Lucas Glenzinski
Exam #2
Section 2:
The model is defined by following the flow of mechanical transference of forces. To start, we
define the initial inputs of an EV vehicle that has a battery stack to provide energy for the
engine. In this case, work is produced by an electric motor proportional to the energy input
which is transferred into rotational motion as both torque and angular velocity. The rotational
motion is then transferred to a transmission that adjusts the torque and speed through gear
ratios which modifies the rotational energy linearly. The transmissions output is connected to the
driveshafts input which is just the transference of the rotational motion to the differential which is
subject to inertia and thus friction as it accumulates torque. The differential then distributes the
rotation to the wheels equally and splits the torque linearly. The wheels transfer rotation to the
road using friction and accumulate rotational energy as it spins. The next interaction is the
interaction between the road and vehicle itself, this completely depends on the input of the road
and the cars’ reaction ex. Slipping, tire stiffness, suspension. The final consideration is the
vehicle itself which has a velocity directly affected by the tire-road interaction which changes
over time depending on the vehicle mass properties.
To simplify:
Engine -> Transmission -> Driveshaft -> Differential -> Wheels -> Tire-Road -> Vehicle Body
Integral and Lag -> Proportional -> Integral with dampening -> Proportional -> Integral with
dampening -> Proportional and Differential -> Integral
To derive the DC motors transfer function, we start by defining its parameters:
1. R_a = armature resistance
2. L_a = armature inductance
3. e_a(t) = applied armature voltage
4. i_a(t) = armature current
5. T(t) = Torque
6. J = Rotor moment of inertia
7. B = damping coefficient
8. Θ(t) = Angular displacement of motor shaft
9. K_t = Torque constant
10. K_b = Back EMF constant
Kirchoffs law is then applied to the armature circuit which yields the equation: e_a(t) =
L_a*(i_a(t)/dt) + R_a*i_a(t) + e_b(t) where e_b(t) = K_b*ω (back EMF proportional to angular
speed). The relation that torque is proportional to armature current yields the equation: T(t) =
K_t * i_a(t) which can the be expanded to: K_t*I_a = J*(dω/dt)+Bω. The laplace of ω(t) must be
taken and divided by e_a(s) in order to get the transfer function. We express e_a in the laplace
domain to acquire this: e_a(s) = R_A*I_a(s) + L_a*s*I_a(s) + K_e*Laplace(ω(s)). Rearrange to
solve for I_a(s): I_a(s) = (e_a(s) – K_e*laplace(ω(s)))/(R_a + L_a*s). Substituting I_a(s) into the
mechanical equation in the laplace domain: K_t( (e_a(s) – K_e*laplace(ω(s)))/(R_a + L_a*s)) =
(J*s + B)*Laplace(ω(s)). Solve for Laplace(ω(s))/e_a(s): Laplace(ω(s))/e_a(s) = K_t/(J*s +
B)*(R_a + L_a*s) + K_t*K_e. The highlighted portion represents the relationship between the
input voltage and the output angular speed in a DC motor. K_t is the engine gain, (J*s + B) is
the Rotors moment of inertia with respect to speed plus the dampening that takes place
resulting in mechanical impedance, (R_a + L_a*s) is the armatures variable inductance plus its
Lucas Glenzinski
Exam #2
resistance for total circuit impedance, and K_t*K_e is the coupling between the electrical and
mechanical portions.
Next, I’ll define the Transmissions transfer function:
1.
2.
3.
4.
T_m(t) = input torque
T_l(t) = load torque on second shaft
ω_1(t) and ω_2(t) = angular velocity of the 2 shafts
n_1 and n_2 = gear ratios
Since it is two gears being used, it is safe to start finding the transfer function using comparative
methods. Starting out, the angular velocities of the shafts will always be proportional to the gear
ratios they are generated from: ω_1(t)/ ω_2(t) = n_2/n_1. The same can be said for torque
related to gear ratios: T_m(t)/T_l(t) = n_1/n_2. After taking the laplace of angular velocity, then
the equality becomes the expression: Laplace(ω_2(s))/Laplace(ω_1(s)) = n_1/n_2. This
expression states that any input is directly related to its output depending on the ratio between
the gears. Revealing that a transmission does not add any integral or derivative elements to the
system.
The driveshaft component of a vehicle system includes damping and integrative properties that
must be included from its parameters:
1.
2.
3.
4.
5.
T_trans = torque from transmission
J_ds = Driveshaft inertia
B_ds = Driveshaft damping coefficient
ω_ds = Driveshaft angular velocity
T_diff = Torque to driveshaft
The final transfer function is related between the velocity and torque input yielding our target
expression of v(s)/T_trans(s) = ?. Now, since the differential is simply distributing the torque
between the two rear wheels in a vehicle, it can be assumed that the differential is a constant
included in the transfer function. Doing so allows the use of the transmission as a start and the
wheels as an end. Modeling the transmission as the laplace of a dynamic transmission yield:
T_trans(s) = J_ds*s*ω_ds(s) + B_ds(s) which can be rearranged to ω_ds(s) = T_trasns(s) /
J_ds*s+B_ds. By then using the linear velocity and wheel angular velocity expression:
ω_w=ω_ds / n (gear ratio), we can relate the wheels to the transmission indicted by the
equation: v(s) = ω_w(s)*r = r * 1/n * T(trans(s) / J_ds*s + B_ds. Simplifying: v(s) / T_trans(s) =
r/N(J_ds*s+B_ds). Since we have already considered the gear ratio of the transmission
separately, we can omit the N in the equation giving: v(s) / T_trans(s) = r/(J_ds*s+B_ds). We
can again omit the r in the equation as that relates to wheel diameter and get: v(s) / T_trans(s) =
1/(J_ds*s+B_ds). J_ds*s + B_ds describes the mechanical dynamics of the driveshaft
susceptible to friction and acceleration response.
By switching the variables around and reinstating the radius of the tire omitted previously, we
can get the transfer function of the wheels without any lengthy steps. This is feasible only
because both components are cylindrical objects subject to friction and a damping effect,
making them similar. Thus, after finding our target expression of V(s)/T_wheel(s). We get:
V(s)/T_wheel(s) = r / J_wheel*s + B_wheel. This represents the tires radius over the dynamic
nature of the wheels.
Lucas Glenzinski
Exam #2
Once the tire has been modeled, we can then analyze its interaction with the road using the
variables:
1.
2.
3.
4.
F_w = force from differential to tire
X = tire and road displacement
B_tire = tire energy loss due to deformation
K_tire = tires resistance to deformation
The interaction between tire and road is similar to a mass-spring-damper system and can be
modeled as such: F_w = B_tire*dx/dt + K_tire*x. A laplace transform can then be applied to get
the equation: F_w(s) = (B_tire*s + K_tire)*X(s). X(s) can be transferred to the other side of the
equation to get the final equation: F_traction(s)/X(s) = B_tire * s + K_tire. Where the variables
are stated above and determines the force transmitted to the road via torque.
The final part is the overall vehicle body dynamics. This is simply a summation of forces which
is characterized by the equation: F_total = m_car*a_car = m_car * dv_car(t)/dt. By using a
transfer function conversion chart, the system is described as: v_car(s) / F_traction(s) =
1/m_car*s. Where m_car is the mass, v_car is the velocity of the vehicle, and F_traction is the
traction force applied by the wheels.