Ch 06 electrical and electromechanical systems

10/27/2013

System Dynamics

6.01

Electrical and Electromechanical Systems

System Dynamics

6.02

Electrical and Electromechanical Systems

§1.Electrical Elements
- Voltage and current are the primary variables used to describe
a circuit’s behavior
- Current: the flow of electrons, the time rate of change of
electrons passing through a defined area

6. Electrical and
Eletromechanical Systems
𝑖=

𝑑𝑄
,
𝑑𝑡

𝑄=

𝑖𝑑𝑡

𝑖: current, 𝐴
𝑄: charge, 𝐶
Electrons are negatively charged, the positive direction of
current flow is opposite to that of the electron flow
HCM City Univ. of Technology, Faculty of Mechanical Engineering

System Dynamics

6.03

Nguyen Tan Tien

Electrical and Electromechanical Systems

§1.Electrical Elements
- Voltage: energy is required to move a charge between two
points in a circuit. The work per unit charge required to do this
is called voltage
Unit of voltage: 𝑉 ≡ 𝐽/𝐶
- Ohm’s Law
𝑣 = 𝑖𝑅
𝑖: current, 𝐴
𝑅: resistance, Ω

battery-lightbulb circuit
HCM City Univ. of Technology, Faculty of Mechanical Engineering

System Dynamics

6.05

Nguyen Tan Tien

Electrical and Electromechanical Systems

§1.Electrical Elements
Electrical quantities

HCM City Univ. of Technology, Faculty of Mechanical Engineering

HCM City Univ. of Technology, Faculty of Mechanical Engineering

System Dynamics

6.04

Nguyen Tan Tien

Electrical and Electromechanical Systems

§1.Electrical Elements
1.Active and passive Elements
- Elements that provide energy are sources, and elements that
dissipate energy are loads
- Circuit elements may be classified as active or passive
• Passive elements: resistors, capacitors, and inductors are
not sources of energy
• Active elements: energy sources that drive the system
- Several types of energy sources

• Chemical: batteries
• Mechanical: generators
• Thermal: thermocouples
• Optical: solar cells
- Active elements are modeled as either ideal voltage sources
or ideal current sources
HCM City Univ. of Technology, Faculty of Mechanical Engineering

System Dynamics

6.06

Nguyen Tan Tien

Electrical and Electromechanical Systems

§1.Electrical Elements
- Power 𝑃: work done per unit time, so the power generated by
an active element, or the power dissipated or stored by a
passive element, can be calculated as
𝑤𝑜𝑟𝑘
𝑤𝑜𝑟𝑘
𝑐ℎ𝑎𝑟𝑔𝑒
𝑝𝑜𝑤𝑒𝑟 =
=
=
= 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 × 𝑐𝑢𝑟𝑟𝑒𝑛𝑡
𝑡𝑖𝑚𝑒 𝑢𝑛𝑖𝑡 𝑐ℎ𝑎𝑟𝑔𝑒
𝑡𝑖𝑚𝑒
2

𝑣
𝑃 = 𝑖𝑣 = 𝑖 2 𝑅 =
𝑅
𝑃: power, 𝑊 ≡ 𝐽/𝑠
𝑖: current, 𝐴
𝑅: resistance, Ω
𝑣: voltage, 𝑉

Nguyen Tan Tien

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Nguyen Tan Tien

1


10/27/2013

System Dynamics

6.07

Electrical and Electromechanical Systems

§1.Electrical Elements
2.Modeling Circuits
- The dynamics of physical systems result from the transfer,
loss, and storage of mass or energy
- Basic laws used to model electrical systems

• Conservation of charge: Kirchhoff ’s current law
• Conservation of energy: Kirchhoff ’s voltage law

System Dynamics

6.08

Electrical and Electromechanical Systems

§1.Electrical Elements
3.Resistances
- Parallel resistances
Kirchhoff’s Current law gives
𝑖 = 𝑖1 + 𝑖2 with 𝑖1 = 𝑣𝑠 /𝑅1, 𝑖2 = 𝑣𝑠 /𝑅2
We have
1
1
1
1
1
1
𝑖=
+
𝑣 = 𝑣,
≡
+
𝑅1 𝑅2 𝑠 𝑅 𝑠
𝑅 𝑅1 𝑅2

and

𝑅2
𝑖,
𝑅1 + 𝑅2
The current-divider rule
𝑖1 =

currents entering the node =
currents leaving the node

System Dynamics

6.09

Nguyen Tan Tien

Electrical and Electromechanical Systems

§1.Electrical Elements9
- Series Resistances
Kirchhoff’s voltage law gives
𝑣𝑠 − 𝑣1 − 𝑣2 = 𝑣𝑠 − 𝑖𝑅1 − 𝑖𝑅2 = 0
or 𝑣𝑠 = 𝑅1 + 𝑅2 𝑖
Current in the loop
1
1
1
1
𝑖=
𝑣 = 𝑣,
≡

𝑅1 + 𝑅2 𝑠 𝑅 𝑠
𝑅 𝑅1 + 𝑅2
and
𝑅1
𝑅2
𝑣1 = 𝑅1 𝑖 =
𝑣,
𝑣2 = 𝑅2 𝑖 =
𝑣
𝑅1 + 𝑅2 𝑠
𝑅1 + 𝑅2 𝑠
𝑣1 𝑅1
=
𝑣2 𝑅2
HCM City Univ. of Technology, Faculty of Mechanical Engineering

6.11

Nguyen Tan Tien

Electrical and Electromechanical Systems

§1.Electrical Elements
4.Capacitance
- Capacitance: the ability of a device to store charge for a
given voltage difference across the element
- A capacitor is designed to store charge. Charge on a
capacitor:
𝑄 = 𝑖𝑑𝑡
- Voltage across the capacitor

𝑄 1
1
𝑣= =
𝑖𝑑𝑡 =
𝐶 𝐶
𝐶

𝑡

𝑖𝑑𝑡 +
0

𝑄0
𝐶

𝐶: capacitance, 𝐹 ≡ 𝐶/𝑉 𝑖: current across the capacitor, 𝐴
𝑄0 : the initial charge on the capacitor at time 𝑡 = 0, 𝐶
1
𝑣=
𝐶

𝑖𝑑𝑡

𝑑𝑣
𝑖=𝐶
𝑑𝑡

HCM City Univ. of Technology, Faculty of Mechanical Engineering

HCM City Univ. of Technology, Faculty of Mechanical Engineering

System Dynamics

6.10

Nguyen Tan Tien

Electrical and Electromechanical Systems

§1.Electrical Elements
- Nonlinear Resistances
• Not all resistance elements have the linear voltage-current
relation 𝑣 = 𝑖𝑅
• An example of a specific diode’s voltage-current
relationship found from experiments is 𝑖 = 0.16(𝑒 0.12𝑣 − 1)
For low voltages, we can approximate this curve with a
straight line whose slope equals the derivative 𝑑𝑖/𝑑𝑣 at 𝑣 =
0
𝑑𝑖
= 0.16(0.12𝑒 0.12𝑣 )
= 0.0192
𝑑𝑣 𝑣=0
𝑣=0
Thus, for small voltages, 𝑖 = 0.0192𝑣, and the resistance is
𝑅 = 𝑣/𝑖 = 1/0.0192 = 52Ω

The voltage-divider rule

System Dynamics

𝑅1
𝑖
𝑅1 + 𝑅2

𝑖1 𝑅2
=
𝑖2 𝑅1

the sum of all the voltage drops
around the loop is equal to zero

HCM City Univ. of Technology, Faculty of Mechanical Engineering

𝑖2 =

Nguyen Tan Tien

HCM City Univ. of Technology, Faculty of Mechanical Engineering

System Dynamics

6.12

Nguyen Tan Tien

Electrical and Electromechanical Systems

§1.Electrical Elements
6.Power and Energy
- The power dissipated by or stored by an electrical element

𝑃 = 𝑖𝑣
𝑃: power, 𝑊 ≡ 𝐽/𝑠
𝑖: current, 𝐴
𝑣: voltage, 𝑉
- Capacitors and inductors store electrical energy as stored
charge and in a magnetic field
• The energy 𝐸 stored in a capacitor
𝑑𝑣
1
𝐸 = 𝑃𝑑𝑡 = 𝑖𝑣𝑑𝑡 =
𝐶
𝑣𝑑𝑡 = 𝐶 𝑣𝑑𝑣 = 𝐶𝑣 2
𝑑𝑡
2
• The energy 𝐸 stored in an inductor
𝑑𝑖
𝐸 = 𝑃𝑑𝑡 = 𝑖𝑣𝑑𝑡 = 𝑖 𝐿
𝑑𝑡 = 𝐿
𝑑𝑡
HCM City Univ. of Technology, Faculty of Mechanical Engineering

1
𝑖𝑑𝑖 = 𝐿𝑖 2
2
Nguyen Tan Tien

2


10/27/2013

System Dynamics

6.13

Electrical and Electromechanical Systems

System Dynamics

6.14

Electrical and Electromechanical Systems

§1.Electrical Elements
Summarizes the voltage-current relations and the energy
expressions for resistance, capacitance, and inductance
elements

§2.Circuit Examples
- Ex.6.2.1
Current and Power in a Resistance Circuit
For the circuit shown in the figure, the applied
voltage is 𝑣𝑠 = 6𝑉 and the resistance is 𝑅 =
10Ω. Determine the current and the power
that must be produced by the power supply
Solution
The current is found from
𝑣𝑠
6
𝑖= =

= 0.6𝐴
𝑅 10
The power is computed from
𝑣𝑠2
62
𝑃=
=
= 3.6𝑊
𝑅
10
Note that we can also compute the power from 𝑃 = 𝑖𝑣𝑠

HCM City Univ. of Technology, Faculty of Mechanical Engineering

HCM City Univ. of Technology, Faculty of Mechanical Engineering

System Dynamics

6.15

Nguyen Tan Tien

Electrical and Electromechanical Systems

§2.Circuit Examples
- Ex.6.2.2

A Summing Circuit
The figure shows a circuit for summing
the voltages 𝑣1 and 𝑣2 to produce 𝑣3 .

Derive the expression for 𝑣3 as a
function of 𝑣1 and 𝑣2 , for the case where
𝑅1 = 𝑅2 = 10Ω and 𝑅3 = 20Ω
Solution
The voltage-current relations
𝑣1 − 𝑣3
𝑣2 − 𝑣3
𝑣3
𝑖1 =
,
𝑖2 =
,
𝑖3 =
𝑅1
𝑅2
𝑅3
Kirchoff’s current law gives 𝑖3 = 𝑖1 + 𝑖2
We obtain
𝑣3 𝑣1 − 𝑣3 𝑣2 − 𝑣3
=
+
⟹ 𝑣3 = 0.4(𝑣1 + 𝑣2 )
𝑅3
𝑅1
𝑅2

HCM City Univ. of Technology, Faculty of Mechanical Engineering

System Dynamics

6.17

Nguyen Tan Tien

Electrical and Electromechanical Systems

§2.Circuit Examples
Resistors 𝑅3 and 𝑅4 are in series 𝑅𝑠 = 2 + 6 = 8Ω
Resistors 𝑅𝑠 and 𝑅2 are in parallel
1
1 1
9
40
=
+ =
⟹ 𝑅𝑝 =
Ω
𝑅𝑝 10 8 40
9

§2.Circuit Examples
- Ex.6.2.3

6.16

Electrical and Electromechanical Systems

Application of the Voltage-Divider Rule
Consider the circuit shown in the
figure. Obtain the voltage 𝑣0 as a

function of the applied voltage 𝑣𝑠
by applying the voltage-divider
rule. Use the values 𝑅1 = 5Ω ,
𝑅2 = 10Ω, 𝑅3 = 6Ω, and 𝑅4 = 2Ω

Solution
Let 𝑣𝐴 be the voltage at the node as shown
The voltage-divider rule applied to
resistors 𝑅3 and 𝑅4 gives
𝑅4
2
1
𝑣0 =
𝑣 =
𝑣 = 𝑣
𝑅3 + 𝑅4 𝐴 6 + 2 𝐴 4 𝐴
HCM City Univ. of Technology, Faculty of Mechanical Engineering

System Dynamics

6.18

Nguyen Tan Tien

Electrical and Electromechanical Systems

§2.Circuit Examples
- Potentiometer: a resistance with a sliding electrical pick-off

Apply the voltage-divider rule

40/9
8
𝑣𝐴 =
𝑣 =
𝑣
5 + 40/9 𝑠 17 𝑠
1
1 8
2
⟹ 𝑣0 = 𝑣𝐴 =
𝑣 =
𝑣
4
4 17 𝑠 17 𝑠

HCM City Univ. of Technology, Faculty of Mechanical Engineering

System Dynamics

Nguyen Tan Tien

• The resistance 𝑅1 between the sliding contact and ground
is a function of the distance 𝑥 of the contact from the end of
the potentiometer
• Potentiometers, commonly called pots, are used as linear
and angular position sensors

