Lectures_12_15_Modeling_of_Electrical_an.pdf

Lectures 12-15
Modeling of Electrical
and Mechanical Systems

Principles of Control Systems Academic year 2013-2014 164
Prof. K. Melhem (Qassim University)
Lecture outline
In these lectures we will learn how to model electric and mechanical systems in terms
of transfer functions and state-space representations.

Electric network transfer functions
Introduction
We model with transfer functions electric systems including passive networks and operational
amplifier circuits. Our guiding principles to develop the transfer functions are Kirchhoff’s laws; We
sum voltages around loops or current at nodes and equate the result to zero. Table below
summarizes the relationships between voltage and current, and between voltage and charge, for the
three passive linear components: resistors, capacitors, and inductors, provided that initial conditions
are zero. From these relationships we can write the differential equations for the network using
Kirchhoff’s laws. Then we can take the Laplace transforms and solve for the transfer function.

Introduction
Below is a technique for simplifying the solution for the transfer function. We take the Laplace
transform for the relationships between voltage and current across a capacitor, resistor, and
inductor to yield
For the capacitor,
V(s) =
1
Cs
I(s)
For the resistor,
V(s) = RI(s)
For the inductor,
V(s) = LsI(s)
We now define the transfer function
V(s)
I(s)
= Z(s)
which is similar to the definition of resistance, that is, the ratio of voltage to current. Unlike the
resistance, this function is applicable to capacitors and inductors and carries information on the
dynamic behavior of the component, since it represents an equivalent differential equation. We call
this particular transfer function impedance. Subsequently, the concept of impedance can be used to
simplify the solution for the transfer function when the Kirchhoff’s voltage law is used.

Introduction
Conversely, let us write the relationships between voltage and current across a capacitor, resistor,
and inductor as
For the capacitor,
I(s) = CsV(s)
For the resistor,
I(s) =
1
R(s)
V(s)
For the inductor,
I(s) =
1
Ls
V(s)
We now define the transfer function
I(s)
V(s)
= Y(s)
which is similar to the definition of conductance (inverse of resistance), that is, the ratio of current
to voltage. Unlike the conductance, this function is applicable to capacitors and inductors and
carries information on the dynamic behavior of the component, since it represents an equivalent
differential equation. We call this particular transfer function admittance. Subsequently, the concept
of admittance can be used to simplify the solution for the transfer function when the Kirchhoff’s
current law is used.

Voltage source and current source
Energy sources are twofold:
• Voltage source: presents a constant voltage to any load (variations in the load do not
appreciably change the voltage).
• Current source: delivers a constant current to any load (variations in the load do not
appreciably change the current).
Note: Practically, a current source can be constructed from a voltage source by placing a large
resistance in series with the voltage source.
For circuit analysis convenience, let us recall of the following theorem:
Norton’s theorem (its dual is called Thvenin’s theorem): a voltage source V(s) in series with an
impedance Z(s) can be replaced by a current source I(s) = V(s)/Z(s) = Y(s)V(s) in parallel with the
admittance Y(s).

Simple circuits via mesh analysis
Transfer functions can be obtained using Kirchhoff’s voltage law in which voltages are summed
around loops or meshes. We call this method loop or mesh analysis. Let us look at an example.
Problem Find the transfer function relating the capacitor voltage VC(s) to the input voltage V(s) in
figure below.
Solution Summing the voltages around the loop, assuming zero initial conditions, yields the
integrodifferential equation for the network as
L
di(t)
dt
+Ri(t)+
1
C
Z t
0
i(τ)dτ = v(t)
Changing variables from current to charge using i(t) = dq(t)/dt yields
L
d2q(t)
dt2
+R
dq(t)
dt
+
1
C
q(t) = v(t)

Since q(t) = CvC(t), we have
LC
d2vC(t)
dt2
+RC
dvC(t)
dt
+vC(t) = v(t)
Taking the Laplace transform assuming zero initial conditions, rearranging terms, and simplifying
yields
(LCs2
+RCs+1)VC(s) = V(s)
Solving for the transfer function VC(s)/V(s), we obtain
VC(s)
V(s)
=
1/LC
s2 + R
L s+ 1
LC
which is shown in figure below.
Let us next demonstrate how the concept of impedance simplifies the solution for the transfer
function.

The Laplace transform of the loop differential equation, assuming zero initial conditions, is
(Ls+R+
1
Cs
)I(s) = V(s)
which is in the form
[Sum of impedances] I(s) = [Sum of applied voltages]
The last form suggests the circuit shown below
in which we add impedances in series as we add resistors in series. We notice that the circuit above
could have been obtained immediately from the original network circuit simply by replacing each
component with its impedance. We call this altered circuit the transformed circuit.

Conversely, let us suppose that we have the transformed circuit
for which we simply apply the Kirchhoff’s voltage law to yield
(Ls+R+
1
Cs
)I(s) = V(s)
Solving for I(s)/V(s), we obtain
I(s)
V(s)
=
1
Ls+R+ 1
Cs
But the voltage across the capacitor is VC(s) = I(s)/Cs, which eventually leads to the same transfer
function
VC(s)
V(s)
=
1/LC
s2 + R
L s+ 1
LC
Conclusion: With the impedance concept we can bypass the differential equation and directly use
the transformed circuit to find the transfer function.

Complex circuits via mesh analysis
To solve complex electrical networks with multiple loops, using mesh analysis, we can perform the
following steps:
1. Replace passive element values with their impedances
2. Replace all sources and time variables with their Laplace transforms
3. Assume a transform current and a current direction in each mesh
4. Write Kirchhoff’s voltage law around each mesh
5. Solve the simultaneous equations for the output
6. Form the transfer function
Let us look at an example.

