07lecture10

Lumped-Element System Dynamics

Joel Voldman*
Massachusetts Institute of Technology
*(with thanks to SDS)

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

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

Outline
> Our progress so far
> Formulating state equations
> Quasistatic analysis
> Large-signal analysis
> Small-signal analysis
> Addendum: Review of 2nd-order system dynamics



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 system
behaves over time (or frequency)



> The Northeastern/ADI RF Switch
> We first lumped the mechanical domain

Fixed support
Images removed due to copyright restrictions.
Figure 11 on p. 342 in: Zavracky, P. M., N. E. McGruer, R. H. Morrison,
I Dashpot b
and D. Potter. "Microswitches and Microrelays with a View Toward Microwave
Applications." International Journal of RF and Microwave Comput-Aided Spring k
Engineering 9, no. 4 (1999): 338-347. + Mass m
V g
z
-
1 µm Cantilever Fixed plate

0.5 µm Pull-down Image by MIT OpenCourseWare.
Silicon electrode Anchor Adapted from Figure 6.9 in Senturia, Stephen D. Microsystem
Design. Boston, MA: Kluwer Academic Publishers, 2001, p.
138. ISBN: 9780792372462.
Image by MIT OpenCourseWare.
Adapted from Rebeiz, Gabriel M. RF MEMS: Theory, Design, and
Technology. Hoboken, NJ: John Wiley, 2003. ISBN: 9780471201694.



> 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
. 1/k
I g

2 + +
Q g
W (Q, g ) = V F m
2εA
C
- -
b



∂W (Q, g ) Q2
> We first analyzed system F= =
∂g Q
2εA
quasistatically ∂W (Q, g ) Qg
V= =
> Saw that there is VERY different ∂Q g
εA
behavior depending on whether
• Charge is controlled dW = VdQ + Fdg
stable behavior at all gaps
• Voltage is controlled dW * = QdV − Fdg
pull-in at g=2/3g0

> Use of energy or co-energy ∂W * εA
depends on what is controlled Q= = Vin
∂V g
g
• Simplifies math
∂W * εAVin 2
F= =
∂g V 2g 2



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?



Outline



Adding dynamics
Fixed support
> Add components to complete Resistor
R I
the system: Dashpot b Spring k
Mass m

+
• Source resistor for the voltage +
source Vin V
- g
-
z
• Inertial mass, dashpot Fixed plate

> This is now our RF switch!
g z 1/k
> System is nonlinear, so we I

R + +
can’t use Laplace to get +
C
F m
Vin V
transfer functions - W(Q,g)
- -

> Instead, model with state b

equations
Adapted from Figure 6.9 in Senturia, Stephen D. Microsystem Design. Boston,
MA: Kluwer Academic Publishers, 2001, p. 138. ISBN: 9780792372462.

Electrical domain Mechanical domain


State Equations
> Dynamic equations for general
system (linear or nonlinear) can be
formulated by solving equivalent
circuit
> In general, there is one state variable
for each independent energy-storage
element (port) Goal:
> Good choices for state variables: ⎡Q ⎤
d ⎢ ⎥ ⎛ functions of ⎞
⎜ ⎟
the charge on a capacitor ⎢ g ⎥ = ⎜ Q,g,g or constants ⎟
⎢g ⎥ ⎝ ⎠
(displacement) and the current in an dt
inductor (momentum) ⎣ ⎦

> For electrostatic transducer, need
three state variables
• Two for transducer (Q,g)
• One for mass (dg/dt)


Formulating state equations
> Start with Q
+ eR - I
> We know that dQ/dt=I R + C
+
> Find relation between I and Vin V
- W(Q,g)

state variables and constants -

KVL : Vin − eR − V = 0
eR = IR
Vin − IR − V = 0
dQ 1
= I = (Vin − V )
dt R Qg
V=
εA
dQ 1 ⎛ Qg ⎞
= ⎜ Vin −
dt R ⎝ εA⎟⎠


> Now we’ll do .
g . 1/k
g z
> We know that dg
=g C + + ek - +
dt
F em m
W(Q,g)
- - eb + -
b
KVL : ek = kz
F − ek − em − eb = 0 F − k ( g 0 − g ) + mg + bg = 0
em = mz
F − kz − mz − bz = 0 eb = bz 1
g = − [F − k ( g 0 − g ) + bg ]
m
dg 1 ⎡ Q2 ⎤
z = g0 − g ⇒ z = − g, z = − g =− ⎢ − k ( g 0 − g ) + bg ⎥
dt m ⎣ 2εA ⎦