Nguyen Tan Tien

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Nguyen Tan Tien

3


10/27/2013

System Dynamics

6.19

Electrical and Electromechanical Systems

§2.Circuit Examples
• Rotational potentiometer

System Dynamics

6.21

Nguyen Tan Tien

Electrical and Electromechanical Systems

§2.Circuit Examples
- Ex.6.2.5
Maximum PowerTransferin a Speaker-AmplifierSystem
A common example of an electrical system is an amplifier and
a speaker

The load is the speaker, which requires current from the
amplifier in order to produce sound. The resistance 𝑅𝐿 is that
of the load. The source supplies a voltage 𝑣𝑠 and a current 𝑖𝑆 ,
and has its own internal resistance 𝑅𝑆 . For optimum efficiency,
we want to maximize the power supplied to the speaker, for
given values of 𝑣𝑆 and 𝑅𝑆 . Determine the value of 𝑅𝐿 to
maximize the power transfer to the load
HCM City Univ. of Technology, Faculty of Mechanical Engineering

System Dynamics

6.23

6.20

Electrical and Electromechanical Systems

§2.Circuit Examples
- Ex.6.2.4
Potentometer
Assuming the potentiometer resistance 𝑅1 is proportional to 𝑥.
Derive the expression for the output voltage 𝑣0 as a function of 𝑥
Solution
The length of the pot is 𝐿 and its total
resistance is 𝑅1 + 𝑅2
From the voltage-divider rule
𝑅1
𝑣𝑜 =
𝑉

𝑅1 + 𝑅2
The resistance 𝑅1 proportional to 𝑥: 𝑅1 = 𝑅1 + 𝑅2 𝑥/𝐿
Substituting the above equation gives
𝑥
𝑣𝑜 = 𝑉 = 𝐾𝑥
𝐿
𝐾 = 𝑉/𝐿: the gain of the pot

• Voltage relation for rotational pot
𝜃
𝑣𝑜 =
𝑉 = 𝐾𝜃
𝜃𝑚𝑎𝑥
• Voltage relation for linear pot
𝑥
𝑣𝑜 = 𝑉 = 𝐾𝑥
𝐿
𝐾: the gain of the pot
HCM City Univ. of Technology, Faculty of Mechanical Engineering

System Dynamics

Nguyen Tan Tien

Electrical and Electromechanical Systems

HCM City Univ. of Technology, Faculty of Mechanical Engineering

System Dynamics

6.22

Electrical and Electromechanical Systems

§2.Circuit Examples
Solution
From Kirchhoff’s voltage law
𝑣𝑠 − 𝑖𝑠 𝑅𝑠 − 𝑖𝑠 𝑅𝐿 = 0
From the voltage-divider rule
𝑅𝐿
𝑣𝐿 =
𝑣
𝑅𝑆 + 𝑅𝐿 𝑆
The power consumed by the load
𝑃𝐿 = 𝑖𝑆2 𝑅𝐿 = 𝑣𝐿2 /𝑅𝐿
Using the relation between 𝑣𝐿 and 𝑣𝑆 we can express 𝑃𝐿 in
terms of 𝑣𝑆 as
𝑅𝐿
𝑃𝐿 =
𝑣2
𝑅𝑆 + 𝑅𝐿 2 𝑆

HCM City Univ. of Technology, Faculty of Mechanical Engineering

System Dynamics

6.24

§2.Circuit Examples
To maximize 𝑃𝐿 for a fixed value of 𝑣𝑆 , we must choose 𝑅𝐿 to

maximize the ratio
𝑅𝐿
𝑟=
𝑅𝑆 + 𝑅𝐿 2
The maximum of 𝑟 occurs where
𝑑𝑟
=0
𝑑𝑅𝐿
or
𝑑𝑟
(𝑅𝑆 + 𝑅𝐿 )2 −2𝑅𝐿 (𝑅𝑆 + 𝑅𝐿 )
=
=0
𝑑𝑅𝐿
(𝑅𝑆 + 𝑅𝐿 )4
⟹ 𝑅𝐿 = 𝑅𝑆
This result for a resistance circuit is a special case of the more
general result known as impedance matching

§2.Circuit Examples
- Ex.6.2.6

HCM City Univ. of Technology, Faculty of Mechanical Engineering

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Nguyen Tan Tien

Nguyen Tan Tien

Nguyen Tan Tien

Electrical and Electromechanical Systems

A Feedback Amplifier
Early in the twentieth century engineers
struggled to design vacuum-tube amplifiers
whose gain remained constant at a
predictable value. The gain is the ratio of
the output voltage to the input voltage. The
vacuum-tube gain 𝐺 can be made large but
is somewhat unpredictable and unreliable
due to heat effects and manufacturing variations. A solution to
the problem is shown in the figure
Derive the input-output relation for 𝑣𝑜 as a function of 𝑣𝑖 .
Investigate the case where the gain 𝐺 is very large

Nguyen Tan Tien

4


10/27/2013

System Dynamics

6.25

Electrical and Electromechanical Systems

§2.Circuit Examples
Solution

Solve for 𝑣𝑜

𝐺
𝑣
1 + 𝐺𝑅2 /(𝑅1 + 𝑅2 ) 𝑖
If 𝐺𝑅2 /(𝑅1 + 𝑅2 ) ≫ 1, then
𝑅1 + 𝑅2
𝑣𝑜 ≈
𝑣𝑖
𝑅2
𝑣𝑜 =

System Dynamics

6.27

6.26

Electrical and Electromechanical Systems

§2.Circuit Examples
1.Loop Currents
Sometimes the circuit equations can be simplified by using the
concept of a loop current, which is a current identified with a
specific loop in the circuit. Use of loop currents usually reduces
the number of unknowns to be found, although when deriving
the circuit equations you must be careful to use the proper

algebraic sum for each element
- Example 6.2.7
Analysis with Loop Currents
Given the values of the voltages and the resistances for the
circuit in the figure. (a) Solve for the
currents 𝑖1 , 𝑖2 and 𝑖3 passing through
the three resistors. (b) Use the loopcurrent method to solve for the
currents

The input voltage to the amplifier
𝑅2
𝑣𝑖 −
𝑣
𝑅1 + 𝑅2 𝑜
Thus the amplifier’s output
𝑅2
𝑣𝑜 = 𝐺 𝑣𝑖 −
𝑣
𝑅1 + 𝑅2 𝑜

HCM City Univ. of Technology, Faculty of Mechanical Engineering

System Dynamics

Nguyen Tan Tien

Electrical and Electromechanical Systems

HCM City Univ. of Technology, Faculty of Mechanical Engineering

System Dynamics

6.28

Nguyen Tan Tien

Electrical and Electromechanical Systems

§2.Circuit Examples
Solution
a.Applying Kirchhoff’s voltage law to the left-hand loop gives
𝑣1 − 𝑅1𝑖1 − 𝑅2 𝑖2 = 0
For the right-hand loop
𝑣2 − 𝑅2 𝑖2 + 𝑅3 𝑖3 = 0
From conservation of charge
𝑖1 = 𝑖2 + 𝑖3
Solving for the solution
𝑅2 + 𝑅3 𝑣1 − 𝑅2 𝑣2
𝑖1 =
𝑅1𝑅2 + 𝑅1 𝑅3 + 𝑅2 𝑅3
𝑅3 𝑣1 + 𝑅1𝑣2
𝑖2 =
𝑅1 𝑅2 + 𝑅1 𝑅3 + 𝑅2 𝑅3
𝑅2 𝑣1 − (𝑅1 + 𝑅2 )𝑣2
𝑖3 =
𝑅1 𝑅2 + 𝑅1 𝑅3 + 𝑅2 𝑅3

§2.Circuit Examples
b.Define the loop currents 𝑖𝐴 and 𝑖𝐵 positive clockwise
Note - voltage drop 𝑅2 𝑖𝐴 due to 𝑖𝐴

- voltage increase 𝑅2 𝑖𝐵 due to 𝑖𝐵
Applying Kirchhoff’s voltage law to the
left-hand loop gives
𝑣1 − 𝑅1𝑖𝐴 − 𝑅2 𝑖𝐴 + 𝑅2 𝑖𝐵 = 0
𝑣2 + 𝑅3 𝑖𝐵 + 𝑅2 𝑖𝐵 − 𝑅2 𝑖𝐴 = 0
The solution
𝑅2 + 𝑅3 𝑣1 − 𝑅2 𝑣2
𝑖𝐴 =
𝑅1 𝑅2 + 𝑅1𝑅3 + 𝑅2 𝑅3
𝑅2 𝑣1 − (𝑅1 + 𝑅2 )𝑣2
𝑖𝐵 =
𝑅1 𝑅2 + 𝑅1 𝑅3 + 𝑅2 𝑅3
⟹ 𝑖1 = 𝑖𝐴 , 𝑖3 = 𝑖𝐵 , 𝑖2 = 𝑖𝐴 + 𝑖𝐵

HCM City Univ. of Technology, Faculty of Mechanical Engineering

HCM City Univ. of Technology, Faculty of Mechanical Engineering

System Dynamics

6.29

Nguyen Tan Tien

Electrical and Electromechanical Systems

System Dynamics

6.30

Nguyen Tan Tien

Electrical and Electromechanical Systems

§2.Circuit Examples
2.Capacitance and Inductance in Circuits
- Ex.6.2.8
Series RC Circuit
The resistor and capacitor in the circuit shown
in the figure are said to be in series because
the same current flows through them. Assume
that the supply voltage 𝑣𝑆 is known. Obtain the
model of the capacitor voltage 𝑣1

§2.Circuit Examples
Solution
From Kirchhoff’s voltage law
𝑣𝑆 − 𝑅𝑖 − 𝑣1 = 0
For the capacitor
1 𝑡
𝑄0
𝑣1 =
𝑖𝑑𝑡 +
𝐶 0
𝐶

HCM City Univ. of Technology, Faculty of Mechanical Engineering

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Nguyen Tan Tien

Differentiate this with respect to 𝑡 to obtain
𝑑𝑣1 1
= 𝑖
𝑑𝑡
𝐶
Then substitute for 𝑖 from the first equation
𝑑𝑣1
1
𝑑𝑣1
=
𝑣 − 𝑣1 ⟹ 𝑅𝐶
+ 𝑣1 = 𝑣𝑆
𝑑𝑡
𝑅𝐶 𝑆
𝑑𝑡
This the required model
Nguyen Tan Tien

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10/27/2013

System Dynamics

6.31

Electrical and Electromechanical Systems

§2.Circuit Examples
- Ex.6.2.9

Pulse Response of a Series RC Circuit
A rectangular pulse input is a positive
step function that lasts a duration 𝐷. One
way of producing a step voltage input is
to use a switch like that shown in the
figure
The battery voltage 𝑉 is constant and the switch is initially
closed at point 𝐵. At 𝑡 = 0 the switch is suddenly moved from
point 𝐵 to point 𝐴. Then at 𝑡 = 𝐷 the switch is suddenly moved
back to point 𝐵. Obtain the expression for the capacitor voltage
𝑣1 (𝑡) assuming that 𝑣1 0 = 0

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Electrical and Electromechanical Systems

§2.Circuit Examples
The solution of this equation for 𝑡 ≥ 𝐷 is simply the free
response with the initial condition 𝑣1 (𝐷)
𝑡−𝐷

𝐷

𝑡−𝐷

𝑣1 𝑡 = 𝑣1 𝐷 𝑒 − 𝑅𝐶 = 𝑉(1 − 𝑒 −𝑅𝐶 )𝑒 − 𝑅𝐶

System Dynamics

6.32

Electrical and Electromechanical Systems

§2.Circuit Examples
Solution
When at 𝑡 = 0 the switch is suddenly
moved from point 𝐵 to point 𝐴, the circuit
is identical to that shown in the figure
with 𝑣𝑆 = 𝑉, and its model is
𝑑𝑣1
𝑅𝐶
+ 𝑣1 = 𝑉
𝑑𝑡
The input voltage is a step function of
magnitude 𝑉. The solution is the forced
response
𝑣1 𝑡 = 𝑉(1 − 𝑒 −𝑡/𝑅𝐶 )
When the switch is moved back to point 𝐵 at time 𝑡 = 𝐷, the
circuit is equivalent to that shown in the figure, whose model is
equation the same above with 𝑉 = 0

HCM City Univ. of Technology, Faculty of Mechanical Engineering

System Dynamics

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Electrical and Electromechanical Systems

§2.Circuit Examples
- Ex.6.2.10
Series 𝑅𝐶𝐿 Circuit
The resistor, inductor, and capacitor in the
circuit shown in the figure are in series because
the same current flows through them. Obtain
the model of the capacitor voltage 𝑣1 with the
supply voltage 𝑣𝑆 as the input
Solution
From Kirchhoff’s voltage law
𝑑𝑖
𝑣𝑆 − 𝑅𝑖 − 𝐿 − 𝑣1 = 0
𝑑𝑡
For the capacitor
1 𝑡
𝑑𝑣1
𝑣1 =
𝑖𝑑𝑡 ⟹ 𝑖 = 𝐶
𝐶 0
𝑑𝑡