Problem Given the network of the figure at the top below, find the transfer function I2(s)/V(s).

Solution We first find the transformed circuit as shown in the central figure. This circuit has two
independent meshes, thus two simultaneous (because of coupling) equations are to be solved for the
transfer function. These equations can be found by summing voltages around Mesh 1 and Mesh 2 as
shown in the figure.
Around Mesh 1 in which current I1(s) flows, we have
R1I1(s)+LsI1(s)−LsI2(s) = V(s)
Around Mesh 2 in which current I2(s) flows, we have
LsI2(s)+R2I2(s)+
1
Cs
I2(s)−LsI1(s) = 0
Combining terms in the last two equations, we find the simultaneous equations in I1(s) and I2(s):
(R1 +Ls)I1(s)−LsI2(s) = V(s)
−LsI1(s)+

Ls+R2 +
1
Cs

I2(s) = 0

Solution (Cont’d) The system of linear equations representing the last two equations has the general
form Ax = B for which unknowns x is given as x = A−1B. If we are interested in a single unknown xk,
Cramer’s rule can be used. To solve for I2(s) Cramer’s rule gives that
I2(s) =
R1 +Ls V(s)
−Ls 0
∆
=
LsV(s)
∆
where
∆ =
R1 +Ls −Ls
−Ls Ls+R2 + 1
Cs
It follows that the transfer function I2(s)/V(s) is
I2(s)
V(s)
=
Ls
∆
=
LCs2
(R1 +R2)LCs2 +(R1R2C +L)s+R1
which is shown in the figure at the bottom before.

Solution (Cont’d) The last equations in I1(s) and I2(s) take the following general form:




Sum of
impedances
around Mesh 1



I1(s)−







Sum of
impedances
common to the
two meshes







I2(s) =




Sum of applied
voltages around
Mesh 1




−







Sum of
impedances
common to the
two meshes







I1(s)+




Sum of
impedances
around Mesh 2



I2(s) =




Sum of applied
voltages around
Mesh 2




Recognizing this form helps us write such equations rapidly from the transformed circuit!

Solution (Cont’d)
Important remark about the number of independent meshes required: In a multi-meshes electric network,
the number of simultaneous differential equations required to describe the system equals the number
of independent meshes. Independent meshes are when we open-circuit one of the meshes, the other
meshes can still have currents flowing through them if there are voltage sources in these meshes. In
our example, there are (sets of) two independent meshes. Determination of the currents flowing in
the independent meshes should give the currents in each electric component in the circuit.

Simple circuits via nodal analysis
Transfer functions can be obtained using Kirchhoff’s current law in which currents are summed at
nodes. We call this method nodal analysis. Let us look at an example.
Problem Repeat the previous problem of RLC network using nodal analysis and without writing a
differential equation.
Solution The transfer function can be obtained by summing currents flowing out of the node whose
voltage is VC(s) (unknown) in the figure below.
We assume currents leaving the node are positive and currents entering the node are negative. The
incoming current is the current flowing through the series resistor and inductor and the outgoing
current is the current through the capacitor. The Kirchhoff’s current law applied at that node gives
VC(s)
1/Cs
+
VC(s)−V(s)
R+Ls
= 0
from which it follows the same transfer function VC(s)/V(s) determined earlier.

Simple circuits via nodal analysis
Solution Notice that the last equation can be rewritten as

1
R+Ls
+Cs

VC(s) =
1
R+Ls
V(s)
which is in the form
[Sum of admittances connected to the node] VC(s) = [Sum of applied currents at the node]
This suggests the following transformed circuit given now with current source and admittances
rather than voltage source and impedances.
1/(R+Ls)V(s) 1/(R+Ls)
Cs VC(s)

Complex circuits via nodal analysis
To solve complex electrical networks with multiple nodes, using nodal analysis, we can perform the
following steps:
1. Replace passive element values with their admittances
2. Replace all sources and time variables with their Laplace transforms
3. Replace transformed voltage sources with transformed current sources
4. Write Kirchhoff’s current law at each node
5. Solve the simultaneous equations for the output
6. Form the transfer function

Problem For the network of the previous problem of R1LR2C network, find the transfer function
I2(s)/V(s), using nodal analysis.
Solution By replacing all the passive elements with their admittances and the voltage sources in series
with an impedance to current sources in parallel with an admittance using Norton’s theorem, we get
the following transformed circuit. Note that G1 = 1/R1 and G2 = 1/R2. We identify two independent
nodes with voltages as the inductor voltage VL(s) and capacitor voltage VC(s), as indicated in the
figure. Now, we apply Kirchhoff’s current law at the two nodes using the concept of admittance.

