Dynamic modeling of physical systems

Dynamic modeling of physical systems

• The main energetic domains encountered in modeling physical systems are:

– mechanic (translational and rotational) – electromagnetic

– hydraulic – thermic (≃)

• These domains are quite different, but they have a very similar dynamic structure. In fact, each energetic domains is mainly characterized by:

– 2 “dynamic” elements D¹ and D² which store the energy (i.e.

capacitors, inductors, masses, springs, etc.);

– 1 “static” element R which dissipates (or generates) the ener- gy(i.e. resistors, frictions, etc.);

– 2 “energy” variables q1(t) and q²(t) which define how much energy is stored within the dynamic elements;

– 2 “power” variables v1(t) and v²(t)which describe how the energy moves within the system;

• The product p(t) = v¹(t)v²(t) has the physical meaning of “power” flowing through the physical section“” characterized by the two power variables v¹(t) and v²(t).

• The two dynamic elements D¹ and D² are “dual”: their differential equa- tions have the same structure and only differ in a reversal of the subscripts

“1” e “2”.

• The dynamic element D¹ is characterized by:

1) the internal variable q¹(t);

2) the input variable v²(t);

3) the output variable v¹(t);

4) the constitutive relation Φ¹(v¹) which links the internal variable q¹ to the output variable v¹:

q1 = Φ¹(v¹) ↔ v1 = Φ⁻¹1 (q¹)

?

1 s

?

Φ⁻¹1 (q¹)

?

v1(t) v2(t)

q1(t)

5) a differential equation which links the internal variable q¹ to the input variable v²:

dq1

dt = v² ↔ dq1 = v²dt

6) the stored energy E¹(q¹) is a function only of the internal energy variable q¹:

E¹ = Z t

0

p(t) dt = Z t

0

v¹v²dt = Z q₁

0

Φ⁻¹1 (q¹) dq¹ = E¹(q¹)

• The input and output power flows of the dynamic element D¹ can be clearly shown using one of the two following graphical representations:

v2a

v1

- 

?

1 s

?

Φ⁻¹1 (q¹)

?

 -

v2b

v1

v2

q1

v1

v2a

 -

6

1 s

6

Φ⁻¹1 (q¹)

6

- 

v1

v2b

v2

q1

In this case the entering power is pe(t) = v²av1 and the exiting power is p_u(t) = v²bv1.

• The dynamic element D² is characterized by:

1) the internal variable q²(t);

2) the input variable v¹(t);

3) the output variable v²(t);

4) the constitutive relation Φ²(v²) which links the internal variable q² to the output variable v²:

q2 = Φ²(v²) ↔ v2 = Φ⁻¹2 (q²)

?

1 s

?

Φ⁻¹2 (q²)

?

v2(t) v1(t)

q2(t)

5) a differential equation which links the internal variable q² to the input variable v¹:

dq2

dt = v¹ ↔ dq2 = v¹dt

6) the stored energy E²(q²) is a function only of the internal energy variable q¹:

E² = Z t

0

p(t) dt = Z t

0

v¹v²dt = Z q₂

0

Φ⁻¹2 (q²) dq² = E²(q²)

• The input and output power flows of the dynamic element D² can be clearly shown using one of the two following graphical representations:

v1a

v2

- 

?

1 s

?

Φ⁻¹2 (q²)

?

 -

v1b

v2

v1

q2

v2

v1a

 -

6

1 s

6

Φ⁻¹2 (q²)

6

- 

v2

v1b

v1

q2

In this case the entering power is pe(t) = v¹av2 and the exiting power is p_u(t) = v¹bv2.

• The static element R is characterized only by a constitutive relation ΦR

that links the two power variables v¹ and v²:

v1 = ΦR(v²) ↔ v2 = ΦG(v¹) The functions ΦR(·) and ΦG(·) are one the inverse of the other:

ΦG(v¹) = Φ⁻¹_R (v¹) ↔ ΦR(v²) = Φ⁻¹_G (v²) This element can be oriented in both ways:

?