Pulse response of a series 𝑅𝐶 circuit
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System Dynamics

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Electrical and Electromechanical Systems

§2.Circuit Examples
Substitute this for 𝑖

HCM City Univ. of Technology, Faculty of Mechanical Engineering

System Dynamics

§2.Circuit Examples
- Ex.6.2.11

𝑑𝑣1
𝑑 2 𝑣1
𝑣𝑆 − 𝑅𝐶
− 𝐿𝐶
− 𝑣1 = 0
𝑑𝑡
𝑑𝑡 2
It can be expressed in the following form

𝑑 2 𝑣1
𝑑𝑣1
𝐿𝐶
+ 𝑅𝐶
+ 𝑣1 = 𝑣𝑆
𝑑𝑡 2
𝑑𝑡

6.36

Nguyen Tan Tien

Electrical and Electromechanical Systems

Parallel 𝑅𝐿 Circuit
𝑅 and 𝐶 in the circuit shown in the figure
are said to be in parallel because they
have the same voltage 𝑣1 . Given supply
current 𝑖𝑆 , Obtain the model of the current
𝑖2 passing through the inductor

Solution
The currents 𝑖1 and 𝑖2 are defined in the figure. Then
𝑑𝑖2
𝑣1 = 𝐿
= 𝑅𝑖1
𝑑𝑡
From conservation of charge, 𝑖1 + 𝑖2 = 𝑖𝑆 , we obtain
𝑑𝑖2
𝐿 𝑑𝑖2

𝐿
= 𝑅(𝑖𝑆 − 𝑖2 ) ⟹
+ 𝑖2 = 𝑖𝑆
𝑑𝑡
𝑅 𝑑𝑡
This is the required model
HCM City Univ. of Technology, Faculty of Mechanical Engineering

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HCM City Univ. of Technology, Faculty of Mechanical Engineering

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Electrical and Electromechanical Systems

§2.Circuit Examples
- Ex.6.2.12

Analysis of a Telegraph Line
Figure shows a circuit representation of

a telegraph line. The resistance 𝑅 is the
line resistance and 𝐿 is the inductance
of the solenoid that activates the
receiver’s clicker. The switch represents
the operator’s key. Assume that when sending a “dot,” the key
is closed for 0.1𝑠. Using the values 𝑅 = 20Ω,𝐿 = 4𝐻, obtain the
expression for the current 𝑖(𝑡) passing through the solenoid

HCM City Univ. of Technology, Faculty of Mechanical Engineering

System Dynamics

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Electrical and Electromechanical Systems

§2.Circuit Examples
- Ex.6.2.13

An 𝑅𝐿𝐶 Circuit with Two Input Voltages
The 𝑅𝐿𝐶 circuit shown in the figure has
two input voltages. Obtain the differential
equation model for the current 𝑖3
Solution
Applying Kirchhoff’s voltage law to the left-hand loop gives
𝑑𝑖3
𝑣1 − 𝑅𝑖1 − 𝐿
=0

𝑑𝑡
For the right-hand loop
1
𝑑𝑖3
𝑣2 −
𝑖2 𝑑𝑡 − 𝐿
=0
𝐶
𝑑𝑡
Differentiate this equation with respect to 𝑡
𝑑𝑣2 1
𝑑 2 𝑖3
− 𝑖2 − 𝐿 2 = 0
𝑑𝑡
𝐶
𝑑𝑡

HCM City Univ. of Technology, Faculty of Mechanical Engineering

System Dynamics

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Electrical and Electromechanical Systems

System Dynamics

6.38

Electrical and Electromechanical Systems

§2.Circuit Examples
Solution
From the voltage law we have
𝑑𝑖
𝐿 + 𝑅𝑖 = 𝑣𝑖 (𝑡)
𝑑𝑡
𝑣𝑖 (𝑡): the input voltage due to the
switch and the 12𝑉 supply
The voltage 𝑣𝑖 (𝑡) can be modeled as an impulsive input
𝑣𝑖 𝑡 = 1.2𝛿(𝑡)
The Laplace transform of equation with 𝑖(0) = 0 gives
1.2
0.3
4𝑠 + 20 𝐼 𝑠 = 1.2 ⟹ 𝐼 𝑠 =
=
4𝑠 + 20 𝑠 + 5
This gives the solution
𝑖 𝑡 = 0.3𝑒 −5𝑡
Note that this solution gives 𝑖(0+) = 0.3, whereas 𝑖(0) = 0.
The difference is due to the impulsive input
HCM City Univ. of Technology, Faculty of Mechanical Engineering

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Electrical and Electromechanical Systems

§2.Circuit Examples
From conservation of charge
𝑖3 = 𝑖1 + 𝑖2
Solve for 𝑖1
1
𝑑𝑖3
𝑖1 =
𝑣 −𝐿
𝑅 1
𝑑𝑡

Then
𝑑𝑣2 1
𝑑 2 𝑖3
− 𝑖3 − 𝑖1 − 𝐿 2 = 0
𝑑𝑡
𝐶
𝑑𝑡
𝑑𝑣2 1
1 1
𝑑𝑖3
𝑑 2 𝑖3
⟹
− 𝑖3 +
𝑣 −𝐿
−𝐿 2 = 0
𝑑𝑡

𝐶
𝐶 𝑅 1
𝑑𝑡
𝑑𝑡
Rearrange this equation to obtain the answer
𝑑 2 𝑖3
𝑑𝑖3
𝑑𝑣2
𝐿𝑅𝐶 2 + 𝐿
+ 𝑅𝑖3 = 𝑣1 + 𝑅𝐶
𝑑𝑡
𝑑𝑡
𝑑𝑡
HCM City Univ. of Technology, Faculty of Mechanical Engineering

System Dynamics

6.42

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Electrical and Electromechanical Systems

§2.Circuit Examples
3.State Variable Models of Circuits
The presence of several current and voltage variables in a
circuit can sometimes lead to difficulty in identifying the
appropriate variables to use for expressing the circuit model.
⟹ Use of state variables can often reduce this confusion
- Ex.6.2.14

State-Variable Model of a Series 𝑅𝐿𝐶 Circuit
Consider the series 𝑅𝐿𝐶 circuit.
Choose a suitable set of state
variables, and obtain the state
variable model of the circuit in matrix
form. The input is the voltage 𝑣𝑠

§2.Circuit Examples
Solution
The energy stored in the capacitor is 𝐶𝑣12 /2 and the energy
stored in the inductor is 𝐿𝑖 2 /2 ⟹ state variables: 𝑣1 and 𝑖
From Kirchhoff’s voltage law
𝑑𝑖
𝑑𝑖 1
1
𝑅
𝑣𝑠 − 𝑅𝑖 − 𝐿 − 𝑣1 = 0 ⟹
= 𝑣 − 𝑣 − 𝑖
𝑑𝑡
𝑑𝑡 𝐿 𝑠 𝐿 1 𝐿
From the capacitor relation
1
𝑑𝑣1 1
𝑣1 =
𝑖𝑑𝑡 ⟹
= 𝑖
𝐶
𝑑𝑡
𝐶

HCM City Univ. of Technology, Faculty of Mechanical Engineering

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Nguyen Tan Tien

𝑑𝑖
𝑅
−
𝑑𝑡 =
𝐿
1
𝑑𝑣1
𝐶
𝑑𝑡

Rewrite in matrix form
1
1
−
𝐿 𝑖 +
𝐿 𝑣𝑠
𝑣1
0
0
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Electrical and Electromechanical Systems

§3.Impedance and Amplifier
- Ex.6.3.1

Coupled 𝑅𝐶 Loops
Determine the transfer function
𝑣𝑜 (𝑠)/𝑣𝑠 (𝑠) of the circuit
Solution
The energy in this circuit is stored
in the two capacitors
The energy stored in a capacitor is expressed by 𝐶𝑣 2 /2
⟹ the state variables are the voltages 𝑣1 and 𝑣𝑜
The capacitance relations are
𝑑𝑣𝑜 1
𝑑𝑣1 1
= 𝑖3 ,
= 𝑖2
𝑑𝑡
𝐶
𝑑𝑡
𝐶
For the right-hand loop
𝑣1 − 𝑣𝑜

𝑑𝑣𝑜
1
𝑖3 =
⟹
=
(𝑣 − 𝑣𝑜 )
𝑅
𝑑𝑡
𝑅𝐶 1

HCM City Univ. of Technology, Faculty of Mechanical Engineering

System Dynamics

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Electrical and Electromechanical Systems

System Dynamics

§3.Impedance and Amplifier
For the left-hand loop

6.44

Electrical and Electromechanical Systems

𝑣𝑠 − 𝑣𝑜

𝑅
From conservation of charge
𝑖2 = 𝑖1 − 𝑖3
1
= (𝑣𝑠 − 2𝑣1 + 𝑣𝑜 )
𝑅
𝑖1 =

𝑑𝑣1
1
=
(𝑣 − 2𝑣1 + 𝑣𝑜 )
𝑑𝑡
𝑅𝐶 𝑠
Transform these equations for zero initial conditions to obtain
1
𝑠𝑉𝑜 𝑠 =
𝑉 𝑠 − 𝑉𝑜 𝑠
𝑅𝐶 1
1
𝑠𝑉1 𝑠 =
[𝑉 𝑠 − 2𝑉1 (𝑠) + 𝑉𝑜 (𝑠)]
𝑅𝐶 𝑠
𝑉𝑜 (𝑠)
1
⟹
=
𝑉𝑠 (𝑠) 𝑅 2𝐶 2 𝑠 2 + 3𝑅𝐶𝑠 + 1
⟹

HCM City Univ. of Technology, Faculty of Mechanical Engineering

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Electrical and Electromechanical Systems

§3.Impedance and Amplifier
1.Impedance
- A resistance resists or “impedes” the flow of current. The
corresponding relation is 𝑣/𝑖 = 𝑅 . Capacitance and
inductance elements also impede the flow of current
- In electrical systems an impedance is a generalization of the
resistance concept and is defined as the ratio of a voltage
transform 𝑉(𝑠) to a current transform 𝐼(𝑠) and thus implies a
current source
- Standard symbol for impedance
𝑉(𝑠)
𝑍(𝑠) ≡
𝐼(𝑠)

§3.Impedance and Amplifier
- The impedance of a resistor is its resistance
𝑍 𝑠 =𝑅
- For a capacitor
1 𝑡
𝐼(𝑠)

𝑣 𝑡 =
𝑖𝑑𝑡 ⟹ 𝑉 𝑠 =
𝐶 0
𝐶 𝑠 𝑠

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Electrical and Electromechanical Systems

The impedance of a capacitor
1
𝑍 𝑠 =
𝐶𝑠
- For an inductor
𝑑𝑖
𝑣 𝑡 = 𝐿 ⟹ 𝑉 𝑠 = 𝐿𝐼 𝑠 𝑠
𝑑𝑡
The impedance of a inductor
𝑍 𝑠 = 𝐿𝑠

System Dynamics

6.48

Nguyen Tan Tien

Electrical and Electromechanical Systems

§3.Impedance and Amplifier
2.Series and Parallel Impedances
- The concept of impedance is useful because the impedances
of individual elements can be combined with series and
parallel laws to find the impedance at any point in the system
- The laws for combining series or parallel impedances are
extensions to the dynamic case of the laws governing series
and parallel resistance elements

§3.Impedance and Amplifier
- Series Impedances
• Two impedances are in series if they have the same
current. If so, the total impedance is the sum of the
individual impedances
𝑍 𝑠 = 𝑍1 𝑠 + 𝑍2 (𝑠)
• Example: a resistor 𝑅 and capacitor 𝐶 in series have the
equivalent impedance
1
𝑅𝐶𝑠 + 1
𝑍 𝑠 =𝑅+
=
𝐶𝑠
𝐶𝑠
𝑉(𝑠)

𝑅𝐶𝑠 + 1
⟹
≡𝑍 𝑠 =
𝐼(𝑠)
𝐶𝑠
and the differential equation model is
𝑑𝑣
𝑑𝑖
𝐶
= 𝑅𝐶 + 𝑖(𝑡)
𝑑𝑡
𝑑𝑡

HCM City Univ. of Technology, Faculty of Mechanical Engineering

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Nguyen Tan Tien

Nguyen Tan Tien

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Electrical and Electromechanical Systems

System Dynamics

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Electrical and Electromechanical Systems

