s

ss

s

s

21

1)(

)(

)(

)(

)(

)(

)(

H

F

X

F

X

F

X

b

m

1/ k x.

F +

-

+

++

- -

-

eb

em

ek

Poles and Zeros

> For 2nd-order system, easy to get

l d f TF

8/8/2019 07 Lecture 10




poles and zeros from TFs

m

k

m

b

m

b−⎟

⎠

⎞⎜⎝

⎛ ±−=

2

1,2 22s

where

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−

−−=

⎥⎥

⎥⎥

⎦

⎤

⎢⎢

⎢⎢

⎣

⎡

++

++=

))((

))((

1

1

1

1)(

21

21

2

2

ssss

sssss

m

mk mbss

s

mk mbss

msH

these are the poles

Spring-mass-dashpot system

> It is a second order system, with222 kb

8/8/2019 07 Lecture 10




two poles

> We conventionally define

• Undamped resonant frequency

• Damping constant• Damped resonant frequency

• Quality factor

b

mQ

js

s

m

b

m

k

00

22

0d

d2,1

0

2

0

2

2,1

0

2

:factor Quality

where

)(systemsdunderdampeFor

2

ω

α

ω

α ω ω

ω α

ω α

ω α α

α

ω

==

=

±−=

<

−±−=

=

=

−

2

0

222 ω α ++=++ ss

m

k sm

bs

> Displays information about dynamics of system

Pole-zero diagram

8/8/2019 07 Lecture 10




97.025.016

1125.02,1 j js ±−=−±−=

function

-2 -1 0 1 2-2

-1.5

-1

-0.5

0

0.5

1

1.5

2Pole-Zero Map

Real Axis

I m a g i n a r y A x i s

pole

zero

15.0)(

22++

=ss

ssH

ωd

SMD-position frequency response

Bode Diagram

8/8/2019 07 Lecture 10




⎟⎟ ⎠

⎞⎜⎜⎝

⎛

−−=∠

+−

=

++−=

22

0

22222

0

2

0

2

2atan)(

4)(

11)(

21 1)(

ω ω

αω ω

ω α ω ω

ω

ω ω α ω ω

j

m

j

jm j

H

H

H

-40

-30

-20

-10

0

10

M a g n

i t u d e ( d B )

10-1

100

101

180

225

270

315

360

P h a s e ( d e g )

Bode Diagram

Frequency (rad/sec)


> Find eigenvalues numerically using MATLAB and A matrix

8/8/2019 07 Lecture 10




[ ] ⎥⎦

⎤⎢⎣

⎡

−−−−==

⎥⎦

⎤⎢⎣

⎡

−−

+−=⎥

⎦

⎤⎢⎣

⎡=Λ

=Λ

⎥⎦

⎤⎢⎣

⎡

−=

j jvv

j

j

68.018.068.018.0

707.0707.0

97.025.00

097.025.0

0

0

)(eig],[

5.00

10

21

2

1

V

AV

A

λ

λ

Documents

07 Lecture 10