**3. HEV configuration and modeling**

A typical update HEV for this research is shown in Figure 9. It is from a very common parallel HEV concept system designed by Daimler Chrysler, namely the P12-Configuration appeared at the 2004 Detroit Motor Show. The model concept includes one (1) conventional ICE and two (2) EM1 and EM2. An automatically controllable conventional friction clutch separates the model drivetrain into two (2) parts: Part 1: ICE with EM1, and Part 2: EM2 and the rest of the powertrain transmission. The EM1 operates as an electrical starter and a supported electrical generator. No torque converter is installed in this configuration. The driven wheel of this model is the rear wheel, which is connected with a conventional automated transferred gearbox and a conventional differential gearbox.

**Figure 9.** Configuration of the parallel hybrid powertrain.

energy available for regenerative braking processes. However, because of the limited charging acceptance of the vehicle batteries, the full ability of regenerative braking process will be never employed unless the deceleration is very slow. Additionally, purchasing a B-ISG system is less cost, which will provide some important advantages over the use of the C-ISG and the D-ISG. From the above overview of HEV configurations, we will select a typical parallel HEV configuration and start the full investigation with this new HEV model in the next section.

A typical update HEV for this research is shown in Figure 9. It is from a very common parallel HEV concept system designed by Daimler Chrysler, namely the P12-Configuration appeared

**3. HEV configuration and modeling**

**Figure 6.** Schematic representation of a B-ISG.

36 New Applications of Electric Drives

**Figure 7.** Schematic representation of a C-ISG.

**Figure 8.** Schematic representation of a D-ISG.

At the only electrical driving for low speeds (below 50 km/h), the clutch between EM1 and EM2 is opened and the vehicle runs in a series hybrid configuration. For this operating mode, only second motor, EM2, drives the vehicle.

Transmission from the series mode to the parallel mode will be occurred at high speed (above 50 km/h) by closing the friction clutch between EM1 and EM2. EM1 will start the ICE for propelling the vehicle. After the ICE already started, the EM1 will turn off. EM2 will turn on if necessary as an electrical generator to charge the vehicle batteries.

In extreme heavy loads and from the control of the driver, both EM1 and EM2 can be simul‐ taneously activated to assist the ICE to run the vehicle in critical driving conditions.

Our research concentrates on the development of reliable control strategies that can control the speeds of these two drivetrain parts and then, to synchronize these parts with a friction clutch to achieve the fast and smooth clutch engagement with reduction of driveline jerk and improvement of the driving comfort.

Figure 10 shows a simplified dynamic model for this hybrid transmission drivetrain. As referred in the references [19] and [20], the first part of the transmission powertrain can be approximately estimated by a lumped inertia of *J*1 where the main shaft holding the ICE, EM1 and one friction clutch plate is modeled as a rigid frame. The second part of this transmission powertrain is also modeled by two inertias of *J*2 and *J*3 connected by a mechanical and damping springs. The conventional transmission gearbox and the differential gearbox are also modeled as a simple transforming torque with a variable term *i* depending on the selected gear transmission ratios.

**Figure 10.** Simplified drivetrain structure.

where *J*<sup>1</sup> is the lumped inertia of ICE and EM1; *J*<sup>2</sup> is the inertia of EM2, and *J*<sup>3</sup> is the lumped inertia of the rest of the transmission powertrain. The transmission powertrain can now be considered as a connection of a spring with the torque stiffness *kθ*, the velocity damping coefficient *kβ*, and the acceleration damping coefficient *kα*. The torques are *MICE*, *MEM* 1, and *MEM* 2 generated from ICE, EM1, and EM2, respectively. Angular angles from shaft1, shaft2, and shaft3 are *θ*1, *θ*2, and *θ*3. Angular velocities *ω*1, *ω*2, and *ω*3 are measured from shaft1, shaft2, and shaft3 respectively. *r* is the vehicle wheel rolling radius.