> State equation for g is easy: dg
=g
dt
> Collect all three nonlinear state equations

⎡ 1⎛ Qg ⎞ ⎤
⎢ ⎜ Vin − ⎟ ⎥
⎡Q ⎤ ⎢ R⎝ εA⎠ ⎥
d ⎢ ⎥ ⎢ ⎥
g⎥= g
dt ⎢ ⎢ ⎥
⎢ g ⎥ ⎢ 1 ⎡ Q2
⎣ ⎦ ⎤⎥
⎢ − m ⎢ 2ε A − k ( g 0 − g ) + bg ⎥ ⎥
⎣ ⎣ ⎦⎦
> Now we are ready to simulate dynamics



Outline



Quasistatic analysis

⎡1 ⎛ Qg ⎞ ⎤
⎢ ⎜V − ⎟ ⎥
d ⎡Q ⎤ ⎢ R ⎝ in εA ⎠ ⎥ Fixed-point
⎢g ⎥ = ⎢g ⎥ analysis
dt ⎢ g ⎥ ⎢ 1 ⎡ Q 2
⎣ ⎦ ⎤⎥
− ⎢
⎢ m 2εA − k ( g 0 − g ) + bg ⎥ ⎥
⎣ ⎣ ⎦⎦ Given static Vin, etc.
State eqns
What is deflection,
(Just now) charge, etc.?
. . 1/k
I g z

R + C + Follow
+ F m causal path
Vin V
- W(Q,g)
(Wed)
- -
b

Equivalent circuit



State equations
For transverse
electrostatic actuator:
State variables ⎡Q ⎤
State variables: x = ⎢g ⎥
Inputs ⎢g ⎥
⎣ ⎦

x = f (x, u) Inputs: u = [Vin ]
y = g (x, u)
⎡1 ⎛ Qg ⎞ ⎤
⎢ ⎜ Vin − ⎟ ⎥
Outputs ⎢R ⎝ εA⎠ ⎥
f (x,u) = ⎢ g ⎥
⎢ ⎥
⎢ 1⎡Q 2
⎤⎥
⎢ − m ⎢ 2ε A − k ( g 0 − g ) + bg ⎥ ⎥
⎣ ⎣ ⎦⎦
Outputs: y = [g ] = g (x, u)


Fixed points
> Definition of a fixed point
• Solution of f(x,u) = 0
• time derivatives 0
> Global fixed point
• A fixed point when u = 0
• Systems can have multiple global fixed points
• Some might be stable, others unstable (consider a
pendulum)

> Operating point
• Fixed point when u is a non-zero constant



Fixed points of the electrostatic actuator
> This analysis is analogous to what we did last time…

Operating point stable
1⎛ Qg ⎞

normalized gap
0 = ⎜Vin − ⎟ 1
R⎝ εA ⎠
0=g
1 ⎡ Q2 ⎤ 0.5
0=− ⎢ − k ( g 0 − g ) + bg ⎥
m ⎣ 2εA ⎦
0
0 0.5 1
Qg normalized voltage
Vin =
εA Last time…
Q 2 Vin 2ε A εAVin
2
= = k ( g0 − g ) g = g0 −
2ε A 2g 2
2kg 2


Outline



Large-signal analysis

⎡1 ⎛ Qg ⎞ ⎤
⎢ ⎜V − ⎟ ⎥
d ⎡Q ⎤ ⎢ R ⎝ in εA ⎠ ⎥ Integrate
⎢g ⎥ = ⎢g ⎥ state eqns
dt ⎢ g ⎥ ⎢ 1 ⎡ Q 2
⎣ ⎦ ⎤⎥
− ⎢
⎢ m 2εA − k ( g 0 − g ) + bg ⎥ ⎥
⎣ ⎣ ⎦⎦ Given a step input
State eqns Vin(t)u(t)

(Earlier) What is g(t), Q(t), etc.?
. . 1/k
I g z

R + C + SPICE
+ F m
Vin V
- W(Q,g)
- -
b

Equivalent circuit



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®

⎡Q ⎤
x = ⎢ g ⎥, u = [Vin ]
(u[1]^2)/(2*e*A) +
⎢ ⎥ + -1/m
Electrostatic force
⎢g ⎥
⎣ ⎦
+
Inertia