§3.Impedance and Amplifier
- Parallel Impedances
• Two impedances are in parallel if they have the same
voltage difference across them. Their impedances combine
by the reciprocal rule
1
1
1
=
+
𝑍(𝑠) 𝑍1 (𝑠) 𝑍2 (𝑠)
• Example: a resistor 𝑅 and capacitor 𝐶 in parallel have the
equivalent impedance
1
1
1
𝑉(𝑠)
𝑅
=
+ ⟹
≡𝑍 𝑠 =
𝑍(𝑠) 1/𝐶𝑠 𝑅

𝐼(𝑠)
𝑅𝐶𝑠 + 1
and the differential equation model is
𝑑𝑣
𝑅𝐶
+ 𝑣 = 𝑅𝑖(𝑡)
𝑑𝑡

§3.Impedance and Amplifier
- Ex.6.3.2
Circuit Analysis Using Impedance
For the circuit shown in the figure, determine the transfer
function between the input voltage 𝑣𝑠 and the output voltage 𝑣𝑜

HCM City Univ. of Technology, Faculty of Mechanical Engineering

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Solution
The equivalent impedance 𝑍(𝑠) for 𝑅 and 𝐶
1
1

1
𝑅
=
+ ⟹𝑍 𝑠 =
𝑍(𝑠) 1/𝐶𝑠 𝑅
𝑅𝐶𝑠 + 1
In this representation we may think of the impedance as a
simple resistance, provided we express the relations in
Laplace transform notation

System Dynamics

6.52

Nguyen Tan Tien

Electrical and Electromechanical Systems

§3.Impedance and Amplifier
Kirchhoff’s voltage law gives
𝑉𝑠 𝑠 − 𝑅1 𝐼 𝑠 − 𝑍 𝑠 𝐼 𝑠 = 0
The output voltage is related to the current by 𝑉𝑜 𝑠 =
𝑍 𝑠 𝐼(𝑠). Eliminating 𝐼(𝑠) from these two relations gives
𝑉𝑜 (𝑠)
𝑉𝑠 𝑠 = 𝑅1
− 𝑉𝑜 𝑠 = 0
𝑍(𝑠)
which yields the desired transfer function
𝑉𝑜 (𝑠)
𝑍(𝑠)

𝑅
=
=
𝑉𝑠 (𝑠) 𝑅1 + 𝑍(𝑠) 𝑅𝑅1 𝐶𝑠 + 𝑅 + 𝑅1
This network is a first-order system whose time constant is
𝑅𝑅1
𝜏=
𝑅 + 𝑅1

§3.Impedance and Amplifier
3.Isolation Amplifier
- A voltage-isolation amplifier is designed to produce an output
voltage that is proportional to the input voltage
- Such an amplifier may be considered to be a voltage source
• if it does not affect the behavior of the source circuit that is
attached to the amplifier input terminals, and
• if the amplifier is capable of providing the voltage
independently of the particular circuit (the “load”) attached
to amplifier output terminals

HCM City Univ. of Technology, Faculty of Mechanical Engineering

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System Dynamics

6.53

§3.Impedance and Amplifier
- Consider the system

𝑉𝑠 𝑠 − 𝐼1 𝑠 𝑍𝑠 𝑠 − 𝑉𝑖 𝑠 = 0,

Nguyen Tan Tien

Electrical and Electromechanical Systems

𝐼1 𝑠 = 𝑉𝑖 (𝑠)/𝑍𝑖 (𝑠)

• 𝑍𝑖 (𝑠) is large ⟶ 𝑖1 is small: the amplifier does not affect the
current 𝑖1 ⟶ the amplifier does not affect the source circuit
• 𝑍𝑖 (𝑠) is large
𝑍𝑖 𝑠
𝑉𝑖 𝑠 =
𝑉 (𝑠) ≈ 𝑉𝑠 (𝑠)
𝑍𝑖 𝑠 + 𝑍𝑠 𝑠 𝑠
⟹ A voltage-isolation amplifier must have a high input impedance
HCM City Univ. of Technology, Faculty of Mechanical Engineering

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Electrical and Electromechanical Systems

§3.Impedance and Amplifier
- Denote the amplifier’s voltage gain as 𝐺. This means that the

amplifier’s output voltage 𝑣𝑜 is 𝑣𝑜 = 𝐺𝑣𝑖 . For the load circuit

𝐺𝑉𝑖 𝑠 − 𝐼𝑜 𝑠 𝑍𝑜 𝑠 − 𝐼𝑜 𝑠 𝑍𝐿 𝑠 = 0 ⟹ 𝐼𝑜 𝑠 =

𝐺𝑉𝑖 (𝑠)
𝑍𝑜 𝑠 + 𝑍𝐿 (𝑠)

𝑍𝐿 𝑠
𝐺𝑉𝑜 (𝑠)
𝑍𝑜 𝑠 + 𝑍𝐿 𝑠
• 𝑍𝑜 (𝑠) is small ⟶ 𝑣𝐿 ≈ 𝐺𝑣𝑜 : the voltage 𝑣𝐿 delivered to the
load is independent of the load

⟹ 𝑉𝐿 𝑠 = 𝑍𝐿 𝑠 𝐼𝑜 𝑠 =

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System Dynamics

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Electrical and Electromechanical Systems

§3.Impedance and Amplifier
4.Operational Amplifier (Op Amp)
- An operational amplifier is a high-gain linear amplifier
- The op-amp approaches the ideal amplifier of the analog
designer’s dreams because it has such ideal characteristics
• Very high open-loop gain: 𝐺 = 100,000 +
• Very high input resistance: 𝑅𝑖𝑛 > 1 𝑀Ω
• Low output resistance:
𝑅𝑜𝑢𝑡 = 50 ÷ 75 Ω
- The output voltage
𝑉𝑜𝑢𝑡 = 𝐺 𝑣2 − 𝑣1
𝑣𝑜𝑢𝑡 : output voltage, 𝑉
𝐺: open-loop gain
𝑣1 : inverting input, 𝑉
𝑣2 : non-inverting input, 𝑉
0.8 × (−𝑉𝑐𝑐 ) ≈ −𝑣𝑠𝑎𝑡 ≤ 𝑣𝑜𝑢𝑡 ≤ +𝑣𝑠𝑎𝑡 ≈ 0.8 × (+𝑉𝑐𝑐 )

§3.Impedance and Amplifier
- Ex.6.3.3
Op-Amp Multiplier
Determine the relation between the input
voltage 𝑣𝑖 and the output voltage 𝑣𝑜 of
the op-amp circuit. Assume that the op
amp has the following properties

• very large gain 𝐺
• 𝑣𝑜 = −𝐺𝑣1
• very large input impedance
Solution
The current drawn by the op amp is very
small 𝑖3 ≈ 0, the input terminal pair can
be represented as an open circuit
𝑣𝑖 − 𝑣1
𝑣1 − 𝑣𝑜
𝑖1 =
,
𝑖2 =
𝑅𝑖
𝑅𝑓

HCM City Univ. of Technology, Faculty of Mechanical Engineering

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Electrical and Electromechanical Systems

§3.Impedance and Amplifier
From conservation of charge, 𝑖1 = 𝑖2 + 𝑖3 ≈ 𝑖2
𝑣𝑖 − 𝑣1 𝑣1 − 𝑣𝑜

⟹
=
𝑅𝑖
𝑅𝑓

System Dynamics

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§3.Impedance and Amplifier
5.Generation Op-Amp Input-Output Relation

Substitute 𝑣1 = −𝑣𝑜 /𝐺 into the equation
𝑣𝑖
𝑣𝑜
𝑣𝑜
𝑣𝑜
+
=−
−
𝑅𝑖 𝑅𝑖 𝐺
𝑅𝑓 𝐺 𝑅𝑓

With very large 𝐺
𝑣𝑖
𝑣𝑜

=−
𝑅𝑖
𝑅𝑓
𝑅𝑓
⟹ 𝑣𝑜 = − 𝑣𝑖
𝑅𝑖
This circuit can be used to multiply a
voltage by the factor 𝑅𝑓 /𝑅𝑖 , and is called an op-amp multiplier.
Note that this circuit inverts the sign of the input voltage
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§3.Impedance and Amplifier

The impedance 𝑍𝑖 (𝑠) of the input elements is defined such that
𝑉𝑖 𝑠 − 𝑉1 𝑠 = 𝑍𝑖 𝑠 𝐼1 (𝑠)
For the feedback elements
𝑉1 𝑠 − 𝑉𝑜 𝑠 = 𝑍𝑓 𝑠 𝐼2 (𝑠)
The high internal impedance: 𝑖3 ≈ 0 ⟹ 𝑖1 ≈ 𝑖2
The amplifier relation:
𝑣𝑜 = −𝐺𝑣1 ⟹ 𝑉𝑜 𝑠 = −𝐺𝑉1 (𝑠)
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§3.Impedance and Amplifier
The gain of this multiplier is 𝑅𝑓 /𝑅𝑖 , with a sign reversal
𝑍𝑓 𝑠
𝑅𝑓
𝑉𝑜 (𝑠)
≈−
=−
𝑉1 (𝑠)
𝑍𝑖 𝑠
𝑅𝑖

𝑍𝑓 𝑠
𝑍𝑖 𝑠
⟹ 𝑉1 𝑠 =
𝑉 𝑠 +
𝑉 (𝑠)
𝑍𝑓 𝑠 + 𝑍𝑖 𝑠 𝑖
𝑍𝑓 𝑠 + 𝑍𝑖 𝑠 𝑜
𝑍𝑓 𝑠
𝑉𝑜 𝑠
𝐺
𝑍𝑖 (𝑠)

⟹
=−
,𝐻 𝑠 =
𝑉1 𝑠
𝑍𝑓 𝑠 + 𝑍𝑖 𝑠 1 + 𝐺𝐻 𝑠
𝑍𝑓 𝑠 + 𝑍𝑖 𝑠

To eliminate the sign reversal, using an inverter, which is a
multiplier having equal resistances

Very large 𝐺 ⟹ |𝐺𝐻(𝑠)| ≫ 1
𝑍𝑓 𝑠
𝑅𝑓
𝑉𝑜 (𝑠)
≈−
=−
𝑉1 (𝑠)
𝑍𝑖 𝑠
𝑅𝑖
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§3.Impedance and Amplifier
- Ex.6.3.4
Integration with Op Amps
Determine
the transfer function
𝑉𝑜 (𝑠)/𝑉𝑖 (𝑠) of the circuit

§3.Impedance and Amplifier
- Ex.6.3.5
Differentiation with Op Amps
Design an op-amp circuit that
differentiates the input voltage

Solution
The impedance of a capacitor is 1/𝐶𝑠 ⟹ the transfer function
of the circuit is found from

𝑍𝑓 𝑠
𝑉𝑜 (𝑠)
1
=−
=−
𝑉𝑖 (𝑠)
𝑍𝑖 𝑠
𝑅𝐶𝑠
In the time domain, the circuit model
1 𝑡
𝑣𝑜 = −
𝑣 𝑑𝑡
𝑅𝐶 0 𝑖

Solution
A differentiator can be created by interchanging the resistance
and capacitance in the integrator circuit
𝑍𝑖 𝑠 = 1/𝐶𝑠,
𝑍𝑓 𝑠 = 𝑅

The circuit is called an op-amp integrator, and is used in many
devices for computing, signal generation, and control
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The input-output relation for this ideal differentiator
𝑍𝑓 𝑠
𝑉𝑜 (𝑠)
=−
= −𝑅𝐶𝑠
𝑉𝑖 (𝑠)
𝑍𝑖 𝑠
The model in the time domain
𝑑𝑣𝑖 (𝑡)
𝑣𝑜 (𝑡) = −𝑅𝐶
𝑑𝑡
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§4.Electric Motors
- Electromechanical systems consist of an electrical subsystem
and a mechanical subsystem with mass and possibly elasticity
and damping
- In some devices, such as motors and speakers, the mass is
driven by a force generated by the electrical subsystem. In
other devices, such as microphones, the motion of the mass

generates a voltage or current in the electrical subsystem
⟹ we must apply electrical principles and Newton’s laws to
develop a model of an electromechanical system
- Often the forces and torques are generated electromagnetically, but other methods are used as well; for example,
piezoelectric devices contain crystals that generate forces
when a voltage is applied to them

§4.Electric Motors
1.Magnetic Coupling
- The majority of electromechanical devices utilize a magnetic field
- Two basic principles apply to a conductor, such as a wire,
carrying a current within a magnetic field
• a force is exerted on the conductor by the field
𝑓 = 𝐵𝐿𝑖
𝑓: force, 𝑁
𝐵: the flux density of the field, 𝑊𝑏/𝑚2
𝐿: the length of the conductor, 𝑚
𝑖: current, 𝐴

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§4.Electric Motors
• if the conductor moves relative to the field, the field induces
a voltage in the conductor that opposes the voltage
producing the current
𝑣𝑏 = 𝐵𝐿𝑣
𝑣𝑏 : the voltage induced in the conductor, 𝑉
𝐵: the flux density of the field, 𝑊𝑏/𝑚2
𝐿: the length of the conductor, 𝑚
𝑣: conductor velocity in the field, 𝑚/𝑠