There are three separated influences to the hybrid vehicle resistances by the air drag; the rolling friction resistance and the combined mechanical torque losses in the transmission gearbox; the differential gearbox and the shaft bearings due to friction. They can be approximately esti‐ mated from the hybrid vehicle velocity resistance torque, *Mv*, as:

$$M\_v = \left(\frac{\rho}{2}c\_w A (r o\_3)^2 + f\_r m g\right)r + a\_0 + a\_1 o\_3 + a\_2 o\_3^2 \tag{1}$$

where drag coefficient; *A*: vehicle frontal area; *ρ*: air density; *cw*; *r*: wheel dynamic radius; *m*: vehicle mass; *f <sup>r</sup>*: resistance coefficient; *ai* : polynomial coefficients; and *g*: natural gravity. The effect of the road dynamics and the road inclination can be considered as the additional disturbances to the system. These additional disturbances can cause some extra accelerating or decelerating torques at the rolling resistances. However, the changes of these effects and the changes of the vehicle load/mass during the vehicle running are not considered in our chapter.

At low speed (less 50 km/h), the powertrain torque model is formulated without the contri‐ bution of the exponential term of *ω*<sup>3</sup> 2 to *Mv*. And the linearization of the resistance torque *Mv* in the above equation (1) is now calculated as:

$$M\_v = M\_{v0} + k\_v \alpha\_3 \tag{2}$$

where *Mv*0 is the initial constant value for the air drag and *kv* is the linear air drag coefficient. Because of the differences in gearbox ratios, the coefficient constant *Mv*0 and *kv* can be varied for each gearbox ratio.

The following torque equations are formulated for the first powertrain part:

$$M\_{1o} = \mathbf{J}\_1 \dot{\alpha}\_1 \tag{3}$$

The driving torque is computed as:

**Figure 10.** Simplified drivetrain structure.

38 New Applications of Electric Drives

and shaft3 respectively. *r* is the vehicle wheel rolling radius.

mated from the hybrid vehicle velocity resistance torque, *Mv*, as:

w

2

æ ö

r

vehicle mass; *f <sup>r</sup>*: resistance coefficient; *ai*

bution of the exponential term of *ω*<sup>3</sup>

for each gearbox ratio.

in the above equation (1) is now calculated as:

chapter.

where *J*<sup>1</sup> is the lumped inertia of ICE and EM1; *J*<sup>2</sup> is the inertia of EM2, and *J*<sup>3</sup> is the lumped inertia of the rest of the transmission powertrain. The transmission powertrain can now be considered as a connection of a spring with the torque stiffness *kθ*, the velocity damping coefficient *kβ*, and the acceleration damping coefficient *kα*. The torques are *MICE*, *MEM* 1, and *MEM* 2 generated from ICE, EM1, and EM2, respectively. Angular angles from shaft1, shaft2, and shaft3 are *θ*1, *θ*2, and *θ*3. Angular velocities *ω*1, *ω*2, and *ω*3 are measured from shaft1, shaft2,

There are three separated influences to the hybrid vehicle resistances by the air drag; the rolling friction resistance and the combined mechanical torque losses in the transmission gearbox; the differential gearbox and the shaft bearings due to friction. They can be approximately esti‐

<sup>3</sup> 0 13 23 ( ) <sup>2</sup> *M c A r f mg r a a a vw r*

where drag coefficient; *A*: vehicle frontal area; *ρ*: air density; *cw*; *r*: wheel dynamic radius; *m*:

effect of the road dynamics and the road inclination can be considered as the additional disturbances to the system. These additional disturbances can cause some extra accelerating or decelerating torques at the rolling resistances. However, the changes of these effects and the changes of the vehicle load/mass during the vehicle running are not considered in our

