# Coordinated Control of Downshift Powertrain of Combined Clutch Transmissions for Electric Vehicles

Authors: Junqiu Li; Han Wei
Journal: Mathematical Problems in Engineering
 (2014)

---

## Abstract

To improve the shift quality of electric vehicles equipped with two-gear automatic transmissions, the coordinated control of the combined clutch and the motor is proposed. The dynamic model of shift process is built up, the dynamic characteristics of each phase of downshift process are analyzed, and linear quadratic optimal control is used to optimize the shift process. As a result, the optimal trajectories of the motor torque and oil pressure of the combined clutch are obtained. Compared to the clutch control only, the simulation results indicate that shift quality is improved remarkably by employing the proposed coordinated control. Specifically, the shift jerk and sliding friction work are decreased by 43% and 44%, respectively, with accelerator pedal angle 50%. In contrast, the reduced percentages are 57% and 89% when accelerator pedal is not depressed.


---

## Body

## 1. Introduction

Electric vehicles (EVs) are paid wide attention to on account of deteriorated environment and energy crisis in recent years. The characteristics of wide working range, constant torque at low speed, and constant power at high speed make the motor suitable for vehicles, and transmissions can be removed in theory. Even so, large EVs still need transmissions with fewer gears to keep climbing performance and high speed performance in balance. Electric buses used in the 2008 Olympic Games adopted an AMT with three gears, whose synchronizers were worn badly [1]. The two-gear combined clutch transmission studied in this paper uses planetary gear set to change the gear ratio, but the torque converter is removed; therefore it integrates the advantages of ATs and AMTs [2, 3].Shift process of conventional automatic transmissions is mostly to engage the oncoming clutch during the process of disengaging the off-going clutch [4–6]. Much different from conventional automatic transmissions, the shift of two-gear transmission is achieved by the combined clutch, which includes a brake and a clutch. Both of them are structurally connected. The shift action is executed by single hydraulic system. Therefore, shift timing problem does not exist any longer and only one hydraulic cylinder is in need, which reduces the complexity of the shift structure and control. Because of these differences, the dynamic model and control strategy of shift process should be studied in detail.Shift control focuses on the control of clutch pressure in initial studies on automatic transmissions [7, 8]. With the increasing demand for shift quality, the coordinated powertrain control has emerged in recent years, the essence of which is coordinated control of the engine and the shift clutch to enhance shift quality and to extend the service life of the clutch simultaneously. With respect to internal combustion engine vehicles (ICEVs), Toyota applied Electric Controlled Transmission-intelligence (ECT-i) to Lexus LS400 [9]. ECT-i can control the automatic transmission A341E and engine concertedly. Ibamoto et al. [10] suggested a way to use the estimated output torque and control of engine ignition advance to optimize the gear shifts of an automatic transmission. A control algorithm that combines speed and torque control of the AMT vehicle powertrain to achieve shift control without using the clutch is proposed [11, 12]. Anna and Govindswamy [13] studied the influence of the engine torque control on the shift quality based on different vehicles. Goetz et al. [14] studied integrated powertrain control of gearshifts on twin clutch transmissions, and a gearshift controller for twin clutch transmissions is developed. The controller incorporates the control of engine variables to achieve synchronization whilst the transfer of engine torque from clutch to clutch is managed by a clutch slip control.The significant difference between EVs and ICEVs is that the driving motor replaces the engine. Good controllability and sensitive response make the driving motor easier to achieve coordinated control of powertrain. Hu et al. and Zhang et al. [15, 16] presented a control strategy of the motor torque and speed to perform the smooth gear shifts in AMTs without releasing the clutch. Gu and Cheng [17] accomplished the coordinated control of upshift power of twin clutch transmission for EVs based on particle swarm optimization (PSO). Zhu et al. [18] studied open-loop control method of DCT shift process under pure electric vehicle system. Both upshift and downshift algorithm were described.Recently, optimal control theory, especially linear quadratic regulator (LQR), has been widely applied in the clutch shift control because of simplicity and engineering advantages. Xue et al. [19] analyzed the oil pressure during the clutch engagement process of CVTs under a variety of work conditions adopting LQR algorithm. Qin and Chen [20] realized the unification of optimal starting control between DCTs and AMTs when the sliding friction work was selected as the minimum object, and the shift jerk was converted to one of constraints.This paper studies the electric vehicle in which the motor is installed in front of the two-speed combined clutch transmission. On the basis of analysis of the shift process, a dynamic model of the shift process is established, and the dynamic characteristics of each phase are analyzed. Linear quadratic optimal control is chosen as the control strategy in which the motor torque and the oil pressure of combined clutch are controlled concertedly to reduce the shift jerk and friction work. As a result, the optimal trajectories of the motor torque and oil pressure of the combined clutch are obtained. The simulation results show that the proposed control strategy can efficiently improve the shift quality.


## 2. Clutch System Description

We consider the powertrain in EVs with a two-gear combined clutch transmission, as schematically shown in Figure1. The powertrain combines the motor characteristics of wide working range, constant torque at low speed, and constant power at high speed with the speed variation of transmission. The combined clutch consists of a brake B and a clutch C, which are structurally connected. Consequently, single hydraulic system can afford to complete the shift action, with oil pressure acting on the brake piston only. Efficient driving and excellent shift quality can be implemented by designing reasonable control strategy.Figure 1
Schematic graph of powertrain with combined clutch transmission.A planetary gear set is adopted as the shift gear and combined clutch is used as the actuator. Power is transferred from the sun gear to the planet carrier. When brake B is engaged and clutch C is disengaged, the powertrain operates on the 1st gear and the speed ratio is given out byi
1
=
1
+
k, where k is the ratio of the teeth number of the ring gear to that of the sun gear. If clutch C is disengaged and brake B is engaged, the vehicle is driven on the 2nd gear with a speed ratio of i
2
=
1.The 2nd to 1st downshift is considered here, which is divided into three phases: clutch C disengagement phase, free phase, and brake B engagement phase. Specifically, firstly hydraulic oil fills the brake cylinder, oil pressure moves the brake piston, and clutch C is disengaged. Then there is a free phase, where both the clutch C and the brake B are disengaged. Lastly, along with the continuous movement of the brake piston, the brake B starts to engage. When the brake B is engaged fully, the downshift action is completed. During the 1st to 2nd upshift, hydraulic system starts to unload and return spring makes the brake B disengaged. After a free phase, clutch C starts to engage until it is fully engaged.