Solution (Cont’d)
Summing currents at the node VL(s) yields
G1[VL(s)−V(s)]+
1
Ls
VL(s)+G2[VL(s)−VC(s)] = 0
Summing currents at the node VC(s) yields
CsVC(s)+G2[VC(s)−VL(s)] = 0
Combining terms, the last two equations can be rewritten as simultaneous equations in VL(s) and
VC(s) as

G1 +G2 +
1
Ls

VL(s)−G2VC(s) = V(s)G1
−G2VL(s)+(G2 +Cs)VC(s) = 0
Solving for the transfer function VC(s)/V(s) yields
VC(s)
V(s)
=
G1G2
C s
(G1 +G2)s2 + G1G2L+C
LC s+ G2
LC

Solution (Cont’d) An advantage of drawing the transformed circuit with admittances and current
source lies in the form of the last equations and its direct relationship to the transformed circuit,
namely


Sum of admittances
connected to Node 1

VL(s)−




Sum of admittances
common to the two
nodes



VC(s) =


Sum of applied
currents at Node 1


−




Sum of admittances
common to the two
nodes



VL(s)+


Sum of admittances
connected to Node 2

VC(s) =


Sum of applied
currents at Node 2


Remarks:
• The above form is not always favorable since transforming voltage sources to current sources is
a difficult task.
• The number of simultaneous equations that must be written is equal to the number of
independent nodes whose voltages are unknown. Determining the voltages of the independent
nodes should lead to determining the voltage across each electric component in the circuit.

A problem-solving technique
In all our examples, we have seen a repeating form in the equations that we can use to our
advantage. If we recognize this form, we need not to write the equations component by component;
we can sum impedances around a mesh in the case of mesh equations or sum admittances at a node
in the case of node equations. Let us look at an example.
Problem (Mesh equation via inspection) Write, but do not solve, the mesh equations for the network
shown in the figure below.

Solution (Cont’d) As seen before, the mesh equations have a predictable form. Actually, the
equations for Meshes 1, 2, and 3 have the following form:




Sum of
impedances
around Mesh 1



I1(s)−







Sum of
impedances
common to
Mesh 1 Mesh 2







I2(s)−







Sum of
impedances
common to
Mesh 1 Mesh 3







I3(s) =




Sum of applied
voltages around
Mesh 1




−







Sum of
impedances
common to
Mesh 1 Mesh 2







I1(s)+




Sum of
impedances
around Mesh 2



I2(s)−







Sum of
impedances
common to
Mesh 2 Mesh 3







I3(s) =




Sum of applied
voltages around
Mesh 2




−







Sum of
impedances
common to
Mesh 1 Mesh 3







I1(s)−







Sum of
impedances
common to
Mesh 2 Mesh 3







I2(s)+




Sum of
impedances
around Mesh 3



I3(s) =




Sum of applied
voltages around
Mesh 3





Solution (Cont’d) Substituting the values from the figure shown before into the last three equations
yields
+(2s+2)I1(s)−(2s+1)I2(s)−I3(s) = V(s)
−(2s+1)I1(s)+(9s+1)I2(s)−4sI3(s) = 0
−I1(s)−4sI2(s)+

4s+1+
1
s

I3(s) = 0
which can be solved simultaneously for any desired transfer function, for example I3(s)/V(s).

Electric network state-space representations
Below, we present a technique for selecting state variables and representing an electric network in
state-space form.
Problem Given the electrical network shown below, find a state-space representation if the output is
the current through the resistor.
Solution First, we label all of the branch currents in the network. These include iL,iR, and iC as
shown in the figure. Second, we choose the state variables as the differentiated quantities in the
derivative equations for all the energy-storage elements. That is, from
C
dvC
dt
= iC
L
diL
dt
= vL

we choose the state variables as vC (the voltage across the capacitor) and iL (the current through the
inductor). Next is to determine the time derivatives of vC and iL as linear combinations of vC and iL
as well as the input v(t). To that end, iC and vL are to be found as such combinations, using
Kirchhoff’s voltage and current laws.
At Node 1, we have
iC = −iR +iL
= −
1
R
vC +iL
which gives iC as a linear combination of state variables vC and iL.
Around the outer loop, we have
vL = −vC +v(t)
which gives vL as a linear combination of the state variable vC and the input v.
Substituting the results found above into the equations determining the derivatives of vC and iL
gives the following state equations
C
dvC
dt
= −
1
R
vC +iL
L
diL
dt
= −vC +v(t)

Or
dvC
dt
= −
1
RC
vC +
1
C
iL
diL
dt
= −
1
L
vC +
1
L
v(t)
Since the output is chosen as iR(t), the output equation is
iR =
1
R
vC
which is a linear combination of state variable vC.
Finally, the state-space representation in matrix form is


v̇C
i̇L

 =


−1/(RC) 1/C
−1/L 0




vC
iL

+


0
1/L

v
iR = [1/R 0]


vC
iL



Operational amplifiers
We now discuss active circuits that can be used to implement controllers and compensators. These
are circuits built around an operational amplifier.
An operational amplifier (or differential amplifier), as pictured in the figure above, is an electronic
amplifier, which has the following characteristics:
1. Differential input, v2(t)−v1(t)
2. High input impedance, Zi = ∞ (ideal)
3. Low output impedance, Zo = 0 (ideal)
4. High constant gain amplification, A = ∞ (ideal)
5. The output vo(t) of an operational amplifier is given by vo(t) = A(v2(t)−v1(t))

Operational amplifiers
The following remarks on operational amplifiers (op-amps) are in order:
• Op-amps are frequently used to amplify signals from sensor circuits. They are also used to
implement analog filters and controllers.
• The amplifier gain A is approximately 105 ∼ 106 for dc signals and ac signals with frequencies
less than approximately 10 Hz. Whereas, the gain A decreases with the signal frequency and
becomes about unity for frequencies of 1 MHz ∼ 50 MHz.
• Since the gain of an op-amp is very high, it is necessary to have a negative feedback from the
output to the input (the inverted input) to make the amplifier stable.
• In an ideal op-amp, no current flows into the input terminals, and the output voltage is not
affected by the load connected to the output terminal. In other words, the input impedance is
infinity and the output impedance is zero. In an actual op-amp, a very small (almost
negligible) current flows into an input terminal and the output cannot be loaded too much. For
the sake of analysis purpose, we make the assumption that the op-amps are ideal.

