# Identification of the Rotor Time Constant in Induction Machines without Speed Sensor

**ABSTRACT** A differential-algebraic method is used to estimate the rotor time constant T_{R} of an induction motor without measurements of the rotor speed/position. The method consists of solving for the roots of a polynomial equation in T_{R} whose coefficients depend only on the stator currents, stator voltages, and their derivatives. Experimental results are presented

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**ABSTRACT:**Traditionally, when approaching controller design with the Youla-Kucˇera parametrization of all stabilizing controllers, the denominator of the rational parameter is fixed to a given stable polynomial, and optimization is carried out over the numerator polynomial. In this note, we revisit this design technique, allowing to optimize simultaneously over the numerator and denominator polynomials. Stability of the denominator polynomial, as well as fixed-order controller design with H<sub>∞</sub> performance are ensured via the notion of a central polynomial and linear matrix inequality (LMI) conditions for polynomial positivity.IEEE Transactions on Automatic Control 10/2005; · 2.72 Impact Factor - SourceAvailable from: John Chiasson[Show abstract] [Hide abstract]

**ABSTRACT:**A method is proposed to estimate the rotor time constant T<sub>R </sub> of an induction motor without measurements of the rotor speed/position. The method consists of solving for the roots of a polynomial equation in T<sub>R</sub> whose coefficients depend only on the stator currents, stator voltages, and their derivatives. Experimental results are presentedIEEE Transactions on Automatic Control 05/2007; · 2.72 Impact Factor

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Identification of the Rotor Time Constant in

Induction Machines without Speed Sensor

M. Li∗, J.N. Chiasson∗, M. Bodson∗∗, L.M. Tolbert∗

∗The University of Tennessee, ECE Department, Knoxville, USA

∗∗The University of Utah, ECE Department, Salt Lake City, USA

Abstract—A differential-algebraic method is used to estimate

the rotor time constant TR of an induction motor without

measurements of the rotor speed/position. The method consists

of solving for the roots of a polynomial equation in TR whose

coefficients depend only on the stator currents, stator voltages,

and their derivatives. Experimental results are presented.

Index Terms—Rotor Time Constant, Sensorless Speed Ob-

server, Induction Motor.

I. INTRODUCTION

Induction motors are very attractive in many applications

owing to their simple structure, low cost, and robust con-

struction. Field-oriented control is now used to obtain high

performance drive of the induction motor because it gives

control characteristics similar to separately excited DC mo-

tors. Implementation of a (rotor-flux) field-oriented controller

requires knowledge of the rotor speed and the rotor time

constant TR to estimate the rotor flux linkages. There has

been considerable work done in the last several years to im-

plement a field-oriented controller without the use of a speed

sensor [1][2][3][4][5][6]. However, many of these methods

still require the value of TR, which can change with time

due to ohmic heating; that is, to be able to update the value

of TR to the controller as it changes is valuable. The work

presented here uses an algebraic approach to identify the rotor

time constant TR without the motor speed information. It is

most closely related to the ideas described in [7][8][9][10][11].

Specifically, it is shown that TRsatisfies a polynomial equation

whose coefficients are functions of the stator currents, the

stator voltages, and their derivatives. A zero of this polynomial

is the value of TR. It is further shown TRis not identifiable

by this technique under steady-state conditions. It is also

true (and shown here) that a standard least-squares approach

cannot identify TR under steady-state conditions. In [4], the

speed ω and TR are identified assuming constant speed but

not (sinusoidal) steady state. In [12], the speed is assumed

constant, but the flux magnitude is perturbed by a small

amplitude sinusoidal signal to identify TR.

The paper is organized as follows. Section II introduces

a space vector model of the induction motor. Section III uses

this model to develop a differential-algebraic equation that TR

must satisfy. Section IV shows that in steady state, TR is not

identifiable by either the differential-algebraic method nor a

standard linear least-squares method. Section V presents the

experimental results, while Section VI gives the conclusions

and future work.

