Examples of physical models

Examples of physical models

1) Dynamics of the rotor of an electric motor.

Consider a mechanical element with inertia J, linear friction coefficient b that rotates at the angular velocity ω to which an external torque c(t) is applied.

c(t)

0 t

? J

ω(t) b

c(t) ω(t)

0 t

?

The differential equation of the system can be obtained from the law of conservation of the angular momentum:

d[Jω(t)]

dt = c(t) − b ω(t) ↔ J ˙ω(t) + b ω(t) = c(t)

Starting from zero initial conditions and applying the Laplace transform one obtains:

J s ω(s) + b ω(s) = C(s) ↔ ω(s) = 1

b + J s C(s) So, the physical system can be described as follows:

C(s) -

c(t)

G(s) 1 b + J s

ω(s)-

ω(t)

where G(s) is the transfer function that describes the physical system:

G(s) = 1 b + J s

2) Mass-spring-damper system.

x b

m F K

• Variables and parameters of the physical system:

x(t) : position m : mass

˙x(t) : velocity K : stiffness of the spring

¨

x(t) : acceleration b : linear friction coefficient F (t) : applied force P (t) : momentum

• Conservation law of the momentum P (t) = m ˙x(t):

d

dt[P (t)] = X

i

Fi(t) → m d

dt[ ˙x(t)] = X

i

Fi(t)

• The following differential equation is then obtained:

d

dt[m ˙x(t)] = F − b ˙x(t) − K x(t) which can be rewritten as follows:

m¨x(t) + b ˙x(t) + K x(t) = F (t)

• Using the Laplace transform (x(0) = ˙x(0) = 0) one obtains:

m s²X(s)+b s X(s)+K X(s) = F (s) ↔ X(s) = F (s)

m s² + b s + K The system can therefore be represented as follows:

-

F (s)

G(s) 1

m s² + b s + K

-

X(s)

3) RL electrical system.

Vi

Ii

R

L Vu

Iu = 0

• Physical law: the variation of the concatenated flux φ_c(t) = LI_i(t) is equal to the voltage V_u(t) applied to the inductance.

d

dt[φ_c(t)] = V_u(t) → L d

dt[I_i(t)] = V_i(t) − R I_i(t)

• Applying the Laplace transform, with zero initial conditions, we have:

L s I_i(s) + R I_i(s) = V_i(s) ↔ I_i(s) = 1 L s + R

| {z } G(s)

V_i(s)

• So, the system can be represented as follows:

-

V_i(s)

G(s) 1 L s + R

-

I_i(s)

• The mathematical relation between the input voltage V_i(t) and the output voltage V_u(t) is described by the following transfer function:

G₁(s) = V_u(s)

V_i(s) = L s L s + R The corresponding differential equation is:

L ˙V_u(t) + R V_u(t) = L ˙V_i(t)

4) Direct Current Electric Motor.

I_a

6

V





L R ⁶

E 

+

− J

ωm b

θ_m

C_m C_e

• POG block diagram of the DC electric motor:

V

I_a

- 

1 R + Ls

?

?

 - -K_e ^-

Cm

E  _

K_e ^{ -}

1 b + Js

6

6

ω_m

-  C_e

The system is described by the following two differential equations:

( L ˙I_a = −RI_a − K_e ω_m + V J ˙ω_m = K_eI_a − b ω_m − C_e

• For the principle of energy conservation, the constant K_e links both the armature current I_a to the motor torque C_m and links the counter motor force E to the angular velocity ω_m:

C_m = K_eI_a, E = K_e ω_m.

• Using the “Mason formula” one obtains the following relation between the output variable ωm(t) and the input variables V (t) and Ce(t):

ωm(s) = G1(s) V (s) + G2(s) Ce(s)

where G₁(s) is the transfer function between input V (t) and output ω_m(t)

G₁(s) = K_e

(R + L s)(b + J s) + K_e²

and G₂(s) is the transfer function between input C_e(t) and output ω_m(t):

G₂(s) = −(R + L s)

(R + L s)(b + J s) + K_e²

• The previous relationship can also be rewritten as follows:

L J s² + (R J + L b)s + R b + K_e²

ωm(s) = KeV (s) − (R + L s)Ce(s) which corresponds to the following second order differential equation:

L J ¨ω_m + (R J + L b) ˙ω_m + (R b + K_e²)ω_m = K_eV − R C_e − L ˙C_e

5) Hydraulic clutch.

Consider the following simplified hydraulic model of a clutch:

P Supply pressure

Q Volumetric flow rate in the valve Kv Constant of prop. of the valve Cm Cylinder hydraulic capacity P¹ Pressure inside the cylinder A Piston section

x Piston position

˙x Speed of the piston mp Piston mass

b Linear friction of the piston Km Spring Stiffness

Fm Spring force on the piston

Valvola (K_v) m_p

b, A K_m

P R = 0 x

P₁ C_m Qx

Q

• A block diagram describing the dynamics of the system is the following:

P

Q

- 

K_v

?

?

 -

 -

1 C_m s

6

6

P₁

-  

Qx

A 

- A ^-F_{x - }

1 b+mps

?

?

˙x

 -

 -

6

1 s

6

K_m

6

F_m

- 

x

0

• The transfer function G(s) which links the input P to the output F_m is:

G(s) = Fm(s)

P (s) = AKmKv

Cmmps³ + (Cmb + Kvmp)s² + (A² + CmKm + Kvb)s + KmKv

which corresponds to the following third order differential equation:

Cmmp

F...m+(Cmb+Kvmp) ¨Fm+(A²+CmKm+Kvb) ˙Fm+KmKvFm = AKmKvP (t)

6) Mechanical transmission system.

Consider the mechanical system sho- wn in figure: an external torque τ is applied to a shaft with inertia J which rotates at speed ω; the shaft moves an elastic gear characterized by a torsional stiffness K₁ and a ra- dius R; the elastic gear moves a mass M which compresses a linear spring having stiffness coefficient K₂.

R

M J

b₁

b₂ K₂

K₁

w t

R

M J

b₁

b₂ K₂

K₁

w t

x x

A block diagram describing the dynamics of the system is the following:

τ

ω

- 

1 b₁ + J s

?

?

 ω -

 -

K1

s

6

6

τ₁

-   1 

R

- 1

R ^{- - }

1 b₂ + M s

?

?

˙x

 -

 -

K2

s

6

6

F₂

- 

F₂

0

The transfer function G(s) that links the input torque τ to the output force F2 can be easily obtained using the Mason formula:

G(s) = K1K2R

a₄s⁴ + a₃s³ + a₂s² + a₁s + a₀ where

a₄ = J M R²

a3 = (b2J + b1M )R²

a₂ = J K₁ + b₁b₂R² + J K₂R² + K₁M R² a₁ = b₁K₁ + b₂K₁R² + b₁K₂R²

a₀ = K₁K₂R²