Inverting operational amplifiers
If v2(t) is grounded, as shown in the figures above, the amplifier is called inverting operational
amplifier, since we have
vo(t) = −Av1(t)
If two impedances are connected to the inverting operational amplifier as shown in the figure on the
right, an important result can be derived as follows. Ia(s) = 0 since input impedance is high.
Therefore, by Kirchhoff’s current law, I1(s) = −I2(s). Also, v1(t) = 0 since gain A is large. Thus,
I2(s) = Vo(s)/Z2(s) by Kirchhoff’s voltage law. But I1(s) = Vi(s)/Z1(s). It follows that
Vo(s)
Vi(s)
= −
Z2(s)
Z1(s)
If Z1(s) = Z2(s), then the inverting op-amp circuit shown acts as a sign inverter.

PID controller using inverting operational amplifiers
A PID controller can be realized by the operational amplifier circuit above. To see this, let us
determine the ratio of the output Eo(s) to the input Ei(s). First, the transfer function E(s)/Ei(s) is
given as
E(s)
Ei(s)
= −
Z2(s)
Z1(s)
where
Z1(s) =
R1
R1C1s+1
, Z2(s) =
R2C2s+1
C2s
Thus
E(s)
Ei(s)
= −
R2C2s+1
C2s
R1C1s+1
R1
= −

R1C1 +R2C2
R1C2
+
1
R1C2s
+C1R2s

PID controller using inverting operational amplifiers
But a PID controller is expressed as
T(s) = Kp +
Ki
s
+Kds
which has no negative sign, compared to the transfer function E(s)/Ei(s) found. For E(s)/Ei(s) to
represent a PID controller we connect our circuit with a sign inverter as shown in the figure. The
overall transfer function Eo(s)Ei(s) is simply
Eo(s)
Ei(s)
=
Eo(s)
E(s)
E(s)
Ei(s)
=

−
R4
R3

−

R1C1 +R2C2
R1C2
+
1
R1C2s
+C1R2s

=
R4(R1C1 +R2C2)
R3R1C2
+
R4
R3R1C2s
+
R4R2C1
R3
s
from which the PID controller gains are to be
Kp =
R4(R1C1 +R2C2)
R3R1C2
Ki =
R4
R3R1C2
Kd =
R4R2C1
R3

PID controller using inverting operational amplifiers - - an example - -
Problem Find the transfer function Vo(s)/Vi(s) for the circuit given below.
Solution Since the admittances of parallel components add, the impedances of cascaded components
add, and the impedance is the reciprocal of admittance, we get
Z1(s) =
1
C1s+ 1
R1
=
1
5.6×10−6s+ 1
360×103
=
360×103
2.016s+1
Z2(s) = R2 +
1
C2s
= 220×103
+
107
s
Substituting Z1(s) and Z2(s) into Vo(s)/Vi(s) = −Z2(s)/Z1(s), we get
Vo(s)
Vi(s)
= −1.232
s2 +45.95s+22.55
s
The aforementioned circuit can be used to implement a PID controller.

Lead or lag compensator using inverting operational amplifiers
The active (or electronic) circuit, as shown in the figure on the left, using an operational amplifier,
has the following transfer function
E(s)
Ei(s)
= −
Z2(s)
Z1(s)
where
Z1(s) =
R1
R1C1s+1
, Z2(s) =
R2
R1C2s+1
Thus
E(s)
Ei(s)
= −
R2
R1
R1C1s+1
R2C2s+1
= −
C1
C2
s+ 1
R1C1
s+ 1
R2C2

Lead or lag compensator using inverting operational amplifiers
The last circuit is sign inverting so that a sign inverter connected at the output or the input of the
circuit may be used as in the figure on the right. The sign inverter transfer function
Eo(s)/E(s) = −R4/R3 makes that the whole transfer function be
Eo(s)
Ei(s)
=
Eo(s)
E(s)
E(s)
Ei(s)
=
R4R2
R3R1
R1C1s+1
R2C2s+1
=
R4C1
R3C2
s+ 1
R1C1
s+ 1
R2C2
= Kcα
Ts+1
αTs+1
= Kc
s+ 1
T
s+ 1
αT
where
Kc =
R4C1
R3C2
, α =
R2C2
R1C1
, T = R1C1
The discussed active network is used to implement the so-called lead compensator when R1C1 R2C2
(or α 1) and lag compensator when R1C1 R2C2 (or α 1).

Operational amplifier circuits used as controllers or compensators
Table below presents a list of operational amplifier circuits that may be used as controllers or
compensators.

A lead-lag compensator can be formed by cascading the lag compensator with the lead
compensator, as shown in figure below.

Passive-circuit realization of compensators
Lag, lead, and lag-lead compensators can also be implemented with passive networks. Table below
summarizes the networks and their transfer functions.
The lag-lead transfer function can be put in the following form:
Gc(s) =

s+ 1
T1

s+ 1
T2

s+ 1
αT1

s+ α
T2

where α 1. The terms with T1 form the lead compensator and the terms with T2 form the lag
compensator. Equation above shows a restriction inherent in using this passive-circuit realization.
Indeed, the dc gain of this network is unity.
However, a lag-lead compensator without this restriction can be implemented with an active
network as previously seen, or with passive networks by cascading the lead and lag networks shown
in the next table. Here, the two networks of lead and lag compensators must be isolated to ensure
that one network does not load the other, as shown in the figure hereafter.