## 3. Combined Clutch Transmission Shift Dynamics Model

The vehicle powertrain may be treated as a multi-rigid-body system. For the purpose of simplification, damping, and elasticity of transmission shaft, bearings and gear mesh for reducing vibration and shock are ignored, and all parts of the powertrain are assumed to exist in the form of concentrated mass [21]. The dynamic model of combined clutch transmission and force analysis of the planet gear are shown in Figures 2 and 3 individually, where T
m is the motor torque. T
f is resistance torque. T
cl
⁡ is the clutch transmission torque. T
br is the brake transmission torque. J
1, J
cl
⁡
1, J
cl
⁡
2, J
br
1, and J
2 are the equivalent moment of inertia of the motor, the driving and driven parts of the clutch, the driving parts of the brake, and the vehicle translational mass, respectively. In addition, ω means the angular velocity. T represents the torque. F shows the force. Parameters relevant to sun gear, ring gear, planet carrier, and planet gears are expressed in subscripts s, r, c, and p.Figure 2
Dynamic model of combined clutch transmission.Figure 3
Force analysis of the planet gear.The dynamic equation of every element in the planetary mechanism during shift process can be presented as [22]
(1)
T
m
-
T
cl
⁡
-
T
s
=
(
J
1
+
J
s
+
J
cl
⁡
1
)
ω
˙
m
T
cl
⁡
-
T
br
-
T
r
=
(
J
cl
⁡
2
+
J
br
1
+
J
r
)
ω
˙
r
T
c
-
T
f
=
(
J
c
+
J
2
)
ω
˙
o
(
F
r
-
F
s
)
R
r
-
R
s
2
=
J
p
ω
˙
p
(
a
)
,
where ω
p
(
a
) is the absolute angular speed of the planet gear:
(2)
ω
p
(
a
)
=
ω
p
(
r
)
+
ω
p
(
e
)
,
where ω
p
(
r
) is the relative angular speed of the planet gear. ω
p
(
r
)
=
(
ω
c
-
ω
s
)
(
2
/
(
k
-
1
)
). ω
p
(
e
) is the transport angular speed of the planet gear, ω
p
(
e
)
=
ω
o.Based on the above equations and according to the planetary gear train kinematics [23],
(3)
ω
p
(
a
)
=
(
ω
o
-
ω
m
)
2
k
-
1
+
ω
o
=
k
+
1
k
-
1
ω
o
-
2
k
-
1
ω
m
ω
m
+
k
ω
r
=
(
1
+
k
)
ω
o
.The principle of virtual work and Lagrange principle can be arranged as follows:(4)
∑
δ
ω
i
·
J
i
ω
˙
i
=
∑
δ
W
i
Z
+
∑
δ
ω
i
·
M
i
e
,
where δ
W
i
Z is virtual work made by the force of constraint, which is zero ignoring the slip of tooth mesh case, and M
i
e is the external torque. As a result, (4) can be transformed into
(5)
∑
i
=
1
n
[
δ
ω
i
(
J
i
ω
˙
i
-
M
i
e
)
]
=
0
,
(6)
δ
ω
s
[
(
J
1
+
J
s
+
J
cl
⁡
1
)
ω
˙
s
+
T
cl
⁡
-
T
m
)
+
δ
ω
r
[
(
J
r
+
J
cl
⁡
2
+
J
br
1
)
ω
˙
r
-
T
cl
⁡
+
T
br
]
+
N
δ
ω
p
×
(
J
p
+
m
p
R
c
2
)
ω
˙
p
(
a
)
+
δ
ω
c
[
(
J
c
+
J
2
)
ω
˙
c
+
T
R
]
=
0
.When (3) are considered, (6) is presented as follows:
(7)
[
(
J
1
+
J
s
+
J
cl
⁡
1
)
+
1
k
2
(
J
r
+
J
cl
⁡
2
+
J
br
1
)
+
4
N
(
k
-
1
)
2
×
(
J
p
+
m
p
R
c
2
)
4
N
(
k
-
1
)
2
1
k
2
]
ω
˙
m
-
[
2
(
k
+
1
)
N
(
k
-
1
)
2
k
+
1
k
2
(
J
r
+
J
cl
⁡
2
+
J
br
1
)
dadadsadsadsdsadasdsad
d
+
2
(
k
+
1
)
N
(
k
-
1
)
2
(
J
p
+
m
p
R
c
2
)
]
ω
˙
o
=
T
m
-
k
+
1
k
T
cl
⁡
+
1
k
T
br
,
[
J
c
+
J
2
+
(
k
+
1
)
2
k
2
(
J
r
+
J
cl
⁡
2
+
J
br
1
)
+
(
k
+
1
)
2
N
(
k
-
1
)
2
×
(
J
p
+
m
p
R
c
2
)
(
k
+
1
)
2
k
2
1
k
2
]
ω
˙
o
-
[
2
(
k
+
1
)
N
(
k
-
1
)
2
k
+
1
k
2
(
J
r
+
J
cl
⁡
2
+
J
br
1
)
gdfgdfgdfgfdgfdgdfgdf
dsd
+
2
(
k
+
1
)
N
(
k
-
1
)
2
(
J
p
+
m
p
R
c
2
)
]
ω
˙
m
=
k
+
1
k
T
cl
⁡
-
k
+
1
k
T
br
-
T
f
.Equation (7) is expressed in matrix form which means the shift dynamic as follows:
(8)
[
J
11
J
12
J
21
J
22
]
[
ω
˙
m
ω
˙
o
]
=
[
1
-
k
+
1
k
1
k
0
0
k
+
1
k
-
k
+
1
k
-
1
]
[
T
m
T
cl
⁡
T
br
T
f
]
=
[
T
m
-
a
11
T
br
-
a
12
T
cl
⁡
-
a
12
T
br
+
a
12
T
cl
⁡
-
T
f
]
,
where
(9)
J
11
=
(
J
1
+
J
s
+
J
cl
⁡
1
)
+
1
k
2
(
J
r
+
J
cl
⁡
2
+
J
br
1
)
+
4
N
(
k
-
1
)
2
(
J
p
+
m
p
R
c
2
)
J
12
=
J
21
=
-
k
+
1
k
2
(
J
r
+
J
cl
⁡
2
+
J
br
1
)
-
2
(
k
+
1
)
N
(
k
-
1
)
2
(
J
p
+
m
p
R
c
2
)
J
22
=
J
c
+
J
2
+
(
k
+
1
)
2
k
2
(
J
r
+
J
cl
⁡
2
+
J
br
1
)
+
(
k
+
1
)
2
N
(
k
-
1
)
2
(
J
p
+
m
p
R
c
2
)
a
11
=
-
1
k
,
a
12
=
k
+
1
k
.
N is the number of the planet gear.T
f depends on road drive resistance, composed of rolling resistance F
f, air resistance F
w, and grade resistance F
i, and can be calculated from (10):
(10)
T
f
=
F
x
r
w
i
0
=
(
F
f
+
F
w
+
F
i
)
r
w
i
0
(11)
F
f
=
m
g
f
cos
⁡
α
(12)
F
i
=
m
g
sin
α
(13)
F
w
=
C
D
A
u
v
2
21.15
,
where m is vehicle mass. g is gravity acceleration. u
v is vehicle speed. α is road ramp. C
D is air resistance coefficient. A is windward area. r
w is tire radius. i
0 is final drive ratio. f is rolling resistance coefficient.During clutch C disengagement phase,T
cl
⁡
≠
0, T
br
=
0, (8) can be transformed into
(14)
[
J
11
J
12
J
21
J
22
]
[
ω
˙
m
ω
˙
o
]
=
[
T
m
-
a
12
T
cl
⁡
a
12
T
cl
⁡
-
T
f
]
.During the free phase,T
cl
⁡
=
0, T
br
=
0, (8) can be transformed into
(15)
[
J
11
J
12
J
21
J
22
]
[
ω
˙
m
ω
˙
o
]
=
[
T
m
-
T
f
]
.During brake B engagement phase,T
cl
⁡
=
0, T
br
≠
0, (8) can be transformed into
(16)
[
J
11
J
12
J
21
J
22
]
[
ω
˙
m
ω
˙
o
]
=
[
T
m
-
a
11
T
br
-
a
12
T
br
-
T
f
]
.


## 4. Optimal Control for Downshift

According to previous studies [24, 25], there are two basic requirements during the engagement of the clutch. One is the comfortability, which means the engagement should be smooth. The other requirement is longer clutch life, which requires the dissipated energy caused by friction to be as small as possible. The shift jerk defined as the vehicle acceleration variation rate is to evaluate the comfortability or smoothness, and the sliding friction work defined as the work produced by friction between the clutch plates is used to evaluate dissipated energy, which has positive correlation with the shift time [26]. The shift jerk and sliding friction work are chosen to establish the object function. The optimal trajectories of oil pressure of the combined clutch and the motor torque are obtained based on linear quadratic optimal control.


### 4.1. Downshift Control Flow

Based on the dynamic analysis of shift and the coordinated control idea of the motor torque and the oil pressure of the combined clutch, the flow chart of the shift control for EVs is proposed as shown in Figure4 [27], where n
r is the speed of ring gear and n
0 is the specified speed value at the end of the free phase. In this paper the downshift coordinated control includes three phases: (1) the coordinated control of the motor torque and the oil pressure of transmission in clutch C disengagement phase; (2) the control of the motor torque in the free phase; (3) the coordinated control of the motor torque and the oil pressure of transmission in brake B engagement phase. After shift, the motor torque and the brake transmission torque should be adjusted to power requirements.Figure 4
Flow chart of downshift coordinated control.


