1.1 CONTROL SYSTEMS A control system is a system which controls the output

response based on the input stimuli. Such a system consists

of subsystems and processes (or plants) assembled for the

purpose of obtaining a desired output and performance,

given a specified input.

The advantage of building control systems include:

1. Power amplification

2. Remote control

3. Convenience of input form

4. Compensation for disturbances

1.2 SYSTEM CONFIGURATIONS A generic open loop system shows that there is no feedback that can be used to compensate for disturbances

and errors.

Such a system consists of:

An input transducer: which converts the form of the input to that used by the controller

Controller: that drives a process or plant

Summing junctions: that add input signals together

The input is sometimes called the reference, while the output is associated with the term controlled variable.

The disadvantages of open loop systems, namely sensitivity to disturbances and inability to correct these

disturbances are overcome in a closed loop system.

Input transducer: converts the form of the input to the form used by the controller

Output transducer: measures the output response and converts it into the form used by the

controller

Feedback path: provides a means of giving feedback from the output transducer

Actuating signal: the input signal (which may contain feedback) that is fed into the controller.

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1.3 ANALYSIS & DESIGN OBJECTIVES Analysis is the process by which a system’s performance is determined, while design is the process of creating

or changing a system’s performance.

A control system is dynamic and responds to

transient responses before reaching a steady

state response that generally resembles the

input. Thus, the objectives of system analysis

and design are to:

produce the desired transient

response

reducing steady state error

achieving stability

robust design

TRANSIENT RESPONSE

The transient response is the sum of the natural and forced responses when the natural response is large. If

we design an elevator with a slow transient response, passengers will become impatient, whereas an

excessively rapid response will cause discomfort.

STEADY-STATE RESPONSE

The steady state response is the sum of the natural and forced responses when the natural response is small.

In essence then, we wish the transient response to decay to zero, otherwise we sacrifice the accuracy of the

steady state response – the quantitative error described as steady state error

STABILITY

The natural response describes the way the system dissipates or acquires energy, and the form of this

response is dependent only on the system and not on the input. However, the form and nature of the forced

response is dependent on the input. We note that the total response for a linear system can be expressed as:

For a control system to be useful:

The natural response must decay to zero, leaving only the forced response

It must oscillate

In some systems which do not meet either of these conditions, the natural response grows without bound,

leading to instability. Such instability often leads to the destruction of a physical device.

OTHER CONSIDERATIONS

Obviously, any engineering task involves the consideration of finances and the need for robust designs.

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1.4 DESIGN PROCESS In general, there are 6 steps to designing a control system:

1. Transform requirements into a physical system

2. Draw a functional block diagram

a. A functional block diagram describes the component parts of the system, such as the input

transducer and controller

3. Transform the system into a schematic

4. Develop a mathematical model

a. Using KVL, KCL, Newton’s Law we can develop a simplified mathematical model. Other tools

used may include the use of linear differential equations, Laplace transform, state space

representation and the transfer function

5. Reduce the block diagram

a. Subsystems models are interconnected to form block diagrams of larger systems, where

each block has a mathematical description.

6. Analyse and design

a. The engineer analyses the system to see if the response specifications and performance

requirements are met by simple adjustments of system parameters, if not, additional

hardware may be designed to achieved the desired performance. Test input signals are used

both analytically and during testing, to verify the design.

1.5 ANTENNA AZIMUTH: INTRODUCTION TO POSITION CONTROL SYSTEMS In the Antenna Azimuth problem, we wish to position

the antenna using a potentiometer. The system must

also be able to adjust for disturbances in the

environment. The detailed layout is shown.

The next step is to derive the functional block

diagram:

We note that the purpose of this system is to have the

azimuth angle output of the antenna track the

input angle of the potentiometer .

The functional block diagram shows that the input command is an angular displacement, which in turn is

converted into a proportional voltage. Similarly, the output angular displacement is converted to a voltage by

the potentiometer in the feedback path. The signal and power amplifiers boost the differential signal between

the input and output voltages such that the actuating signal can drive the plant.

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Note that if we increase the gain of the signal amplifier,

the actuating signal may drive the motor too hard, the

transient response is overdamped, and causes the motor

to overshoot the final value and the system is forced to

make corrections. This indicates the possibility of damped

oscillations about the steady state value.

In most systems the steady state error – the difference

between the output and the input after the transients

have disappeared will still be non-zero; for these systems

a simple gain adjustment to regulate the transient response is either not effective or leads to a trade off

between the desired transient response and the desired steady state accuracy. To solve this problem we can

implement a filter as a compensator.

Finally, we seek to transform the system into a schematic.