⎡1 ⎛ Qg ⎞ ⎤ 1
⎢ ⎜Vin − ⎟ ⎥ s 3
⎢R ⎝ εA ⎠ ⎥ Spring k b Damping Velocity gdot
f(x, u) = ⎢ g ⎥
⎢ ⎥
⎢ 1 ⎡ Q2 ⎤⎥ Sum2
1
s 2
⎢− m ⎢ 2εA − k ( g 0 − g ) + bg ⎥ ⎥ _
+ g
⎣ ⎣ ⎦⎦ Position

At-rest gap g _ 0 1/(e*A) *

Qg
_
1/R 1
1 + s 1

V_in 1/R Charge Q

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
<
g_min
Accel > 0 ? >
+ +
g > g_min?
(u[1]^2)/(2*e*A) +
+ -1/m
Electrostatic force + Inertia 1
s
Switch Velocity
3
0
Spring k b Damping gdot
Zero

1
Sum2 s 2
_
+ g
Position

At-rest gap g _ 0 1/(e*A) *

Qg
_
1/R 1 1
1 + s
1/R Q
V_in Charge

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

7 10 8 1.6

6 10 8 Drive (scaled) 1.4
Charge Release
1.2
5 10 8
1.0
4 10 8

Position
Pull-in
0.8
3 10 8
0.6
2 10 8
0.4
1 10 8 0.2
0 10 0 0.0
0 50 100 150 200 250 300 0 50 100 150 200 250 300
Time Time

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

1.0
1.4

0.5 1.2
Release Drive
1.0
0.0 Release
Velocity

Position
Pull-in 0.8

-0.5 0.6 Discharge
0.4
-1.0
0.2
Pull-in
-1.5 0.0
0 50 100 150 200 250 300 0 50 100 150 200 250 300
Time Time
Adapted from Figure 7.11 in Senturia, Stephen D. Microsystem Design. Boston, MA: Kluwer Academic Publishers, 2001, p. 177. ISBN: 9780792372462.



Outline



Small-signal analysis

Given O.P.

Linearized Time What is g(t)
Linearize Integrate
State eqns state eqns response due to small
(Jacobians) changes in
Take LT Vin(t)?

Ei aly Form TFs
An
g e si
nf s
ct Transfer SSS Frequency
n
functions response
poles & Given O.P.
form TFs zeros
How fast
Natural
Equivalent Linearize Linearized can I wiggle
system
circuit circuit tip?
dynamics




Given O.P.

Linearize Integrate
Vin(t)?

Given O.P.

How fast
Equivalent can I wiggle
circuit tip?



Linearization about a fixed point
> 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
> Using Taylor’s theorem, a system can x = f (x, u)
be linearized about any fixed point x (t ) = X 0 + δx (t )
u(t ) = U0 + δu(t )
> We can do this in one dimension or
many Operating point
d (δx )
f(x) X0 + = f (X 0 + δx, U0 + δu)
df/dx dt
~ f(X0+δ x) Multi-dimensional Taylor
f( X0)=Y0 ⎛ ∂f ⎞ ⎛ ⎞
d (δx ) ⎟δx (t ) + ⎜ ∂f i
X0 + = f (X 0 , U0 ) + ⎜ i
⎜ ∂x ⎟ ⎜ ∂u
⎟δu(t )
⎟
dt
⎝ j X 0 ,U 0 ⎠ ⎝ j X 0 ,U 0 ⎠
δx
x Cancel
X0
d (δxi (t ) ) ⎛ ∂f i ⎞ ⎛
⎟δx (t ) + ⎜ ∂f i
⎞
=⎜
⎜ ∂x ⎟ i ⎜ ∂u
⎟δu (t )
⎟ i
dt
df ⎝ j X 0 ,U 0 ⎠ ⎝ j X 0 ,U 0 ⎠
f ( X 0 + δx ) ≈ f ( X 0 ) + δx J2
dx X0
J1