### 4.2. Linear Quadratic Optimal Control

The control law worked out by linear quadratic optimal control is the linear function of state variables, which endows it with significance in engineering. In the shift control, disturbance matrix is out of consideration generally. However, it exists in this paper for further accurate control model.The linear time-varying system can be written into the following state space model:(17)
X
˙
(
t
)
=
A
X
(
t
)
+
B
U
(
t
)
+
V
,
where V is the disturbance matrix. We look for a control which minimizes the performance index:
(18)
J
=
1
2
∫
0
t
f
(
X
(
t
)
T
Q
1
X
(
t
)
+
U
(
t
)
T
Q
2
U
(
t
)
)
d
t
.The Hamiltonian dynamic equation governing the clutch engagement can be constructed and presented in the following form:(19)
H
=
1
2
(
X
(
t
)
T
Q
1
X
(
t
)
+
U
(
t
)
T
Q
2
U
(
t
)
)
+
λ
T
(
A
X
(
t
)
+
B
U
(
t
)
+
V
)
,
where λ is lagrangian multiplier.According to the maximum principle, (19) can be acquired.One has(20)
∂
H
∂
u
=
Q
2
U
(
t
)
+
B
T
λ
(
t
)
=
0
;
equally,
(21)
U
(
t
)
=
-
Q
2
-
1
B
T
λ
(
t
)
.Using the Hamiltonian approach and the adjoint equation, The following differential equations can be obtained(22)
λ
˙
(
t
)
=
-
∂
H
∂
x
=
-
Q
X
(
t
)
-
A
T
λ
(
t
)
,
X
˙
(
t
)
=
A
X
(
t
)
-
B
Q
2
-
1
B
T
λ
(
t
)
+
V
,
with the set of conditions
(23)
X
(
t
0
)
=
x
0
,
(24)
N
[
X
(
t
f
)
,
t
f
]
=
0
,
(25)
λ
(
t
f
)
=
∂
N
T
∂
X
(
t
f
)
v
,
where t
0 is initial time, t
f is the terminal time, v is the unknown constant, and N is the terminal constraint.Considering the effect of the perturbation matrix and the terminal constraint [28], define
(26)
λ
(
t
)
=
P
(
t
)
X
+
M
(
t
)
v
+
h
(
t
)
(27)
ψ
=
K
(
t
)
X
+
L
(
t
)
v
+
r
(
t
)
,
where ψ is the terminal function. P
(
t
) is the matrix that solves the Riccati equation. h
(
t
) has been introduced in order to compensate for the presence of the disturbance vector V. M
(
t
) is used to solve the problem of the unknown constant v. By computing (26)-(27) at t
f, the P
(
t
f
), M
(
t
f
), h
(
t
f
), K
(
t
f
), L
(
t
f
), and r
(
t
f
) can be obtained.By substituting (26)-(27) in (22), after simple algebraic manipulations one obtains the following three matrix differential equations:
(28)
P
˙
(
t
)
+
P
(
t
)
A
+
A
T
P
(
t
)
-
P
(
t
)
B
Q
2
-
1
B
T
P
(
t
)
+
Q
1
=
0
M
˙
(
t
)
-
P
(
t
)
B
Q
2
-
1
B
T
M
(
t
)
+
A
T
M
(
t
)
=
0
h
˙
(
t
)
-
P
(
t
)
B
Q
2
-
1
B
T
h
(
t
)
+
A
T
h
(
t
)
+
P
(
t
)
V
=
0
.These differential equations can be solved backward in time from the above terminal conditions. Now, by differentiating (27) we have
(29)
K
˙
(
t
)
X
+
K
(
t
)
X
˙
+
L
˙
(
t
)
v
+
r
˙
(
t
)
=
0
,
and, by using (22)
(30)
[
K
˙
(
t
)
+
K
(
t
)
A
1
-
K
(
t
)
B
Q
2
-
1
B
T
P
(
t
)
]
X
(
t
)
+
[
L
(
t
)
-
K
(
t
)
B
Q
2
-
1
B
T
M
(
t
)
]
v
+
r
˙
-
K
B
Q
2
-
1
B
T
h
+
K
V
=
0
,
that holds for any X
(
t
) and v. Therefore
(31)
K
˙
(
t
)
+
K
(
t
)
A
-
K
(
t
)
B
Q
2
-
1
B
T
P
(
t
)
=
0
L
˙
(
t
)
-
K
(
t
)
B
Q
2
-
1
B
T
M
(
t
)
=
0
r
˙
(
t
)
-
K
(
t
)
B
Q
2
-
1
B
T
h
(
t
)
+
K
(
t
)
V
=
0
,
which determine K
(
t
), L
(
t
), and r
(
t
). From (23), since v is a constant, one obtains
(32)
v
=
L
-
1
(
t
0
)
[
ψ
-
K
(
t
0
)
X
(
t
0
)
+
r
(
t
0
)
]
.Finally, the optimal control variablesU
*
(
t
) and state variables X
*
(
t
) can be obtained.


### 4.3. Optimal Control Model for Downshift

#### 4.3.1. Clutch C Disengagement Phase

Considering the shift jerk and sliding friction work of downshift process and the coordinated control idea, the following state variables, control variables, and transformations are introduced below:(33)
x
1
=
ω
m
-
ω
r
,
x
2
=
T
m
,
x
3
=
T
cl
⁡
,
u
1
=
d
T
m
d
t
,
u
2
=
d
T
cl
⁡
d
t
.Substitution of transformations in (33) into (14) results in the equation of motion of the dynamic system in terms of the state variables as follows:
(34)
X
˙
=
A
1
X
+
B
1
U
+
V
1
,
where
(35)
X
=
[
x
1
x
2
x
3
]
T
,
U
=
[
u
1
u
2
]
T
,
A
1
=
[
0
a
12
(
J
21
+
J
22
)
J
11
J
22
-
J
12
J
21
-
a
12
2
(
J
11
+
2
J
12
+
J
22
)
J
11
J
22
-
J
12
J
21
0
0
0
0
0
0
]
,
B
1
=
[
0
0
1
0
0
1
]
,
V
1
=
[
a
12
(
J
11
+
J
21
)
J
11
J
22
-
J
12
J
21
T
f
0
0
]
.When clutch C is disengaged completely, its torque must be zero; that is, the terminal constraint is(36)
N
[
x
(
t
m
)
,
t
m
]
=
x
3
(
t
m
)
=
0
,
where t
m is the end time of clutch C disengagement.In this phase, the shift jerkj and the sliding friction work W are expressed below:
(37)
j
=
d
a
d
t
=
r
w
i
0
d
2
ω
o
d
t
2
=
r
w
i
0
(
J
11
J
22
-
J
21
J
12
)
×
[
a
12
(
J
11
+
J
21
)
u
2
-
J
21
u
1
]
,
W
=
∫
0
t
m
T
cl
⁡
(
ω
m
-
ω
r
)
d
t
=
∫
0
t
m
x
1
x
3
d
t
.It is shown through experiments that the shift jerk is inversely proportional to friction work and to time as well. Therefore, the shift jerk may be reduced by increasing the frictional time but cause short service life of the clutch at the expense of the plate excessive wear. For a compromise of the two evaluation criterions in contradictory, an objective function is proposed for the optimal control and expressed below:(38)
J
=
1
2
∫
0
t
m
(
W
+
η
j
2
)
d
t
=
1
2
∫
0
t
m
{
x
1
x
3
+
η
[
a
12
(
J
11
+
J
21
)
u
2
-
J
21
u
1
]
2
}
d
t
=
1
2
∫
0
t
m
(
x
T
Q
1
x
+
u
T
Q
2
u
)
d
t
,
where η
(
0
<
η
<
1
) is weight coefficient of the shift jerk and the larger the η is, the more the shift jerk is considered [29]:
(39)
Q
1
=
[
0
0
0.5
0
0
0
0.5
0
0
]
,
Q
2
=
η
[
J
21
1
0
-
2
a
12
(
J
11
+
J
21
)
J
21
a
12
2
(
J
11
+
J
21
)
2
]
.Obviously, the optimal coordinated control is equivalent to seeking the optimal trajectories of the motor torque and the clutch C transmission torque to minimize the value of the object function.By computing (26)-(27) at t
m, one obtains
(40)
P
(
t
m
)
=
[
0
0
0
0
0
0
0
0
0
]
,
M
(
t
m
)
=
[
0
0
1
]
,
222222222222
h
(
t
m
)
=
[
0
0
0
]
K
(
t
m
)
=
[
0
0
1
]
,
L
(
t
m
)
=
0
,
r
(
t
m
)
=
0
.The optimal trajectories of motor torque and clutch C transmission torque in the clutch C disengagement phase can be derived from solving the differential equations as described above with different weight coefficient.


