University of Misan

College of Engineering CONTROL | Electrical Engineering

Department Hisham Altai

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Definitions:

1- Control Systems: A system is a combination of components that

act together to perform a specific goal.

2- Reference input: It is the actual signal input to the control

system.

3- Controlled variable (output): the quantity that must be

maintained at prescribed value.

4- Disturbance: An unwanted input signal that affects the output

signal.

Open-Loop control system: A system in which the output has no

effect on the input action. In other words, the output is neither

measured nor fed back for comparison with the input. One practical

example is a washing machine.

Figure 1 Open-Loop Control System.

Reference signal

Plant Controlled

variable

(Output)

Disturbance

Input

Summing point

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1- Closed-Loop Control System: A system in which the output has

an effect on the input quantity in a way that can maintain the

desired output value. An example is a room temperature control

system.

Figure 2 Closed-loop control system

Figure 3 Multiloop feedback system with an inner loop and an outer loop.

Closed-Loop versus Open-Loop Control Systems.

Closed-Loop:

The use of feedback makes the system response insensitive to

external disturbances and internal variations in system parameters.

More complicated and more expensive comparing with Open-Loop.

Open-Loop

The open-loop control system is easier to build because system

stability is not a major problem.

Reference Plant Controller

Sensor

Controlled

variable

(Output)

Measured

Error

Signal

Forward Path

Feedback Path

Contro

l

Signal

Desired output

response

Outer loop

Inner loop

Controller 2 Controller 1

Sensor 1

Sensor 2

Actual output

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It is sensitive to external disturbances.

2- Plants: The device, process, or system that need to be controlled.

3- Control unit (dynamic element): the unit that reacts to an actuating

signal to produce a desired output. This unit does the work of

controlling the output and thus may be a power amplifier.

4- Feedback control system: The unit that provides the means for

feeding back the output quantity, or a function of the output, in order

to compare it with the reference input.

5- Actuating signal: The signal that is difference between the

reference input and the feedback signal if actuates the control unit

in order to maintain the output of the desired value.

6- The sensor or measuring element is a device that converts the

output variable into another suitable variable, such as a

displacement, pressure, voltage, etc.

7- The actuator is a power device that produces the input to the plant

according to the control signal so that the output signal will

approach the reference input signal.

8- Automatic Controllers. An automatic controller compares the actual

value of the plant output with the reference input (desired value), determines

the deviation, and produces a control signal that will reduce the deviation to

zero or to a small value.

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Figure 4 Block diagram of an industrial control system, which consists of an automatic controller, an actuator, a plant, and a sensor

The controller detects the actuating error signal, which is usually at

a very low power level, and amplifies it to a sufficiently high level

Examples of control systems

Room temperature control system

The output signal from a temperature sensing device such as a

thermocouple or a resistance thermometer is compared with the desired

temperature. Any difference or error causes the controller to send a

control signal to the gas solenoid valve which produces a linear

movement of the valve stem, thus adjusting the flow of gas to the

burner of the gas fire. The desired temperature is usually obtained from

manual adjustment of a potentiometer.

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Figure 5 Room temperature control system

Figure 6 Block diagram of room temperature control system.

Steady conditions will exist when the actual and desired

temperatures are the same, and the heat input exactly balances the heat

loss through the walls of the building.

The system can operate in two modes:

a) Proportional control: Here the linear movement of the valve

stem is proportional to the error. This provides a continuous

modulation of the heat input to the room producing very precise

temperature control. This is used for applications where

temperature control, of say better than l °C, is required (i.e.

hospital operating theatres, industrial standards rooms, etc.)

where accuracy is more important than cost.

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b) On-off control: Also called thermostatic or bang-bang control,

the gas valve is either fully open or fully closed, i.e. the heater is

either on or off. This form of control produces an oscillation of

about 2 or 3 °C, of the actual temperature about the desired

temperature, but is cheap to implement and is used for low-cost

applications (i.e. domestic heating systems).

