The research objects are a PS-HEB equipped with an SPGS (shown in Figure 1) and a two-speed clutchless AMT. The SPGS is used as power-split device, and the engine is connected to the planetary carrier while the motor/generator (MG) 1 is connected to the sun gear. The power of the SPGS is transmitted to the output shaft of the AMT by ring gear. The AMT is used to change the speed ratio of the drive motor (MG2) to the output shaft.
2.1 Engine Model
A single hidden layer back propagation neural network (BPNN) with 35 neurons is selected to establish the dynamic model of the engine in Ref. [28]. On flat and straight roads, the driver controls the pedal to allow the vehicle speed to frequently switch between high and low, thereby resulting in AMT frequent shift. The engine speed, torque, and throttle opening data during the shift are recorded as the training sample data of the engine neural network model. The expected output of the BPNN is engine torque, and the input of which includes engine speed and throttle opening data obtained from actual driving conditions.
Some other experimental data are collected to verify the generalization ability of engine model. The comparison between the torque output by the engine model and the actual engine torque is shown in Figure 2. Sudden change in torque is observed during engine start-up and shutdown phases. The torque output by the engine model has a slight deviation from the actual torque, but the output can follow actual torque in the remaining time. This finding shows that the BPNN model trained from engine running data can truly reflect the engine dynamic characteristics and has a good generalization ability.
2.2 Motor Model
In this section, the parameter identification principle of the least square method is adopted to identify the parameters of the first-order inertial model and finally obtain the dynamic motor model. The corresponding transfer function of the continuous system is as follows:
$$G_{m} (s) = \frac{{T_{m,act} (s)}}{{T_{m,req} (s)}} = \frac{K}{T \cdot s + 1},$$
(1)
where \(T_{m,req}\) and \(T_{m,act}\) are the act torque and required torque of MG, \(K\) represents the scaling factor to be identified, and \(T\) is the response time constant of the motor to be identified. When \(s \approx \frac{2}{{T_{s} }} \cdot \frac{{1 - Z^{ - 1} }}{{1 + Z^{ - 1} }}\) and \(T_{s}\) is the sampling time, Eq. (1) can be discretized by bilinear transformation as follows:
$$T_{m,act} (t) = \alpha [T_{m,req} (t) + T_{m,req} (t - 1)] - \beta T_{m,act} (t - 1),$$
(2)
where \(\alpha { = }K/(2T/T_{s} + 1),\) \(\beta = (1 - 2T/T_{s} )/(2T/T_{s} + 1).\)
In the running process, MG is affected by the rotary resistance moment, including bearing friction and ventilation resistance. Therefore, the transfer function of the motor is not strictly a first-order process, and an inevitable steady state error is found between \(T_{m,req}\) and \(T_{m,act}\). Compared with the actual output torque of the motor, the rotation resistance moment is small and generally not considered. An improved parameter identification method, which can identify the torque increase and torque decrease process and eliminate the adverse influence of rotation resistance moment on motor parameter identification as far as possible, is proposed to establish a more accurate motor transfer function model.
In accordance with the operating condition of the motor, the corresponding transfer function is selected, and the final fitting effect is shown in Figure 3. The figure shows that, in the selected working conditions, the first-order inertia model of the motor can better reflect the actual torque output characteristics.
2.3 Two-Speed AMT Model
The gear shifting actuator (GSA) uses direct current motor (DCM) as the power source, and ball screw or worm gear is used to convert the DCM torque into the flat power required for gear shifting. Thus, the gear shifting control accuracy is relatively high, and the cost is relatively low. The typical electric controlled AMT gearbox with high precision ball screw can achieve rapid and efficient shift. When the GSA receives the shift command, the DCM drives the screw and the connected shift fork to move in a straight line. When the speed of the input shaft synchronizes with the speed of the synchronizer, the transmission is shifted into the target gear. The mechanism model of the main components of the two-speed AMT gearbox is established in this section, including the gear shifting actuator, the synchronizer, the input, and output shaft of gearbox.
2.3.1 GSA Model
The GSA comprises DCM and ball screw, and the DCM output shaft is fixedly connected with a ball screw nut. In the DCM, the torque balance relationship is as follows:
$$T_{s\_m} = I_{s\_m} \frac{{{\text{d}}\omega_{s\_m} }}{{{\text{d}}t}} + K_{f} \omega_{s\_m} + T_{d} ,$$
(3)
where \(T_{s\_m}\) is the output torque of DCM, \(I_{s\_m}\) is the sum of the inertia of DCM and ball screw, \(K_{f}\) is the rotational damping coefficient of DCM, and \(T_{d}\) is the load torque of DCM.
The output torque of DCM is proportional to the motor current, as follows:
$$T_{s\_m} = K_{m} I,$$
(4)
where \(K_{m}\) is the motor torque constant.
In accordance with Eqs. (3) and (4), the DCM model is built in MATLAB/Simulink.