1.6 INPUT TEST SIGNALS Input Function Description Sketch Use

Impulse

Transient response modeling

Step

Transient response & steady state error

Ramp

Steady state error

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Parabola

Steady state error

Sinusoid

Transient response modeling, steady state error

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2.1 LAPLACE TRANSFORM The Laplace transform is defined as:

Where 𝑠 = 𝜎 + 𝑗𝜔. The inverse Laplace transform is given by:

The following table summarizes some of the most common Laplace pairs:

𝒇(𝒕) 𝑭(𝒔) 𝜹(𝒕) 𝟏 𝒖(𝒕) 𝟏

𝒔

𝒕𝒖(𝒕) 𝟏𝒔𝟐

𝒕𝒏𝒖(𝒕) 𝒏!𝒔𝒏+𝟏

𝒆−𝒂𝒕𝒖(𝒕) 𝟏𝒔 + 𝒂

𝐬𝐢𝐧(𝝎𝒕)𝒖(𝒕) 𝝎𝒔𝟐 + 𝝎𝟐

𝐜𝐨𝐬(𝝎𝒕)𝒖(𝒕) 𝒔𝒔𝟐 + 𝝎𝟐

EXAMPLE 2.1

: Find the Laplace transform of 𝑓(𝑡) = 𝐴𝑒−𝑎𝑡𝑢(𝑡)

𝐹(𝑠) = � 𝐴𝑒−(𝑎+𝑠)𝑡 𝑑𝑡∞

0=

𝐴𝑠 + 𝑎

In many cases, we need not work from first principles to find the Laplace transform of a function. Instead, we can make use of the Laplace theorems to assist us.

ℒ{𝑓(𝑡)} = 𝐹(𝑠) = � 𝑓(𝑡)𝑒−𝑠𝑡 𝑑𝑡∞

0−

ℒ−1{𝐹(𝑠)} =1

2𝜋� 𝐹(𝑠)𝑒𝑠𝑡 𝑑𝑠 = 𝑓(𝑡)𝑢(𝑡)𝜎+𝑗𝜔

𝜎−𝑗𝜔

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2.2 PROPERTIES OF LAPLACE TRANSFORMS Theorem Name Linearity ℒ{𝑘𝑓1(𝑡) + 𝑐𝑓2(𝑡)} = 𝑘𝐹1(𝑠) + 𝑐𝐹2(𝑠) Frequency Shift ℒ{𝑒−𝑎𝑡𝑓(𝑡)} = 𝐹(𝑠 + 𝑎) Time shift ℒ{𝑓(𝑡 − 𝑇)} = 𝑒−𝑠𝑇𝐹(𝑠) Scaling ℒ{𝑓(𝑎𝑡)} =

1𝑎𝐹 �

𝑠𝑎�

Differentiation ℒ �𝑑𝑛𝑓𝑑𝑡𝑛

� = 𝑠𝑛𝐹(𝑠) − 𝑠𝑛−1𝑓(0−) … .−𝑠𝑓(0−)

Integration ℒ �� 𝑓(𝜏)𝑑𝜏

1

0−1� =

𝐹(𝑠)𝑠

Final value1 𝑓(∞) = lim𝑠→0

𝑠𝐹(𝑠) Initial value2 𝑓(0+) = lim

𝑠→∞𝑠𝐹(𝑠)

We can make use of these Laplace transform theorems and known Laplace transform pairs since we can:

• Convert a linear differential equation into the Laplace domain • Use partial fractions to reduce the equation in the Laplace domain into known Laplace transform such

that we can obtain the inverse Laplace transform

EXAMPLE 2.2

Conversion into the Laplace domain yields:

𝑠2𝑌(𝑠) + 12𝑠𝑌(𝑠) + 32𝑌(𝑠) =32𝑠

𝑌(𝑠)[𝑠2 + 12𝑠 + 32] =32𝑠

𝑌(𝑠) =32

𝑠(𝑠2 + 12𝑠 + 32)

: Find the Laplace transform 𝑌(𝑠) given that all initial conditions are zero for:

𝑑2𝑦𝑑𝑡2

+ 12𝑑𝑦𝑑𝑡

+ 32𝑦 = 32𝑢(𝑡)

EXAMPLE 2.3

1 For this theorem to work, all roots of the denominator of 𝐹(𝑠) must have negative real parts, and no more than one can be at the origin.

: Find the Laplace transform, 𝑋(𝑠) if the initial conditions are 𝑥(0) = 1, �̇� = −1 for:

𝑑2𝑥𝑑𝑡2

+ 2𝑑𝑥𝑑𝑡

+ 3𝑥 = 𝑟(𝑡)

𝑠2𝑋(𝑠) − 𝑠(−1) − 1 + 2𝑠𝑋(𝑠) − 2 + 3𝑋(𝑠) = 𝑅(𝑠)

(𝑠2 + 2𝑠 + 3)𝑋(𝑠) = 𝑅(𝑠) + 3 − 𝑠

2 For this theorem to work, 𝑓(𝑡) must be continuous or have a step discontinuity at 𝑡 = 0

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Find 𝑦(𝑡) if the initial conditions are zero for:

𝑑2𝑦𝑑𝑡2

+ 12𝑑𝑦𝑑𝑡

+ 32𝑦 = 32𝑢(𝑡)

𝑌(𝑠) =32

𝑠(𝑠2 + 12𝑠 + 32)=

32𝑠(𝑠 + 4)(𝑠 + 8) =

𝐴𝑠

+𝐵

𝑠 + 4+

𝐶𝑆 + 8

𝐴 =32

(0 + 4)(0 + 8) = 1

𝐵 =32

(−4)(−4 + 8) = −2

𝐶 =32

(−8)(−8 + 4) = 1

∴ 𝑌(𝑠) =1𝑠−

2𝑠 + 4

+1

𝑠 + 8

EXAMPLE 2.4

Using the Laplace transform table then:

𝑦(𝑡) = 𝑢(𝑡) − 2𝑒−4𝑡𝑢(𝑡) + 𝑒−8𝑡𝑢(𝑡)

Find the inverse Laplace transform of 3𝑠(𝑠2+2𝑠+5)

using partial fractions with complex roots.