Linearization About an Operating Point
> The resulting set of δx (t ) = J1δx + J2δu(t)
equations are linear, and ⎡ ∂f1 ⎤
∂f1 ∂f1
have dynamics described ⎢ ⎥
⎢ ∂Q O.P. ∂g O. P.
∂g O. P. ⎥
by the Jacobians of f(x,u) ⎢ ∂f ∂f 2 ∂f 2 ⎥
J1 = ⎢ 2 ⎥
evaluted at the fixed point. ⎢ ∂Q O.P. ∂g O. P.
∂g O. P. ⎥
⎢ ∂f 3 ∂f 3 ∂f 3 ⎥
> These describe how much ⎢ ⎥
⎢ ∂Q O.P.
⎣ ∂g O.P.
∂g O. P. ⎥
⎦
a small change in one state
⎛ ⎞
variable affects itself or ⎜ g0ˆ Q0 ⎟
⎜ − − 0 ⎟
another state variable ⎜ RεA RεA ⎟ ⎛1⎞
⎛ δQ ⎞ ⎜ ⎟⎛ δQ ⎞ ⎜ ⎟
The O.P. must be evaluated d ⎜ δg ⎟ = ⎜ 0 0 1 ⎟⎜ ⎟ ⎜ R ⎟
> ⎜ ⎟ δg +
⎟⎜ ⎟ ⎜ 0 ⎟ in
(δV )
to use the Jacobian dt ⎜ ⎟ ⎜ ⎜ ⎟ ⎜ ⎟
⎝ δg ⎠ ⎜ Q0 k b ⎟⎝ δg ⎠ ⎜ 0 ⎟
⎜− − − ⎟ ⎝ ⎠
> Example – linearization of ⎜ mεA m m⎟
the voltage-controlled ⎜ ⎟ J2
⎝ J1 ⎠
electrostatic actuator


State Equations for Linear Systems
> Normally expressed with:
• x: a vector of state variables
• u: a vector of inputs x = Ax + Bu
• y: a vector of outputs
y = Cx + Du
• 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 arbitrary
SMALL inputs
• Remember, this is only valid for small deviations from O.P.



Direct Integration in Time
Impulse Response

> Can integrate via Simulink® 1000
model (as before) or

Charge
Charge (Q)
500
MATLAB®

> First define system in 0
MATLAB® 2
• using ss(J1,J2,C,D) or

Displacement
alternate method 0
> Can use MATLAB® -2
commands step, initial,
impulse etc. -4
4
2
> Response of electrostatic
0
Velocity

actuator to impulse of voltage
-2
• Parameters from text (pg -4
167)
-6
0 5 10 15 20 25
Time (sec)



Given O.P.

Linearize Integrate
Take LT Vin(t)?

Form TFs

Transfer SSS Frequency
functions response
poles & Given O.P.
zeros
How fast
circuit tip?



Solve via Laplace transform
> Use Laplace Transforms
to solve in frequency x = Ax + Bu
domain ⇓ Unilateral Laplace
• Transform DE to algebraic sX ( s ) − x (0) = AX( s ) + BU( s )
equations
sIX ( s ) − x (0) = AX( s ) + BU( s )
• Use unilateral Laplace to
allow for non-zero IC’s (sI − A )X( s) = x(0) + BU( s)



Transfer Functions
> 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. ⎡ Q( s ) ⎤
formulations ⎢ ⎥
⎢ Vin ( s ) ⎥
> For our actuator, we would get three transfer ⎢ g( s) ⎥
H(s) = ⎢ ⎥
functions ⎢ Vin ( s ) ⎥
⎢ g( s) ⎥
⎢ ⎥
⎣ Vin ( s ) ⎦


Sinusoidal Steady State
> When a LTI system is
driven with a sinusoid, the u (t ) = U 0 cos(ωt )
steady-state response is a
sinusoid at the same Y ( jω ) = H ( jω )U ( jω )
frequency
> The amplitude of the y sss (t ) = Y0 cos(ωt + θ )
response is |H(jω)|
> The phase of the response
relative to the drive is the
angle of H(jω) Y0 = H ( jω ) U 0
> A plot of log magnitude vs Im{H ( jω )}
tanθ =
Re{H ( jω )}
log frequency and angle
vs log frequency is called
a Bode plot



Bode plot of electrostatic actuator
Bode Diagram
> Use Matlab®
command 40
20
bode with previously 0

Magnitude (dB)
-20
defined system sys
-40
-60
> Evaluate only one of TFs -80
-100
g( s)
H(s) = -120
Vin ( s ) -140
180
> This tells us how quickly 135

we can wiggle tip! 90

Phase (deg)
• At a certain OP! 45

0

-45

-90
-2 -1 0 1 2 3
10 10 10 10 10 10
Frequency (rad/sec)




Given O.P.