#### 4.3.2. Free Phase

During this phase, both the clutch C and the brake B are disengaged. When the motor torque is constant, the shift jerk and the sliding friction work will keep being zero, which is an ideal state. However, in order to accomplish the fast synchronization of the driving and driven parts of the brake and to decrease the sliding friction work in the brake B engagement phase, the motor torque is supposed to be controlled under the condition that the shift jerk is less than the recommended value, 10 m/s3.Define(41)
x
1
=
ω
r
,
x
2
=
T
m
,
u
=
d
T
m
d
t
.The state equation has the following expression, resembling (15):
(42)
X
˙
=
A
2
X
+
B
2
U
+
V
2
,
where
(43)
X
=
[
x
1
x
2
]
T
,
U
=
[
u
1
]
T
A
2
=
[
0
a
11
J
22
-
a
12
J
21
J
11
J
22
-
J
12
J
21
0
0
]
,
B
2
=
[
0
1
]
,
V
2
=
[
(
a
11
J
12
-
a
12
J
11
)
T
f
J
11
J
22
-
J
12
J
21
0
]
.At the end of free phase, the angular velocity of the ring gear is required to be less than 30 rad/s; that is, the terminal constraint is(44)
N
[
x
(
t
n
)
,
t
n
]
=
x
1
(
t
n
)
≤
30
,
where t
n is the end time of free phase.During this phase, the shift jerk is expressed below:(45)
j
=
r
w
i
0
d
d
t
(
J
21
T
m
J
11
J
22
-
J
21
J
12
)
=
J
21
r
w
i
0
(
J
11
J
22
-
J
21
J
12
)
u
.As a result, the object function can be given as follows:(46)
J
=
1
2
∫
0
t
n
u
2
d
t
=
1
2
∫
0
t
n
U
T
Q
2
U
d
t
,
where Q
1
=
0, Q
2
=
[
1
].The optimal torque of the motor control law in free phase can be derived from solutions of the above differential matrix equations.During this phase, ignoring the frictional resistance in sealing ring and splines, the equation of motion of the brake piston is established as(47)
m
c
x
¨
c
+
c
x
˙
c
+
k
s
(
x
0
+
x
c
)
=
p
A
c
,
where m
c is the equivalent piston mass. x
0 is the initial compression amount of the return spring. x
c is piston displacement. c is the piston friction coefficient. k
s is the return spring stiffness. p is the oil pressure.According to (47), the oil pressure of combined clutch in free phase can be acquired.


#### 4.3.3. Brake B Engagement Phase

Define(48)
x
1
=
ω
r
,
x
2
=
T
m
,
x
3
=
T
br
,
u
1
=
d
T
m
d
t
,
u
2
=
d
T
br
d
t
.The state equation (51) can be derived from (14):
(49)
X
˙
=
A
3
X
+
B
3
U
+
V
3
,
where
(50)
X
=
[
x
1
x
2
x
3
]
T
,
U
=
[
u
1
u
2
]
T
,
A
3
=
[
0
a
11
J
22
-
a
12
J
21
J
11
J
22
-
J
12
J
21
2
a
11
a
12
J
12
-
a
11
2
J
22
-
a
12
2
J
11
J
11
J
22
-
J
12
J
21
0
0
0
0
0
0
]
,
B
3
=
[
0
0
1
0
0
1
]
,
V
3
=
[
a
11
J
12
-
a
12
J
11
J
11
J
22
-
J
12
J
21
T
f
0
0
]
.At the end of the brake engagement phase, the speed difference of the driving and driven parts of the brake B is zero; that is, the terminal constraint is(51)
N
[
x
(
t
f
)
,
t
f
]
=
x
1
(
t
f
)
=
0
,
where t
f is the end time of brake B engagement phase.In this phase, the shift jerkj and the sliding friction work W are expressed below:
(52)
j
=
r
w
i
0
(
J
11
J
22
-
J
21
J
12
)
×
[
(
a
11
J
21
-
a
12
J
11
)
u
2
-
J
21
u
1
]
W
=
∫
0
t
f
T
br
ω
r
d
t
=
∫
0
t
f
x
1
x
3
d
t
.Similar to the clutch C disengagement phase, the objective function can be expressed as follows:(53)
J
=
1
2
∫
0
t
f
(
W
+
η
j
2
)
d
t
=
1
2
∫
0
t
f
{
x
1
x
3
+
[
(
a
11
J
21
-
a
12
J
11
)
u
2
-
J
21
u
1
]
2
}
d
t
=
1
2
∫
0
t
f
(
X
T
Q
1
X
+
U
T
Q
2
U
)
d
t
,
where
(54)
Q
1
=
[
0
0
0.5
0
0
0
0.5
0
0
]
,
Q
2
=
η
[
J
21
2
0
-
2
J
21
(
a
11
J
21
-
a
12
J
11
)
(
a
11
J
21
-
a
12
J
11
)
2
]
.The optimal trajectories of the motor torque and brake B transmission torque in brake engagement phase can be acquired with reference to the solving process in clutch disengagement phase.The correlation of the clutch transmission torque and the oil pressure is given by (55) [30]. Apparently, the optimal trajectory of the oil pressure can be transformed from the optimal trajectory of the clutch/brake transmission torque:
(55)
T
=
μ
Z
2
(
R
2
3
-
R
1
3
)
3
(
R
2
2
-
R
1
2
)
|
F
s
-
p
A
c
|
,
where T is the clutch/brake transmission torque. F
s is the return spring force. μ is the friction coefficient of clutch/brake plate, treated as a constant in this paper. R
1 and R
2 are clutch/brake plate inner and outer radius, respectively. Z is the plate number of the clutch/brake. A
c is the brake piston area.In reality, the shift time is very short, which makes the real-time online calculation of the differential matrix equation impractical. A solution to this problem is to calculate the optimal trajectory under different working conditions offline and then to conduct function fitting. At last, the fitting coefficients are stored in the memory of vehicle control system. We can get the optimal trajectory quickly through the look-up table and interpolation way to get fit coefficients in online control.


## 4.1. Downshift Control Flow

Based on the dynamic analysis of shift and the coordinated control idea of the motor torque and the oil pressure of the combined clutch, the flow chart of the shift control for EVs is proposed as shown in Figure4 [27], where n
r is the speed of ring gear and n
0 is the specified speed value at the end of the free phase. In this paper the downshift coordinated control includes three phases: (1) the coordinated control of the motor torque and the oil pressure of transmission in clutch C disengagement phase; (2) the control of the motor torque in the free phase; (3) the coordinated control of the motor torque and the oil pressure of transmission in brake B engagement phase. After shift, the motor torque and the brake transmission torque should be adjusted to power requirements.Figure 4
Flow chart of downshift coordinated control.