II. MATHEMATICAL MODEL OF INDUCTION MOTOR

The starting point of the analysis is a space vector model

of the induction motor given by (see e.g., pp. 568 of [13])

d

dtiS=

d

dtψR= −1

dω

dt=npM

β

TR(1 − jnPωTR)ψR− γiS+

TR(1 − jnPωTR)ψR+M

n

where iS, iSa+ jiSb, ψR, ψRa+ jψRb, and uS, uSa+

juSb. Here, θ is the position of the rotor, ω = dθ/dt is the

rotor speed, npis the number of pole pairs, iSa, iSbare the

(two-phase equivalent) stator currents, ψRa, ψRbare the (two-

phase equivalent) rotor flux linkages, RS, RR are the stator

and rotor resistances, respectively, M is the mutual inductance,

LSand LRare the stator and rotor inductances, respectively, J

is the moment of inertia of the rotor, and τLis the load torque.

LR

RR, σ = 1 −

RS

σLS

TR

TRis referred to as the rotor time constant, while σ is called

the total leakage factor.

1

σLSuS

(1)

TRiS

(2)

JLR

ImiSψ∗

R

o

−τL

J,

(3)

The symbols TR =

M2

LSLR, β =

M

σLSLR,

γ =

+βM

have been used to simplify the expressions.

III. DIFFERENTIAL-ALGEBRAIC APPROACH TO TR

ESTIMATION

The idea of the differential-algebraic approach is to solve

(1) and (2) for TR[14][15]. However, equations (1) and (2)

are only four equations while there are six unknowns, namely

ψRa, ψRb, dψRa/dt, dψRb/dt, ω, and TR. Equation (3) is not

used because it introduces the additional unknown τL. To find

two more independent equations, equation (1) is differentiated

to obtain

d2

dt2iS

=

β

TR(1 − jnPωTR)d

−γd

dtψR− jnPβψR

dω

dt

dtiS+

1

σLS

d

dtuS.

(4)

Using the (complex-valued) equations (1) and (2), one can

solve for ψRandd

dtψRin terms of ω, iSand uSand substitute

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the resulting expressions into (4) to obtain

d2

dt2iS= −

1

TR(1 − jnPωTR)

+βM

T2

R

jnPTR

1 − jnPωTR

µd

dtiS+ γiS−

1

σLSuS

1

σLS

¶dω

¶

(1 − jnPωTR)iS− γd

µd

dtiS+

d

dtuS

−

dtiS+ γiS−

1

σLSuS

dt.

(5)

Solving (5) for dω/dt gives

dω

dt= −(1 − jnPωTR)2

βM

T2

R

jnPT2

R

+1 − jnPωTR

jnPTR

×

(1 − jnPωTR)iS− γd

d

dtiS+ γiS−

dtiS+

1

σLS

d

dtuS−d2

dt2iS

1

σLSuS

.

(6)

The left-hand side of (6) is real, so the right-hand side must

also be real. Note by (1) that diS/dt + γiS− uS/(σLS) =

β

TR(1 − jnPωTR)ψRso that the right-hand side of (6) is

singular if and only if

¯¯¯ψR

real part has the form

dω

dt

+a0(uSa,uSb,iSa,iSb).

¯¯¯ψR

¯¯¯ = 0. Other than at startup,

¯¯¯ 6= 0 in normal operation of the motor. Separating the

right-hand side of (6) into its real and imaginary parts, the

=a2(uSa,uSb,iSa,iSb)ω2+ a1(uSa,uSb,iSa,iSb)ω

(7)

The expressions for a2(uSa,uSb,iSa,iSb), a1(uSa,uSb,iSa,

iSb), and a0(uSa,uSb,iSa,iSb) are lengthy in terms of uSa,

uSb, iSa, iSb, and their derivatives as well as of the machine

parameters including TR. As a consequence, they are not

explicitly presented here. Their steady-state expressions are

given in [6].

On the other hand, the imaginary part of the right-hand

side of (6) must be zero. In fact, the imaginary part of (6) is

a second degree polynomial equation in ω of the form

q(ω)

,

q2(uSa,uSb,iSa,iSb)ω2+ q1(uSa,uSb,iSa,iSb)ω

+q0(uSa,uSb,iSa,iSb)