Linearize Integrate
Take LT Vin(t)?

Form TFs

Transfer SSS Frequency
functions response
poles & Given O.P.
zeros
How fast
Natural
system
circuit tip?
dynamics



Poles and Zeros
> 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
−Q0
g( s) ε ARm
H(s) = =
Vin ( s ) 3 ⎛ 1 b⎞ 2 ⎛ 1 b k⎞ ⎛ 1 k Q02 1 ⎞
s +⎜ + ⎟s +⎜ + ⎟s+⎜ − 2 2 ⎟
⎝ RC0 m ⎠ ⎝ RC0 m m ⎠ ⎝ RC0 m ε A Rm ⎠
εA
where C0 =
ˆ
g0
> MATLAB solution for poles is VERY long


Pole-zero diagram
> Displays information 1
Pole-Zero Map

tip velocity
about dynamics of 0.5
pole
ωd

Imaginary Axis
zero
system function 0

• Matlab command -0.5
pzmap
-1
-9 -8 -7 -6 -5 -4 -3 -2 -1 0

> Useful for examining Real A i
R l axis

dynamics, stability, 1
Pole-Zero Map

gap
etc. 0.5
Imaginary Axis
0

-0.5

-1
-9 -8 -7 -6 -5 -4 -3 -2 -1 0
Real Axis
Real axis




Given O.P.

Linearize Integrate
Take LT Vin(t)?

Ei aly Form TFs
An
g e si
nf s
n
functions response
poles & Given O.P.
zeros
How fast
Natural
system
circuit tip?
dynamics



Eigenfunction Analysis
> For an LTI system, we can find the eigenvalues and
eigenvectors of the J1 (or A) matrix describing the internal
dynamics

For scalar 1st-order system: If we try solution:
dx
= λx ⇒ x(t ) = K 0 e λt + K1 x (t ) = Ke λt
dt
Plug into DE:

λx = Ax
Our linear (or linearized)
homogeneous systems look like:
•This is an eigenvalue
δx (t ) = J1δx + J2δu(t) equation
dx d (δx ) •If we find λ we can find
= Ax = J1δx natural frequencies of
dt dt system



> 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
> We can use either λi or si to
determine natural frequencies
of system

> As we increase applied
voltage
• Stable damped resonant
frequency decreases Increasing voltage

jω
> Plotting poles as system
changes is a root-locus plot

σ



Spring softening
> 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
1.00

Damped resonant frequency
0.90
0.80

ε AV 2 0.70

k'=k− 3
0.60
g 0.50

0.40

This is called 0.30
0 0.01 0.02 0.03 0.04 0.05 0.06
spring softening Voltage
Adapted from Figure 7.5 in Senturia, Stephen D. Microsystem Design.
Boston, MA: Kluwer Academic Publishers, 2001, p. 169. ISBN: 9780792372462.




Given O.P.

Linearize Integrate
Take LT Vin(t)?

Ei aly Form TFs
An
g e si
nf s
n
functions response
poles & Given O.P.
form TFs zeros
How fast
Natural
system
dynamics



Linearized Transducers
> Can we directly linearize
our equivalent circuit? I
1/k U
R
YES! +
C
+ +
> This is perhaps the most Vin(t) + F Fout m
- V
common analysis in the -
W - -
b
literature
Source Transducer Load
> First, choose what is load Find OP
and what is transducer
Linearize
• Here we include spring
δI δU
with transducer R
+ +
+ Linearized
δVin(t) δV δFout m
- Transducer
- -
b

