transfer function of electrical and mechanical systems

Polytechnic UniversityMechanical Engineering Department

System Dynamics and Controls - ME 5050 - 24

Profesora: Sandra L. Ordóñez E.

In general, a physical system that can be represented by a linear time-invariant differential equation can be also represented by a transfer function. In this chapter sections we are going to determine the transfer function of electrical and mechanical systems. Remember that the general procedure to determine the transfer function of a system is:

- Determine the differential equations that describe the behavior of the system- Find the Laplace transform of such differential equations.- Represent the system as a transfer function, described by: the output (as an s

polynomial) / the input (as an s polynomial).

The differential equations of a system correspond to its mathematical model. This is determined using the physics laws like Newton’s and Kirchhoff’s.

Transfer function of electrical systems

1. Passive circuits (that use only passive elements such as resistances, inductances and capacitances):When we have an electrical system represented by the electrical circuit in Figure 1, we should first identify its input signal and its possible output signals.

Figure 1. Electrical system, RLC circuit

a) If the system’s input is v(t), and its output is i(t), the differential equation that describe the system dynamics is:

Then, the Laplace transform of this equation (when the initial conditions are zero) is:

Factorizing I(s):

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+_v(t) vc(t)

+

_C

LR

i(t)

Now we represent the system by its transfer function:

b) If the system’s input is v(t), and its output is vc(t), the differential equation that describe the system dynamics is:

Then, the Laplace transform of this equation (when the initial conditions are zero) is:

but,

Now we represent the system by its transfer function:

We can find the transfer function directly without determine the differential equations of the system using the Laplace transform of each element and then determine the KVL equations.

The voltage drop in each element of a RLC circuit is:- For the inductor: VL(s)=LsI(s)- For the Resistor: VR(s)=RI(s)- For the Capacitor: VC(s) = I(s)/Cs

V(s)=VL(s)+VR(s)+VC(s) = LsI(s)+RI(s)+I(s)/Cs, That is the same that we had before (above).

Since we now that the impedance Z is defined as: V(s)/I(s), then, - ZI(s)=Ls

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LCs

L

Rs

LC1

/1

2

Vc(s)V(s)

- ZR(s)=R- ZC(s) = 1/Cs

c) If we want to determine the transfer function of the same circuit using nodes, then:

That is identical to the one we found before (above).

2. Active circuits (It is an electronic circuit which uses active devices such as transistors, operational amplifiers, or integrated circuits for its operation and which requires a power source for operation).

An operational-amplifier (op-amp) is a device that amplifies analog signals, with a gain A. The op-amp has two inputs and one output. The ideal op-amp is showed in Figure 2.

va = vb

i1 = 0, i2 = 0, Ro = 0, Ri = vo = vb - va

Figure 2. Ideal Operational Amplifer

a) Transfer function of an Inverter amplifier (Figure 3).

Figure 3. Inverter amplifier

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va

_

+

Vo(s)

Vi(s)

Z2

Z1

vb

To find the transfer function of the amplifier we can use nodal analysis:

Since we know that Vb = Va and Va is connected to ground, then, Vb = Va = 0, and the equation will become:

Which means that the gain A of the inverter amplifier will be (-Z2/Z1), and at the output Vo(s) I will have the input multiplied by that gain A and inverted.

Example 1:Remembering that the impedances for elements such as L, R and C are:

- ZI(s)=Ls- ZR(s)=R- ZC(s) = 1/Cs

Then we can determine the transfer function of the system showed in Figure 4, as follows:

Figure 4.

We first determine the transfer function using the impedances (same as above).

Then, we calculate the values of Z2 and Z1 in terms of R and C and their Laplace representation.

We can start with Z2, where R2 and C2 are in series, then, Z2 is the algebraic sum of their Laplace transforms:

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Z2

Z1

Vi(s) vb

va

_

+

Vo(s)

R1

R2C1

C2

And, Z1 will be the parallel between R1 and C1:

Then, the transfer function of the circuit shown in Figure 4 will be:

b) Transfer function of an Non-Inverter amplifier (Figure 5).