Then, the ball screw model is established. The driving torque on the screw is the load torque (\(T_{d}\)) on the DCM in Eq. (3). According to the principle of ball screw drive, the flat power generated by the driving torque on the nut is as follows:
$$F_{l} = \frac{{2\uppi }}{h}T_{d} ,$$
(5)
where \(h\) is the lead of the ball screw.
Considering the friction loss between the nut and the screw, on the premise of ignoring the damping force, the final output shifting force \(F_{s}\) of the screw nut is obtained as follows:
$$F_{s} = (1 - f)F_{l} .$$
(6)
In accordance with this relationship, the ball screw model was built, and the DCM and ball screw models were integrated to obtain the gear shift actuator model.
2.3.2 Synchronizer Model
The speed synchronization process of the synchronizer is an important link in the shifting process. In most articles, the contact reaction force generated between the locking ring and the gear ring during the synchronization stage is a constant value, and the resulting friction torque, namely, the synchronization torque, is a constant value [29, 30]. However, in fact, contact reaction force during synchronization is a complex process that gradually increases from 0 to shift force [31, 32]. To reduce the difficulty of modeling, the contact reaction force at the synchronization stage is assumed to increase linearly with time from 0, reach the maximum value after time \(t_{s}\), and remain stable, as shown in Figure 4.
The kinetic model of the synchronizer in the synchronization stage is established, as shown in Figure 5.
When the contact reaction force of the synchronizer is \(F_{tch}\), the synchronization torque of the locking ring acting on the gear ring is as follows:
$$T_{syn1} = \frac{{\mu_{s} F_{tch} R_{c} }}{\sin \alpha },$$
(7)
where \(T_{syn1}\) is the synchronizing torque acting on the gear ring of AMT, \(\mu_{s}\) is the friction coefficient of cone surface, \(R_{c}\) is the average effective radius of friction of the cone, and \(\alpha\) is the half cone angle.
2.3.3 Input Shaft Model
The main function of the AMT input shaft model is to calculate the motion state of the input shaft according to the internal dynamics and kinematics of AMT. When the synchronizer and gear ring are completely engaged, the speed of the MG2 input shaft is obtained as follows:
$$n_{a\_in} = \omega_{r} i_{g} ,$$
(8)
where \(n_{a\_in}\) is the speed of the input shaft, \(\omega_{r}\) is the speed of carrier, and \(i_{g}\) is the ratio of AMT.
Assuming that the input shaft speed remains unchanged during the process of synchronizer separation and re-engagement, and the input shaft speed in the process of re-engagement is as follows:
$$n_{a\_in} = n_{0} + \int_{0}^{{t_{d} }} {(T_{syn1} /i_{g} + T_{m} )/(I_{m} + I_{a} )} {\text{d}}t,$$
(9)
where \(n_{0}\) represents the speed of input shaft before shifting, and \(t_{d}\) is the duration of gear shifting process, \(T_{m}\) is MG2 torque, \(I_{a}\) represents the moment of inertia of the AMT equivalent to the input shaft, which can be expressed as =:
$$I_{a} = I_{1} + \frac{{I_{2} }}{{i_{12} }} + \frac{{I_{3} }}{{i_{12} i_{23} }},$$
(10)
where \(I_{a}\) is the equivalent moment of inertia of the AMT; \(I_{1}\), \(I_{2}\), and \(I_{3}\) are moment of inertia of input shaft, intermediate shaft, and output shaft, respectively; \(i_{12}\) is the ratio of input shaft to intermediate shaft; \(i_{23}\) is the ratio of intermediate shaft to output shaft.
When the gearbox shifts from low-speed gear to high, the speed of the high gear on the output shaft is higher than that of the synchronizer. The synchronous torque hinders the rotation of MG2 and the input shaft of the gearbox during the synchronization process. On the contrary, during downshifts, the synchronous torque accelerates the rotation of MG2 and gearbox input shaft.
2.3.4 Output Shaft Model
The gearbox output shaft model focuses on the external output characteristics of the AMT and is used to calculate the output torque. When not shifting, the MG2 torque is the output torque of the gearbox after overcoming the torque consumed by MG2 inertia and gearbox inertia. In the shifting process, the output torque of AMT is \(T_{syns2}\), the synchronous torque acting on the output shaft, the same magnitude, and opposite direction with \(T_{syns1}\).
When not shifting, the torque of the AMT output shaft is calculated using Eq. (11).
$$T_{a\_out} { = }i_{g} (T_{m} - (I_{m} + I_{a} )\dot{\omega }_{m} ),$$
(11)
where, \(\dot{\omega }_{m}\) denotes the angular acceleration of MG2.
When shifting, the torque of the AMT output shaft is calculated as Eq. (12).
$$T_{a\_out} { = }i_{g} (T_{m} - (I_{m} + I_{a} )\dot{\omega }_{m} ) + T_{syn2} .$$
(12)
2.3.5 Model Integration and Validation
The AMT model is obtained by integrating the models mentioned above, and the accuracy of the model is further verified by simulation. With the driver’s pedals as input, partial signals of the vehicle and transmission during acceleration are obtained, as shown in Figure 6.