## 4.2. Linear Quadratic Optimal Control

The control law worked out by linear quadratic optimal control is the linear function of state variables, which endows it with significance in engineering. In the shift control, disturbance matrix is out of consideration generally. However, it exists in this paper for further accurate control model.The linear time-varying system can be written into the following state space model:(17)
X
˙
(
t
)
=
A
X
(
t
)
+
B
U
(
t
)
+
V
,
where V is the disturbance matrix. We look for a control which minimizes the performance index:
(18)
J
=
1
2
∫
0
t
f
(
X
(
t
)
T
Q
1
X
(
t
)
+
U
(
t
)
T
Q
2
U
(
t
)
)
d
t
.The Hamiltonian dynamic equation governing the clutch engagement can be constructed and presented in the following form:(19)
H
=
1
2
(
X
(
t
)
T
Q
1
X
(
t
)
+
U
(
t
)
T
Q
2
U
(
t
)
)
+
λ
T
(
A
X
(
t
)
+
B
U
(
t
)
+
V
)
,
where λ is lagrangian multiplier.According to the maximum principle, (19) can be acquired.One has(20)
∂
H
∂
u
=
Q
2
U
(
t
)
+
B
T
λ
(
t
)
=
0
;
equally,
(21)
U
(
t
)
=
-
Q
2
-
1
B
T
λ
(
t
)
.Using the Hamiltonian approach and the adjoint equation, The following differential equations can be obtained(22)
λ
˙
(
t
)
=
-
∂
H
∂
x
=
-
Q
X
(
t
)
-
A
T
λ
(
t
)
,
X
˙
(
t
)
=
A
X
(
t
)
-
B
Q
2
-
1
B
T
λ
(
t
)
+
V
,
with the set of conditions
(23)
X
(
t
0
)
=
x
0
,
(24)
N
[
X
(
t
f
)
,
t
f
]
=
0
,
(25)
λ
(
t
f
)
=
∂
N
T
∂
X
(
t
f
)
v
,
where t
0 is initial time, t
f is the terminal time, v is the unknown constant, and N is the terminal constraint.Considering the effect of the perturbation matrix and the terminal constraint [28], define
(26)
λ
(
t
)
=
P
(
t
)
X
+
M
(
t
)
v
+
h
(
t
)
(27)
ψ
=
K
(
t
)
X
+
L
(
t
)
v
+
r
(
t
)
,
where ψ is the terminal function. P
(
t
) is the matrix that solves the Riccati equation. h
(
t
) has been introduced in order to compensate for the presence of the disturbance vector V. M
(
t
) is used to solve the problem of the unknown constant v. By computing (26)-(27) at t
f, the P
(
t
f
), M
(
t
f
), h
(
t
f
), K
(
t
f
), L
(
t
f
), and r
(
t
f
) can be obtained.By substituting (26)-(27) in (22), after simple algebraic manipulations one obtains the following three matrix differential equations:
(28)
P
˙
(
t
)
+
P
(
t
)
A
+
A
T
P
(
t
)
-
P
(
t
)
B
Q
2
-
1
B
T
P
(
t
)
+
Q
1
=
0
M
˙
(
t
)
-
P
(
t
)
B
Q
2
-
1
B
T
M
(
t
)
+
A
T
M
(
t
)
=
0
h
˙
(
t
)
-
P
(
t
)
B
Q
2
-
1
B
T
h
(
t
)
+
A
T
h
(
t
)
+
P
(
t
)
V
=
0
.These differential equations can be solved backward in time from the above terminal conditions. Now, by differentiating (27) we have
(29)
K
˙
(
t
)
X
+
K
(
t
)
X
˙
+
L
˙
(
t
)
v
+
r
˙
(
t
)
=
0
,
and, by using (22)
(30)
[
K
˙
(
t
)
+
K
(
t
)
A
1
-
K
(
t
)
B
Q
2
-
1
B
T
P
(
t
)
]
X
(
t
)
+
[
L
(
t
)
-
K
(
t
)
B
Q
2
-
1
B
T
M
(
t
)
]
v
+
r
˙
-
K
B
Q
2
-
1
B
T
h
+
K
V
=
0
,
that holds for any X
(
t
) and v. Therefore
(31)
K
˙
(
t
)
+
K
(
t
)
A
-
K
(
t
)
B
Q
2
-
1
B
T
P
(
t
)
=
0
L
˙
(
t
)
-
K
(
t
)
B
Q
2
-
1
B
T
M
(
t
)
=
0
r
˙
(
t
)
-
K
(
t
)
B
Q
2
-
1
B
T
h
(
t
)
+
K
(
t
)
V
=
0
,
which determine K
(
t
), L
(
t
), and r
(
t
). From (23), since v is a constant, one obtains
(32)
v
=
L
-
1
(
t
0
)
[
ψ
-
K
(
t
0
)
X
(
t
0
)
+
r
(
t
0
)
]
.Finally, the optimal control variablesU
*
(
t
) and state variables X
*
(
t
) can be obtained.


## 4.3. Optimal Control Model for Downshift

### 4.3.1. Clutch C Disengagement Phase

Considering the shift jerk and sliding friction work of downshift process and the coordinated control idea, the following state variables, control variables, and transformations are introduced below:(33)
x
1
=
ω
m
-
ω
r
,
x
2
=
T
m
,
x
3
=
T
cl
⁡
,
u
1
=
d
T
m
d
t
,
u
2
=
d
T
cl
⁡
d
t
.Substitution of transformations in (33) into (14) results in the equation of motion of the dynamic system in terms of the state variables as follows:
(34)
X
˙
=
A
1
X
+
B
1
U
+
V
1
,
where
(35)
X
=
[
x
1
x
2
x
3
]
T
,
U
=
[
u
1
u
2
]
T
,
A
1
=
[
0
a
12
(
J
21
+
J
22
)
J
11
J
22
-
J
12
J
21
-
a
12
2
(
J
11
+
2
J
12
+
J
22
)
J
11
J
22
-
J
12
J
21
0
0
0
0
0
0
]
,
B
1
=
[
0
0
1
0
0
1
]
,
V
1
=
[
a
12
(
J
11
+
J
21
)
J
11
J
22
-
J
12
J
21
T
f
0
0
]
.When clutch C is disengaged completely, its torque must be zero; that is, the terminal constraint is(36)
N
[
x
(
t
m
)
,
t
m
]
=
x
3
(
t
m
)
=
0
,
where t
m is the end time of clutch C disengagement.In this phase, the shift jerkj and the sliding friction work W are expressed below:
(37)
j
=
d
a
d
t
=
r
w
i
0
d
2
ω
o
d
t
2
=
r
w
i
0
(
J
11
J
22
-
J
21
J
12
)
×
[
a
12
(
J
11
+
J
21
)
u
2
-
J
21
u
1
]
,
W
=
∫
0
t
m
T
cl
⁡
(
ω
m
-
ω
r
)
d
t
=
∫
0
t
m
x
1
x
3
d
t
.It is shown through experiments that the shift jerk is inversely proportional to friction work and to time as well. Therefore, the shift jerk may be reduced by increasing the frictional time but cause short service life of the clutch at the expense of the plate excessive wear. For a compromise of the two evaluation criterions in contradictory, an objective function is proposed for the optimal control and expressed below:(38)
J
=
1
2
∫
0
t
m
(
W
+
η
j
2
)
d
t
=
1
2
∫
0
t
m
{
x
1
x
3
+
η
[
a
12
(
J
11
+
J
21
)
u
2
-
J
21
u
1
]
2
}
d
t
=
1
2
∫
0
t
m
(
x
T
Q
1
x
+
u
T
Q
2
u
)
d
t
,
where η
(
0
<
η
<
1
) is weight coefficient of the shift jerk and the larger the η is, the more the shift jerk is considered [29]:
(39)
Q
1
=
[
0
0
0.5
0
0
0
0.5
0
0
]
,
Q
2
=
η
[
J
21
1
0
-
2
a
12
(
J
11
+
J
21
)
J
21
a
12
2
(
J
11
+
J
21
)
2
]
.Obviously, the optimal coordinated control is equivalent to seeking the optimal trajectories of the motor torque and the clutch C transmission torque to minimize the value of the object function.By computing (26)-(27) at t
m, one obtains
(40)
P
(
t
m
)
=
[
0
0
0
0
0
0
0
0
0
]
,
M
(
t
m
)
=
[
0
0
1
]
,
222222222222
h
(
t
m
)
=
[
0
0
0
]
K
(
t
m
)
=
[
0
0
1
]
,
L
(
t
m
)
=
0
,
r
(
t
m
)
=
0
.The optimal trajectories of motor torque and clutch C transmission torque in the clutch C disengagement phase can be derived from solving the differential equations as described above with different weight coefficient.