At low speed (less 50 km/h), the powertrain torque model is formulated without the contri‐

where *Mv*0 is the initial constant value for the air drag and *kv* is the linear air drag coefficient. Because of the differences in gearbox ratios, the coefficient constant *Mv*0 and *kv* can be varied

w

*MM k v vv* = +0 3

= + ++ + ç ÷

2 2

w

è ø (1)

 w

: polynomial coefficients; and *g*: natural gravity. The

to *Mv*. And the linearization of the resistance torque *Mv*

(2)

$$M\_{1o} = M\_{\text{ICE}} + M\_{M1} - M\_{\text{C}} \tag{4}$$

The friction torque *MC* transmitted by the clutch can be divided into two engagement modes: The locked mode when the friction torque (*MC*) exceeds the static friction capacity (*M <sup>f</sup>* max *Static* ):

$$\boldsymbol{M}\_{\complement} = \frac{2}{3} \boldsymbol{r}\_{\complement} \boldsymbol{F}\_{\complement \complement \complement \complement \complement \textbf{w} \textbf{then} \left( \boldsymbol{M}\_{\complement \complement \textbf{C}} = \boldsymbol{M}\_{\complement / \text{max}}^{\text{Static}} \right) \tag{5}$$

where *FNC*: normal force exerted on the clutch; *rC*: clutch corresponding radius; and *μS* : clutch static friction coefficient.

And the slipping mode:

$$M\_{\mathbb{C}} = r\_{\mathbb{C}} F\_{\mathbb{NC}} \text{sign}(a\_1 - a\_2) \mu\_{\mathbb{K}} \text{ when } (M\_{\mathbb{C}} < M\_{\text{/max}}^{\text{Static}})$$

where *μK* : clutch slipping kinetic friction coefficient.

The following torque equations are applied for the second part:

$$\begin{aligned} \mathbf{M}\_{2o} &= k\_o \theta\_2 + \frac{k\_o}{\dot{\mathbf{i}}} \theta\_3 + k\_v o \mathbf{o}\_3 & \quad \mathbf{a} \\ \text{or} \\ \mathbf{M}\_{2o} &= J\_2 \dot{o} \dot{o}\_2 \mathbf{i} + J\_3 \dot{o}\_3 + k\_v o \mathbf{o}\_3 & \quad \mathbf{b} \end{aligned} \tag{6}$$

And

$$\dot{M}\_{2o} = J\_2 \ddot{\alpha}\_2 \dot{\imath} + k\_\alpha \left(\frac{\dot{\alpha}\_2}{\dot{\imath}} - \dot{\alpha}\_3\right) + k\_\beta \left(\frac{\alpha\_2}{\dot{\imath}} - \alpha\_3\right) \tag{7}$$

The torque *M*2*<sup>o</sup>* is computed:

$$M\_{2o} = \left(M\_{EM2} + M\_c\right) \eta \dot{\imath} - M\_{v0} \tag{8}$$

with *η*: efficiency of gearbox and differential.

The above torque equations can be transformed to the following dynamic equations:

$$
\dot{\theta}\_1 = \alpha\_1 \tag{9}
$$

$$\dot{o}\_1 = -\frac{k\_{\rho1}a\_1}{J\_1} + \frac{M\_{\text{kCE}}}{J\_1} + \frac{M\_{M1}}{J\_1} + \frac{-M\_\odot}{J\_1} \tag{10}$$

where *kβ*1 is friction coefficient in shaft 1.

$$
\dot{\theta}\_2 = \alpha\_2 \tag{11}
$$

$$\dot{\rho}\_2 = -\frac{k\_{\rho\_2}\alpha\_3}{J\_2 i} - \frac{J\_3 \dot{\alpha}\_3}{J\_2 i} - \frac{\eta M\_{M2}}{J\_2} + \frac{\eta M\_\odot}{J\_2} - \frac{M\_{v0}}{J\_2 i} \tag{12}$$

where *kβ*2 is friction coefficient in shaft 2.

$$
\dot{\theta}\_3 = \alpha\_3 \tag{13}
$$

$$
\dot{\alpha}\_3 = \frac{k\_{\mu 3} \alpha\_3}{J\_3} + \mathcal{M}\_{v0} \tag{14}
$$

where *kβ*3 is friction coefficient in shaft 3.