A lead-lag compensator can be formed by cascading the passive-circuit lag compensator with the
passive-circuit lead compensator with isolation, as shown in figure below.

Active- and passive-circuit realization of compensators
Controllers and compensators can be realized either by active networks, using operational amplifier
circuits, or by passive networks using only resistors and capacitors (for electric networks).
• Compensators that use pure integration (PI controllers) for improving steady-state error or
pure differentiation (PD controllers) for improving transient response are defined as ideal
compensators. Ideal compensators must be implemented with active networks, which, in the
case of electric networks, require the use of active amplifiers with additional power sources for
their operation. An advantage of ideal integral compensators is that steady-state error is
reduced to zero. Electromechanical ideal compensators, such as tachometers, are often used to
improve transient response, since they can be conveniently interfaced with the plant.
• Compensators that can be implemented with only passive elements (lag, lead, and lag-lead
compensators) such as resistors and capacitors do not use pure integration or differentiation
and are not ideal compensators. Advantages of passive-circuit compensators are that they are
less expensive and do not require additional power sources for their operation. Their
disadvantage is that the steady-state error is not driven to zero.
• The choice between an active- or passive-circuit compensator revolves around cost, weight,
desired performance, transfer function, and the interface between the compensator and the
other hardware.

Translational mechanical system transfer functions
Next we see how to model a translational mechanical system by a transfer function. Like electrical
networks, mechanical systems have three passive linear components. Two of them, the spring and
the mass, are energy-storage elements, and one of them, the viscous damper, dissipates energy.

The concept of impedance
Like electric networks, impedances can be defined for mechanical systems, which can be used to
write the equations of motion and find the transfer function without bypassing by the differential
equations.
Taking the Laplace transform of force-displacement column of the previous table, we obtain
For the spring,
F(s) = KX(s)
For the viscous damper,
F(s) = fvsX(s)
For the mass,
F(s) = Ms2
X(s)
If we define impedance for mechanical components as
ZM(s) =
F(s)
X(s)
and apply the definition to the last equation above, we arrive at the impedances of each component
as summarized in the table. The impedance concept allows us to write the force-displacement
relationships for mass and viscous damper algebraically as in the case of spring.

To find transfer functions for translational mechanical systems, we can perform the following steps:
1. Assume a positive direction of motion, for example, to the right, for each moving body.
2. Draw a free-body diagram for each moving body, placing on it all the forces that act on the
body either in the direction of motion or opposite to it.
3. Use Newton’s law to form a differential equation of motion for each moving body, by summing
forces and setting the sum equal to zero.
4. Finally, assuming zero initial conditions, we take the Laplace transform of each differential
equation, separate variables, and arrive at the transfer function required.

Transfer function - - One degree of freedom - -
Problem Find the transfer function X(s)/F(s) for the system shown below.
Solution We begin with drawing the free-body diagram and placing on it the forces felt by the mass.
We assume the mass is traveling toward the right.

Transfer function - - One degree of freedom - -
Solution We now write the differential equation of motion using Newton’s second law to sum to zero
the different forces acting on the mass and to yield
M
d2x(t)
dt2
+ fv
dx(t)
dt
+Kx(t) = f(t)
Taking the Laplace transform, assuming zero initial conditions,
Ms2
X(s)+ fvsX(s)+KX(s) = F(s) or (Ms2
+ fvs+K)X(s) = F(s)
Solving for the transfer function yields
G(s) =
X(s)
F(s)
=
1
Ms2 + fvs+K
Using the concept of impedance, we notice that the Laplace-transformed equation of motion is of
the form
[Sum of impedances connected to the motion at x] X(s) = [Sum of applied forces at x]

Transfer function - - Multiple degrees of freedom - -
In order to work a problem with multiple degrees of freedom, we draw the free-body diagram for
each point of motion and then use superposition. For each free-body diagram, we begin by holding
all other points of motion still and finding the forces acting on the body due only to its own motion.
Then we hold the body still and activate the other points of motion one at a time, placing on the
original body the forces created by the adjacent motion.

Transfer function - - Two degrees of freedom - -
Problem Find the transfer function X2(s)/F(s) for the system shown below.
Solution The system has two independent degrees of freedom, since each mass can be moved while
the other is held still. Thus, two simultaneous equations of motion will be required to describe the
system. The two equations come from the free-body diagrams of each mass.

If we hold M2 still and move M1 to the right, we see the forces shown in figure at the top on the left.
If we hold M1 still and move M2 to the right, we see the forces shown in the figure at the top on the
right. The total forces acting on M1 are the superposition, as shown in the figure at the bottom.
Now the Laplace transform of the equations of motion for M1 can be written as

M1s2
+( fv1 + fv3 )s+(K1 +K2)

X1(s)−( fv3 s+K2)X2(s) = F(s)

Similarly, the Laplace transform of the equation of motion for M2 is
−( fv3 s+K2)X1(s)+

M2s2
+( fv2 + fv3 )s+(K2 +K3)

X2(s) = 0
From the last two simultaneous equations of motion, the transfer function X2(s)/F(s) is

X2(s)
F(s)
= G(s) =
fv3 s+K2
∆
where
∆ =
M1s2 +( fv1 + fv3 )s+K1 +K2 −( fv3 s+K2)
−( fv3 s+K2) M2s2 +( fv2 + fv3 )s+K2 +K3
Notice that the last two simultaneous equations of motion are in the general form:




Sum of impedances
connected to
the motion at x1



X1(s)−




Sum of impedances
between
x1 and x2



X2(s) =




Sum of applied
forces
at x1




−




Sum of impedances
between
x1 and x2



X1(s)+




Sum of impedances
connected to
the motion at x2



X2(s) =




Sum of applied
forces
at x2




Note: If we become familiar with this form, we can find the equations of motion by inspection,
without drawing the free-body diagrams.