Source Load


Linearized Transducer Model
> First, find O.P.
ˆ
V0 , g 0 , Q0 > Recast in terms of port variables
δI
> Next, generate matrix to ⎡δ Q⎤ ⎡ s ⎤
⎢ ⎥
relate incremental port ⎢δ g ⎥ = ⎢δ U ⎥
⎣ ⎦
variables to each other ⎣ s⎦
• Start from energy and force > Define intermediate variables
relations εA Q0
C0 = , V0 =
V=
Qg
ˆ
g0 C0
εA
Q2
Fout = − k ( g0 − g ) > Final expression
2ε A
• Linearize (take partials…) ⎡ 1 V0 ⎤
ˆ
⎡ g0 Q0 ⎤ ⎡δ V ⎤ ⎢ sC0 sg 0 ⎥ ⎡δ I ⎤
ˆ
⎡δ V ⎤ ⎢ ε A ε A ⎥ ⎡δ Q ⎤ ⎢δ F ⎥ = ⎢ V ⎥⎢ ⎥
k ⎥ ⎣δ U ⎦
⎢δ F ⎥ = ⎢ Q ⎥⎢ ⎥ ⎣ out ⎦ ⎢ 0
⎣ out ⎦ ⎢ 0 k ⎥⎣ ⎦
δg ⎢ sg s ⎥
⎢ε A
⎣ ⎥
⎦ ⎣ ˆ0 ⎦


> Now we want to convert this relation into a circuit
> Many circuit topologies are consistent with this matrix relation
> THIS IS NOT UNIQUE!
i i' u' u i 1:Γ/κ2 u
1:Γ
+ + + + + +
-1/k' 1/k ∗
Co

{
1/k*
υ υ' F' F υ F
Co 1/k
_ _ _ _ _ _

i u i u
1:Γ
+ + + +
Γ2/κ∗ Co 1
1_Γ k_Γ/C o
υ F υ Co F
Co
_ Γ
_ _ _

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



> This is the one used in the text
⎡ δ V ⎤ ⎡ Z EB ϕ Z EB ⎤ ⎡ δ I ⎤
⎢δ F ⎥ = ⎢ϕ Z Z MO ⎥ ⎢δ U ⎥
⎣ out ⎦ ⎣ EB ⎦⎣ ⎦
⎛ ϕ 2 Z EB ⎞
Z MS = Z MO ⎜1 − ⎟
⎝ Z MO ⎠
> Uses a transformer
• Transforms port variables
• Doesn’t store energy
> What we want to do now is
identify ZEB, ZMS and ϕ, and
figure out what they mean…
⎛ e2 ⎞ ⎛ ϕ 0 ⎞⎛ e1 ⎞
⎜ ⎟=⎜
⎜ f ⎟ ⎜0
⎟
− 1 ⎟⎜ f ⎟
⎜ ⎟
⎝ 2⎠ ⎝ ϕ ⎠⎝ 1 ⎠



⎡ δ V ⎤ ⎡ Z EB ϕ Z EB ⎤ ⎡ δ I ⎤
⎢δ F ⎥ = ⎢ϕ Z Z MO ⎥ ⎢δ U ⎥
⎣ out ⎦ ⎣ EB ⎦⎣ ⎦
⎛ ϕ 2 Z EB ⎞
Z MS = Z MO ⎜1 − ⎟
⎝ Z MO ⎠
⎡ 1 V0 ⎤
⎡δV ⎤ ⎢ sC0 sg 0 ⎥ ⎡δI ⎤
ˆ
⎢δF ⎥ = ⎢ V ⎥
k ⎥ ⎢δU ⎥
⎣ ⎦ ⎢ 0 ⎣ ⎦
⎢ sg 0
⎣ ˆ s ⎥ ⎦
⎛ 2 1 ⎞
k ⎜ ⎛ Q0 ⎞ sC0 ⎟ k ⎛ ⎛ Q0 ⎞ 1 ⎞
2
⎜ ⎟
Z MS ⎜ g ⎟ k ⎟ = s ⎜1 − ⎜ g ⎟ C k ⎟
= ⎜1 − ⎜ ⎟
s ⎜ ⎝ ˆ0 ⎠ ⎜ˆ ⎟
⎝ s ⎟
⎠ ⎝ ⎝ 0⎠ 0 ⎠ C0V0
ϕ=
k⎛ Q02 ⎞ Q02 ˆ
g0
= ⎜1 − ⎟ k'= k −
s ⎜ εAkg 0 ⎟
⎝ ˆ ⎠ εAg ˆ 0




> C0 represents the
capacitance of the
structure seen from the
electrical port
> It is simply the capacitance
εA
at the gap given by the C0 =
ˆ
g0
operating point
> As Vin increases, C0 will
increase until the structure
pulls in
> This is a tunable capacitor