$$\ddot{o}\_{3} = \frac{k\_{\rho2}o\_{2}}{J\_{3}\dot{i}} - \frac{(k\_{\rho2}I\_{2}\dot{i}^{2} + k\_{a}k\_{v})o\_{3}}{J\_{2}J\_{3}\dot{i}^{2}} - \left(\frac{k\_{v} + k\_{a}}{J\_{3}} + \frac{k\_{a}}{J\_{2}\dot{i}^{2}}\right)\dot{o}\_{3} + \frac{k\_{a}\eta(M\_{M2} + M\_{\subset})}{J\_{2}J\_{3}\dot{i}} - \frac{k\_{a}M\_{v0}}{J\_{2}J\_{3}\dot{i}^{2}}\tag{15}$$

Replace the torque generated by a DC motor in the following formula:

$$\mathbf{M}\_{\text{DC\\_MOTOR}} = \frac{k\_T}{R\_I}V - \frac{k\_E k\_T}{R\_I} \alpha \tag{16}$$

#### Modeling and Control Strategy for Hybrid Electrical Vehicle http://dx.doi.org/10.5772/61415 41

where *MDC*\_*MOTOR*: torque by DC Motor; *kT* : motor torque constant, *kT* = *MTorque ICurrent* (Nm/A); *kE*: motor electromotive force (EMF) constant (V/ (rad/s)), *kE* =*kT* ; *RI* : terminal resistance (Ohm), *V* : power supply (Volts), and *ω*: angular velocity (rpm).

Then a new set of vehicle dynamics is installed as:

*M M M iM* 22 0 *o EM c* ( ) *<sup>v</sup>* = +-

The above torque equations can be transformed to the following dynamic equations:

1 1 q w

1 1 1

2 2 q w

3 3 q w

3 3 3 0 3

*J* b w

=- - + + - ç ÷ è ø

*k kk M V*

*k*

w

 w

Replace the torque generated by a DC motor in the following formula:

\_

*DC MOTOR*

*v*

*M*

2 2 22 3 2 0 3 2 2 3 2 3 2 3 3 2 2 3 2 3 ( ) ( ) *<sup>v</sup> <sup>v</sup> MC v k k Ji kk k k k k M M kM J i JJi J J i JJi JJi*

+ æ ö + +

aa

 w

&& & (15)

*T ET*

*I I*

*R R* = -

wh

2 3 33 2 0

 h

=- - - + - & & (12)

 a

w

h

2 2 2 22 *M Cv k J M MM Ji Ji J J Ji*

1 111 *ICE M <sup>C</sup> k MM M JJJJ*

with *η*: efficiency of gearbox and differential.

40 New Applications of Electric Drives

1

w

where *kβ*1 is friction coefficient in shaft 1.

2

w

where *kβ*2 is friction coefficient in shaft 2.

where *kβ*3 is friction coefficient in shaft 3.

 ba

b

w

w

2

b w

b w h

(8)

= & (9)

= & (11)

= & (13)

a

(16)

& = + (14)


$$
\dot{\theta}\_1 = \begin{bmatrix} 0 + & \alpha\_1 + & 0 + & 0 + & 0 + & 0 + & 0 \end{bmatrix} + \begin{bmatrix} 0 + & 0 + & 0 + & 0 + & 0 \end{bmatrix} \tag{17}
$$

$$
\rho\_1 \dot{\rho}\_1 = \begin{bmatrix} -\left(k\_{\rho1} + \frac{k\_{\Gamma1}k\_{\Gamma1}}{R\_{\Gamma1}}\right)\rho\_1\\ 0 + \frac{-\left(k\_{\rho1} + \frac{k\_{\Gamma1}}{R\_{\Gamma1}}\right)\rho\_1}{J\_1} + \begin{bmatrix} 0 & 0 & 0 \end{bmatrix} + \begin{bmatrix} \frac{M\_{\Gamma1}}{\rho\_1} + \frac{k\_{\Gamma1}V\_1}{R\_{\Gamma1}} + \begin{bmatrix} 0 & -M\_c\\ 0 & 0 \end{bmatrix} + \begin{bmatrix} 0 & 0 \end{bmatrix} \end{bmatrix} \tag{18}
$$