3𝑠(𝑠2 + 2𝑠 + 5) =

𝐴𝑠

+𝐵𝑠 + 𝐶

𝑠2 + 2𝑠 + 5

3 =35

(𝑠2 + 2𝑠 + 5) + 𝐵𝑠2 + 𝐶𝑠

= �35

+ 𝐵� 𝑠2 + �65

+ 𝐶� 𝑠 + 3

𝐴 =35

,𝐵 = −35

, 𝐶 = −65

∴ 3

5𝑠−

35 𝑠 + 6

5𝑠2 + 2𝑠 + 5

=3

5𝑠−

35 𝑠 + 6

5(𝑠 + 1)2 + 22

=3

5𝑠−

35�

𝑠 + 1(𝑠 + 1)2 + 22

+12

2(𝑠 + 1)2 + 22

�

ℒ−1 �3

5𝑠−

35�

𝑠 + 1(𝑠 + 1)2 + 22

+12

2(𝑠 + 1)2 + 22

�� =35−

35𝑒−𝑡 �𝑐𝑜𝑠2𝑡 +

12𝑠𝑖𝑛2𝑡�

EXAMPLE 2.5

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2.3 TRANSFER FUNCTION The transfer function shows the relationship between the input and output of a system. Suppose we have an n-th order linear, time invariant differential equation:

𝑎𝑛𝑑𝑛𝑐(𝑡)𝑑𝑡𝑛

+ 𝑎𝑛−1 𝑑𝑛−1𝑐(𝑡)𝑑𝑡𝑛−1

+ ⋯+ 𝑎0𝑐(𝑡) = 𝑏𝑚𝑑𝑚𝑟(𝑡)𝑑𝑡𝑚

+ 𝑏𝑚−1 𝑑𝑚−1𝑟(𝑡)

𝑑𝑚+ ⋯+ 𝑏0𝑟(𝑡)

Taking the Laplace transform and rearranging for 𝐶(𝑠)/𝑅(𝑠) yields the transfer function:

Note that when we took the Laplace transform, we evaluated it with zero initial conditions.

Knowing the transfer function now, allows us to draw the block diagram of a subsystem as shown.

Find the transfer function 𝐶(𝑠)/𝑅(𝑠) for:

𝑑3𝑐𝑑𝑡3

+ 3𝑑2𝑐𝑑𝑡2

+ 7𝑑𝑐𝑑𝑡

+ 5𝑐 =𝑑2𝑟𝑑𝑡2

+ 4𝑑𝑟𝑑𝑡

+ 3𝑟

EXAMPLE 2.6

𝑠3𝐶(𝑠) + 3𝑠2𝐶(𝑠) + 7𝑠𝐶(𝑠) + 5𝐶(𝑠) = +𝑠2𝑅(𝑠) + 4𝑠𝑅(𝑠) + 3𝑅(𝑠)

𝐶(𝑠)𝑅(𝑠) =

𝑠2 + 4𝑠 + 3𝑠3 + 3𝑠2 + 7𝑠 + 5

Find the differential equation corresponding to the transfer function 𝐺(𝑠) = 2𝑠+1𝑠2+6𝑠+2

EXAMPLE 2.7

𝑠2𝐶(𝑠) + 6𝑠𝐶(𝑠) + 2𝐶(𝑠) = 2𝑠𝑅(𝑠) + 𝑅(𝑠)

𝑑2𝑐𝑑𝑡2

+ 6𝑑𝑐𝑑𝑡

+ 2𝑐 = 2𝑑𝑟𝑑𝑡

+ 𝑟

𝐺(𝑠) =𝑏𝑚𝑠𝑚 + 𝑏𝑚−1𝑠𝑚−1 + ⋯+ 𝑏0𝑎𝑛𝑠𝑛 + 𝑎𝑛−1𝑠𝑛−1 + ⋯+ 𝑎0

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Find the ramp response for a system whose transfer function is 𝐺(𝑠) = 𝑠(𝑠+4)(𝑠+8)

EXAMPLE 2.8

The Laplace transform of a ramp response is 𝑅(𝑠) = 1𝑠2

𝐶(𝑠)𝑅(𝑠) =

𝑠(𝑠 + 4)(𝑠 + 8)

𝐶(𝑠) =1

𝑠(𝑠 + 4)(𝑠 + 8)

=1

32𝑠−

116(𝑠 + 4) −

132(𝑠 + 8)

𝑐(𝑡) =1

32−

116

𝑒−4𝑡 −1

32𝑒−8𝑡

2.4 ELECTRICAL NETWORK TRANSFER FUNCTIONS In this section we apply the transfer function to the mathematical modeling of electric circuits. We present 3 of the most basic passive linear components used in electric circuit modeling.