The test results show that the synchronizing torque acting on MG2 is 0 prior to shifting. The AMT begins to shift from low-speed gear to high-speed gear at 20 s. At this time, the MG2 torque changes to 0, the synchronizing torque acting on the input shaft increases in reverse, and the speed of MG2 declines. Meanwhile, the synchronizing torque has a driving effect on the vehicle, and the output torque of AMT increases gradually with the synchronizing torque. When the MG2 speed declines to the target speed, the synchronizing torque disappears, and the input shaft engages with the output shaft to complete the shift. At 25 s, the gearbox starts to shift from high-speed gear to low-speed gear, and the synchronizing torque increases positively. The synchronizing torque acting on input shaft is smaller than that of the up-shift process due to the larger transmission ratio. Thus, the MG2 speed synchronization time is increased. The synchronizing torque is the resistance torque for the driving wheel during the process, the gearbox outputs the negative torque until the end of the upshift process. The change in system output torque has insignificant impact on the vehicle speed due to the short duration of gear shifting.
The simulation results show that the AMT model can reflect the internal dynamic relationship and external output characteristics of the transmission. Thus, the expected design goal is achieved, and the requirements of simulation accuracy of the entire powertrain are satisfied.
2.4 Longitudinal Dynamics Model
Ignoring the rotational viscous damping, elastic deformation of bearing, and meshing deformation of gear, the basic dynamic relations of SPGS are shown in Eqs. (13)–(15).
$$I_{r} \dot{\omega }_{r} = F \cdot R - T_{r} ,$$
(13)
$$(I_{e} + I_{c} )\dot{\omega }_{e} = T_{e} - F \cdot R - F \cdot S,$$
(14)
$$(I_{g} + I_{s} )\dot{\omega }_{g} = F \cdot S - T_{g} ,$$
(15)
where \(T_{r}\), \(T_{e}\), and \(T_{g}\) denote the ring torque, engine torque, and MG1 torque, respectively; \(I_{r}\), \(I_{c}\), and \(I_{s}\) represent the moment of inertia of ring gear, planetary carrier, and sun gear, respectively; \(I_{e}\) and \(I_{g}\) are moment of inertia of engine and MG1, respectively; \(\omega_{r}\), \(\omega_{e}\), and \(\omega_{g}\) are speed of ring gear, engine, and MG1, respectively; \(R\) and \(S\) are the radii of the ring gear and sun gear, respectively; \(F\) is the internal force of SPGS.
The characteristic parameter (the ratio of ring gear teeth to sun gear teeth) of SPGS is defined as \(k_{p}\), as follows:
$$k_{p} { = }\frac{R}{S}.$$
(16)
According to Eqs. (13), (14), and (16), \(T_{r}\) can be expressed as follows:
$$T_{r} { = }\frac{{k_{p} }}{{1 + k_{p} }}T_{e} - I_{r} \dot{\omega }_{r} - \frac{{k_{p} }}{{1 + k_{p} }}(I_{e} + I_{c} )\dot{\omega }_{e} .$$
(17)
Then, according to the AMT model, the AMT dynamic equation can be obtained using Eq. (18), as follows:
$$i_{g}^{2} (I_{m} + I_{a} )\dot{\omega }_{r} = T_{m} i_{g} + T_{syn2} - T_{a} .$$
(18)
The output torques of ring gear and AMT are coupled to the drive wheel to overcome vehicle driving resistance. If only longitudinal dynamics of the vehicle is considered, then the following relationship can be obtained:
$$\left( {\frac{{n_{t} I_{t} }}{{R_{t} }} + MR_{t} } \right)\frac{{\dot{\omega }_{r} R_{t} }}{{i_{0} }} = (T_{r} + T_{a} )i_{0} - T_{f} ,$$
(19)
where \(T_{f}\) is the driving resistance of the bus, \(I_{t}\) is the wheel inertia, \(n_{t}\) is the number of wheel, \(M\) is the vehicle mass, \(a\) is longitudinal acceleration, \(R_{t}\) is the wheel radius, and \(i_{0}\) is the final drive ratio.
On the basis of this analysis, the longitudinal dynamic equation between SPGS and driving wheels can be obtained by combining Eqs. (17), (18), and (19), as follows:
$$\begin{aligned} & \left( {\frac{{k_{p} }}{{1 + k_{p} }}T_{e} + T_{m} i_{g} + T_{syn} } \right)i_{0} - T_{f} \\ & \quad = \frac{{k_{p} }}{{1 + k_{p} }}(I_{e} + I_{c} )\dot{\omega }_{e} i_{0} + \left[ {i_{g}^{2} i_{0}^{2} (I_{m} + I_{a} ) + I_{r} i_{0}^{2} + n_{t} I_{t} + MR_{t}^{2} } \right]\frac{{\dot{\omega }_{r} }}{{i_{0} }}. \\ \end{aligned}$$
(20)