$$
\dot{\theta}\_2 = \begin{bmatrix} 0+ & 0+ & 0+ & a\_2+ & 0+ & 0+ & 0 \end{bmatrix} + \begin{bmatrix} 0+ & 0+ & 0+ & 0+ & 0+ \end{bmatrix} \tag{19}
$$

$$\begin{aligned} \dot{\alpha}\_2 &= \begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} -\left(k\_{\mu 2} + \frac{k\_{\varepsilon 2}k\_{\tau 2}}{R\_{l\_2}}\right)a\_3 & -\frac{1}{I\_2 i}\dot{a}\_3\\ & & & & \frac{-I\_3}{I\_2 i} \end{bmatrix} \\\\ +\left[0 + \quad 0 + \quad \frac{-\eta k\_{\tau 2}V\_2}{R\_{l\_2}I\_2} + \quad \frac{\eta M\_c}{I\_2} + \quad \frac{-M\_{z0}}{I\_2 i}\right] \end{aligned} \tag{20}$$

$$
\dot{\theta}\_3 = \begin{bmatrix} 0 + & 0 + & 0 + & 0 + & 0 + & 0 \end{bmatrix} + \begin{bmatrix} 0 + & 0 + & 0 + & 0 + & 0 \end{bmatrix} \tag{21}
$$

$$
\dot{\rho}\_3 = \begin{bmatrix} 0 + & 0 + & 0 + & 0 + & 0 + & \frac{k\_{\rho 3} \rho\_3}{I\_3} + & 0 \end{bmatrix} + \begin{bmatrix} 0 + & 0 + & 0 + & 0 + & M\_{r0} \end{bmatrix} \tag{22}
$$

$$\begin{aligned} \partial\_{3} &= \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & \frac{-\left(k\_{g2} + \frac{k\_{z2}k\_{r2}}{R\_{12}}\right)a\_{2}}{I\_{3}i} & 0 & 0 & \frac{-\left(k\_{g2}I\_{2}i^{2} + k\_{a}k\_{v}\right)a\_{3}}{I\_{2}I\_{3}i^{2}} & -\left(\frac{k\_{v} + k\_{a}}{I\_{3}} + \frac{k\_{a}}{I\_{2}i^{2}}\right)\partial\_{3} \\\\ 0 & 0 & \frac{-k\_{a}\eta k\_{r2}V\_{2}}{R\_{12}I\_{2}I\_{3}i} + \frac{k\_{a}\eta M\_{c}}{I\_{2}I\_{3}i} + \frac{-k\_{a}M\_{c0}}{I\_{2}I\_{3}i^{2}} \end{bmatrix} \end{aligned} \tag{23}$$

#### 42 New Applications of Electric Drives

If we define the state variables as *x*<sup>0</sup> <sup>=</sup> *<sup>θ</sup>*<sup>1</sup> *<sup>ω</sup>*<sup>1</sup> *<sup>θ</sup>*<sup>2</sup> *<sup>ω</sup>*<sup>2</sup> *<sup>θ</sup>*<sup>2</sup> *<sup>ω</sup>*<sup>3</sup> *<sup>ω</sup>*˙ <sup>3</sup> ' for the positions, angular velocities, and acceleration on vehicle shafts 1, 2, 3 respectively, and the input variables as *<sup>u</sup>*<sup>0</sup> <sup>=</sup> *MICE <sup>V</sup>*<sup>1</sup> *<sup>V</sup>*<sup>2</sup> *MC Mv*<sup>0</sup> ' for the torque for ICE, voltage for EM1; voltage for EM2, torque on clutch, and the initial air-drag torque load, a space state form of the vehicle dynamics is set up:

( ) 1 1 1 1 1 2 2 2 2 3 3 0 2 2 3 3 3 2 2 2 2 2 2 2 2 2 3 3 2 3 2 01000 0 0 0 000 0 0 00010 0 0 00000 00000 1 0 00000 0 000 0 *E T E T E T v v k k <sup>k</sup> R J k k <sup>k</sup> <sup>R</sup> <sup>J</sup> <sup>x</sup> J i J i k J k k <sup>k</sup> R k Ji kk kk k J i JJi J J i* b b b b b a a a w w <sup>é</sup> <sup>ù</sup> <sup>ê</sup> <sup>ú</sup> æ ö <sup>ê</sup> <sup>ú</sup> - + ç ÷ <sup>ê</sup> <sup>ú</sup> è ø <sup>ê</sup> <sup>ú</sup> <sup>ê</sup> ê ê <sup>ê</sup> æ ö <sup>ê</sup> - + ç ÷ <sup>ê</sup> è ø - <sup>=</sup> <sup>ê</sup> ê ê ê ê ê ê <sup>ê</sup> æ ö <sup>ê</sup> - + ç ÷ <sup>ê</sup> è ø - + æ ö <sup>+</sup> - + ç ÷ <sup>ê</sup> ë è øû & &<sup>0</sup> 1 1 11 1 2 0 22 2 2 2 2 223 23 2 3 00 0 0 0 1 1 0 0 00 0 0 0 <sup>1</sup> 0 0 00 0 0 0 00 0 0 1 0 0 *T T T x k J RJ J k <sup>u</sup> RJ J Ji kk k k RJJi JJi JJi* a aa h h h h ú ú ú ú ú ú ú ú ú ú ú ú ú ú ú ú ú é ù ê ú - - <sup>+</sup> ë û (24)

The new dynamic modeling in (24) allows having a deep viewing on the acceleration *ω*˙ 3 and jerk *ω*¨ 3 of HEVs. This is also one of the mail contributions of this study.

When the vehicle travels in low speed (below 50 km/h), only EM2 is running. Then, the inputs: *MICE* =0, *V*<sup>1</sup> =0, *MC* =0 and the state variables: *θ*<sup>1</sup> =0, *ω*<sup>1</sup> =0, the system becomes:

$$\begin{aligned} \dot{\mathbf{x}}\_{p} &= \boldsymbol{\mu}\_{p}\mathbf{x}\_{p} + \boldsymbol{\mathcal{D}}\_{p}\boldsymbol{u}\_{p} \\ \boldsymbol{y}\_{p} &= \boldsymbol{\mathcal{C}}\_{p}\mathbf{x}\_{p} + \boldsymbol{\mathcal{D}}\_{p}\boldsymbol{u}\_{p} \end{aligned} \text{ with} \\ \begin{bmatrix} 0 & \mathbf{1} & \mathbf{0} & \mathbf{0} \\ 0 & -\frac{\boldsymbol{k}\_{p2} + \frac{\mathbf{k}\_{p2}}{R\_{f2}}}{R\_{f2}} & \mathbf{0} & \mathbf{0} \\ \mathbf{0} & \mathbf{0} & \mathbf{0} & \mathbf{1} \\ \mathbf{0} & \mathbf{0} & \mathbf{0} & -\frac{\left(\mathbf{k}\_{p3} - \mathbf{k}\_{p}\right)}{I\_{3}} \end{bmatrix}; \boldsymbol{B}\_{p} = \begin{bmatrix} \mathbf{0} & \mathbf{0} \\ \frac{\mathbf{k}\_{p2}}{R\_{f2}} & \mathbf{0} \\ \mathbf{0} & \mathbf{0} \\ \mathbf{0} & -\mathbf{1} \end{bmatrix} \\ \mathbf{C}\_{p} = \begin{bmatrix} 0 & \mathbf{0} & \mathbf{0} & \mathbf{1} \\ \mathbf{k}\_{\phi} & \mathbf{0} & \frac{\mathbf{k}\_{\phi}}{I} & \mathbf{0} \end{bmatrix}; \boldsymbol{D}\_{p} = \begin{bmatrix} \mathbf{0} & \mathbf{0} \\ \mathbf{0} & \mathbf{1} \end{bmatrix} \end{aligned} \tag{25}$$

If we define the state variables as *x*<sup>0</sup> <sup>=</sup> *<sup>θ</sup>*<sup>1</sup> *<sup>ω</sup>*<sup>1</sup> *<sup>θ</sup>*<sup>2</sup> *<sup>ω</sup>*<sup>2</sup> *<sup>θ</sup>*<sup>2</sup> *<sup>ω</sup>*<sup>3</sup> *<sup>ω</sup>*˙ <sup>3</sup> '