### 4.3.2. Free Phase

During this phase, both the clutch C and the brake B are disengaged. When the motor torque is constant, the shift jerk and the sliding friction work will keep being zero, which is an ideal state. However, in order to accomplish the fast synchronization of the driving and driven parts of the brake and to decrease the sliding friction work in the brake B engagement phase, the motor torque is supposed to be controlled under the condition that the shift jerk is less than the recommended value, 10 m/s3.Define(41)
x
1
=
ω
r
,
x
2
=
T
m
,
u
=
d
T
m
d
t
.The state equation has the following expression, resembling (15):
(42)
X
˙
=
A
2
X
+
B
2
U
+
V
2
,
where
(43)
X
=
[
x
1
x
2
]
T
,
U
=
[
u
1
]
T
A
2
=
[
0
a
11
J
22
-
a
12
J
21
J
11
J
22
-
J
12
J
21
0
0
]
,
B
2
=
[
0
1
]
,
V
2
=
[
(
a
11
J
12
-
a
12
J
11
)
T
f
J
11
J
22
-
J
12
J
21
0
]
.At the end of free phase, the angular velocity of the ring gear is required to be less than 30 rad/s; that is, the terminal constraint is(44)
N
[
x
(
t
n
)
,
t
n
]
=
x
1
(
t
n
)
≤
30
,
where t
n is the end time of free phase.During this phase, the shift jerk is expressed below:(45)
j
=
r
w
i
0
d
d
t
(
J
21
T
m
J
11
J
22
-
J
21
J
12
)
=
J
21
r
w
i
0
(
J
11
J
22
-
J
21
J
12
)
u
.As a result, the object function can be given as follows:(46)
J
=
1
2
∫
0
t
n
u
2
d
t
=
1
2
∫
0
t
n
U
T
Q
2
U
d
t
,
where Q
1
=
0, Q
2
=
[
1
].The optimal torque of the motor control law in free phase can be derived from solutions of the above differential matrix equations.During this phase, ignoring the frictional resistance in sealing ring and splines, the equation of motion of the brake piston is established as(47)
m
c
x
¨
c
+
c
x
˙
c
+
k
s
(
x
0
+
x
c
)
=
p
A
c
,
where m
c is the equivalent piston mass. x
0 is the initial compression amount of the return spring. x
c is piston displacement. c is the piston friction coefficient. k
s is the return spring stiffness. p is the oil pressure.According to (47), the oil pressure of combined clutch in free phase can be acquired.


### 4.3.3. Brake B Engagement Phase

Define(48)
x
1
=
ω
r
,
x
2
=
T
m
,
x
3
=
T
br
,
u
1
=
d
T
m
d
t
,
u
2
=
d
T
br
d
t
.The state equation (51) can be derived from (14):
(49)
X
˙
=
A
3
X
+
B
3
U
+
V
3
,
where
(50)
X
=
[
x
1
x
2
x
3
]
T
,
U
=
[
u
1
u
2
]
T
,
A
3
=
[
0
a
11
J
22
-
a
12
J
21
J
11
J
22
-
J
12
J
21
2
a
11
a
12
J
12
-
a
11
2
J
22
-
a
12
2
J
11
J
11
J
22
-
J
12
J
21
0
0
0
0
0
0
]
,
B
3
=
[
0
0
1
0
0
1
]
,
V
3
=
[
a
11
J
12
-
a
12
J
11
J
11
J
22
-
J
12
J
21
T
f
0
0
]
.At the end of the brake engagement phase, the speed difference of the driving and driven parts of the brake B is zero; that is, the terminal constraint is(51)
N
[
x
(
t
f
)
,
t
f
]
=
x
1
(
t
f
)
=
0
,
where t
f is the end time of brake B engagement phase.In this phase, the shift jerkj and the sliding friction work W are expressed below:
(52)
j
=
r
w
i
0
(
J
11
J
22
-
J
21
J
12
)
×
[
(
a
11
J
21
-
a
12
J
11
)
u
2
-
J
21
u
1
]
W
=
∫
0
t
f
T
br
ω
r
d
t
=
∫
0
t
f
x
1
x
3
d
t
.Similar to the clutch C disengagement phase, the objective function can be expressed as follows:(53)
J
=
1
2
∫
0
t
f
(
W
+
η
j
2
)
d
t
=
1
2
∫
0
t
f
{
x
1
x
3
+
[
(
a
11
J
21
-
a
12
J
11
)
u
2
-
J
21
u
1
]
2
}
d
t
=
1
2
∫
0
t
f
(
X
T
Q
1
X
+
U
T
Q
2
U
)
d
t
,
where
(54)
Q
1
=
[
0
0
0.5
0
0
0
0.5
0
0
]
,
Q
2
=
η
[
J
21
2
0
-
2
J
21
(
a
11
J
21
-
a
12
J
11
)
(
a
11
J
21
-
a
12
J
11
)
2
]
.The optimal trajectories of the motor torque and brake B transmission torque in brake engagement phase can be acquired with reference to the solving process in clutch disengagement phase.The correlation of the clutch transmission torque and the oil pressure is given by (55) [30]. Apparently, the optimal trajectory of the oil pressure can be transformed from the optimal trajectory of the clutch/brake transmission torque:
(55)
T
=
μ
Z
2
(
R
2
3
-
R
1
3
)
3
(
R
2
2
-
R
1
2
)
|
F
s
-
p
A
c
|
,
where T is the clutch/brake transmission torque. F
s is the return spring force. μ is the friction coefficient of clutch/brake plate, treated as a constant in this paper. R
1 and R
2 are clutch/brake plate inner and outer radius, respectively. Z is the plate number of the clutch/brake. A
c is the brake piston area.In reality, the shift time is very short, which makes the real-time online calculation of the differential matrix equation impractical. A solution to this problem is to calculate the optimal trajectory under different working conditions offline and then to conduct function fitting. At last, the fitting coefficients are stored in the memory of vehicle control system. We can get the optimal trajectory quickly through the look-up table and interpolation way to get fit coefficients in online control.