Figure 5. Non-Inverter amplifier

To find the transfer function of the amplifier we can use nodal analysis:

Since we know that Vb = Va and Va is our input Vi, then, Vb = Va = Vi, and the equation will become:

Which means that the gain A of the inverter amplifier will be (1+Z2/Z1), and at the output Vo(s) it will be the input multiplied by that gain A and with the same polarity.

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va

_

+

Vo(s)

Vi(s)

Z2

Z1

vb

Example 2:Remembering that the impedances for elements such as L, R and C are:

- ZI(s)=Ls- ZR(s)=R- ZC(s) = 1/Cs

Then we can determine the transfer function of the system showed in Figure 6, as follows:

Figure 6.

We first determine the transfer function using the impedances (same as above).

Then, we calculate the values of Z2 and Z1 in terms of R and C and their Laplace representation.

We can start with Z1, where R1 and C1 are in parallel, and that will be in series with R1:

And, Z1 will be calculated in the same way that we calculated Z1. R4 and C2 are in parallel, and that will be in series with R3:

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C2Z2

Z1

Vi(s)

vb

va

_

+

Vo(s)

R2

R4

C1 R1

R3

Then, my transfer function will be:

Transfer function of mechanical systems

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a) Mathematical model of a springIf lo is the natural length of the spring (when it is not compressed or tensioned), and ∆l is the elongation, then, in equilibrium (when all movement has ceased), the static analysis of the system would be:

Figure 7. Static analysis of a spring-mass system

T = Mg = K∆l, where K is the static constant of the springIn equilibrium, the force created by the spring equals the gravitational force (weight) of the mass and there is no movement.

Dynamic analysis: What should happen if an external force F is applied altering the equilibrium point?

Newton’s Laws:- - The sum of all external forces acting over a body is equal to the

If m0, then F ma. For example, this is the case of a space ship.

Then for our system, the sum of the vertical forces should be:

Figure 8. Dynamic analysis of a spring-mass system

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T

Mg

M

M

lo lo

∆lMg

x(t)M

T

Mg F

And since Mg = K∆l,

Then, the second order differential equation that describes the dynamics of the system around the equilibrium point is:

- Mathematical model of Dampers, shock absorbers or viscous friction.This is a force that is opposing to movement and it is experimented by any body that is moving in the presence of a fluid medium (liquid or gas). The force created by a shock absorber opposes to movement and it is proportional to the velocity of movement. The shock absorber only dissipates energy.

Figure 9. mathematical model of a damper

Combining both elements in the same system:

Example 3: Find the transfer function of the system in Figure 10.

Figure 10. Mechanical system that combines spring-damper mechanical elements

Applying the Laplace transform to this second order differential equation, and assuming initials conditions equal to zero, we have:

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2

2 )())()(

dt

txdmtKxlKlKtF

oil

x(t)F

F(t)

x(t)

M

K B

The transfer function of the system represented in block diagrams would be:

Example 4Determine the differential equations that describe the dynamics of the system in and determine their Laplace transform.

Figure 11

Since there are three masses, and there will be three independent movements (three degrees of freedom), then the dynamic of the system will be represented by three differential equations (one per mass).

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KBsMs 2

1 X(s)F(s)

F(t)

x3(t)

M3

K4

B3

M2

B2

K3

K2

M1

B1K1

x2(t)

x1(t)

The first differential equation, for mass M1 will be:

The second equation, for mass M2:

And the third equation, for mass M3:

If we apply the Laplace transform in each of those differential equations (assuming initial conditions equal to zero), and reorganizing the terms when have:

Using these equations we can find the transfer function X3(s)/F(s), or X2(s)/F(s), or X1(s)/F(s), by solving the equations for the chosen variable (X1(s), X2(s), X3(s)).

b) Rotary systems

According to Newton’s second law,

Where J = Inertia, = rotation angle, = angular velocity, = angular acceleration

- Torsion bar.When a torque (T) is applied to a torsion bar, there will be a reaction torque (TK) in the other end of the bar.