Laplace transform

In order to compute the time response of a dynamic system, it is

necessary to solve the differential equations (system mathematical

model) for given inputs. Laplace transform is one of the favored ways

by control engineers to do this. This technique transforms the problem

from the time (or t) domain to the Laplace (or s) domain. The advantage

in doing this is that complex time domain differential equations

become relatively simple s domain algebraic equations. When a

suitable solution is arrived at, it is inverse transformed back to the time

domain.

ℒ[𝑓(𝑡)] = ∫ 𝑓(𝑡)𝑒−𝑠𝑡𝑑𝑡 = 𝐹(𝑠)∞

0

Where 𝑠 is complex variable 𝜎 ± 𝑗𝜔 and is called the Laplace operator.

Example: Find Laplace transform for

1- 𝒇(𝒕) = 𝟏

2- 𝒇(𝒕) = 𝒆−𝒂𝒕

1-

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𝓛[𝒇(𝒕)] = ∫ 𝟏𝒆−𝒔𝒕𝒅𝒕∞

𝟎

= [−𝟏

𝒔(𝒆−𝒔𝒕)]

𝟎

∞

= [−𝟏

𝒔(𝟎 − 𝟏)] =

𝟏

𝒔

2-

𝓛[𝒇(𝒕)] = 𝑭(𝒔) = ∫ 𝒆−𝒂𝒕𝒆−𝒔𝒕𝒅𝒕∞

𝟎

∫ 𝒆−(𝒔+𝒂)𝒕𝒅𝒕∞

𝟎

[−𝟏

𝒔 + 𝒂(𝒆−(𝒔+𝒂)𝒕)]

𝟎

∞

[−𝟏

𝒔 + 𝒂(𝟎 − 𝟏)]

𝟏

𝒔 + 𝒂

Derivatives: the Laplace transform of a time derivative is

𝒅𝒏

𝒅𝒕𝒏𝒇(𝒕) = 𝒔𝒏𝑭(𝑺) − 𝒇(𝟎)𝒔𝒏−𝟏 − 𝒇′(𝟎)𝒔𝒏−𝟐 − ⋯

Where 𝑓(0), 𝑓′(0) are the initial conditions, or the values of

𝑓(𝑡),𝑑

𝑑𝑡𝑓(𝑡) etc. 𝑎𝑡 𝑡 = 0.

Example: Find the Laplace transform of the following differential

equation given:

(a) Initial conditions 𝑥𝑜 = 4,𝑑𝑥0

𝑑𝑡= 3

(b) Zero initial conditions

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𝒅𝟐𝒙𝒐

𝒅𝒕𝟐+ 𝟑

𝒅𝒙𝒐

𝒅𝒕+ 𝟐𝒙𝒐 = 𝟓

Solution

(a) Initial conditions 𝒙𝒐 = 𝟒,𝒅𝒙𝟎

𝒅𝒕= 𝟑

(𝒔𝟐𝑿𝒐(𝒔) − 𝟒𝒔 − 𝟑) + 𝟑(𝒔𝑿𝒐(𝒔) − 𝟒) + 𝟐𝑿𝒐(𝒔) =𝟓

𝒔

𝒔𝟐𝑿𝒐(𝒔) + 𝟑𝒔𝑿𝒐(𝒔) + 𝟐𝑿𝒐(𝒔) =𝟓

𝒔+ 𝟒𝒔 + 𝟑 + 𝟏𝟐

(𝒔𝟐 + 𝟑𝒔 + 𝟐)𝑿𝒐(𝒔) =𝟓 + 𝟒𝒔𝟐 + 𝟏𝟓𝒔

𝒔

𝑿𝒐(𝒔) =𝟒𝒔𝟐 + 𝟏𝟓𝒔 + 𝟓

𝒔(𝒔𝟐 + 𝟑𝒔 + 𝟐)