ΦR(v²)

?

v2(t)

v1(t)

ΦG(v¹)

6

6

v2(t)

v1(t)

• The input and output power flows of the static element R can be clearly shown using one of the two following graphical representations:

v2a

v1

- 

ΦR(v²)

?

?

 -

v2b

v1

v2

v2

v1a

 -

ΦG(v¹)

6

6

- 

v2

v1b

v1

• The power p(t) dissipated (or generated) by the static element R is equal to the product of the two power variables:

p(t) = v¹(t)v²(t) = ΦR(v²) v² = v¹Φ⁻¹_R (v¹)

• Electromagnetic domain:

Name Constitutive Rel. Linear case Differentioal Eq.

D₁ C Capacitor

q1 Q Charge Q = ΦC(V ) Q = C V d Q dt = I v1 V Voltage

D₂ L Inductor

q² φ Flux φ = ΦL(I) φ = L I d φ dt = V v2 I Current

R R Resistance V = ΦR(I) V = R I

Linear dynamic elements

• Inductor (φ = L I):

L

V1 V2

I I

d

dt [L I] = V¹ − V2

V1 ^{- } 1 L s

?

?

I

 -

V2

• Capacitor (Q = C V ):

C

I1 I2

V

I^{1 - }

1 C s

?

?

 -

I²

• Mechanic Translational domain:

Name Constitutive Rel. Linear case Differentioal Eq.

D₁ M Mass

q¹ P Momentum P = ΦM( ˙x) P = M ˙x d P

dt = F v1 ˙x Velocity

D₂ E String

q2 x Displacement x = ΦE(F ) x = E F d x dt = ˙x v2 F Force

R b Friction F = Φb( ˙x) F = b ˙x

• Mechanic Rotational domain:

Name Constitutive Rel. Linear case Differentioal Eq.

D₁ J Inertia

q1 P Ang. Momentum P = ΦJ(ω) P = J ω d P dt = τ v1 ω Ang. Velocity

D₂ E Rot. Spring

q2 θ Ang. Displacement θ = ΦE(τ ) θ = E τ d θ

dt = ω v2 τ Torque

R b Rot. Friction τ = Φb(ω) τ = b ω

• Hydraulic domain:

Name Constitutive Rel. Linear case Differentioal Eq.

D₁ C_I Hyd. Capacitor

q1 V Volume V = ΦC(P ) V = CI P d V

dt = Q v1 P Pressure

D₂ L_I Hyd. Inductor

q2 φ_I Hyd. Flux φ_I = ΦL(Q) φI = LI Q d φ_I

dt = P v2 Q Volume flow rate

R R Hyd. Resistor P = ΦR(Q) P = RI Q

• Hydraulic Inductor:

L_I

P1 P2

Q Q

d

dt [LIQ] = P¹ − P2

P1 ^{- } 1 L_Is

?

?

Q

 -

P2

• Hydraulic Capacitor:

C_I

Q1 Q2

P

d

dt [CIP] = Q¹ − Q2

Q1 ^{- } 1 C_Is

?

?

P

 -

Q2

• Hydraulic Resistance:

R_I P1 P2

Q Q

P1 ^{- } 1 R_I

?

P2

Examples

• Mass-dumper system.

Differential equation:

d(m ˙x)

d t = F − b ˙x Using the Laplace transform:

X˙ (s) = 1

m s + b F(s)

one obtains the transfer function:

G(s) = 1 m s + b

x

m F b

0

 -

b

6

6 - 

- 

1 m s

?

?

˙x

 -

F

• Parallel spring-dumper system:

R R

˙x¹

˙x²

b

K x2

x1

R

˙x¹

 -

b+ ^K_s

6

6 - 

R

˙x²

• Series spring-dumper system:

R R

˙x¹

˙x³

˙x²

b K

x2

x3

x1

R

˙x²

 

1 b

?

?

- -

 -

K s

6

6 - 

R

˙x¹