§4.Electric Motors

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- The circuit represents the electrical
behavior of the conductor and the
mass 𝑚 represents the mass of the

conductor and any attached mass
The power generated by the circuit:
𝑣𝑏 𝑖
The power applied to the mass 𝑚 by the force 𝑓: 𝑓𝑣
Neglecting any energy loss due to resistance in the
conductor or friction or damping acting on the mass
𝑣𝑏 𝑖 = 𝑓𝑣 = 𝐵𝐿𝑖𝑣
In addition, from Newton’s law
𝑚𝑣 = 𝑓
Therefore
𝑚𝑣 = 𝐵𝐿𝑖
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§4.Electric Motors
2.The D’Arsonval Meter

§4.Electric Motors
- Ex.6.4.1

The current to be measured is passed through a coil to which
a pointer is attached. The coil is positioned within a magnetic
field and is wrapped around an iron core to strengthen the
effects of the field. The core thus acts like an inductor. The
interaction between the current and the field produces a torque
that tends to rotate the coil and pointer. This rotation is
opposed by a torsional spring of stiffness 𝑘 𝑇
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Electrical and Electromechanical Systems

A Model of the D'Arsonval Meter
Derive a model of a D’Arsonval meter
in terms of the coil angular
displacement 𝜃 and the coil current 𝑖.
The input is the applied voltage 𝑣𝑖 .
Discuss the case where there are 𝑛
coils around the core

Solution
The torque acting on both sides of the coil
𝐿
𝑇 = 𝑓𝑟 = 2𝐵 𝑖 𝑟 = (𝐵𝐿𝑟)𝑖
2
The equation of motion of the core/coil unit is
𝑑2𝜃
𝑑𝜃
𝐼 2 +𝑐
+ 𝑘 𝑇 𝜃 = 𝑇 = (𝐵𝐿𝑟)𝑖
𝑑𝑡
𝑑𝑡
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§4.Electric Motors
The rotation of the coil induces a voltage 𝑣𝑏 in the coil that is
proportional to the coil’s linear
velocity 𝑣 such that 𝑣𝑏 = 𝐵𝐿𝑣. The
linear velocity is related to the coil’s
angular velocity 𝜃 by 𝑣 = 𝑟𝜃. Thus
𝑑𝜃

𝑣𝑏 = 𝐵𝐿𝑣 = 𝐵𝐿𝑟
𝑑𝑡
The coil circuit is represented in the figure, where
𝑅 represents the resistance of the wire in the coil
Kirchhoff’s voltage law applied to the coil circuit gives
𝑑𝑖
𝑣𝑖 − 𝐿 − 𝑅𝑖 − 𝑣𝑏 = 0
𝑑𝑡
𝑑𝑖
𝑑𝜃
⟹ 𝐿 + 𝑅𝑖 + 𝐵𝐿𝑟
= 𝑣𝑖
𝑑𝑡
𝑑𝑡

§4.Electric Motors
If there are 𝑛 coils
• the resulting torque expression: 𝑇 = 𝑛(𝐵𝐿𝑟)𝑖
• the induced voltage expression: 𝑣𝑏 = 𝑛𝐵𝐿𝑟𝜃
the model is
𝑑2𝜃
𝑑𝜃
𝐼 2 +𝑐
+ 𝑘 𝑇 𝜃 = 𝑇 = 𝑛(𝐵𝐿𝑟)𝑖
𝑑𝑡
𝑑𝑡
𝑑𝑖
𝑑𝜃
𝐿 + 𝑅𝑖 + 𝑛𝐵𝐿𝑟
= 𝑣𝑖

𝑑𝑡
𝑑𝑡
If the applied constant voltage 𝑣𝑖 , the system will reach a
steady-state in which the pointer comes to rest. At steadystate, 𝜃 = 𝑑𝑖/𝑑𝑡 = 0, the above equations become
𝑣𝑖
𝑛𝐵𝐿𝑟𝑖 𝑛𝐵𝐿𝑟𝑣𝑖
𝑖= ,
𝜃=
=
𝑅
𝑘𝑇
𝑅𝑘 𝑇
This equation can be used to calibrate the device by relating the
pointer displacement 𝜃 to either the measured current 𝑖 or voltage 𝑣𝑖

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§4.Electric Motors
3.DC Motors
- There are many types of electric

motors, but the two main categories
are direct current (dc) motors and
alternating current (ac) motors
- Within the dc motor category there
are the armature-controlled motor
and the field-controlled motor
- Basic motor elements: stator, rotor, armature, commutator
• Stator: stationary and provides the magnetic field
• Rotor: an iron core that is supported by bearings
• Armature: the coils are attached to the rotor
• Commutator: a pair of electrically conducting, spring-loaded
carbon sticks (called brushes) that slide on the armature
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§4.Electric Motors
4.Model of an Armature Controlled DC Motors
Model
- input: the armature voltage 𝑣𝑎
- output: the armature current 𝑖𝑎
motor speed 𝜔

The torque on the armature
𝑇 = 𝑛𝐵𝐿𝑖𝑎 𝑟 = 𝑛𝐵𝐿𝑟 𝑖𝑎 = 𝐾𝑇 𝑖𝑎
𝐾𝑇 ≡ 𝑛𝐵𝐿𝑟: the motor’s torque constant given by the
manufacturer of motor
The voltage in the armature, back emf, is proportional to the speed
𝑣𝑏 = 𝑛𝐵𝐿𝑣 = 𝑛𝐵𝐿𝑟 𝜔 = 𝐾𝑏 𝜔
𝐾𝑏 ≡ 𝑛𝐵𝐿𝑟: the motor’s back emf constant or the voltage
constant. Note that 𝐾𝑏 ≡ 𝐾𝑇
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§4.Electric Motors

The back emf is a voltage drop in the armature
circuit. Thus, Kirchhoff’s voltage law gives
𝑑𝑖𝑎
𝑣𝑎 − 𝑅𝑎 𝑖𝑎 − 𝐿𝑎
− 𝐾𝑏 𝜔 = 0
𝑑𝑡
Apply Newton’s law
𝑑𝜔
𝐼
= 𝑇 − 𝑐𝜔 − 𝑇𝐿
𝑑𝑡
𝑑𝜔
⟹𝐼
= 𝐾𝑇 𝑖𝑎 − 𝑐𝜔 − 𝑇𝐿
𝑑𝑡
The above two equations constitute the system model

§4.Electric Motors
Motor Transfer Functions
𝐼𝑎 (𝑠) 𝐼𝑠 + 𝑐
𝐼𝑎 (𝑠)
𝐾𝑏
=
,
=
𝑉𝑎 (𝑠)
𝐷(𝑠)
𝑇𝐿 (𝑠) 𝐷(𝑠)
Ω(𝑠)
𝐾𝑇

Ω(𝑠) 𝐿𝑎 𝑠 + 𝑅𝑎
=
,
=
𝑉𝑎 (𝑠) 𝐷(𝑠)
𝑇𝐿 (𝑠)
𝐷(𝑠)
2
with 𝐷(𝑠) ≡ 𝐿𝑎 𝐼𝑠 + 𝑅𝑎 𝐼 + 𝑐𝐿𝑎 𝑠 + 𝑐𝑅𝑎 + 𝐾𝑏 𝐾𝑇
State-Variable Form of the Motor Model
𝑑𝑖𝑎
1
= (𝑣𝑎 − 𝑅𝑎 𝑖𝑎 − 𝐾𝑏 𝜔)
𝑑𝑡
𝐿𝑎
𝑑𝜔 1
= (𝐾𝑇 𝑖𝑎 − 𝑐𝜔 − 𝑇𝐿 )
𝑑𝑡
𝐼

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§4.Electric Motors
5.Field Controlled Motors
- Require two power supplies
• for the armature circuit
• for the field circuit
- Requite a control circuit to
maintain a constant armature
current in the presence of the
back emf, which varies with motor speed and field strength
- In general, the field strength 𝐵(𝑖𝑓 ) is a nonlinear function of
the field current 𝑖𝑓 . The torque on the armature
𝑇 = 𝑛𝐵 𝑖𝑓 𝐿𝑖𝑎 𝑟 = 𝑛𝐿𝑟𝑖𝑎 𝐵 𝑖𝑓 = 𝑇(𝑖𝑓 )
Often the linear approximation 𝑇 − 𝑇𝑟 ≈ 𝐾𝑇 (𝑖𝑓 − 𝑖𝑓𝑟 ) is used
𝑇𝑟 ,𝑖𝑓𝑟 : torque, current at a reference operating equilibrium
𝐾𝑇 : the slope of the 𝑇(𝑖𝑓 ) curve at the reference condition

§4.Electric Motors
- Ex.6.4.2
Model of a Field-Controlled dc Motor
Develop a model of the field-controlled motor

Solution
The voltage 𝑣𝑓 is applied to the
field circuit, whose inductance
and resistance are 𝐿𝑓 and 𝑅𝑓 .
No back emf exists in the field
circuit because it does not move within the field, and
Kirchhoff’s voltage law applied to the field circuit gives
𝑑𝑖𝑓
𝑣𝑓 = 𝑅𝑓 𝑖𝑓 + 𝐿𝑓
𝑑𝑡
For the inertia 𝐼
𝑑𝜔
𝐼
= 𝑇 − 𝑐𝜔 − 𝑇𝐿 = 𝐾𝑇 𝑖𝑓 − 𝑐𝜔 − 𝑇𝐿
𝑑𝑡

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§5.Analysis of Motor Performance
1.Steady-State Motor Response

The steady-state operating conditions can be obtained by
applying the final value theorem, 𝑥 ∞ = lim 𝑠𝑋(𝑠), to the 𝑇𝐹s
𝑠→0

• The steady-state value of 𝑖𝑎
to applied voltage 𝑉𝑎
1
(𝐼𝑠 + 𝑐) 𝑉𝑎
𝑐𝑉𝑎
𝑠 ⟹ 𝑖 = lim 𝑠𝐼 𝑠 =
𝐼𝑎 𝑠 =
𝑎
𝑎
𝑠→0
𝐷(𝑠)
𝑐𝑅𝑎 + 𝐾𝑏 𝐾𝑇
where 𝐷(𝑠) ≡ 𝐿𝑎 𝐼𝑠 2 + 𝑅𝑎 𝐼 + 𝑐𝐿𝑎 𝑠 + 𝑐𝑅𝑎 + 𝐾𝑏 𝐾𝑇
to applied torque 𝑇𝐿
1
𝐾𝑏 𝑇𝐿
𝐾𝑏 𝑇𝐿
𝑠 ⟹ 𝑖 = lim 𝑠𝐼 𝑠 =
𝐼𝑎 𝑠 =
𝑎
𝑎
𝑠→0
𝐷(𝑠)
𝑐𝑅𝑎 + 𝐾𝑏 𝐾𝑇
𝑐𝑉𝑎 + 𝐾𝑏 𝑇𝐿
⟹ 𝑖𝑎 =

𝑐𝑅𝑎 + 𝐾𝑏 𝐾𝑇
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§5.Analysis of Motor Performance
• The steady-state value of 𝜔
to applied voltage 𝑉𝑎
1
𝐾𝑇 𝑉𝑎
𝐾𝑇 𝑉𝑎
𝑠 ⟹ 𝜔 = lim 𝑠Ω 𝑠 =
Ω(𝑠) =
𝑠→0
𝐷(𝑠)
𝑐𝑅𝑎 + 𝐾𝑏 𝐾𝑇
where 𝐷(𝑠) ≡ 𝐿𝑎 𝐼𝑠 2 + 𝑅𝑎 𝐼 + 𝑐𝐿𝑎 𝑠 + 𝑐𝑅𝑎 + 𝐾𝑏 𝐾𝑇
to applied torque 𝑇𝐿
1
𝐿𝑎 𝑠 + 𝑅𝑎 𝑇𝐿
𝑠 ⟹ 𝜔 = lim 𝑠Ω 𝑠
Ω 𝑠 =−

𝑠→0
𝐷 𝑠
−𝑅𝑎 𝑇𝐿
=
𝑐𝑅𝑎 + 𝐾𝑏 𝐾𝑇
𝐾𝑇 𝑉𝑎 − 𝑅𝑎 𝑇𝐿
⟹𝜔=
𝑐𝑅𝑎 + 𝐾𝑏 𝐾𝑇
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§5.Analysis of Motor Performance