(8)

and, if ω is the speed of the motor, then q(ω) = 0. The

qi are functions of uSa, uSb, iSa, iSb, and their derivatives

as well as of the machine parameters including TR. The

expressions for q2(uSa,uSb,iSa,iSb), q1(uSa,uSb,iSa,iSb),

and q0(uSa,uSb,iSa,iSb) are also lengthy and not explicitly

presented here. (Their steady-state expressions are given in

[6].) If the speed was measured, then (8) would be equal to

zero and could then be solved for TR. However, in the problem

being considered, ω is not known. To eliminate ω, q(ω) in (8)

is differentiated to obtain

d

dtq(ω) = (2q2ω + q1)dω

where dq(ω)/dt ≡ 0 if ω is equal to the motor speed. Next,

dω/dt in (9) is replaced by the right-hand side of (7) so that

dt+ ˙ q2ω2+ ˙ q1ω + ˙ q0

(9)

(9) may be written as

dq(ω)

dt

=g(ω) , 2q2a2ω3+ (2q2a1+ q1a2+ ˙ q2)ω2

+(2q2a0+ q1a1+ ˙ q1)ω + q1a0+ ˙ q0.

(10)

g(ω) is a third-order polynomial equation in ω for which the

speed of the motor is one of its zeros. Dividing1g(ω) in (10)

by q(ω)2in (8), g(ω) may be rewritten as

³

+r1(uSa,uSb,iSa,iSb)ω + r0(uSa,uSb,iSa,iSb)

g(ω) =1

q2

(2q2a2ω + 2q2a1− q1a2+ ˙ q2)q(ω)

´

(11)

r1(uSa,uSb,iSa,iSb)

,

2q2

+q2

2a0− q2q1a1+ q2˙ q1− 2q2q0a2

1a2− q1˙ q2

q2q1a0+ q2˙ q0− 2q2q0a1

+q0q1a2− q0˙ q2.

(12)

r0(uSa,uSb,iSa,iSb)

,

(13)

If ω is equal to the speed of the motor, then both g(ω) = 0

and q(ω) = 0, and one obtains

r(ω) , r1(uSa,uSb,iSa,iSb)ω + r0(uSa,uSb,iSa,iSb) = 0.

(14)

This is now a first-order polynomial equation in ω which

uniquely determines the motor speed ω as long as r1 (the

coefficient of ω) is nonzero. (It is shown in Appendix VII-A

that r1 6= 0 in steady state.) Solving for the motor speed ω

using (14), one obtains

ω = −r0/r1.

(15)

Next, replace ω in (8) by the expression in (15) to obtain

q2r2

0− q1r0r1+ q0r2

1≡ 0.

(16)

The expressions for qi, ri are in terms of motor parameters

(including TR) as well as the stator currents, voltages, and

their derivatives. Expanding the expressions for q0, q1, q2, r0,

and r1, one obtains a twelfth-order polynomial equation in

TR, which can be written as

X

Solvingequation(17)gives

Ci(uSa,uSb,iSa,iSb) of (17) contain third-order derivatives

of the stator currents and second-order derivatives of the stator

voltages making noise a concern. For short time intervals in

which TRdoes not vary, (17) must hold identically with TR

constant. In order to average out the effect of noise on the

Ci, (17) is integrated over a time interval [t1,t2] to obtain

µ

12

i=0

Ci(uSa,uSb,iSa,iSb)Ti

R= 0.

(17)

TR.

Thecoefficients

12

X

1Given

ng,deg{q(ω)} = nq, the Euclidean division algorithm ensures that there

are polynomials γ(ω),r(ω) such that g(ω) = γ(ω)q(ω) + r(ω) and

deg{r(ω)} ≤ deg{q(ω)} − 1 = nq − 1. Consequently if, for example,

ω0is a zero of both g(ω) and q(ω), then it must also be a zero of r(ω).

2q2 6= 0 if ω and the stator electrical frequency ωS are nonzero, which

hold under normal operating conditions. See [6][16].

i=0

1

t2− t1

Zt2

t1

Ci(uSa,uSb,iSa,iSb)dt

¶

Ti

R= 0. (18)

thepolynomials

g(ω),q(ω)

in

ω

with

deg{g(ω)}=

Page 3

3

There are 12 solutions satisfying (18). However, simulation

results have always given 10 conjugate solutions. The remain-

ing two solutions include the correct value of TR while the

other one was either negative or close to zero. The method is to

compute the coefficients

t2−t1

roots of (18). Among the positive real roots is the correct value

of TR. Experimental results using this method are presented

in Section V.