(b) Zero initial conditions

At 𝒕 = 𝟎, 𝒙𝒐 = 𝟎𝒅𝒙𝒐

𝒅𝒕= 𝟎

𝒔𝟐𝑿𝒐(𝒔) + 𝟑𝒔𝑿𝒐(𝒔) + 𝟐𝑿𝒐(𝒔) =𝟓

𝒔

𝑿𝒐(𝒔) =𝟓

𝒔(𝒔𝟐 + 𝟑𝒔 + 𝟐)

Table 1 Common Laplace transform pairs

Time function 𝒇 ( 𝒕 ) Laplace transform 𝓛[𝒇(𝒕)] = 𝑭(𝒔)

1- Unit impulse 𝟏

2- Unit step 1 𝟏

𝒔

3- Unit ramp 𝒕 𝟏

𝒔𝟐

4- 𝒕𝒏 𝒏!

𝒔𝒏+𝟏

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5- 𝒆−𝒂𝒕 𝟏

𝒔+𝒂

6- 𝟏 − 𝒆−𝒂𝒕 𝟏

𝒔+𝒂

7- 𝐬𝐢𝐧 𝝎𝒕 𝝎

𝒔𝟐+𝝎𝟐

8- 𝐜𝐨𝐬 𝝎𝒕 𝒔

𝒔𝟐+𝝎𝟐

9- 𝒆−𝒂𝒕𝐬𝐢𝐧 𝝎𝒕 𝝎

(𝒔+𝝎)𝟐+𝝎𝟐

10- 𝒆−𝒂𝒕 (𝐜𝐨𝐬 𝝎𝒕 −𝒂

𝝎𝐬𝐢𝐧 𝝎𝒕)

𝒔

(𝒔+𝝎)𝟐+𝝎𝟐

Inverse transform

𝑓(𝑡) = ℒ−1[𝐹(𝑠)] =1

2𝜋𝑗∫ 𝐹(𝑠)𝑒𝑠𝑡𝑑𝑠

𝜎+𝑗𝜔

𝜎−𝑗𝜔

In practice, inverse transformation is most easily achieved by using

partial fractions to break down solutions into standard components,

and then use tables of Laplace transform pairs

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Table 2 Common partial fraction pairs

Example: Find Laplace inverse transform for:

3𝑠+2

𝑠(𝑠+1)

Solution

𝟑𝒔 + 𝟐

𝒔(𝒔 + 𝟏)=

𝑨

𝒔+

𝑩

𝒔 + 𝟏

𝟑𝒔 + 𝟐 = 𝑨(𝒔 + 𝟏) + 𝑩𝒔

𝟑𝒔 + 𝟐 = (𝑨 + 𝑩)𝒔 + 𝑨

𝑨 = 𝟐

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(𝟐 + 𝑩) = 𝟑

𝑩 = 𝟏

𝓛−𝟏 [𝟐

𝒔+

𝟏

𝒔 + 𝟏] = 𝟐 + 𝒆−𝒕

Transfer function

The function of a linear time invariant differential equation system

is defined as the ratio of Laplace transform of the output( response

function) to the Laplace transform of the input(drive function) under

the assumption that all initial conditions are zero.

Where:

𝒙 is the input

𝒚 is the output

𝐓𝐫𝐚𝐧𝐬𝐟𝐞𝐫 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 = 𝑮(𝒔) = |𝓛[𝒐𝒖𝒕𝒑𝒖𝒕]

𝓛[𝒊𝒏𝒑𝒖𝒕]| 𝒛𝒆𝒓𝒐 𝒊𝒏𝒕𝒊𝒂𝒍 𝒄𝒐𝒏𝒅𝒊𝒕𝒊𝒐𝒏𝒔

=𝒀(𝒔)

𝑿(𝒔)=

𝒃𝟎𝒔𝒎 + 𝒃𝟏𝒔𝒎−𝟏 + ⋯ + 𝒃𝒎−𝟏𝒔 + 𝒃𝒎

𝒂𝟎𝒔𝒏 + 𝒂𝟏𝒔𝒏−𝟏 + ⋯ + 𝒃𝒏−𝟏𝒔 + 𝒃𝒏

Block Diagrams. A block diagram of a system is a pictorial

representation of the functions performed by each component and of

the flow of signals.