*<sup>u</sup>*<sup>0</sup> <sup>=</sup> *MICE <sup>V</sup>*<sup>1</sup> *<sup>V</sup>*<sup>2</sup> *MC Mv*<sup>0</sup> '

42 New Applications of Electric Drives

1 1

*E T*

1 1

00000

000 0

1

b

*k k <sup>k</sup> R J*

up:

0

ê ê

ê ê ê ê ê ê

1

*T*

*k*

0 0

1 11 1

*J RJ J*

2

*T*

*k*


h

00 0 0 0 00 0 0 1

00 0 0 0 <sup>1</sup> 0 0

00 0 0 0 1 1 0 0

é ù ê ú -

2

*T*

ë û

a

h

22 2 2

h

*<sup>u</sup> RJ J Ji*

223 23 2 3

 aa

 h

*kk k k RJJi JJi JJi*

velocities, and acceleration on vehicle shafts 1, 2, 3 respectively, and the input variables as

on clutch, and the initial air-drag torque load, a space state form of the vehicle dynamics is set

for the torque for ICE, voltage for EM1; voltage for EM2, torque

( )

*R k Ji kk kk k J i JJi J J i*

 a

3 3 2 3 2

2 2

*E T*

*k k <sup>k</sup>*

0

2

jerk *ω*¨ 3 of HEVs. This is also one of the mail contributions of this study.

*MICE* =0, *V*<sup>1</sup> =0, *MC* =0 and the state variables: *θ*<sup>1</sup> =0, *ω*<sup>1</sup> =0, the system becomes:


b

00000 1 0

& &<sup>0</sup>

*<sup>R</sup> <sup>J</sup> <sup>x</sup> J i J i*

<sup>ê</sup> æ ö <sup>ê</sup> - + ç ÷ <sup>ê</sup> è ø - <sup>=</sup> <sup>ê</sup>

00000 0

01000 0 0

0 000 0 0

<sup>é</sup> <sup>ù</sup> <sup>ê</sup> <sup>ú</sup> æ ö <sup>ê</sup> <sup>ú</sup> - + ç ÷ <sup>ê</sup> <sup>ú</sup> è ø <sup>ê</sup> <sup>ú</sup> <sup>ê</sup>

00010 0 0

2 2 2 2 2

The new dynamic modeling in (24) allows having a deep viewing on the acceleration *ω*˙ 3 and

When the vehicle travels in low speed (below 50 km/h), only EM2 is running. Then, the inputs:

<sup>ê</sup> æ ö <sup>ê</sup> - + ç ÷ <sup>ê</sup> è ø - + æ ö <sup>+</sup> - + ç ÷ <sup>ê</sup> ë è øû

3 3 3

w

2 2

*E T*

2 3 3

2 2

a a

w

*x*

(24)

ú ú ú ú ú ú ú ú ú ú ú ú ú ú ú ú ú

*v v*

2 2

2

b

*k J*

b

b

*k k <sup>k</sup>*

for the positions, angular

where *xp* = *θ*<sup>2</sup> *ω*<sup>2</sup> *θ*<sup>3</sup> *ω*<sup>3</sup> ' , *up* <sup>=</sup> *<sup>V</sup>*<sup>2</sup> *Mv*<sup>0</sup> ' , the outputs, *yp* <sup>=</sup> *<sup>ω</sup>*<sup>3</sup> *TTorque*<sup>3</sup> ' , are the vehicle velocity (measured) *ω*3 and the vehicle torque (unmeasured) *TTorque*3 generated on shaft 3. In this case, the torsional rigidity (Torque/angle): *k<sup>θ</sup>* = *MTorque <sup>φ</sup>* <sup>=</sup> *GJ <sup>l</sup>* , where *<sup>φ</sup>* is the angle of twist *<sup>φ</sup>* <sup>=</sup>*θ*<sup>2</sup> <sup>−</sup> *θ*3 *i* (rad). *G* is the shear modulus or modulus of rigidity of mild carbon steel, *G* =81500.10<sup>6</sup> (N/m). *l* is the length of the shaft where the torque is being applied, in this case, we assumed: *l* =1.5 (m) for the vehicle drivetrain length. *J* is the moment of inertia, *J* = *J*<sup>2</sup> + *J*3 (m4 ).