1

Rt2

t1Cidt and then compute the

IV. IDENTIFIABILITY OF TRIN STEADY STATE

A. Differential-algebraic approach

The polynomial (18) is now considered with the machine in

steady state so that, in particular, the speed is constant. That

is, uSa+ juSb = USejωStand iSa+ jiSb = ISejωStare

substituted into (8) and (14). In steady state, the motor speed

in (15) becomes (see Appendix VII-A and [16])

ω = −r0

r1

=ωS(1 − S)

np

(19)

where S , (ωS−npω)/ωSis the normalized slip and ωSis the

electrical frequency. Substituting the steady-state expressions

for q2, q1, and q0as well as the expression (19) for ω into

(8), one obtains q2ω2+ q1ω + q0=

n2

pT2

R|IS|4ω2

σ(1 + S2ω2

npωS|IS|4LS(1 − σ)2³

µωS(1 − S)

That is, in steady state (8) and (14) hold independent of the

value of TRand thus so does (17) making TRunidentifiable

in steady state by this method.

SLS(1 − σ)2(1 − S)

ST2

R)

µωS(1 − S)

ST2

np

¶2

+

1 − ω2

ST2

SLS(1 − σ)2(1 − S)

σ (1 + S2ω2

R(1 − S)2´

σ(1 + S2ω2

¶

R)

×

np

−|IS|4ω2

ST2

R)

≡ 0.

B. Linear least-squares approach

Vélez-Reyes et al [3][4] have used least-squares methods for

simultaneous parameter and speed identification in induction

machines. In the approach used herein, dω/dt is taken to be

zero so that a linear (in the parameters) regressor model can be

obtained. Specifically, consider the mathematical model of the

induction motor in (5). Assuming constant speed, dω/dt = 0

so that this equation reduces to

d2

dt2iS

+βM

T2

R

=−1

TR(1 − jnPωTR)

(1 − jnPωTR)iS− γd

µd

dtiS+ γiS−

1

σLSuS

1

σLS

¶

dtiS+

d

dtuS

(20)

where iS= iSa+ jiSband uS= uSa+ juSb. Decomposing

equation (20) into its real and imaginary parts gives

d2iSa

dtTR

dt

µ

−γdiSa

dtσLS

=

1

µ

−diSa

−diSb

−RS

σLSiSa+

RS

σLSiSb+

duSa

dt

1

σLSuSa

1

σLSuSb

¶

+npω

dt

−

¶

+

1

(21)

and

d2iSb

dt

=

1

TR

µ

−diSb

µ

+

dt

−diSa

−RS

σLSiSb+

RS

σLSiSa+

duSb

dt

1

σLSuSb

¶

−npω

−γdiSb

dt

−

1

σLSuSa

¶

dt

1

σLS

.

(22)

The goal here is to estimate TRwithout knowledge of ω. So, it

is now assumed the motor parameters are all known except for

TR. The set of equations (21) and (22) may then be rewritten

in regressor form as

y(t) = W (t)K

(23)

where K ∈ R2, y ∈ R2, and W ∈ R2×2are given by

K ,

The regressor system (23) is linear in the parameters. The

standard linear least-squares approach is to let (i.e., collect

data at) t = 0,T,2T,··· ,NT, multiply (23) on the left by

WT(nT), sum WT(nT)y(nT) = WT(nT)W (nT)K from

t = 0 to t = NT, and finally compute the solution to

·

1/TR

npω

¸

,

y(t),

duSa

dt

− σLSd2iSa

dt

− RSdiSa

dt

duSb

dt

− σLSd2iSb

dt

− RSdiSb

σLSdiSb

dt

,

W (t) ,

LSdiSa

dt

− uSa+ RSiSa

dt

− uSb+ RSiSb

LSdiSb

dt

− uSb+ RSiSb

−σLSdiSa

dt

+ uSa− RSiSa

.

RWK = RY W

(24)

where

RW,

N

X

n=0

WT(nT)W(nT),RY W,

N

X

n=0

WT(nT)y(nT).

A unique solution to (24) exists if and only if RWis invertible.

However, RW is never invertible in steady state as is now

shown. To proceed, define

·

In steady state where uSa+ juSb = USejωStand iSa+

jiSb = ISejωSt, det(D(t)) = i2

D(t)TD(t) = |IS|2I2×2. Multiply both sides of (23) on the

left by D(t) to obtain

D(t) =

iSb(t)

iSa(t)

−iSa(t)

iSb(t)

¸

.