## 4.3.1. Clutch C Disengagement Phase

Considering the shift jerk and sliding friction work of downshift process and the coordinated control idea, the following state variables, control variables, and transformations are introduced below:(33)
x
1
=
ω
m
-
ω
r
,
x
2
=
T
m
,
x
3
=
T
cl
⁡
,
u
1
=
d
T
m
d
t
,
u
2
=
d
T
cl
⁡
d
t
.Substitution of transformations in (33) into (14) results in the equation of motion of the dynamic system in terms of the state variables as follows:
(34)
X
˙
=
A
1
X
+
B
1
U
+
V
1
,
where
(35)
X
=
[
x
1
x
2
x
3
]
T
,
U
=
[
u
1
u
2
]
T
,
A
1
=
[
0
a
12
(
J
21
+
J
22
)
J
11
J
22
-
J
12
J
21
-
a
12
2
(
J
11
+
2
J
12
+
J
22
)
J
11
J
22
-
J
12
J
21
0
0
0
0
0
0
]
,
B
1
=
[
0
0
1
0
0
1
]
,
V
1
=
[
a
12
(
J
11
+
J
21
)
J
11
J
22
-
J
12
J
21
T
f
0
0
]
.When clutch C is disengaged completely, its torque must be zero; that is, the terminal constraint is(36)
N
[
x
(
t
m
)
,
t
m
]
=
x
3
(
t
m
)
=
0
,
where t
m is the end time of clutch C disengagement.In this phase, the shift jerkj and the sliding friction work W are expressed below:
(37)
j
=
d
a
d
t
=
r
w
i
0
d
2
ω
o
d
t
2
=
r
w
i
0
(
J
11
J
22
-
J
21
J
12
)
×
[
a
12
(
J
11
+
J
21
)
u
2
-
J
21
u
1
]
,
W
=
∫
0
t
m
T
cl
⁡
(
ω
m
-
ω
r
)
d
t
=
∫
0
t
m
x
1
x
3
d
t
.It is shown through experiments that the shift jerk is inversely proportional to friction work and to time as well. Therefore, the shift jerk may be reduced by increasing the frictional time but cause short service life of the clutch at the expense of the plate excessive wear. For a compromise of the two evaluation criterions in contradictory, an objective function is proposed for the optimal control and expressed below:(38)
J
=
1
2
∫
0
t
m
(
W
+
η
j
2
)
d
t
=
1
2
∫
0
t
m
{
x
1
x
3
+
η
[
a
12
(
J
11
+
J
21
)
u
2
-
J
21
u
1
]
2
}
d
t
=
1
2
∫
0
t
m
(
x
T
Q
1
x
+
u
T
Q
2
u
)
d
t
,
where η
(
0
<
η
<
1
) is weight coefficient of the shift jerk and the larger the η is, the more the shift jerk is considered [29]:
(39)
Q
1
=
[
0
0
0.5
0
0
0
0.5
0
0
]
,
Q
2
=
η
[
J
21
1
0
-
2
a
12
(
J
11
+
J
21
)
J
21
a
12
2
(
J
11
+
J
21
)
2
]
.Obviously, the optimal coordinated control is equivalent to seeking the optimal trajectories of the motor torque and the clutch C transmission torque to minimize the value of the object function.By computing (26)-(27) at t
m, one obtains
(40)
P
(
t
m
)
=
[
0
0
0
0
0
0
0
0
0
]
,
M
(
t
m
)
=
[
0
0
1
]
,
222222222222
h
(
t
m
)
=
[
0
0
0
]
K
(
t
m
)
=
[
0
0
1
]
,
L
(
t
m
)
=
0
,
r
(
t
m
)
=
0
.The optimal trajectories of motor torque and clutch C transmission torque in the clutch C disengagement phase can be derived from solving the differential equations as described above with different weight coefficient.


## 4.3.2. Free Phase

During this phase, both the clutch C and the brake B are disengaged. When the motor torque is constant, the shift jerk and the sliding friction work will keep being zero, which is an ideal state. However, in order to accomplish the fast synchronization of the driving and driven parts of the brake and to decrease the sliding friction work in the brake B engagement phase, the motor torque is supposed to be controlled under the condition that the shift jerk is less than the recommended value, 10 m/s3.Define(41)
x
1
=
ω
r
,
x
2
=
T
m
,
u
=
d
T
m
d
t
.The state equation has the following expression, resembling (15):
(42)
X
˙
=
A
2
X
+
B
2
U
+
V
2
,
where
(43)
X
=
[
x
1
x
2
]
T
,
U
=
[
u
1
]
T
A
2
=
[
0
a
11
J
22
-
a
12
J
21
J
11
J
22
-
J
12
J
21
0
0
]
,
B
2
=
[
0
1
]
,
V
2
=
[
(
a
11
J
12
-
a
12
J
11
)
T
f
J
11
J
22
-
J
12
J
21
0
]
.At the end of free phase, the angular velocity of the ring gear is required to be less than 30 rad/s; that is, the terminal constraint is(44)
N
[
x
(
t
n
)
,
t
n
]
=
x
1
(
t
n
)
≤
30
,
where t
n is the end time of free phase.During this phase, the shift jerk is expressed below:(45)
j
=
r
w
i
0
d
d
t
(
J
21
T
m
J
11
J
22
-
J
21
J
12
)
=
J
21
r
w
i
0
(
J
11
J
22
-
J
21
J
12
)
u
.As a result, the object function can be given as follows:(46)
J
=
1
2
∫
0
t
n
u
2
d
t
=
1
2
∫
0
t
n
U
T
Q
2
U
d
t
,
where Q
1
=
0, Q
2
=
[
1
].The optimal torque of the motor control law in free phase can be derived from solutions of the above differential matrix equations.During this phase, ignoring the frictional resistance in sealing ring and splines, the equation of motion of the brake piston is established as(47)
m
c
x
¨
c
+
c
x
˙
c
+
k
s
(
x
0
+
x
c
)
=
p
A
c
,
where m
c is the equivalent piston mass. x
0 is the initial compression amount of the return spring. x
c is piston displacement. c is the piston friction coefficient. k
s is the return spring stiffness. p is the oil pressure.According to (47), the oil pressure of combined clutch in free phase can be acquired.


## 4.3.3. Brake B Engagement Phase

Define(48)
x
1
=
ω
r
,
x
2
=
T
m
,
x
3
=
T
br
,
u
1
=
d
T
m
d
t
,
u
2
=
d
T
br
d
t
.The state equation (51) can be derived from (14):
(49)
X
˙
=
A
3
X
+
B
3
U
+
V
3
,
where
(50)
X
=
[
x
1
x
2
x
3
]
T
,
U
=
[
u
1
u
2
]
T
,
A
3
=
[
0
a
11
J
22
-
a
12
J
21
J
11
J
22
-
J
12
J
21
2
a
11
a
12
J
12
-
a
11
2
J
22
-
a
12
2
J
11
J
11
J
22
-
J
12
J
21
0
0
0
0
0
0
]
,
B
3
=
[
0
0
1
0
0
1
]
,
V
3
=
[
a
11
J
12
-
a
12
J
11
J
11
J
22
-
J
12
J
21
T
f
0
0
]
.At the end of the brake engagement phase, the speed difference of the driving and driven parts of the brake B is zero; that is, the terminal constraint is(51)
N
[
x
(
t
f
)
,
t
f
]
=
x
1
(
t
f
)
=
0
,
where t
f is the end time of brake B engagement phase.In this phase, the shift jerkj and the sliding friction work W are expressed below:
(52)
j
=
r
w
i
0
(
J
11
J
22
-
J
21
J
12
)
×
[
(
a
11
J
21
-
a
12
J
11
)
u
2
-
J
21
u
1
]
W
=
∫
0
t
f
T
br
ω
r
d
t
=
∫
0
t
f
x
1
x
3
d
t
.Similar to the clutch C disengagement phase, the objective function can be expressed as follows:(53)
J
=
1
2
∫
0
t
f
(
W
+
η
j
2
)
d
t
=
1
2
∫
0
t
f
{
x
1
x
3
+
[
(
a
11
J
21
-
a
12
J
11
)
u
2
-
J
21
u
1
]
2
}
d
t
=
1
2
∫
0
t
f
(
X
T
Q
1
X
+
U
T
Q
2
U
)
d
t
,
where
(54)
Q
1
=
[
0
0
0.5
0
0
0
0.5
0
0
]
,
Q
2
=
η
[
J
21
2
0
-
2
J
21
(
a
11
J
21
-
a
12
J
11
)
(
a
11
J
21
-
a
12
J
11
)
2
]
.The optimal trajectories of the motor torque and brake B transmission torque in brake engagement phase can be acquired with reference to the solving process in clutch disengagement phase.The correlation of the clutch transmission torque and the oil pressure is given by (55) [30]. Apparently, the optimal trajectory of the oil pressure can be transformed from the optimal trajectory of the clutch/brake transmission torque:
(55)
T
=
μ
Z
2
(
R
2
3
-
R
1
3
)
3
(
R
2
2
-
R
1
2
)
|
F
s
-
p
A
c
|
,
where T is the clutch/brake transmission torque. F
s is the return spring force. μ is the friction coefficient of clutch/brake plate, treated as a constant in this paper. R
1 and R
2 are clutch/brake plate inner and outer radius, respectively. Z is the plate number of the clutch/brake. A
c is the brake piston area.In reality, the shift time is very short, which makes the real-time online calculation of the differential matrix equation impractical. A solution to this problem is to calculate the optimal trajectory under different working conditions offline and then to conduct function fitting. At last, the fitting coefficients are stored in the memory of vehicle control system. We can get the optimal trajectory quickly through the look-up table and interpolation way to get fit coefficients in online control.