- Rotary viscous friction.It occurs when a body is moving within a fluid medium and it is proportional to the velocity.

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1 2T TK

K

1 2T TK

B

Combining both elements in the same system:

Example 5: Find the transfer function of the system in Figure 12.

Figure 12. Simple rotary system

Applying the Laplace transform to this second order differential equation, and assuming initials conditions equal to zero, we have:

Example 6Determine the differential equations that describe the system shown in Figure 13, and their Laplace transform assuming initial conditions equal to zero.

Figure 13

Since there are three inertias, and there will be three independent movements (three degrees of freedom), then the dynamic of the system will be represented by three differential equations (one per inertia).

The first differential equation, for inertia J1 will be:

The second equation, for inertia J2:

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KB

JT

KJ1

T1 2

D1

J2 J3

D2 D3

3

And the third equation, for inertia J3:

If we apply the Laplace transform in each of those differential equations (assuming initial conditions equal to zero), and reorganizing the terms when have:

Using these equations we can find the transfer function 3(s)/T(s), or 2(s)/T(s), or 1(s)/T(s), by solving the equations for the chosen variable (1(s), 2(s), 3(s)).

c) Systems with gearsThe gears are used to transmit a torque from one axis to another at different velocities. This concept is analog to an electrical transformer.

Where N1, r1 and N2, r2 are the number of teeth and radius of the gears 1 and 2 respectively. The gears circumference is:2r1 = N1 and 2r2 = N2 where is the proportionality constant that relates the number of teeth of the gear to its circumference. Since the gears 1 and 2 are mechanically connected, has the same value in both relations. Then,

And its tangential movement is described by:

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T1 1

T2 2

r2, N2

r1, N1

2

1

2

1

2

1

2

1

2

2

1

1

2

222

N

N

r

r

N

N

r

r

N

r

N

r

The relation between the torques T1 and T2 will be:

Reflecting the input/outputWhen a rotational system has gears, usually it is necessary to reflect one or more parts of the system to another one, to find the transfer function.

For example, the differential equation that describes the behavior of the system of Figure 14 in terms of T2 and 2 is:

Figure 14

Applying the Laplace transform:

Having in mind that:

If we want to express the systems in terms of T1 instead of T2, and eliminate the gears, we should have the system shown in Figure 15:

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Ftangential

1T1

2T2

r1

r2

r1, N1

r2, N2

K

J

D

T1 1

T2 2

K

J

D

T1(N2/N1) 2

Figure 15

And expressing it in terms of 1 instead of 2, will result in the equivalent system of Figure 16:

Figure 16

Generalizing, the mechanical rotational impedances can be reflected through the gears multiplying by:

Example 7. Find the transfer function 2/T1 for the system of Figure 17.

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T1 1 2

2

1

N

NJ

2

2

1

N

NK

2

2

1

N

ND

r1, N1

r2, N2

K

J2

D2

T1 1

T2 2

J1

D1

Figure 17

When we look at this system we initially think that the system is described for two differential equations since the system has two inertias. However, since those inertias are jointed by gears, there will be only one independent movement so there will be only one differential equation. Since we need to have the system in some familiar form (only inertias, dampers and torsion bars), we need to find an equivalent system using the relation mentioned above.

For an equivalent system (shown in Figure 18), the differential equation that describes its behavior would be:

Figure 18

And since

Then,

Since the equivalent of elements in the same axis will be its algebraic sum, and the equivalent of the elements from other axis are the same elements multiplied by the relation of the gears, then, the equivalent inertia Jeq, equivalent damper Deq and equivalent torsion bar Keq of the system will be:

Figure 19

Gears in cascade:

For a system with gears in cascade as shown in Figure 20, the rotation angles can be expressed in terms of the number of teeth of all the gears as follows:

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Keq

Jeq

Deq

T2 2

T1(N2/N1) 2

2

2

1

21 J

N

NJ

K

2

2

1

21 D

N

ND

N2

N1

1

2

N4

N3 3

N6

N5 4

Figure 20

Example 8Find the equivalent elements (Jeq, Deq) of the system shown in Figure 21, if you reflect all the elements to axis 1.