Sa(t) + i2

Sb(t) = |IS|2,

D(t)y(t) = D(t)W (t)K

or

·

·

RSωS|IS|2− ωSP

σLSω2

−ωSLS|IS|2+ Q

RS|IS|2− P

S|IS|2− ωSQ

¸

=

RS|IS|2− P

σLSωS|IS|2− Q

¸

K (25)

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4

where P , uSaiSa+ uSbiSband Q , uSbiSa− uSaiSbare

the real and reactive powers, respectively, whose steady-state

expressions are given by (30) and (31) in the Appendix. Using

(30) and (31) to replace P and Q in (25), one obtains

¯D

,

D(t)W (t)

−|IS|2(1 − σ)ωSLS

1 + S2ω2

=

ST2

R

·

S2ω2

SωSTR

ST2

R

SωSTR

1

¸

(26)

¯Y

,

D(t)y(t)

−ωS|IS|2(1 − σ)ωSLS

1 + S2ω2

=

ST2

R

·

SωSTR

1

¸

.

(27)

That is, in steady state,¯D , D(t)W (t) ∈ R2×2and¯Y ,

D(t)y(t) ∈ R2are constant matrices. Further, it is easily

seen that the determinant of¯D , D(t)W (t) is zero. Also,

RDW

,

N

X

|IS|2

n=1

(D(nT)W (nT))T(D(nT)W (nT)))

=

N

X

n=1

WT(nT)W(nT) = |IS|2RW.

RDWis singular because D(t)W (t) is constant and singular.

It then follows that RWis also singular using steady-state data.

Further,

RDWY

,

N

X

|IS|2

n=1

(D(nT)W (nT))T(D(nT)y(nT)))

=

N

X

n=1

WT(nT)y(nT) = |IS|2RY W.

Thus RW and RY W are given by

RW

=RDW/|IS|2= N¯DT¯D/|IS|2

N |IS|2(1 − σ)2ω2

1 + S2ω2

=

SL2

S

ST2

R

·

S2ω2

SωSTR

ST2

R

SωSTR

1

¸

(28)

RY W

=RDWY/|IS|2= N¯DT¯Y /|IS|2

ωSN |IS|2(1 − σ)2ω2

1 + S2ω2

=

SL2

S

ST2

R

·

SωSTR

1

¸

,

(29)

where again¯D and¯Y are from (26) and (27), respectively.

By inspection of (28) and (29), K = [0 ωS]Tis one

solution to (24). The null space of RW is generated by

[−1/TRSωS]Tso that all possible solutions are given by

[0 ωS]T+ α[−1/TRSωS]Tfor some α ∈ R. In summary,

solving (24) using steady-state data leads to an infinite set

of solutions so that TR is not identifiable using the linear

regressor (23) with steady-state data.

V. EXPERIMENTAL RESULTS

To demonstrate the viability of the speed sensorless estima-

tor (18) for TR, experiments were performed. A three-phase,

0.5 hp, 1735 rpm (np = 2 pole-pair) induction motor was

driven by an ALLEN-BRADLEY PWM inverter to obtain the

data. Given a speed command to the inverter, it produces PWM

voltages to drive the induction motor to the commanded speed.

Here a step speed command was chosen to bring the motor

from standstill up to the rated speed of 188 rad/s. The stator

currents and voltages were sampled at 10 kHz. The real-time

computing system RTLAB from OPAL-RT with a fully inte-

grated hardware and software system was used to collect data

[17]. Filtered differentiation (using digital filters) was used

for the derivatives of the voltages and currents. Specifically,

the signals were filtered with a third-order Butterworth filter

whose cutoff frequency was 100 Hz. The voltages and currents

were put through a 3 − 2 transformation to obtain their two-

phase equivalent values.

Using the data {uSa,uSb,iSa,iSb} collected between

0.84sec to 0.91sec, which includes the time the motor ac-

celerates, the quantities duSa/dt, duSa/dt, diSa/dt, diSb/dt,

d2iSa/dt2, d2iSb/dt2, d3iSa/dt3, d3iSb/dt3are calculated

and used to evaluate the coefficients Ci, i = 1,2,··· ,12 in

equation (18). Solving (18), one obtains the 12 solutions

TR1= +0.1064

TR3= −0.0576 + j0.0593

TR5= −0.0037 + j0.0166

TR7= −0.0072 + j0.0103

TR9= +0.0125 + j0.0077

TR11= +0.0065 + j0.0018

TR2= −0.0186

TR4= −0.0576 − j0.0593

TR6= −0.0037 − j0.0166

TR8= −0.0072 − j0.0103

TR10= +0.0125 − j0.0077

TR12= +0.0065 − j0.0018.