Component Voltage-Current Current-voltage Voltage-charge Impedance 𝒁(𝒔) = 𝑽(𝒔)

𝑰(𝒔)

Admittance: 𝒀(𝒔) =

𝟏𝒁(𝒔)

Capacitor 𝑣𝑐 =

1𝐶� 𝑖𝑐(𝑡) 𝑑𝑡𝑡

0 𝑖𝑐(𝑡) = 𝐶

𝑑𝑣(𝑡)𝑑𝑡

𝑣(𝑡) =𝑞(𝑡)𝐶

1𝐶𝑠

𝐶𝑠

Resistor 𝑣(𝑡) = 𝑅𝑖(𝑡) 𝑖(𝑡) =

𝑣(𝑡)𝑅

𝑣(𝑡) = 𝑅𝑑𝑞(𝑡)𝑑𝑡

𝑅 1

𝑅

Inductor 𝑣𝐿(𝑡) = 𝐿

𝑑𝑖(𝑡)𝑑𝑡

𝑖(𝑡) =1𝐿� 𝑣𝐿(𝑡)𝑑𝑡𝑡

0 𝑣(𝑡) = 𝐿

𝑑2𝑞(𝑡)𝑑𝑡2

𝐿𝑠 1

𝐿𝑠

MESH ANALYSIS Mesh analysis is the electric circuit analysis technique which aims at using voltage loops (KVL). When we are given a circuit with capacitors and inductors, the resulting voltage loops will be differential equations. We can transform the circuit into the Laplace domain by:

• Redrawing the original network showing all the time variables as Laplace transform 𝑉(𝑠), 𝐼(𝑠),𝑉𝐶(𝑠). • Replace the component values with their impedance values.

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Find the transfer function relating the capacitor voltage 𝑉𝐶(𝑠) to the input voltage 𝑉(𝑠).

Example 2.9

Taking a KVL loop:

−𝑉(𝑠) + 𝐿𝑠𝐼(𝑠) + 𝑅𝐼(𝑠) +𝐼(𝑠)𝐶𝑠

= 0

But:

𝑉𝑐(𝑠) =𝐼(𝑠)𝐶𝑠

→ 𝐼(𝑠) = 𝐶𝑠𝑉𝑐(𝑠)

∴ 𝐿𝐶𝑠2𝑉𝑐(𝑠) + 𝑅𝐶𝑠𝑉𝑐(𝑠) + 𝑉𝐶(𝑠) = 𝑉(𝑠)

𝑉𝐶(𝑠)𝑉(𝑠) =

1𝐿𝐶𝑠2 + 𝑅𝐶𝑠 + 1

In more complex circuits, mesh analysis involves multiple loops and nodes. Again we:

1. Replace passive element values with their impedances 2. Replace all sources and time variables with their Laplace transform 3. Assume a transform current and a current direction in each mesh 4. Write Kirchoff’s voltage law around each mesh 5. Solve the simultaneous equations for the output

Find the transfer function 𝐼2(𝑠)/𝑉(𝑠)

Example 2.10

For each mesh loop:

𝑅1𝐼1(𝑠) + 𝐿𝑠�𝐼1(𝑠) − 𝐼2(𝑠)� = 𝑉(𝑠)

𝐿𝑠�𝐼2(𝑠) − 𝐼1(𝑠)� + 𝑅2𝐼2(𝑠) +𝐼2(𝑠)𝐶𝑠

= 0

We then rearrange the equations:

(𝑅1 + 𝐿𝑠)𝐼1(𝑠) − 𝐿𝑠𝐼2(𝑠) = 𝑉(𝑠)

−𝐿𝑠𝐼1(𝑠) + �𝐿𝑠 + 𝑅2 +1𝐶𝑠� 𝐼2(𝑠) = 0

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By the Cramer rule:

𝐼2(𝑠) =�𝑅1 + 𝐿𝑠 𝑉(𝑠)−𝐿𝑠 0

�

Δ=𝐿𝑠𝑉(𝑠)Δ

Where:

Δ = �𝑅1 + 𝐿𝑠 −𝐿𝑠

−𝐿𝑠 𝐿𝑠 + 𝑅2 +1𝐶𝑠�

∴𝐼2(𝑠)𝑉(𝑠) =

𝐿𝐶𝑠2

(𝑅1 + 𝑅2)𝐿𝐶𝑠2 + (𝑅1𝑅2𝐶 + 𝐿)𝑠 + 𝑅1

NODAL ANALYSIS Like in the time domain, nodal analysis makes use of Kirchoff’s current law and sums the currents at the nodes. This also applies in the Laplace domain. The general procedure is to:

1. Replace passive element values with their admittances 2. Replace all sources and time variables with their Laplace transform 3. Replace transformed voltage sources with transformed current sources 4. Write Kirchoff’s current law at each node 5. Solve the simultaneous equations for the output 6. Form the transfer function

Find the transfer function 𝑉𝐶(𝑠)/𝑉(𝑠)of the above circuit using nodal analysis.