Figure 21

This system is the resumed version of the system shown in Figure 22.

We can start with the equivalent elements in the third axis where there are only two inertias in series:

Then, we can calculate the equivalent elements of the second axis:

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N2

N1, J2,D2

1

2

N4

N3, J3 3

J5

J4

J2,D2

T1

And then finally we can calculate the equivalent elements of the first axis:

Then,

Another representation of the same system is:

Figure 22

d) Rack and pinion(converts radial motion to linear motion)If a pinion with radius r has an angular movement , then, the linear displacement x will be:

x = r

Transfer function of a DC motorA DC motor is used to move loads and is called an actuator. An actuator is a device that provides the motive power to the process. DC motors are used widely in control applications such as robotics manipulators, tape transport mechanisms, disk drives, machine tools and others.

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T1

J5

3

J4N4

N2N3

2

D2

J3J2D2

J2

1

N1

The DC motor has two parts, the armature and the field. A model of the DC motor is shown in Figure 23.

Figure 23. DC motor equivalent circuit

The flux is defined as: = Kf if

The motor torque is defined as: Tm = K1 ia(t) = K1Kfif(t)ia(t)Then, to control the motor one current must be maintained constant while the other current becomes the input current. In a field current controlled motor, the armature current is maintained constant. In the armature controlled motor, the field current is the one that is maintained constant.

Field current controlled motor (constant ia)

Tm(t) = K1 ia(t) = K1Kf ia(t) if(t) Tm(s) = Km If(s)

constant

In the motor circuit:Vf(s) = (Rf +Lf s) If(s) If(s) = Vf(s)/(Rf +Lf s) Tm(s) = TL(s) + Td(s) Tm(s) = TL(s) = Km If(s) =Km Vf(s)/(Rf+Lfs)

But, TL(s) = Js2(s)+bs (s) = (Js+b)s (s)

Then, the DC motor transfer function when the field is controlled is:

And can be represented by the block diagram shown in Figure 24

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T. Load T. disturbance (often negligible)

(s)Km

s

1

ff RsL 1

Field

bJs 1

Load

Speed Tm(s)+ Position, (s)

Output

If(s) Vf(s) TL(s)

Td(s)-

°

°

°

°

if

ia

Lf

Rf

Ra

La

J

Inertia. Friction b

Armature

Field

vf

+

-

,

Figure 24. Field controlled DC motor block diagram

Armature controlled motor (constant if)

This is the most common used method to control a DC motor.

Figure 25. DC motor equivalent circuit

Va(s) = (Ra +La s) Ia(s)+Vb(s) Vb(s) = Kb (s)

Ia(s) = (Va(s)- Kb (s)) /(Ra +La s)

Tm(s) = TL(s) + Td(s) Tm(s) = TL(s) = Km Ia(s)

But, TL(s) = Js2(s)+bs (s) = (Js+b)s (s), so,

TL(s) = Km Ia(s) = Km (Va(s)- Kb (s)) /(Ra +La s)

(Js+b)s (s) = Km (Va(s)- Kb (s)) /(Ra +La s), and since (s) = s (s),

(Js+b)s (s) = Km (Va(s)- Kb s (s)) /(Ra +La s)

(Ra +La s) (Js+b)s (s) + Km Kb s (s) = Km Va(s)

((Ra +La s) (Js+b)+ Km Kb )s (s) = Km Va(s)

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Back electromotive force (back emf) voltage. Proportional to the motor speed

Armature

°

°

ia

LaRa

J

Inertia. Friction bvf

+

-

,

°ifRf

Lf Field

°

And can be represented by the block diagram shown in Figure 26.

Figure 26. Armature controlled DC motor block diagram

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Ia(s)

Td(s)-

(s)

Kms

1

aa RsL 1

Armature

bJs 1

Load

Speed Tm(s)+ Position, (s)

Output

Va(s) TL(s)

KbBack emf

+

-

transfer function of electrical and mechanical systems

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