- Ex.6.5.1
No-Load Speed and Stall Torque
Motor: 𝐾𝑇 = 𝐾𝑏 = 0.05𝑁𝑚/𝐴 , 𝑐 = 10−4 𝑁𝑚𝑠/𝐴 , 𝑅𝑎 = 0.5Ω ,
𝜔𝑚𝑎𝑥 = 3000𝑟𝑝𝑚, 𝑖𝑚𝑎𝑥 = 30𝐴. Compute the no-load speed,
the no-load current, and the stall torque. Determine whether
the motor can be used with an applied voltage of 𝑣𝑎 = 10𝑉
Solution
𝑐𝑉𝑎 + 𝐾𝑏 𝑇𝐿
10−4 × 10 + 0.05𝑇𝐿
𝑖𝑎 =
=
= 0.392 + 19.61𝑇𝐿
𝑐𝑅𝑎 + 𝐾𝑏 𝐾𝑇 10−4 × 0.5 + 0.05 × 0.05
𝐾𝑇 𝑉𝑎 − 𝑅𝑎 𝑇𝐿
0.05 × 10 − 0.5𝑇𝐿
ω=
=
= 196.1 − 196.1𝑇𝐿
𝑐𝑅𝑎 + 𝐾𝑏 𝐾𝑇 10−4 × 0.5 + 0.05 × 0.05
No load, 𝑇𝐿 = 0
𝑖𝑎 = 0.392𝐴 < 𝑖𝑚𝑎𝑥
𝜔 = 196.1𝑟𝑎𝑑 = 1872𝑟𝑝𝑚 < 𝜔𝑚𝑎𝑥
Stall torque, 𝜔 = 0 𝑇𝐿 = 1𝑁𝑚
𝑖𝑎 = 20𝐴 < 𝑖𝑚𝑎𝑥

§5.Analysis of Motor Performance
2.Motor Dynamics Response
Ex.6.5.2
Response of an Armature-Controlled dc Motor
The parameter values for a certain motor are

𝐾𝑇 = 𝐾𝑏 = 0.05𝑁𝑚/𝐴
𝑐 = 10−4 𝑁𝑚𝑠/𝐴
𝑅𝑎 = 0.5Ω
𝐿𝑎 = 2 × 10−3 𝐻
𝐼 = 9 × 10−5𝑘𝑔𝑚2
where 𝐼 includes the inertia of the armature and that of the
load. The load torque 𝑇𝐿 is zero. Obtain the step response of
𝑖𝑎 (𝑡) and 𝜔(𝑡) if the applied voltage is 𝑣𝑎 = 10𝑉

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§5.Analysis of Motor Performance

Solution
Substituting the given parameter values into motor’s 𝑇𝐹s
𝐼𝑎 (𝑠)
9 × 10−5 𝑠 + 10−4
=
𝑉𝑎 (𝑠) 18 × 10−8 𝑠 2 + 4.52 × 10−5 𝑠 + 2.55 × 10−3
Ω(𝑠)
0.05
=
𝑉𝑎 (𝑠) 18 × 10−8 𝑠 2 + 4.52 × 10−5 𝑠 + 2.55 × 10−3
If 𝑣𝑎 is a step function of magnitude 10𝑉
5 × 103𝑠 + 5.555 × 104
𝐶1
𝐶2
𝐶3
𝐼𝑎 𝑠 =
= +
+
𝑠 𝑠 + 165.52 𝑠 + 85.59
𝑠 𝑠 + 165.52 𝑠 + 85.59
6
2.777 × 10
𝐷1
𝐷2
𝐷3
Ω 𝑠 =
=
+
+
𝑠 𝑠 + 165.52 𝑠 + 85.59

𝑠
𝑠 + 165.52 𝑠 + 85.59
Evaluating the partial-fraction coefficients
𝑖𝑎 𝑡 = 0.39 − 61𝑒 −165.52𝑡 + 61.74𝑒 −85.59𝑡
𝜔 𝑡 = 196.1 + 210𝑒 −165.52𝑡 − 406𝑒 −85.59𝑡

§5.Analysis of Motor Performance
Step response of an armature-controlled dc motor

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- The large overshoot in 𝑖𝑎 , which is caused by the numerator
dynamics. The plot shows that the steady-state calculation of
𝑖𝑎 = 0.39𝐴 greatly underestimates the maximum required
current, which is approximately 15𝐴
- In practice, a pure step input is impossible, and thus the
required current will not be as high as 15𝐴. The real input
would take some time to reach 10𝑉

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Electrical and Electromechanical Systems

§5.Analysis of Motor Performance
3.The Effect of Armature Inductance
- The denominator of motor transfer functions
𝐷(𝑠) ≡ 𝐿𝑎 𝐼𝑠 2 + 𝑅𝑎 𝐼 + 𝑐𝐿𝑎 𝑠 + 𝑐𝑅𝑎 + 𝐾𝑏 𝐾𝑇
If setting 𝐿𝑎 = 0, the model reduces to a first-order model
- Another reason 𝐿𝑎 is sometimes neglected is that it is difficult
to calculate or to measure
- With 𝐿𝑎 = 0 the characteristic equation and the motor
differential equations are still only second order and thus are
manageable. Be careful in using the approximation 𝐿𝑎 = 0

§5.Analysis of Motor Performance
4.Determining Motor Parameters
- Motor parameter values can be applied from
• the motor manufacturer
• the experiment
- The voltage and torque constants 𝐾𝑏 , 𝐾𝑇 : cannot be
calculated from 𝑛𝐵𝐿𝑟 because the value of the magnetic
field parameter 𝐵 might be difficult to determine
- The armature inertia 𝑖𝑎 can be
• calculated from the formula for the inertia of a cylinder
• measured by suspending it with a metal wire and
measuring the torsional oscillation frequency 𝑓𝑛 𝐻𝑧 as the

armature twists on the wire, then
𝑖𝑎 = 𝑘 𝑇 /(2𝜋𝑓𝑛 )2
𝑘 𝑇 : the torsional spring constant of the wire

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§5.Analysis of Motor Performance

- Some parameters can be measured with static (steady-state)
tests
• By slowly increasing the load torque 𝑇𝐿 until the motor stalls
and measuring the resulting stall current, we can compute
𝐾𝑇 from 𝐾𝑇 = 𝑇𝐿 /𝑖𝑎
• Knowing the voltage 𝑉𝑎 , we can compute the armature
resistance from 𝑅𝑎 = 𝑉𝑎 /𝑖𝑎
• By measuring the no-load speed 𝜔 and the resulting
current 𝑖𝑎 , and knowing 𝑉𝑎 , 𝑅𝑎 , and 𝐾𝑇 , we can compute 𝑐
from the steady-state relations with 𝑇𝐿 = 0
- Because 𝑐 is difficult to determine precisely, its value is rarely
reported by motor manufacturers. However, in motors with
good bearings the damping can be slight and is often taken
to be zero

§5.Analysis of Motor Performance
- The inductance 𝐿𝑎 is
• difficult to determine ⟹ special instruments such as an
impedance meter can be used to measure 𝐿𝑎
• often assumed to be very small and therefore is often taken
to be zero. This is sometimes a good approximation, but
not always
- Units for 𝐾𝑇 and 𝐾𝑏
𝑇
𝑁𝑚
𝐾𝑇 =
=
𝑖𝑎
𝐴
𝑣𝑏

𝑉
𝑊
𝑁𝑚/𝑠
𝑁𝑚
𝐾𝑏 =
=
= 𝑉𝑠 = 𝑠 =
𝑠=
𝜔
1/𝑠
𝐴
𝐴
𝐴

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System Dynamics

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Electrical and Electromechanical Systems

§5.Analysis of Motor Performance
5.The Trapezoidal Profile and Motion Control

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§5.Analysis of Motor Performance
• Average required current, 𝑖𝑟𝑚𝑠 , and motor torque, 𝑇𝑟𝑚𝑠 . The
average usually stated is the rms average, which stands for
root mean square. For torque, it is calculated as follows
𝑇𝑟𝑚𝑠 =

1
𝑡𝑓

𝑡𝑓

𝑇 2 𝑡 𝑑𝑡 ,

0

𝑖𝑟𝑚𝑠 =

𝑇𝑟𝑚𝑠
𝐾𝑇

• Maximum speed error: the maximum difference between
the desired speed given by the profile and the actual speed
• Average speed error: the average speed error is commonly

computed as the rms value
• Displacement error: the difference between the specified
displacement and the actual displacement
• System response time: the system must be able to respond
fast enough to follow the profile
- Note: not all these criteria are important for every application
HCM City Univ. of Technology, Faculty of Mechanical Engineering

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§5.Analysis of Motor Performance
6.Motor and Amplifier Performance
- In evaluating the performance of a motion-control system,
the following are important
• Energy consumption per cycle: the sum of the energy loss
in the motor resistance and in the damping

- The trapezoidal speed profile shown in the figure is used in
many motion-control applications. The constant-speed phase
of the profile is called the slew phase
- Aplications
• Speed control: conveyor of assembly line
• Position control: robot arm moving a specified angle

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𝑡𝑓

𝐸=

𝑅𝑖 2 𝑡 𝑑𝑡 +

0

𝑡𝑓

𝑐𝜔2 𝑡 𝑑𝑡

0

𝑡𝑓 : the duration of the cycle
𝜔: the speed of the system at the location of the damping
• Maximum required current, 𝑖𝑚𝑎𝑥 , and maximum motor
torque, 𝑇𝑚𝑎𝑥
• Maximum required motor speed, 𝜔𝑚𝑎𝑥
• Maximum required voltage, 𝑣𝑚𝑎𝑥
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§5.Analysis of Motor Performance
- In the following, assume that the damping constant 𝑐 is zero
- Obtain the following motor model
𝑑𝑖
𝑣 = 𝑅𝑖 + 𝐿 + 𝐾𝑏 𝜔
𝑑𝑡
𝑑𝜔
𝐼
= 𝐾𝑇 𝑖 − 𝑇𝑑
𝑑𝑡
𝜔:
the motor speed, 𝑟𝑎𝑑/𝑠
𝜔𝐿 = 𝜔/𝑁: the load speed, 𝑟𝑎𝑑/𝑠
𝑁:
the speed reduction ratio
𝑇𝑑 = 𝑇𝐿 /𝑁: the torque 𝑇𝑑 opposing the motor torque 𝑇 is due
to the torque 𝑇𝐿 acting on the load, 𝑁𝑚
𝐼:
the inertia includes the motor inertia 𝐼𝑚 and the
load inertia 𝐼𝐿 reflected to the motor shaft, 𝑘𝑔𝑚2
𝐼 = 𝐼𝑚 + 𝐼𝐿 /𝑁 2
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§5.Analysis of Motor Performance
- Energy lost
With 𝑐 = 0, the expression for the energy loss per cycle
𝑡𝑓

𝐸=

𝑅𝑖 2 𝑡 𝑑𝑡 = 𝑅

0

𝑡𝑓
0

𝑑𝑡

Assuming that 𝑇𝑑 is constant, and expanding this expression
𝑅𝑖 2 𝑡𝑓 2
2𝑅𝐼𝑇𝑑 𝑡𝑓
𝑅𝑇𝑑2 𝑡𝑓

𝐸= 2
𝜔 𝑑𝑡 +
𝜔 𝑑𝑡 + 2
𝑑𝑡
𝐾𝑇 0
𝐾𝑇2 0
𝐾𝑇 0
With the assumption that the motion begins and ends at rest
so that 𝜔(0) = 𝜔(𝑡𝑓 ), the second integral is zero
𝑅𝑇𝑑2 𝑡𝑓
𝑅𝐼2 𝑡𝑓 2
𝐸= 2
𝜔 𝑑𝑡 +
𝐾𝑇 0
𝐾𝑇2
Note that the first term on the right depends on the profile,
while the second term depends on the disturbance torque 𝑇𝑑
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§5.Analysis of Motor Performance
0 ≤ 𝑡 ≤ 𝑡1 ,
𝜔 = 𝜔𝑚𝑎𝑥 /𝑡1 = 𝜃𝑓 /𝑡1 𝑡2