𝑉𝐿(𝑠) − 𝑉(𝑠)𝑅1

+𝑉𝐿(𝑠)𝐿𝑠

+�𝑉𝐿(𝑠) − 𝑉𝐶(𝑠)�

𝑅2= 0

𝐶𝑠𝑉𝑐(𝑠) +𝑉𝑐(𝑠) − 𝑉𝐿(𝑠)

𝑅2= 0

Example 2.11

Rearranging we get:

�𝐺1 + 𝐺2 +1𝐿𝑠� 𝑉𝐿(𝑠) − 𝐺2𝑉𝐶(𝑠) = 𝑉(𝑠)𝐺1

−𝐺2�𝑉𝐿(𝑠)� + (𝐺2 + 𝐶𝑠)𝑉𝑐(𝑠) = 0

Again using Cramer’s rule, solving yields:

𝑉𝐶(𝑠)𝑉(𝑠) =

�𝐺1𝐺2𝑠𝐶 �

(𝐺1 + 𝐺2)𝑠2 + 𝐺1𝐺2𝐿 + 𝐶𝐿𝐶 + 𝐺2

𝐿𝐶

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2.4 OPAMPS Recall that an ideal operational amplifier has the following characteristics:

• Differential input • Infinite input impedance • Zero output impedance • Infinite gain amplification

Like in normal circuit analysis in the time domain, we can use the same techniques to find the transfer function of operational amplifiers in the Laplace domain.

Find the transfer function 𝑉𝑜(𝑠)/𝑉𝑖(𝑠) for the circuit shown.

Example 2.12

The components in the feedback loop give an impedance of:

𝑍1 =1𝑠𝐶2

+ 𝑅2 =10 × 106

𝑠+ 220 × 103

The components at the input give an impedance of:

1𝑍2

=1𝑅1

+ 𝑠𝐶1 =1

360000+ (5.6 × 10−6)𝑠

∴ 𝑍2 =1

1360000 + 5.6 × 10−6𝑠

=360 × 103

1 + 2.016𝑠

Nodal analysis at the inverting terminal yields:

𝑉𝑜(𝑠)𝑍1

= −𝑉𝑖(𝑠)𝑍2

𝑉𝑜(𝑠)𝑉𝑖(𝑠) = −�

10 × 106

𝑠+ 220 × 103� �

1 + 2.016𝑠360 × 103

�

= −10 × 106 + 20.16 × 106𝑠 + 220 × 103 + 443520𝑠2

360 × 103𝑠

= −1.232𝑠2 + 49.95𝑠 + 22.55

𝑠

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The previous circuit is an inverting configuration because the transfer function is negative. In the non-inverting configuration, we find that the transfer function is positive.

Find the transfer function 𝑉𝑜(𝑠)/𝑉𝑖(𝑠) for the circuit given.

Example 2.13

For the series connection the impedance is:

𝑍1(𝑠) =1𝐶1𝑠

+ 𝑅1

And for the parallel components:

𝑍2(𝑠) =

𝑅2𝐶2𝑠

𝑅2 + 1𝑠𝐶2

=𝑅2

𝑅2𝐶2𝑠 + 1

Then by nodal analysis:

𝑉𝑜(𝑠) − 𝑉𝑖(𝑠)𝑍2(𝑠) =

𝑉𝑖(𝑠)𝑍1(𝑠)

𝑉𝑜(𝑠) = 𝑉𝑖(𝑠) �1 +𝑍2(𝑠)𝑍1(𝑠)�

𝑉𝑜(𝑠)𝑉𝑖(𝑠) = 1 +

𝑅2𝐶1𝑠(𝑅2𝐶2𝑠 + 1)(1 + 𝑅1𝐶1𝑠)

= 1 +𝑅2𝐶1𝑠

𝑅2𝐶2𝑠 + 𝑅2𝑅1𝐶1𝐶2𝑠2 + 𝑅1𝐶1𝑠 + 1

=𝐶2𝐶1𝑅2𝑅1𝑠2 + (𝐶2𝑅2 + 𝐶1𝑅2 + 𝐶1𝑅2)𝑠 + 1

𝐶2𝐶1𝑅2𝑅1𝑠2 + (𝐶2𝑅2 + 𝐶1𝑅1)𝑠 + 1

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2.5 TRANSLATIONAL MECHANICAL SYSTEM TRANSFER FUNCTIONS So far we have looked at modeling electric circuits in the Laplace domain. We can do the same thing for mechanical systems – both translational and rotational.

There are many parallels between mechanical and electrical networks – mechanical systems also have three passive, linear components, notably, the spring and the mass, which are energy storage elements and the viscous damper which dissipates energy. The energy storage elements are similar to the capacitor and inductor in electric networks, while the viscous damper parallels the passive resistor element.

To begin with we define:

𝑘 = 𝑆𝑝𝑟𝑖𝑛𝑔 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝑓𝑣 = 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑣𝑖𝑠𝑐𝑜𝑢𝑠 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛

𝑀 = 𝑀𝑎𝑠𝑠

The following table presents a force-velocity, force-displacement and impedance relationship for each component.

Component Force-Velocity Force-Displacement Impedance 𝒁 = 𝑭𝑿

𝐹 = 𝑘� 𝑣(𝑡) 𝑑𝑡𝜏

0

𝐹 = 𝑘𝑥(𝑡)

𝑘

𝐹 = 𝑓𝑣𝑣(𝑡)

𝐹 = 𝑓𝑣𝑑𝑥(𝑡)𝑑𝑡

𝑓𝑣𝑠

𝐹 = 𝑀𝑑𝑣(𝑡)𝑑𝑡

𝐹 = 𝑀𝑑2𝑥(𝑡)𝑑𝑡2

𝑀𝑠2

Many mechanical systems are similar to multiple loop and multiple node electrical networks, where more than one simultaneous differential equation is required to describe the system.