Transfer function - - Equations of motion by inspection - -
Problem Write, but do not solve, the equations of motion for the translational mechanical system
shown below.
Solution The system has three degrees of freedom, since each of the three masses can be moved
independently while the others are held still.

Solution (Cont’d) The form of the Laplace-transformed equations of motion for masses M1,M2, and
M3 are







Sum of
impedances
connected to
the motion at x1







X1(s)−







Sum of
impedances
between
x1 and x2







X2(s)−







Sum of
impedances
between
x1 and x3







X3(s) =




Sum of
applied forces
at x1




−







Sum of
impedances
between
x1 and x2







X1(s)+







Sum of
impedances
connected to
the motion at x2







X2(s)−







Sum of
impedances
between
x2 and x3







X3(s) =




Sum of
applied forces
at x2




−







Sum of
impedances
between
x1 and x3







X1(s)−







Sum of
impedances
between
x2 and x3







X2(s)+







Sum of
impedances
connected to
the motion at x3







X3(s) =




Sum of
applied forces
at x3





Solution (Cont’d) M1 has two springs, two viscous dampers, and mass associated with its motion.
There is one spring between M1 and M2, and one viscous damper between M1 and M3. Thus, the
equation of motion for M1 is
[M1s2
+( fv1 + fv3 )s+(K1 +K2)]X1(s)−K2X2(s)− fv3 sX3(s) = 0
Similarly, the equations of motion for M2 and M3 are
−K2X1(s)+[M2s2
+( fv2 + fv4 )s+K2]X2(s)− fv4 sX3(s) = F(s)
− fv3 sX1(s)− fv4 sX2(s)+[M3s2
+( fv3 + fv4 )s]X3(s) = 0
Note: We can solve the last equations for any displacements X1(s),X2(s), or X3(s), or transfer
function.

Translational mechanical systems
A state-space representation
Problem Find the state equations for the translational mechanical system shown in the figure below.
Solution First, the differential equations of motion are
M1
d2x1
dt2
+D
dx1
dt
+Kx1 −Kx2 = 0
−Kx1 +M2
d2x2
dt2
+Kx2 = f(t)
Now, let v1 = dx1/dt and v2 = dx2/dt, and then select x1,v1,x2, and v2 as the state variables (normal
choice for mechanical systems).

Translational mechanical systems
A state-space representation
It follows that the state equations of the system are simply given by
dx1
dt
= v1
dv1
dt
= −
K
M1
x1 −
D
M1
v1 +
K
M1
x2
dx2
dt
= v2
dv2
dt
=
K
M2
x1 −
K
M2
x2 +
1
M2
f(t)
In matrix form, we have







ẋ1
v̇1
ẋ2
v̇2







=







0 1 0 0
−K/M1 −D/M1 K/M1 0
0 0 0 1
K/M2 0 −K/M2 0














x1
v1
x2
v2







+







0
0
0
1/M2







f(t)

Rotational mechanical system transfer functions
Rotational mechanical systems are handled the same way as translational mechanical systems
except that torque replaces force and angular displacement replaces translational displacement.
Similarly, the mechanical components for rotational systems are the same as those for translational
systems except that the components undergo rotation instead of translation.

Problem Find the transfer function θ2(s)/T(s) for the rotational mechanical system shown below.
The rod is supported by bearings at each end and is undergoing torsion. A torque is applied at the
left, and the displacement is measured at the right.
We approximate the system by assuming that the torsion acts like a spring concentrated at one
particular point in the rod, with an inertia J1 to the left and an inertia J2 to the right. We also
assume that the damping inside the flexible shaft is negligible. The schematic of the system is
shown in the figure at the right. The system has 2 degrees of freedom so that it will take two
simultaneous equations to solve.

Next, we draw a free-body diagram of J1, using superposition. Figure at the left shows the torques
on J1 if J2 is held still and J1 rotates. Figure at the center shows the torques on J1 if J1 is held still
and J2 rotates. Figure at the right shows the final free-body diagram for J1, from which the equation
of motion, after summing torques, is
(J1s2
+D1s+K)θ1(s)−Kθ2(s) = T(s)

Similarly, the free-body diagram of J2 is shown above, from which the equation of motion, summing
all the forces acting on J2, is
−Kθ1(s)+(J2s2
+D2s+K)θ2(s) = 0

Using the last two simultaneous equations of motion yields the transfer function
θ2(s)
T(s)
=
K
∆
where
∆ =
J1s2 +D1s+K −K
−K J2s2 +D2s+K
Notice that the last two simultaneous equations of motion are in the general form:




Sum of impedances
connected to
the motion at θ1



θ1(s)−




Sum of impedances
between
θ1 and θ2



θ2(s) =




Sum of applied
torques
at θ1




−




Sum of impedances
between
θ1 and θ2



θ1(s)+




Sum of impedances
connected to
the motion at θ2



θ2(s) =




Sum of applied
torques
at θ2




Note: If we become familiar with this form, we can find the equations of motion by inspection,
without drawing the free-body diagrams.

Problem Write, but do note solve, the Laplace transform of the equations of motion for the system
shown below.