> k’ represents the effective
spring
> A combination of the
mechanical spring k and
the electrical spring
> This is an electrically Q02
k'= k −
tunable spring! εAg 0
ˆ
• Spring softening shows up
in k’

> As Vin increases, k’ will
decrease from k (at Vin=0)
to 0 (at Vin=Vpi)


> ϕ represents the
electromechanical
coupling

> Represents how much the
capacitance changes with ϕ=
C0V0 Q0
=
ˆ
g0 ˆ
g0
gap

> A measure of sensitivity ∂C ∂ εA
ϕ = −V0 = −V0
∂g O. P.
∂g g O. P.
εA
= V0
ˆ2
g0
C0V0
=
ˆ
g0



Transfer Functions
> Can use linearized
R 1:ϕ 1/k’ δU
circuit to construct H(s)
m
using complex δVin +
impedances - C0 b

> Usually helpful to
“eliminate” transformer
ϕ2/k’ δU
R
> Transformer changes m/ϕ2
δVin
impedances + C0
Z1 - b/ϕ
2

Z2 =
ϕ2
1:ϕ

Can now get any transfer
Z1 Z2 function using standard
circuit analysis



Linearized Transducer Models
> Now we can understand Nguyen’s filter!

Image removed due to copyright restrictions. Image removed due to copyright restrictions.

Figure 9 on p. 17 in Nguyen, C. T.-C. "Vibrating RF MEMS Figure 12 on p. 62 in: Nguyen, C. T.-C. "Micromechanical

Overview: Applications to Wireless Communications." Filters for Miniaturized Low-power Communications."
Proceedings of SPIE Int Soc Opt Eng 5715 (January 2005): 11-25. Proceedings of SPIE Int Soc Opt Eng 3673 (July 1999): 55-66.



Small-signal analysis summary

Given O.P.

Linearize Integrate
Take LT Vin(t)?

Ei aly Form TFs
An
g e si
nf s
n
functions response
poles & Given O.P.
form TFs zeros
How fast
Natural
system
dynamics



Conclusions
> 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

. 1/k
x

+ ek - +
F
+ em m
-
- eb + -
b

d ⎡ x⎤ ⎡ x ⎤ State x = Ax + Bu
⎢ x ⎥ = ⎢ 1 (F − kx − bx )⎥
dt ⎣ ⎦ ⎢ m ⎣ ⎥
⎦
eqns y = Cx + Du



Direct Integration in Time
> Example: Spring-mass-dashpot step response
• k=m=1;b=0.5;

Step Response
1.5

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

0.5
Velocity

0

-0.5
0 5 10 15 20 25
Tim e (sec)



Transfer Functions
> Can get TFs from A,B,C matrices

Y ( s ) = CX ( s ) + DU( s )
[
Y ( s ) = C (sI − A ) x (0) + (sI − A ) BU( s ) + DU( s )
−1 −1
]
Assume transient has died out (XZIR=0)
No feed-through (D=0)

Y( s) = ⎡C ( sI − A )−1 B ⎤ U( s )
⎣ ⎦
Y( s ) = H ( s )U( s )

H ( s ) = ⎡C ( sI − A ) B ⎤
−1
⎣ ⎦


Transfer Functions
> Let’s do analytically & via MATLAB ⎡ X( s ) ⎤ ⎡ 1 ⎤
⎢ F( s) ⎥ ⎢ 2
⎡ s 0⎤ ⎡ 0 1 ⎤
⎢
⎥
⎥ = ⎢ ms + sb + k ⎥
sI − A = ⎢ ⎥ − ⎢− k
0 s ⎦ ⎢ m − b m⎥
⎥ H( s) =
⎢ X( s ) ⎥ ⎢
⎣ ⎣ ⎦ s ⎥
⎢ ⎥ ⎢ ms 2 + sb + k ⎥
⎡ s −1 ⎤ ⎣ F( s) ⎦ ⎣ ⎦
= ⎢k
⎢ m s+b ⎥ ⎡ 1 ⎤
⎣ m⎥⎦ ⎢ s 2 + 0.5s + 1⎥
H( s) = ⎢ ⎥
1 ⎡ s + b m 1⎤
(sI − A ) = ⎢ k
−1
⎥ ⎢ s ⎥
Δ⎢− s⎥ ⎢ s + 0.5s + 1⎥
⎣ 2
⎦
⎣ m ⎦
Δ = s ( s + b m) + k m = s 2 + s b m + k m >> [n,d]=ss2tf(A,B,C,D)