Linear independence implies that a point of motion in a system can still move if all other points of motion are held still. The number of linearly independent motions is commonly referred to as the degrees of freedom.

In mechanical systems, the number of equations of motion required is equal to the number of linearly independent motions

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In our analysis of mechanical systems, we make use of linear independence – for each free body diagram we superimpose the forces acting on the body when every other point of motion is held still and let that point move, and with the reverse situation whereby all other points are in motion while the point of interest is held still. We can do this because of Newton’s Laws of motion.

Find the transfer function 𝑋2(𝑠)/𝐹(𝑠).

Example 2.14

We first focus on 𝑀1’s motion, holding every other motion still:

𝐹(𝑠) = 𝑓𝑣1𝑠𝑋1(𝑠) + 𝐾1𝑋1(𝑠) + 𝐾2𝑋1(𝑠) + 𝑓𝑣3𝑠𝑋1(𝑠) + 𝑀1𝑠2𝑋1(𝑠)

We then hold 𝑀1 still and let all other bodies move and analyse the forces on 𝑀1:

−𝑓𝑣3𝑠𝑋2(𝑠) − 𝐾2𝑋2(𝑠)

Superimposing, we get:

[𝑀1𝑠2 + (𝑓𝑣1 + 𝑓𝑣2)𝑠 + (𝐾1 + 𝐾2)]𝑋1(𝑠) − [𝑓𝑣3𝑠 + 𝐾2]𝑋2(𝑠) = 𝐹(𝑠)

For mass 𝑀2, we do the same thing, and this yields:

−[𝑓𝑣3𝑠 + 𝐾2]𝑋1(𝑠) + [𝑀2𝑠2 + (𝑓𝑣2 + 𝑓𝑣3)𝑠 + 𝐾2 + 𝐾3]𝑋2(𝑠) = 0

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By Cramer’s Rule:

𝑋2(𝑠) =�𝑀1𝑠2 + (𝑓𝑣1 + 𝑓𝑣2)𝑠 + (𝐾1 + 𝐾2) 𝐹(𝑠)

−(𝑓𝑣3𝑠 + 𝐾2) 0 �

Δ=𝐹(𝑠)(𝑓𝑣3𝑠 + 𝐾2)

Δ

𝑋2(𝑠)𝐹(𝑠) =

𝑓𝑣3𝑠 + 𝐾2Δ

Where:

Δ = �𝑀1𝑠2 + (𝑓𝑣1 + 𝑓𝑣2)𝑠 + (𝐾1 + 𝐾2) 𝑓𝑣3𝑠 + 𝐾2−(𝑓𝑣3𝑠 + 𝐾2) [𝑀2𝑠2 + (𝑓𝑣2 + 𝑓𝑣3)𝑠 + 𝐾2 + 𝐾3]�

Write, but do not solve, the equations of motion for the mechanical network shown.

Example 2.15

There are three degrees of freedom, since each of the three masses can be moved independently while the others are held still. We focus on mass 𝑀1 first:

[𝑀1𝑠2 + (𝑓𝑣1 + 𝑓𝑣3)𝑠 + (𝐾1 + 𝐾2)]𝑋1(𝑠) − 𝐾2𝑋2(𝑠) − 𝑓𝑣3𝑠𝑋3(𝑠) = 0

Then 𝑀2:

−𝐾2𝑋1(𝑠) + [𝑀2𝑠2 + (𝑓𝑣2 + 𝑓𝑣4)𝑠 + 𝑘2]𝑋2(𝑠) − 𝑓𝑣4𝑠𝑋3(𝑠) = 𝐹(𝑠)

Then 𝑀3:

−𝑓𝑣3𝑋1(𝑠) − 𝑓𝑣4𝑋2(𝑠) + (𝑀3𝑠2 + (𝑓𝑣3 + 𝑓𝑣4)𝑠)𝑋3(𝑠) = 0

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2.6 ROTATIONAL MECHANICAL SYSTEMS In rotational mechanical systems, we deal with torque and angular displacement rather than forces and translational displacement. We define:

𝐾 = 𝑠𝑝𝑟𝑖𝑛𝑔 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝐷 = 𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑣𝑖𝑠𝑐𝑜𝑢𝑠 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛

𝐽 = 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑖𝑛𝑒𝑟𝑡𝑖𝑎

Component Torque-angular velocity Torque-angular displacement

Impedance 𝒁𝑴(𝒔) =𝑻(𝒔)𝜽(𝒔)

𝐾� 𝜔(𝑡)𝑑𝜏𝑡

0

𝑇(𝑡) = 𝐾𝜃(𝑡) 𝐾

𝑇(𝑡) = 𝐷𝜔(𝑡) 𝑇(𝑡) = 𝐷

𝑑𝜃(𝑡)𝑑𝑡

𝐷𝑠

𝑇(𝑡) = 𝐽𝑑𝜔(𝑡)𝑑𝑡

𝑇(𝑡) = 𝐽𝑑2𝜃(𝑡)𝑑𝑡2

𝐽𝑠2

The concept of degrees of freedom carries over to rotational systems, except that we test a point of motion by rotating it while holding still all other points of motion. The number of points of motion that can be rotated while all others are held still equals the number of equations of motion required to describe the system.