𝑡1 ≤ 𝑡 ≤ 𝑡2 , 𝜔 = 0
𝑡2 ≤ 𝑡 ≤ 𝑡𝑓 ,
𝜔 = −𝜔𝑚𝑎𝑥 /𝑡1 = −𝜃𝑓 /𝑡1 𝑡2
𝑡𝑓

⟹

𝑡1

𝜔2 𝑑𝑡 =

0

𝜃𝑓
𝑡1 𝑡2

0

𝑡2

𝑑𝑡 +

𝑡𝑓

0𝑑𝑡 +

−

0

2

2

𝜃𝑓
𝑡1 𝑡2

=

2

𝜃𝑓
𝑡1 𝑡2

𝑡1 + 𝑡𝑓 − 𝑡2 =

𝜃𝑓
𝑡1 𝑡2

6.92

Electrical and Electromechanical Systems

§5.Analysis of Motor Performance
- Maximum Motor Speed

2

𝐼𝜔 + 𝑇𝑑
𝐾𝑇

System Dynamics

2

𝑑𝑡

2

2𝑡1

Evaluate the integral in
previous equation for the
trapezoidal profile, which is
specified in terms of the
motor speed 𝜔(𝑡)
The total angular displacement is the area under the
trapezoidal profile shown in the figure. Assuming that 𝜃(0) =
0, and because 𝑡𝑓 − 𝑡2 = 𝑡1 , we have
𝑡𝑓
1
𝜃 𝑡𝑓 = 𝜃𝑓 =
𝜔𝑑𝑡 = 2 𝜔𝑚𝑎𝑥𝑡1 + 𝜔𝑚𝑎𝑥 𝑡2 − 𝑡1 = 𝜔𝑚𝑎𝑥𝑡2
2
0
𝜃𝑓
⟹ 𝜔𝑚𝑎𝑥 =
𝑡2
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§5.Analysis of Motor Performance
- Maximum Motor Torque
Maximum required acceleration for the trapezoidal profile
𝜃𝑓
𝜔𝑚𝑎𝑥
𝛼𝑚𝑎𝑥 =
=
𝑡1
𝑡1 𝑡2
The motor torque
𝑑𝜔
𝑇 = 𝐾𝑇 𝑖 = 𝐼
+ 𝑇𝑑 = 𝐼𝛼 + 𝑇𝑑
𝑑𝑡
The maximum required motor torque
𝜃𝑓
𝜔𝑚𝑎𝑥
𝑇𝑚𝑎𝑥 = 𝐼𝛼𝑚𝑎𝑥 + 𝑇𝑑 = 𝐼
+ 𝑇𝑑 = 𝐼
+ 𝑇𝑑
𝑡1
𝑡1 𝑡2

Since 𝑡𝑓 − 𝑡2 = 𝑡1
𝐸=

𝑅𝐼2
𝐾𝑇2

𝜃𝑓
𝑡1 𝑡2

2

2𝑡1 +

𝑅𝑇𝑑2 𝑡𝑓
𝑅 2𝐼2 𝜃𝑓2
= 2
+ 𝑇𝑑2 𝑡𝑓
𝐾𝑇2
𝐾𝑇 𝑡1 𝑡22

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§5.Analysis of Motor Performance

- RMS Motor Torque
1 𝑡𝑓 2
1
2
𝑇𝑟𝑚𝑠
=
𝑇 𝑡 𝑑𝑡 =
𝑡𝑓 0
𝑡𝑓
2
⟹ 𝑇𝑟𝑚𝑠
=

1
𝑡𝑓

𝑡1

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𝑡𝑓

6.96

Electrical and Electromechanical Systems

(𝐼𝛼 + 𝑇𝑑 )2 𝑑𝑡

0

𝑡2

(𝐼𝛼𝑚𝑎𝑥 + 𝑇𝑑 )2 𝑑𝑡 +

0

(0 + 𝑇𝑑 )2 𝑑𝑡

𝑡1
𝑡𝑓

2
𝑇𝑟𝑚𝑠
=

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Nguyen Tan Tien

§5.Analysis of Motor Performance

+
This reduces to

HCM City Univ. of Technology, Faculty of Mechanical Engineering

(−𝐼𝛼𝑚𝑎𝑥 + 𝑇𝑑 )2 𝑑𝑡

𝑡2

1
2
2𝐼2 𝛼𝑚𝑎𝑥
𝑡1 + 𝑇𝑑2 (𝑡1 + 𝑡2 )
𝑡𝑓

Since 𝛼𝑚𝑎𝑥 = 𝜃𝑓 /𝑡1 𝑡2 and 𝑡1 + 𝑡2 = 𝑡𝑓
𝑇𝑟𝑚𝑠 =

2𝐼2 𝜃𝑓2
𝑡𝑓 𝑡1 𝑡22

+ 𝑇𝑑2

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Electrical and Electromechanical Systems

§5.Analysis of Motor Performance
- Amplifier Requirements
Using the motor current equation 𝑖 = 𝑇/𝐾𝑇 , the maximum
current and the rms current required
𝑇𝑚𝑎𝑥
𝑇𝑟𝑚𝑠
𝑖𝑚𝑎𝑥 =
,
𝑖𝑟𝑚𝑠 =
𝐾𝑇
𝐾𝑇
The motor voltage equation
𝑑𝑖
𝑣
𝐿 𝑑𝑖 𝐾𝑏
𝑣 = 𝑅𝑖 + 𝐿 + 𝐾𝑏 𝜔 ⟹ = 𝑖 +
+ 𝜔
𝑑𝑡
𝑅
𝑅 𝑑𝑡 𝑅
If the electrical time constant 𝐿/𝑅 is very small, neglect the

second term on the right to obtain
𝑣
𝐾𝑏
=𝑖+ 𝜔
𝑅
𝑅
The maximum voltage required is given approximately by
𝑣𝑚𝑎𝑥 = 𝑅𝑖𝑚𝑎𝑥 + 𝐾𝑏 𝜔𝑚𝑎𝑥

§5.Analysis of Motor Performance
- Ex.6.5.3
Calculating Motor-Amplifier Requirements
The trapezoidal profile requirements for a specific application
are given in the following table, along with the load and
motor data

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§5.Analysis of Motor Performance
Solution

The total inertia 𝐼
𝐼𝐿
4 × 10−3
𝐼 = 𝐼𝑚 + 2 = 10−3 +
= 2 × 10−3 𝑘𝑔𝑚2
𝑁
22
The reduction ratio is 𝑁 = 2, the required motor displacement
is 𝑁𝜃𝐿𝑓 = 2 10𝜋 = 20𝜋 𝑟𝑎𝑑, and the load torque as felt at
the motor shaft is 𝑇𝑑 = 𝑇𝐿 /𝑁 = 0.1/2 = 0.05 𝑁𝑚
The motor’s energy consumption per cycle
2 2(4 × 10−6)(20𝜋)2
𝐸=
+ (0.05)20.6 = 22𝐽/𝑐𝑦𝑐𝑙𝑒
(0.3)2
0.2(0.4)2
The power consumption is 22/𝑡𝑓 = 37𝐽/𝑠 = 37𝑊
The maximum speed for the trapezoidal profile
𝜔𝑚𝑎𝑥 = 𝜃𝑓 /𝑡2 = 50𝜋 𝑟𝑎𝑑/𝑠 = 1500 𝑟𝑝𝑚
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Determine the motor and amplifier requirements

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6.100

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Electrical and Electromechanical Systems

§5.Analysis of Motor Performance
The rms torque
𝑇𝑟𝑚𝑠 =

2(4 × 10−6 )(20𝜋)2
+ (0.05)2 = 1.28𝑁𝑚
0.6(0.2)(0.4)2

The maximum required torque
2 × 10−3 (20𝜋)
𝑇𝑚𝑎𝑥 =
+ 0.05 = 1.57 + 0.05 = 1.62𝑁𝑚
0.2(0.4)
Note that the load torque contributes little to 𝑇𝑚𝑎𝑥 . Most of
the required torque is needed to accelerate the inertia
The amplifier requirements
1.62
𝑖𝑚𝑎𝑥 =
5.4𝐴
0.3
1.28

𝑖𝑟𝑚𝑠 =
= 4.27𝐴
0.3
𝑣𝑚𝑎𝑥 = 2 5.4 + 0.3 50𝜋 = 10.8 + 47.1 = 57.9𝑉
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§6.Sensors and Electroacoustic Devices
1.Tachometer
Equipment used to measure linear and rotational velocity
Consider the circuit equation for
an armature-controlled motor
𝑑𝑖𝑎
𝑣𝑎 − 𝑖𝑎 𝑅𝑎 − 𝐿𝑎
− 𝐾𝑏 𝜔 = 0
𝑑𝑡
With the tachometer
• there is no applied voltage 𝑣𝑎 = 0
• at steady-state 𝑑𝑖𝑎 /𝑑𝑡 = 0
⟹ −𝑖𝑎 𝑅𝑎 − 𝐾𝑏 𝜔 = 0
Denote the voltage 𝑖𝑎 𝑅𝑎 across the resistor by 𝑣𝑡
𝑣𝑡 = 𝐾𝑏 𝜔
Measure the voltage 𝑣𝑡 ⟹ Determine the velocity 𝜔

§6.Sensors and Electroacoustic Devices
2.Accelerometer
Apply the Newton’s law
𝑘
𝑘
𝑚𝑥 = −𝑐 𝑥 − 𝑧 − 𝑥 − 𝑧 − 𝑥 − 𝑧
2
2
or 𝑚𝑦 + 𝑐𝑦 + 𝑘𝑦 = −𝑚𝑧
The transfer function between the input 𝑧
and the output 𝑦
𝑌(𝑠)
−𝑚𝑠 2
=
𝑍(𝑠) 𝑚𝑠 2 + 𝑐𝑠 + 𝑘
The device can be used
- as a vibrometer to measure the amplitude of a sinusoidal
displacement 𝑧 = 𝐴𝑠𝑖𝑛𝜔𝑡, or
- as an accelerometer to measure the amplitude of the
acceleration 𝑧 = −𝐴𝜔2 sin𝜔𝑡

HCM City Univ. of Technology, Faculty of Mechanical Engineering

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Nguyen Tan Tien

Nguyen Tan Tien

17


10/27/2013

System Dynamics

6.103

Electrical and Electromechanical Systems

§6.Sensors and Electroacoustic Devices
3.Strain Gauge Accelerometers

System Dynamics

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Nguyen Tan Tien

Electrical and Electromechanical Systems

§6.Sensors and Electroacoustic Devices
- Ex.6.6.1
An Electromagnetic Speaker
Develop a model of the electromagnetic
speaker shown in the figure, and obtain the
transfer function relating the diaphragm
displacement 𝑥 to the applied voltage 𝑣
Solution

Consider the simplified model
𝑚: the mass of the diaphragm and the coil
𝑘: spring constant 𝑘
𝑐: damping constant
𝑓: the magnetic force, 𝑓 = 𝑛𝐵𝐿𝑖 = 𝐾𝑓𝑖
𝐾𝑓 ≡ 𝑛𝐵𝐿

𝑛: the number of turns in the coil
HCM City Univ. of Technology, Faculty of Mechanical Engineering

System Dynamics

6.107

6.104

Electrical and Electromechanical Systems

§6.Sensors and Electroacoustic Devices
5.Electro Acoustic Devices
- Speakers and microphones are common
examples of a class of electromechanical
devices called electroacoustic
• Speaker
electrical energy ⟹ mechanical energy
• Microphone
mechanical energy ⟹ electrical energy
- A radio amplifier produces a current in a coil
that is attached to a diaphragm in the cone.
This causes the coil and diaphragm to move relative to the

permanent magnet. The motion of the diaphragm produces air
pressure waves, which is sound

4.Piezoelectric Devices

HCM City Univ. of Technology, Faculty of Mechanical Engineering

System Dynamics

Nguyen Tan Tien

Electrical and Electromechanical Systems

HCM City Univ. of Technology, Faculty of Mechanical Engineering

System Dynamics

6.106

Nguyen Tan Tien

Electrical and Electromechanical Systems

§6.Sensors and Electroacoustic Devices
From Newton’s law
𝑑2𝑦
𝑑𝑥
𝑚 2 = −𝑐
− 𝑘𝑥 + 𝐾𝑓 𝑖
𝑑𝑡

𝑑𝑡
Consider the electrical subsystem
𝐿: the coil’s inductance
𝑅: the coil’s resistance
The coil experiences a back emf, 𝐾𝑏 𝑥 ,
because it is a current conductor moving
in a magnetic field
From Kirchhoff’s voltage law
𝑑𝑖
𝑑𝑥
𝑣 = 𝐿 + 𝑅𝑖 + 𝐾𝑏
𝑑𝑡
𝑑𝑡
𝑣: the voltage signal from the amplifier
HCM City Univ. of Technology, Faculty of Mechanical Engineering

System Dynamics

6.108

Nguyen Tan Tien

Electrical and Electromechanical Systems

§6.Sensors and Electroacoustic Devices
Transforming the equation and solving for 𝑋(𝑠) gives
𝐾𝑓
𝑋 𝑠 =
𝐼(𝑠)
𝑚𝑠 2 + 𝑐𝑠 + 𝑘

Transforming the equation and solving for 𝐼(𝑠) gives
1
𝐼 𝑠 =
[𝑉 𝑠 − 𝐾𝑏 𝑠𝑋(𝑠)]
𝐿𝑠 + 𝑅
Eliminating 𝐼(𝑠) from the above two equations, obtain the
desired transfer function
𝐾𝑓
𝑋(𝑠)
=
𝑉(𝑠) 𝑚𝐿𝑠 3 + 𝑐𝐿 + 𝑚𝑅 𝑠 2 + 𝑘𝐿 + 𝑐𝑅 + 𝐾𝑓 𝐾𝑏 𝑠 + 𝑘𝑅