At the high speed (more than 50 km/h), EM1 starts ICE while the clutch is still open, the dynamic equations in the first part become:

$$
\dot{\theta}\_1 = \begin{bmatrix} 0 + & \alpha\_1 \end{bmatrix} + \begin{bmatrix} 0 + & 0 \end{bmatrix} \tag{26}
$$

$$
\phi\_1 = \left[ 0 \quad -\frac{\left( k\_{\rho 1} + \frac{k\_{E1} k\_{T1}}{R\_{11}} \right)}{J\_1} \alpha\_1 \right] + \left[ \frac{\mathcal{L} k\_{T1}}{R\_{11} J\_1} V\_1 + \quad \frac{1}{J\_1} M\_{\text{kCE}} \right] \tag{27}
$$

where *ς* is a new coefficient added to the EM1 as a compensated load to the starting period. Then the system becomes:

$$\begin{aligned} \begin{bmatrix} \dot{\boldsymbol{x}}\_{\epsilon} = \boldsymbol{A}\_{\epsilon} \boldsymbol{x}\_{\epsilon} + \boldsymbol{B}\_{\epsilon} \boldsymbol{u}\_{\epsilon} \\ \boldsymbol{y}\_{\epsilon} = \boldsymbol{C}\_{\epsilon} \boldsymbol{x}\_{\epsilon} + \boldsymbol{D}\_{\epsilon} \boldsymbol{u}\_{\epsilon} \end{bmatrix} \text{with} \\ \begin{bmatrix} \boldsymbol{0} & \mathbf{1} \\ \boldsymbol{A}\_{s} = \begin{bmatrix} \boldsymbol{k}\_{\rho\_{1}} + \frac{\boldsymbol{k}\_{T1}^{2}}{\boldsymbol{R}\_{11}} \\ \boldsymbol{0} - \frac{\boldsymbol{k}\_{\rho\_{1}} + \frac{\boldsymbol{k}\_{T1}^{2}}{\boldsymbol{R}\_{11}}}{\boldsymbol{J}\_{1}} \end{bmatrix} \boldsymbol{j} \boldsymbol{B}\_{s} = \begin{bmatrix} \boldsymbol{0} & \mathbf{0} \\ \frac{\boldsymbol{\mathcal{E}} \boldsymbol{k}\_{T1}}{\boldsymbol{R}\_{11} \boldsymbol{J}\_{1}} & \frac{\boldsymbol{1}}{\boldsymbol{J}\_{1}} \end{bmatrix} \\ \boldsymbol{C}\_{\epsilon} = \begin{bmatrix} \boldsymbol{0} & \mathbf{1} \\ \boldsymbol{0} & \frac{\boldsymbol{k}\_{\rho\_{1}} + \frac{\boldsymbol{k}\_{T1} \boldsymbol{k}\_{T1}}{\boldsymbol{R}\_{1}}}{\boldsymbol{J}\_{1}} \end{bmatrix} \boldsymbol{j} \boldsymbol{D}\_{s} = \begin{bmatrix} \boldsymbol{0} & \mathbf{0} \\ \mathbf{0} & \mathbf{0} \end{bmatrix} \end{aligned} \tag{28}$$

where, *xe* <sup>=</sup> *<sup>θ</sup>*<sup>1</sup> *<sup>ω</sup>*<sup>1</sup> ' , *ue* <sup>=</sup> *<sup>V</sup>*<sup>1</sup> *MICE* ' , *ye* <sup>=</sup> *<sup>ω</sup>*<sup>1</sup> *TTorque*<sup>1</sup> ' . *TTorque*1 is the output torque (unmeasured) on shaft 1.

Using comprehensive HEV modeling equations from (24) to (28), we can develop MPC controllers to this HEV in the next section.