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Figure 7 Element of a block diagram.

Summing Point: A circle with a cross is the symbol that indicates a

summing operation. The plus or minus sign at each arrowhead indicates

whether that signal is to be added or subtracted.

Figure 8 Summing point

Branch Point: A branch point is a point from which the signal from a

block goes concurrently to other blocks or summing points.

Open-Loop Transfer Function and Feedforward Transfer Function

The ratio of the feedback signal B(s) to the actuating error signal E(s) is

called the open-loop transfer function. That is,

𝑶𝒑𝒆𝒏 − 𝒍𝒐𝒐𝒑 𝒕𝒓𝒂𝒏𝒔𝒇𝒆𝒓 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏 =𝑩(𝒔)

𝑬(𝒔) = 𝑮(𝒔)𝑯(𝒔)

The ratio of the output C(s) to the actuating error signal E(s) is called the

feedforward transfer function, so that

𝑭𝒆𝒆𝒅𝒇𝒐𝒓𝒘𝒂𝒓𝒅 𝒕𝒓𝒂𝒏𝒔𝒇𝒆𝒓 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏 =𝑪(𝒔)

𝑬(𝒔) = 𝑮(𝒔)

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If the feedback transfer function H(s) is unity, then the open-loop transfer

function and the feedforward transfer function are the same.

Closed-Loop Transfer Function. For the system shown above, the

output C(s) and input R(s) are related as follows: since

𝑪(𝒔) = 𝑮(𝒔)𝑬(𝒔)

𝑬(𝒔) = 𝑹(𝒔) − 𝑩(𝒔)

= 𝑹(𝒔) − 𝑯(𝒔)𝑪(𝒔)

Eliminating E(s) from these equations gives

𝑪(𝒔) = 𝑮(𝒔) [𝑹(𝒔) − 𝑯(𝒔)𝑪(𝒔)]

𝑪(𝒔)

𝑹(𝒔)=

𝑮(𝒔)

𝟏 + 𝑯(𝒔)𝑪(𝒔)

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Closed-Loop System Subjected to a Disturbance.

Figure 9 Closed-loop system subjected to disturbance

In examining the effect of the disturbance D(s), we may assume that the

reference input is zero; we may then calculate the response CD(s) to the

disturbance only. This response can be found from

𝑪𝑫(𝒔)

𝑫(𝒔)=

𝑮𝟐(𝒔)

𝟏 + 𝑮𝟏(𝒔)𝑮𝟐(𝒔)𝑯(𝒔)

On the other hand, in considering the response to the reference input

R(s), we may assume that the disturbance is zero. Then the response

CR(s) to the reference input R(s) can be obtained from

𝑪𝑹(𝒔)

𝑹(𝒔)=

𝑮𝟏(𝒔)𝑮𝟐(𝒔)

𝟏 + 𝑮𝟏(𝒔)𝑮𝟐(𝒔)𝑯(𝒔)

𝑪(𝒔) = 𝑪𝑫(𝒔) + 𝑪𝑹(𝒔)

=𝑮𝟐(𝒔)

𝟏 + 𝑮𝟏(𝒔)𝑮𝟐(𝒔)𝑯(𝒔)[𝑮𝟏𝑹(𝒔) + 𝑫(𝒔)]

References

1- Ogata, Katsuhiko, and Yanjuan Yang. "Modern control engineering." (1970): 1.

2- Burns, Roland. Advanced control engineering. Butterworth-Heinemann, 2001.

3- Dorf, Richard C. Modern control systems. Addison-Wesley Longman Publishing

Co., Inc., 1995.