Solution (Cont’d) The form of the Laplace-transformed equations of motion for the rotational masses
J1,J2, and J3 are given as







Sum of
impedances
connected to
the motion at θ1







θ1(s)−







Sum of
impedances
between
θ1 and θ2







θ2(s)−







Sum of
impedances
between
θ1 and θ3







θ3(s) =




Sum of
applied torques
at θ1




−







Sum of
impedances
between
θ1 and θ2







θ1(s)+







Sum of
impedances
connected to
the motion at θ2







θ2(s)−







Sum of
impedances
between
θ2 and θ3







θ3(s) =




Sum of
applied torques
at θ2




−







Sum of
impedances
between
θ1 and θ3







θ1(s)−







Sum of
impedances
between
θ2 and θ3







θ2(s)+







Sum of
impedances
connected to
the motion at θ3







θ3(s) =




Sum of
applied torques
at θ3





Solution (Cont’d) As the figure shows, the last equations are given as
(J1s2
+D1s+K)θ1(s)−Kθ2(s)−0θ3(s) = T(s)
−Kθ1(s)+(J2s2
+D2s+K)θ2(s)−D2sθ3(s) = 0
−0θ1(s)−D2sθ2(s)+(J3s2
+D3s+D2s)θ3(s) = 0
Note: We can solve for any transfer function, for example, θ3(s)/T(s).

Transfer functions for systems with gears
Rotational mechanical systems, especially those driven by motors, are rarely seen without
associated gears driving the load. Gears provide mechanical advantages to rotational systems. For a
car going uphill we shift to provide more torque and less speed. On the straightaway, we shift to
provide more speed and less torque.
The distance traveled along each gear’s circumference is the same:
r1θ1 = r2θ2 or
θ2
θ1
=
r1
r2
=
N1
N2
Assuming that gears do not absorb or store energy, the energy into Gear 1 equals the energy out
from Gear 2, that is
T1θ1 = T2θ2 or
T2
T1
=
θ1
θ2
=
N2
N1

Transfer functions for systems with gears - - Example 1 - -
What happens to mechanical systems that are driven by gears? In other words, can the mechanical
impedances be reflected from one side to the other in order to eliminate the gears?
T1 can be reflected to the output by multiplying by N2/N1, from which the equation of motion is:
(Js2
+Ds+K)θ2(s) = T1(s)
N2
N1
which is the equivalent system at the output after reflection of input torque.

By converting θ2(s) into an equivalent θ1(s), the equivalent system at the input, shown in figure
above, has the equation of motion
(Js2
+Ds+K)
N1
N2
θ1(s) = T1(s)
N2
N1
After simplification,
J

N1
N2
2
s2
+D

N1
N2
2
s+K

N1
N2
2
#
θ1(s) = T1(s)
Conclusion Rotational mechanical impedance can be reflected through the gear by multiplying the
mechanical impedance by the ratio of the number of teeth of gear on destination shaft to the
number of teeth of gear on source shaft.

Problem Find the transfer function θ2(s)/T1(s) for the system shown below.
Solution The two inertias do not undergo linearly independent motions, since they are tied together
by the gears. Thus, this is a one degree of freedom system. Now, we want to find the equivalent
system at the output after reflection of impedances.

Solution (Cont’d) Impedances J1 and D1 are reflected to the output by multiplying with N2/N1 while
torque T1 is reflected to the output by N2/N1. This gives the equivalent system of the figure at the
right, from which the equation of motion is
(Jes2
+Des+Ke)θ2(s) = T1(s)
N2
N1
where
Je = J1

N2
N1
2
+J2; De = D1

N2
N1
2
+D2; Ke = K2
Solving for θ2(s)/T1(s), the transfer function is found to be
θ2(s)
T1(s)
=
N2/N1
Jes2 +Des+Ke

Additional problems
Problem Write the nodal equations for the network shown in the figure below.
Solution We first notice that this network is difficult to be transformed into a circuit with current
source and admittances as the voltage source comes with no impedance in series. For this reason,
nodal equations will be determined based upon the transformed circuit with voltage source and
impedances.

Additional problems
Solution We write the Kirshooff’s current law at the nodes whose voltages are V1(s) and Vo(s) as
shown in the figure below. As can be seen, this network has three independent nodes (i.e., three
independent loops) whose voltages are sufficient to determine the voltage across any component in
the network. We can choose two nodes as stated before and the third one is that of source voltage
V(s). As the third node has a known voltage, we need effectively only two independent nodes to find
the nodal equations.

Additional problems
Solution From the figure, the nodal equations at the chosen nodes are simply
V1(s)−V(s)+
V1(s)
1+2s
+
V1(s)−Vo(s)
2+3s
= 0
Vo(s)−V1(s)
2+3s
+
Vo(s)
4
+
Vo(s)−V(s)
5/s
= 0
or
6s2 +12s+5
6s2 +7s+2
V1(s)−
1
3s+2
Vo(s) = V(s)
−
1
3s+2
V1(s)+
1
20
12s2 +23s+30
3s+2
Vo(s) =
s
5
V(s)

Additional problems
In order to clarify the representation of electric networks in state-space form, we look at the
following example.
Problem Find the state and output equations for the electrical network shown in the figure below if
the outputs are the voltage vR2 across resistor R2 and the current iR2 through the same resistor.
Solution Immediately notice that this network has a voltage-dependent current source.