## 5. Simulation Results and Analysis

Based on MATLAB software, a simulation model is built up to analyze the above discussed shift control strategy and to compare the shift quality employing coordinated control with that adopting oil pressure of combined clutch only. The focus is put on uphill downshift in this paper, and the road ramp studied is 5%. The shift vehicle speed is set as 25 km/h. The vehicle parameters are shown in Table1.Table 1
The vehicle parameters.
Parameters
Value
Vehicle mass fully equippedm/kg
15000
Air resistance coefficientC
D
0.6
Rolling resistance coefficientf
0.0015
Vehicle frontal areaA/m2
7.82
Tire radiusr
w/m
0.478
Final drive ratioi
0
6.5
Transmission ratio (2 gears)
[3.5; 1]
Ratio of the number of teeth of ring gear and sun geark
2.5
Motor
Permanent magnet synchronous motor
Rated/peak power/(kW)
130/170
Rated/peak torque/(Nm)
653/850
Plate number of the clutch/brakeZ
6/8
Equivalent piston massm
c/kg
6.82
Brake piston friction coefficientc/(kg/s)
0.5
Return spring stiffnessk
s/(N/mm)
2520
Return spring initial compression amountx
0/mm
4.2
Brake piston areaA
c/mm2
32.371
Friction coefficient of clutch/brake discμ
0.13
J
11
/
J
12
/
J
21
/
J
22
0.63/−0.044/−0.044/81.2
Weight coefficient of the shift jerkη
0.5Figure5 indicates simulation results of the coordinated control of powertrain and the control of oil pressure of the combined clutch only when the accelerator pedal angle is 50%, consisting of four graphs, which is similar to Figure 6, with the accelerator pedal angle 0: (a) shows the optimal trajectories of the motor torque, (b) depicts the hydraulic oil pressure of the combined clutch, (c) shows the speeds of the motor, the ring gear, and the transmission output shaft, and (d) pictures the shift jerk in downshift. Table 2 represents the maximum sliding friction work during downshift.Table 2
The maximum sliding friction work.
Coordinated control
Combined clutch control only
Accelerator pedal angle/%
Sliding friction work/KJ
Accelerator pedal angle/%
Sliding friction work/KJ
50
1.34
50
2.4
0
1.53
0
14.5Downshift simulation results with the accelerator pedal angle 50%.
(a)
The motor torque curves
(b)
The oil pressure curves
(c)
The speed curves
(d)
The shift jerk curveDownshift simulation results with the accelerator pedal angle 0.
(a)
The motor torque curves
(b)
The oil pressure curves
(c)
The speed curves under coordinated control
(d)
The speed curves under clutch control only
(e)
The shift jerk curvesIn the case where the coordinated control of powertrain is employed, the motor torque is not determined by the accelerator pedal any more, the signal of which just represents power requirement from the driver. As can be seen from the profiles of the motor torque in Figure5, the motor torque is kept roughly constant in clutch disengagement phase, which means that the changes of the motor torque can be neglected to simplify the control process, and at the end of this phase, the clutch transmission torque goes to zero. In free phase, the motor torque keeps increasing to regulate the ring gear rotational speed to the specified range and then to decrease the shift time and the sliding friction work. In brake engagement phase, the motor torque is supposed to reduce to make sure that the vehicle speed is almost constant. At the point of synchronization of the driving and driven parts of the brake, actual transition from stick to slip at the brake is accomplished. Namely, the brake transmission torque becomes static friction torque that does not depend on the oil pressure, which leads to a big shift jerk. At the end of brake engagement phase, the motor torque depends on motor speed and accelerator pedal when combined clutch is controlled only. The absence of motor torque control makes a further bigger shift jerk.The combined clutch is largely different from the conventional wet clutch in structure, which results in the tremendous difference in oil pressure between two kinds of clutches. From the profiles of the oil pressure in Figures5 and 6, the oil pressure curves of combined clutch consist of five periods. Firstly, the combined clutch is filled with oil quickly, completed in an instant, so that the oil pressure increases rapidly, aiming to make the actual transmission torque of the clutch equal to its dynamic friction torque to mitigate the shift jerk at the beginning of the clutch disengagement. Secondly, the oil pressure keeps increasing slowly to surpass the initial value of return spring, to separate friction plates of the clutch, and to decrease the clutch transmission torque to zero. Thirdly, the increased oil pressure is used to eliminate the surplus space of the brake until the friction plates are connected. During this period, both the clutch and brake transmission torque are zeros. In the following period, the brake plates clearance has been eliminated. Oil pressure is rising continuously until the driving and driven parts of brake B reach the same speed. This is main stage to control the shift jerk and sliding friction work. At the end of this period, the actual transition is from slip to stick at the brake. After that, the brake transmission torque has nothing to do with the oil pressure. During the last period, in order to produce torque reserve for the bake in case the brake slips, the oil pressure is raised to the line pressure, which is the main pressure in the hydraulic system. It must be stressed that the rapid increase of oil pressure will not affect the driving comfort since the driving and driven parts of the brake already achieved synchronization.With the accelerator pedal angle 0, the control of oil pressure of the combined clutch only means there is power interrupt during downshift. It is shown from Figure6 that there is little variation in the speed of the motor, the ring gear, and the transmission output shaft during the clutch C disengagement phase and free phase. By contrast, the coordinated control strategy adjusts the three speeds through increasing the motor torque in the two phases.It is shown from Figures5 and 6 that there are three significant differences between the two accelerator pedal angles when coordinated control of powertrain is employed: the changing trend of combined clutch oil pressure during brake B engagement phase, the speed variation during the free phase, and the shift jerk at the end of brake B engagement phase. Ultimately, it comes down to the initial value of motor torque. In other words, the terminal value of motor torque must be zero with combined clutch control only.Compared to the combined clutch control only, the downshift time is 0.45 s with coordinated control of powertrain, 0.2 s less than that in the control strategy without the motor. In addition, the adjustment of the ring gear rotational speed in free phase makes the sliding friction work down from 2.4 KJ to 1.34 KJ when the accelerator pedal angle is 50% and from 14.5 KJ to 1.53 KJ when the accelerator pedal angle is 0. In other words, the sliding friction work is decreased by 44% and 89% separately. During the brake engagement phase when the accelerator pedal angle is 50%, the maximum value of the shift jerk is almost equal to that in the control strategy without the motor because the motor torque is supposed to decrease to make sure that the vehicle speed is almost constant. But the shift jerk is from 8.1 m/s3 down to 1.6 m/s3, which improves the shift comfortability at the time of synchronization of the driving and driven parts of the brake B. In the whole shift process, the coordinated control of powertrain makes the shift jerk change from 8.1 m/s3 to 4.6 m/s3 with the accelerator pedal angle 50% and from 13.1 m/s3 down to 5.6 m/s3 with the accelerator pedal angle 0, which means the shift jerk is decreased by 43% and 57% under different accelerator pedal angles.


## 6. Conclusions

(1)
Based on two-gear combined clutch transmission, the dynamic model of shift process is established by the principle of virtual work and Lagrange principle. On this basis, the shift process is divided into three phases and each phase has different dynamics equation.
(2)
The dynamic coordinated system for EVs regards the motor and transmission as a whole. Through coordinated control of the motor torque and the oil pressure of the combined clutch, the shift jerk and sliding friction work are reduced by a large margin no matter how much the accelerator pedal angle is. At the same time, the shift time can be shortened by adjusting the speed difference of the driving and driven parts of brake in free phase. In a word, an effective method is provided to solve the problem of automatic transmission shift control in this paper.



---

*Source: 963725-2014-06-01.xml*