§7.Matlab Applications
Step Response from Transfer Functions
𝐼𝑎 (𝑠)
𝐼𝑠 + 𝑐
=
𝑉(𝑠) 𝐿𝑎 𝐼𝑠 2 + 𝑅𝑎 𝐼 + 𝑐𝐿𝑎 𝑠 + 𝑐𝑅𝑎 + 𝐾𝑏 𝐾𝑇
Ω(𝑠)
𝐾𝑇
=
𝑉(𝑠) 𝐿𝑎 𝐼𝑠 2 + 𝑅𝑎 𝐼 + 𝑐𝐿𝑎 𝑠 + 𝑐𝑅𝑎 + 𝐾𝑏 𝐾𝑇
The motor parameters
Matlab % Program motor_par.m (Motor parameters in SI units)
global KT Kb La Ra Im cm
KT = 0.05;Kb = KT;
La = 2e-3;Ra = 0.5;
Im = 9e-5;cm = 1e-4

HCM City Univ. of Technology, Faculty of Mechanical Engineering

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Nguyen Tan Tien

Nguyen Tan Tien

18


10/27/2013

System Dynamics

6.109

Electrical and Electromechanical Systems

System Dynamics

6.110

Electrical and Electromechanical Systems

§7.Matlab Applications
LTI models based on the motor transfer functions
Matlab % Program motor_tf.m (TFs for voltage input)
I = Im; c = cm;
current=tf([I,c],[La*I,Ra*I+c*La,c*Ra+Kb*KT]); %current
speed=tf(KT,[La*I,Ra*I+c*La,c*Ra+Kb*KT]); %speed

Computes and plots the step response for an input of 10𝑉
Matlab % Program motor_step.m (Motor step response)
motor_par
motor_tf
[current, tc] = step(current);
[speed, ts] = step(speed);
subplot(2,1,1),plot(tc,10*current),...
xlabel('t (s)'),ylabel('Current (A)')
subplot(2,1,2),plot(ts,10*speed),...
xlabel('t (s)'),ylabel('Speed (rad/s)')

§7.Matlab Applications
Modified Step Response
A more realistic model of a suddenly applied voltage input is
𝑣(𝑡) = 10(1 − 𝑒 −𝑡/𝜏 ), where 𝜏 is no larger than the system’s
dominant time constant
Computes motor response to this modified step input for 𝜏 = 0.01𝑠
Matlab % Program motor_mod (Response with modified step)
motor_par
motor_tf
mod_step
% Program mod_step.m
% Motor simulation with modified step input
t = (0:0.0001:0.07);
v = 10*(1-exp(-t/0.01));
ia = lsim(current,v,t);
plot(t,ia,t,v)

HCM City Univ. of Technology, Faculty of Mechanical Engineering

HCM City Univ. of Technology, Faculty of Mechanical Engineering

System Dynamics

6.111

Nguyen Tan Tien

Electrical and Electromechanical Systems

System Dynamics

6.112

Nguyen Tan Tien

Electrical and Electromechanical Systems

§7.Matlab Applications
Step Response from State Variable Model
𝑑𝑖𝑎 𝑣 − 𝑖𝑎 𝑅𝑎 − 𝐾𝑏 𝜔
=
𝑑𝑡
𝐿𝑎
𝑑𝜔 𝐾𝑇 𝑖𝑎 − 𝑐𝜔 − 𝑇𝑑 𝐾𝑇 𝑖𝑎 − 𝑐𝜔 − 𝑇𝐿 /𝑁
=
=
𝑑𝑡
𝐼
𝐼

𝑁: the gear ratio
The appropriate state and input matrices
𝑣
𝑖
𝒙 = 𝑎 ,𝒖 = 𝑇
𝜔
𝐿
𝑅𝑎
𝐾𝑏
1
−
−
0
𝐿𝑎
𝐿𝑎
𝐿
1 0
0 0
𝑨=
,𝑩 = 𝑎
,𝑪 =
,𝑫 =
𝐾𝑇
𝑐
0 0
0 0
1
−
0 −
𝐼

𝐼
𝑁𝐼

§7.Matlab Applications
The parameter values for the load
Matlab % load_par.m Load parameters
IL = 0;
cL = 0;
N = 1;
TL = 0;
Inertia, damping, the matrices and the state space model
Matlab % motor_mat.m Motor state matrices
I = Im +IL/N^2;
c = cm + cL/N^2;
A = [-Ra/La,-Kb/La;KT/I,-c/I;];
B = [1/La,0;0, -1/(N*I)];
C = [1,0;0,1];
D = [0,0;0,0];
sysmotor = ss(A,B,C,D);

HCM City Univ. of Technology, Faculty of Mechanical Engineering

HCM City Univ. of Technology, Faculty of Mechanical Engineering

System Dynamics

6.113

Nguyen Tan Tien

Electrical and Electromechanical Systems

System Dynamics

6.114

Nguyen Tan Tien

Electrical and Electromechanical Systems

§7.Matlab Applications
Computes and plots the response due to a step voltage of
magnitude 10𝑣, with the load torque 𝑇𝐿 = 0
Matlab % state_step.m (Motor step response with state
model)
motor_par
load_par
motor_mat
[y, t] = step(sysmotor);
subplot(2,1,1),plot(t,10*y(:,1)),...
xlabel('t (s)'),ylabel('Current (A)')
subplot(2,1,2),plot(t,10*y(:,2)),...
xlabel('t (s)'),ylabel('Speed (rad/s)')

§7.Matlab Applications
Trapezoidal Response
The applied voltage is the following trapezoidal function, which
is shown in the figure for the case where 𝑣𝑚𝑎𝑥 = 20𝑉, 𝑡1 =
0.3𝑠, 𝑡2 = 0.9𝑠, 𝑡𝑓 = 1.2𝑠, 𝑡3 = 1.5𝑠
𝑣𝑚𝑎𝑥

𝑡
0 ≤ 𝑡 ≤ 𝑡1
𝑡1
𝑣𝑚𝑎𝑥
𝑡1 < 𝑡 < 𝑡2
𝑣 𝑡 = 𝑣𝑚𝑎𝑥
(𝑡𝑓 − 𝑡) 𝑡2 ≤ 𝑡 ≤ 𝑡𝑓
𝑡1
0
𝑡𝑓 < 𝑡 ≤ 𝑡3

HCM City Univ. of Technology, Faculty of Mechanical Engineering

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Nguyen Tan Tien

Nguyen Tan Tien

19


10/27/2013

System Dynamics

6.115

Electrical and Electromechanical Systems

System Dynamics

6.116

Electrical and Electromechanical Systems

§7.Matlab Applications
Creates the voltage array 𝑣
Matlab % Program trapezoid.m (Trapezoidal voltage profile)
t1 = 0.3; t2 = 0.9; tfinal = 1.2; t3 = 1.5; v_max = 20;
dt = t3/1000; t = (0:dt:t3);
for k = 1:1001
if t(k) <= t1
v(k) = (v_max/t1)*t(k);
elseif t(k) <= t2
v(k) = v_max;
elseif t(k) <= tfinal
v(k) = (v_max/t1)*(tfinal-t(k));
else
v(k) = 0;
end
end

§7.Matlab Applications
Computes the performance measures relating to energy
consumption, maximum current, voltage, and torque required
and rms current and torque
Matlab % Program performance.m
% Computes motor performance measures.
ia = y(:,1);

speed = y(:,2);
E = trapz(t,Ra*ia.^2)+trapz(t,c*speed.^2)
i_max = max(ia)
i_rms = sqrt(trapz(t,ia.^2)/t3)
T_max = KT*i_max
T_rms = KT*i_rms
speed_max = max(speed);
v_max = Ra*i_max+Kb*speed_max

HCM City Univ. of Technology, Faculty of Mechanical Engineering

HCM City Univ. of Technology, Faculty of Mechanical Engineering

System Dynamics

6.117

Nguyen Tan Tien

Electrical and Electromechanical Systems

System Dynamics

6.118

Nguyen Tan Tien

Electrical and Electromechanical Systems

§7.Matlab Applications

Uses performance and trapezoid to compute the response
Matlab % Program trapresp.m (Motor trapezoidal response)
motor_par
load_par
motor_mat
trapezoid
u = [v', TL*
ones(size(v'))];
y = lsim(sysmotor,u,t);
subplot(2,1,1),plot(t,y(:,1)),...
xlabel('t (s)'),ylabel('Current (A)')
subplot(2,1,2),plot(t,y(:,2)),...
xlabel('t (s)'),ylabel('Speed (rad/s)')
performance

§7.Matlab Applications
Nonlinear Damping
Investigate the effects of a nonlinear damping torque 𝑐𝑁 𝜔2 on
the speed and current, for a step input voltage
𝑐𝑁 = 5 × 10−6 𝑁𝑚𝑠 2 /𝑟𝑎𝑑 2
𝜔 = 0 ÷ 2000𝑟𝑝𝑚
A linear approximation with the
coefficient
𝑐 = 10−3 𝑁𝑚𝑠/𝑟𝑎𝑑
The nonlinear motor equation
with 𝑇𝑑 = 0
𝑑𝜔 𝐾𝑇 𝑖𝑎 − 𝑐𝑁 𝜔2
=
𝑑𝑡
𝐼

HCM City Univ. of Technology, Faculty of Mechanical Engineering

HCM City Univ. of Technology, Faculty of Mechanical Engineering

System Dynamics

6.119

Nguyen Tan Tien

Electrical and Electromechanical Systems

§7.Matlab Applications
Creat the nlmotor.m function
Matlab function xdot = nlmotor(t,x)
% nonlinear damping in motor
global KT Kb La Ra I cN v
xdot = [(-Ra*x(1)-Kb*x(2)+v)/La; (KT*x(1)cN*x(2).^2)/I];
Computes response for nonlinear damping
Matlab Program motor.m
clear
global I cN v
motor_par
load_par
cm = 1e-3;
motor_mat
HCM City Univ. of Technology, Faculty of Mechanical Engineering

System Dynamics

6.120

Nguyen Tan Tien

Electrical and Electromechanical Systems

§7.Matlab Applications
cN = 5e-6; v = 10;
[y1, t1]= step(sysmotor);
y1 = v*y1;
speed = (0:2000)*(2*pi)/60;
[t2, y2] = ode23(@nlmotor, [0, 0.05], [0, 0]);
subplot(2,1,1), plot(t2,y2(:,1),'--',t1,y1(:,1)),...
ylabel('i_a(t) (A)'),...
axis([0 0.05 0 20]), gtext('Linear'),gtext('Nonlinear'),
subplot(2,1,2), ...
plot(t2,y2(:,2)*60/(2*pi),'--',t1,y1(:,2)*60/(2*pi)),...
xlabel('t (s)'),axis([0 0.05 0 2000]),...
ylabel('\omega(t) (rpm)'),gtext('Linear'),...
gtext('Nonlinear')

Nguyen Tan Tien

HCM City Univ. of Technology, Faculty of Mechanical Engineering

Nguyen Tan Tien

20

10/27/2013

System Dynamics

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Electrical and Electromechanical Systems

System Dynamics

6.122

Electrical and Electromechanical Systems

§8.Simulink Applications
Simulation with a Pulse Input
Use a single pulse of amplitude 12𝑉 and duration 0.02𝑠 as an
input to the transfer function of a 𝑅𝐶 circuit, 𝑅 = 104 Ω and 𝐶 =
10−6 𝐹
𝑉𝑜 (𝑠)
1
1
=
=
𝑉𝑖 (𝑠) 𝑅𝐶𝑠 + 1 0.01𝑠 + 1
Simulink model

§8.Simulink Applications
Torque Limitation in Motors

Use Simulink to examine the effects of torque limits on the step
response for a motor with the following model
1
Ω 𝑠 =
[𝑇 𝑠 − 𝑇𝑑 (𝑠)]
𝐼𝑚 𝑠 + 𝑐𝑚
1
𝑇 𝑠 = 𝐾𝑇 𝐼𝑎 𝑠 = 𝐾𝑇
𝑉 𝑠
𝐿𝑎 𝑠 + 𝑅𝑎 𝑎
𝑉𝑎 𝑠 = 𝑉 𝑠 − 𝐾𝑏 Ω(𝑠)
Numerical values
𝐾𝑇 = 𝐾𝑏 = 0.05𝑁𝑚/𝐴 𝐿𝑎 = 2 × 10−3 𝐻
𝑅𝑎 = 0.5Ω
𝐼𝑚 = 9 × 10−5 𝑘𝑔𝑚2
𝑐𝑚 = 10−4 𝑁𝑚𝑠/𝐴
Limit motor torque:
±0.4𝑁𝑚
Disturbance torque:
𝑇𝑑 = 0.01𝑁𝑚
Step voltage input:
𝑉𝑎 = 10𝑉

HCM City Univ. of Technology, Faculty of Mechanical Engineering

HCM City Univ. of Technology, Faculty of Mechanical Engineering

System Dynamics

6.123

Nguyen Tan Tien

Nguyen Tan Tien

Electrical and Electromechanical Systems

§8.Simulink Applications
The Simulink model

HCM City Univ. of Technology, Faculty of Mechanical Engineering

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21