Additional problems
Solution (Cont’d) First, we label all the branch currents on the network as shown in the figure.
Second, we choose the current through the inductor iL and the voltage across the capacitor vC as the
states variables whose time derivatives are given as
L
diL
dt
= vL
C
dvC
dt
= iC
Since the time derivatives of state variables should be expressed as linear combinations of the state
variables themselves as well the input, we proceed to find vL and iC as linear combinations of iL,vC,
and i(t).
Around the mesh containing L and C, we have
vL = vC +vR2 = vC +iR2 R2
But at Node 2, iR2 = iC +4vL. Substituting the last equation for iR2 into the expression vL yields
vL = vC +(iC +4vL)R2
Solving for vL, we get
vL =
1
1−4R2
(vC +iCR2)

Additional problems
Solution (Cont’d) Now, we need to find iC in terms of state variables and input. Thus, at Node 1 we
can write the sum of the currents as
iC = i−iR1 −iL
= i−
vR1
R1
−iL
= i−
vL
R1
−iL
Last equation are simultaneous relating vL and iC in terms of the state variables iL and vC, as well
the input i(t). These equations can be rewritten as
(1−4R2)vL −R2iC = vC
−
1
R1
vL −iC = iL −i(t)
Solving simultaneously for vL and iC, using Cramer’s rule, yields
vL =
1
∆
[R2iL −vC −R2i(t)]
and
iC =
1
∆
[(1−4R2)iL +
1
R1
vC −(1−4R2)i(t)]

Additional problems
Solution (Cont’d) where
∆ = −

(1−4R2)+
R2
R1

Substituting the expressions found of vL and iC into the time derivatives of state variables,
simplifying, and writing the result in matrix form renders the following state equation:


i̇L
v̇C

 =


R2/(L∆) −1/(L∆)
(1−4R2)/(C∆) 1/(R1C∆)




iL
vC

+


−R2/(L∆)
−(1−4R2)/(C∆)

i(t)
Now, we derive the output equations. Since the specified outputs are vR2 and iR2 , we notice that the
mesh containing C,L, and R2,
vR2 = −vC +vL
while at Node 2,
iR2 = iC +4vL
Substituting the expressions of vL and iC found in terms of state variables and input into the last
two equations, the output equation in matrix form is


vR2
iR2

 =


R2/∆ −(1+1/∆)
1/∆ (1−4R1)/(R1∆)




iL
vC

+


−R2/∆
−1/∆

i(t)

Additional problems
Problem Find the transfer function g(s) = X2(s)/F(s) for the translational mechanical system shown
in the figure below.

Additional problems
Solution
Writing the Laplace-transformed equations of motion,
(s2
+3s+1)X1(s)−(3s+1)X2(s) = F(s)
−(3s+1)X1(s)+(s2
+4s+1)X2(s) = 0
Solving for X2(s),
X2(s) =
s2 +3s+1 F(s)
−(3s+1) 0
s2 +3s+1 −(3s+1)
−(3s+1) s2 +4s+1
=
(3s+1)F(s)
s(s3 +7s2 +5s+1)
Hence,
X2(s)
F(s)
=
3s+1
s(s3 +7s2 +5s+1)

Additional problems
Problem Find the transfer function G(s) = θ2(s)/T(s) for the rotational mechanical system as shown
in figure below.

Additional problems
Solution Reflecting impedances to θ2 gives

200+3

50
5
2
+200

5
25
×
50
5
2
#
s2
+

1000

5
25
×
50
5
2
#
s+

250+3

50
5
2
#
=
50
5
T(s)
Thus,
θ2(s)
T(s)
=
10
1300s2 +4000s+550

Additional problems
Problem Find the transfer function G(s) = θ4(s)/T(s) for the rotational mechanical system as shown
in figure below.
Solution Reflecting impedances and applied torque to respective sides of the spring yields the
following equivalent circuit:

Additional problems
Solution (Cont’d)
Writing the equations of motion,
θ2(s)−θ3(s) = 4T(s)
−θ2(s)+(s+1)θ3(s) = 0
Solving for θ3(s),
θ3(s) =
1 4T(s)
−1 0
1 −1
−1 s+1
=
4T(s)
s
Hence θ3(s)/T(s) = 4/s. But theta4(s) = 1/5θ3(s). Thus, θ4(s)/T(s) = 4/(5s).

Additional problems
Problem Find the transfer function G(s) = θL(s)/T(s) for the rotational mechanical system as shown
in figure below.
Solution Reflecting impedances and applied torque to respective sides of the viscous damper yields
the following equivalent circuit:

Additional problems
Solution (Cont’d) Writing the equations of motion,
(s2
+s)θ2(s)−sθ3(s) = 10T(s)
−sθ2(s)+(s+1)θ3(s)−θ4(s) = 0
−θ3(s)(s+1)θ4(s) = 0
Solving for θ4(s),
θ4(s) =
s(s+1) −s 10T(s)
−s s+1 0
0 −1 0
s(s+1) −s 0)
−s s+1 −1
0 −1 s+1
=
10sT(s)
s2(s+1)2
=
10T(s)
s(s+1)2
Thus, θ4(s)/T(s) = 10/s(s+1)2. But θL(s) = 5θ4(s). Hence,
θL(s)
T(s)
=
50
s(s+1)2

Suggested problems
Students are suggested to solve the following problems from the textbook:
E 2.20, E 2.28, P 2.13, P 2.44, E 3.12, E 3.15, E 3.20, P 3.1, P 3.6, AP 3.2
Students are encouraged to solve the assigned problems by hand before seeking help
from classmates or the teacher. Subsequently, the accompanying solutions can be
checked for confirmation.

Lectures_12_15_Modeling_of_Electrical_an.pdf

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