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

Full text

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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)}=

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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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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

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0.850.90.9511.05
−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
õ(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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