⎡1 0⎤ 1 ⎡ s + b m 1⎤ ⎡0 ⎤ n= s2 s1 s0
C(sI − A ) B = ⎢
−1
⎥Δ⎢ k ⎥⎢ ⎥
1 0 -0.0000 1.0000
⎣0 1 ⎦ ⎢ − m s ⎥ ⎢ m ⎥
⎣ ⎦⎣ ⎦ 0 1.0000 -0.0000
⎡1 ⎤
1 ⎢ m⎥
= d=
Δ ⎢s ⎥ 1.0000 0.5000 1.0000
⎣ m⎦


Transfer Functions
> Can also construct H(s) directly
X( s ) 1
using complex impedances and = H2 ( s ) =
F( s ) Z( s )
circuit model
.
x 1/k 1
=
b + ms + k s
+ ek - + =
s

F
+ em m
ms 2 + bs + k
-
- eb + - X( s ) sX( s )
=
b F( s ) F( s )
F − ek − em − eb = 0
X( s ) 1
k = H1 ( s ) =
ek = kx = x F( s ) ms 2 + bs + k
s
eb = bx
em = mx = msx


Poles and Zeros
> For 2nd-order system, easy to get
poles and zeros from TFs

⎡ 1 ⎤
1 ⎢s + sb m + k m⎥
2

H(s) = ⎢ ⎥
m⎢ s ⎥ where
⎢ s2 + s b m + k m ⎥
⎣ ⎦
⎡ 1 ⎤ b ⎛ b ⎞ k
2

⎢ ⎥
1 ⎢ ( s − s1 )( s − s2 ) ⎥ s1,2 = − ± ⎜ ⎟ −
= 2m ⎝ 2m ⎠ m
m⎢ s ⎥
⎢ ( s − s )( s − s ) ⎥ these are the poles
⎣ 1 2 ⎦



Spring-mass-dashpot system
> It is a second order system, with
two poles s2 + b s+k = s 2 + 2αs + ω 0
2
m m
> We conventionally define k
ω0 =
• Undamped resonant frequency m
• b
Damping constant α=
2m
• Damped resonant frequency
s1, 2 = −α ± α 2 − ω 0
2
• Quality factor
For underdamped systems (α < ω 0 )
s1, 2 = −α ± jω d
where
ω d = ω 02−α 2
Quality factor :
ω 0 mω 0
Q= =
2α b



Pole-zero diagram
> Displays information about dynamics of system
function s
H2 ( s ) = 2
s + 0.5s + 1
s1, 2 = −0.25 ± j 1 − 1 = −0.25 ± j 0.97
16
Pole-Zero Map
2

1.5
pole
1
Imaginary Axis

0.5
zero ωd
0

-0.5

-1

-1.5

-2
-2 -1 0 1 2
Real Axis



SMD-position frequency response

Bode Diagram
1 1 10
H( jω ) =
m − ω 2 + 2αjω + ω 0
2 0

Magnitude (dB)
1 1 -10
H( jω ) =
m (ω 0 − ω 2 ) 2 + 4α 2ω 2
2 -20

-30
⎛ 2αω ⎞
∠H( jω ) = − atan⎜ 2
⎜ ω −ω 2 ⎟
⎟ -40
⎝ 0 ⎠ 360

315

Phase (deg) 270

225

180
-1 0 1
10 10 10
Frequency (rad/sec)



> Find eigenvalues numerically using MATLAB and A matrix

⎡0 1 ⎤
A=⎢
⎣ 0 − 0.5⎥ ⎦
[ V , Λ ] = eig( A)
⎡λ1 0 ⎤ ⎡− 0.25 + 0.97 j 0 ⎤
Λ=⎢ =⎢
⎣ 0 λ2 ⎥ ⎣
⎦ 0 − 0.25 − 0.97 j ⎥
⎦
⎡ 0.707 0.707 ⎤
V = [v1 v2 ] = ⎢ ⎥
⎣− 0.18 − 0.68 j − 0.18 − 0.68 j ⎦



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