TR must be a real positive number, so TR = 0.1064 is the

only possible choice. This value compares favorably with the

value of TR= 0.11 obtained using the method of Wang et al

[18], which requires a speed sensor.

To illustrate the identified TR, a simulation of the induction

motor model was carried out using the measured voltages as

input. Then the simulation’s output [stator currents computed

according to (1) and (2)] are used to compare with the

measured (stator currents) outputs. Figure 1 shows the sampled

two-phase equivalent current iSb and its simulated response

iSb−sim. The phase a current iSa is similar, but shifted by

π/(2np). The resulting phase b current iSb−sim from the

simulation corresponds well with the actual measured current

iSb. Note that in equation (1) γ =

on TR.

RS

σLS

+βM

TR

also depends

VI. CONCLUSIONS AND FUTURE WORK

This paper presented a differential-algebraic approach to

the estimation of the rotor time constant of an induction

motor without using a speed sensor. The experimental results

demonstrated the practical viability of this method. Though the

method is not applicable in steady state, neither is a standard

linear least-squares approach. Future work includes studying

an on-line implementation of the estimation algorithm and

using such an online estimate in a speed sensorless field-

oriented controller.

VII. APPENDIX: STEADY-STATE EXPRESSIONS

In the following, ωS denotes the stator frequency and S

denotes the normalized slip defined by S , (ωS− npω)/ωS.

With uSa+juSb= USejωStand iSa+jiSb= ISejωSt, it is

Page 5

5

0.85 0.90.951 1.05

−20

−15

−10

−5

0

5

10

15

20

25

Time in Seconds

Currents in Amps

iSb

iSb−sim

Fig. 1. Phase b current iSband its simulated response iSb−sim.

shown in [19] that under steady-state conditions, the complex

phasors USand ISare related by (Sp,

RR

σωSLR=

1

σωSTR)

IS

=

US

RS+ jωSLS

³³

1 + jS

Sp

US

´

´

/

³

1 + j

S

σSp

´´

=

³

RS+

(1−σ)Sω2

1+S2ω2

SLSTR

ST2

R

+ j

ωSLS(1+σS2ω2

1+S2ω2

ST2

R)

ST2

R

,

and straightforward calculations (see [6]) give

P

,

uSaiSa+ uSbiSb= Re(USI∗

µ

uSbiSa− uSaiSb= Im(USI∗

|IS|2ωSLS

1 + S2ω2

S)

=|IS|2

RS+(1 − σ)Sω2

SLSTR

ST2

R

1 + S2ω2

¶

(30)

Q

,

S)

=

¡1 + σS2ω2

ST2

R

¢

ST2

R

.

(31)

A. Steady-State Expression for r1and r0

It is now shown that the steady-state value of r1in (12) is

nonzero. Substituting the steady-state values of q2, q1, q0, a2,

a1, and a0shown in [6] (noting that ˙ q1≡ 0 and ˙ q2≡ 0 in

steady state) into (12) gives

µ

ω3

S

1 + T2

µ

ω4

S(1 − S)

where

Ãµ(1 − σ)

µ(1 − σ)

r1

=−|IS|6

³

|IS|6

1

1 + S2ω2

S(1 − S)2´2

1

1 + S2ω2

³

ST2

R

¶3n4

p(1 − σ)6L2

σ4

1

den

p(1 − σ)6L2

σ4

R× (1 − S)2´2 1

S

×

Rω2

r0

=

ST2

R

¶3n3

ST2

S

×

1 + ω2

den

den

,

npTR|IS|4

σTR

1 + S2ω2

ST2

R− Sω2

ST2

ST2

R

1 + S2ω2

¶2!

R

¶2

+

σ

ωS

1 + S2ω2

ST2

R

.

(32)

Recall from Section III [following (6)] that den = 0 if and

only if

¯¯¯ψR

¯¯¯ = 0. It is then seen that r16= 0 in steady state.

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