Again, in our analysis, we make use of the superposition of torques – we rotate a body while holding all other points still and place on its free body diagram all torques due to the body’s own motion. Then, holding the body still, we rotate adjacent points of motion one at a time and add the torques due to the adjacent motion to the free body diagram. Note that the torques will sum up to zero due to an equilibrium system. This process is repeated for each point of motion.

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Find the transfer function 𝜃2(𝑠)/𝑇(𝑠), for the rotational system shown. The rod is supported by bearings at either end and is undergoing torsion. A torque is applied at the left, and the displacement is measured at the right.

Example 2.16

There are 2 degrees of freedom, sine each inertia can be rotated while the other is held still. There are thus 2 equations of motion:

For the free body diagram of 𝐽1, we show the torques due to its own motion, due to the motion of the other bodies and then superimpose:

(𝐽1𝑠2 + 𝐷1𝑠 + 𝐾)𝜃1(𝑠) − 𝐾𝜃2(𝑠) = 𝑇(𝑠)

Then for the free body diagram of 𝐽2:

−𝐾𝜃1(𝑠) + (𝐽2𝑠2 + 𝐷2𝑠 + 𝐾)𝜃2(𝑠) = 0

Using Cramer’s rule:

𝜃2(𝑠) =�𝐽1𝑠

2 + 𝐷1𝑠 + 𝐾 𝑇(𝑠)−𝐾 0

�

Δ=𝐾𝑇(𝑠)𝛥

Where:

Δ = �𝐽1𝑠2 + 𝐷1𝑠 + 𝐾 −𝐾−𝐾 𝐽2𝑠2 + 𝐷2𝑠 + 𝐾

�

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Find the transfer function 𝜃2(𝑠)/𝑇(𝑠) for the system shown.

Example 2.17

There are 2 degrees of freedom – the cylinder and the point between the 2 springs.

For the cylinder:

(𝑠2 + 𝑠 + 1)𝜃1(𝑠) − (𝑠 + 1)𝜃2(𝑠) = 𝑇(𝑠)

For the point 𝑥2:

−(𝑠 + 1)𝜃1(𝑠) + (2𝑠 + 2)𝜃2(𝑠) = 0

𝜃2(𝑠) =�𝑠2 + 𝑠 + 1 𝑇(𝑠)−(𝑠 + 1) 0 �

�s2 + s + 1 −(s + 1)−(s + 1) 2s + 2 �

=(s + 1)T(s)

2s3 + 3s2 + 2s + 1

𝜃2(𝑠)𝑇(𝑠) =

12𝑠2 + 𝑠 + 1

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2.7 TRANSFER FUNCTIONS FOR SYSTEMS WITH GEARS Gears provide mechanical advantage to rotational systems. This because gears allow you to match the drive system and the load; a tradeoff between speed and torque. In this section we ignore the effect of backlash – the situation in which the drive gear rotates through a small angle before making contact with the meshed gear.

The ratio of the angular displacement of the gears is inversely proportional to the ratio of the number of teeth. In an idealized case, the energy transferred between gears is conserved. Thus:

Derive a relationship for the transfer function 𝑇1(𝑠)/𝜃1(𝑠)

Example 2.18

We note that 𝑇1 can be reflected to the output by multiplying by 𝑁2/𝑁1 to give us 𝑇1′. This results in:

(𝐽𝑠2 + 𝐷𝑠 + 𝐾)𝜃2(𝑠) = 𝑇1(𝑠) 𝑁2𝑁1

Now: 𝜃2 = 𝑁1𝑁2

𝜃1

(𝐽𝑠2 + 𝐷𝑠 + 𝐾) �𝑁1𝑁2� 𝜃1(𝑠) = 𝑇1(𝑠) �

𝑁2𝑁1�

�𝐽 �𝑁1𝑁2�2

𝑠2 + 𝐷 �𝑁1𝑁2�2

𝑠 + 𝐾 �𝑁1𝑁2�2

� 𝜃1(𝑠) = 𝑇1(𝑠)

We generalize the result by showing that:

Rotational mechanical impedances can be reflected through gear trains by multiply the mechanical impedance by the ratio:

Where the impedance to be reflected is attached to the source shaft and is being reflected to the destination shaft.

𝑇1𝑇2

=𝜃2𝜃1

=𝑟1𝑟2

=𝑁1𝑁2

�𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡𝑒𝑒𝑡ℎ 𝑜𝑓 𝑔𝑒𝑎𝑟 𝑜𝑛 𝑑𝑒𝑠𝑡𝑖𝑛𝑎𝑡𝑖𝑜𝑛 𝑠ℎ𝑎𝑓𝑡𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡𝑒𝑒𝑡ℎ 𝑜𝑓 𝑔𝑒𝑎𝑟 𝑜𝑛 𝑠𝑜𝑢𝑟𝑐𝑒 𝑠ℎ𝑎𝑓𝑡

�2

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Find 𝜃2(𝑠)/𝑇1(𝑠)

Example 2.19

When we reflect the input shaft to the output shaft, we clearly see that there is only 1 degree of freedom:

Reflecting the inertia and the viscous damper, we get:

𝐽1′ = 𝐽1 �𝑁2𝑁1�2

, 𝐷1′ = 𝐷1 �𝑁2𝑁1�2

Thus we have:

𝐽𝑒 = 𝐽1 �𝑁2𝑁1�2

+ 𝐽2

𝐷𝑒 = 𝐷1 �𝑁2𝑁1�2

+ 𝐷2

And analysis of the resulting free-body shows that:

(𝐽𝑒𝑠2 + 𝐷𝑒𝑠 + 𝐾2)𝜃2(𝑠) = 𝑇1(𝑠) 𝑁2𝑁1

𝜃2(𝑠)𝑇1(𝑠) =

𝑁2𝑁1

𝐽𝑒𝑠2 + 𝐷𝑒𝑠 + 𝐾2

With a gear train, we can continually reflect the angular displacement by multiplying through by the ratio:

𝜃2𝜃1

=𝑁1𝑁2

Thus:

𝜃4 =𝑁1𝑁3𝑁5𝑁2𝑁4𝑁6

𝜃1

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Find 𝜃2(𝑠)/𝑇(𝑠), for the rotational mechanical system with gears.

Example 2.20

We shall reflect everything to the input shaft noting that the gears are not lossless since we need to take into account their inertia and viscous friction.

Reflecting the third shaft to the middle shaft we get:

�𝑁3𝑁4�2

(𝐽4 + 𝐽5)

And the middle shaft (including the load of the bottom shaft):

�𝑁1𝑁2�2

�𝐽2 + 𝐽3 + �𝑁3𝑁4�2

(𝐽4 + 𝐽5)� + �𝑁1𝑁2�2

𝐷2

Thus, the total inertia in the system is:

𝐽𝑒 = 𝐽1 + (𝐽2 + 𝐽3) �𝑁1𝑁2�2

+ (𝐽4 + 𝐽5) �𝑁1𝑁3𝑁2𝑁4

�2

And total viscous friction:

𝐷𝑒 = 𝐷1 + �𝑁1𝑁2�2

𝐷2

The transfer function becomes:

(𝐽𝑒𝑠2 + 𝐷𝑒𝑠)𝜃1(𝑠) = 𝑇1(𝑠)

𝜃1(𝑠)𝑇1(𝑠) =

1𝐽𝑒𝑠2 + 𝐷𝑒𝑠

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ELECTROMECHANICAL SYSTEMS We focus on deriving the transfer function 𝜃𝑚(𝑠)

𝐸𝑎(𝑠) of an electric motor from first principles. A motor is an

electromechanical component that yields a displacement output for a voltage input.

The basic motor schematic is shown in which an external magnetic field causes the interaction of the magnetic field generated through the armature coils carrying a current of 𝑖𝑎(𝑡). The armature feels a force of 𝐵𝑙𝑖𝑎(𝑡) when it is perpendicular to the external magnetic field, and the resulting torque turns the rotor:

𝑇𝑚(𝑠) = 𝐾𝑡𝐼𝑎(𝑠)

Since the armature cuts through magnetic flux, Lenz’s Law and Faraday’s Law suggest the existence of a voltage and its associated magnetic which opposes the change which caused it - the back emf, which is given by:

𝑣𝑏(𝑡) = 𝐾𝑏𝑑𝜃𝑚(𝑡)𝑑𝑡

Where 𝐾𝑏 is a back emf constant and 𝑑𝜃𝑚(𝑡)𝑑𝑡

is the angular velocity of the motor. The Laplace transform of it

yields:

𝑉𝑏(𝑠) = 𝐾𝑏𝑠𝜃𝑚(𝑠)

The resulting loop equation thus gives us:

𝑅𝑎𝐼𝑎(𝑠) + 𝐿𝑎𝑠𝐼𝑎(𝑠) + 𝑉𝑏(𝑠) = 𝐸𝑎(𝑠)

(𝑅𝑎 + 𝐿𝑎𝑠)𝑇𝑚(𝑠)𝐾𝑡

+ 𝐾𝑏𝑠𝜃𝑚(𝑠) = 𝐸𝑎(𝑠)

The typical mechanical loading on a motor is shown in the figure, with 𝐽𝑚 being equivalent inertia at the armature and includes both the armature inertia and the load inertia reflected to the armature. 𝐷𝑚 is the equivalent viscous damping at the armature.

𝑇𝑚(𝑠) = (𝐽𝑚𝑠2 + 𝐷𝑚𝑠)𝜃𝑚(𝑠)

∴ (𝑅𝑎 + 𝐿𝑎𝑠)(𝐽𝑚𝑠2 + 𝐷𝑚𝑠)𝜃𝑚(𝑠)

𝐾𝑡+ 𝐾𝑏𝑠𝜃𝑚(𝑠) = 𝐸𝑎(𝑠)

Since the armature inductance is small, then we approximate the equation to be:

𝜃𝑚(𝑠) �𝑅𝑎𝐾𝑡

(𝐽𝑚𝑠2 + 𝐷𝑚𝑠) + 𝐾𝑏� = 𝐸𝑎(𝑠)

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Where:

𝐽𝑚 = 𝐽𝑎 + 𝐽𝐿 �𝑁1𝑁2�2

, 𝐷𝑚 = 𝐷𝑎 + 𝐷𝐿 �𝑁1𝑁2�2

To find the electrical constants in the transfer function, we perform a dynamometer test which measures the torque and speed of a motor under the condition of a constant applied voltage. From this we can measure 𝑇𝑠𝑡𝑎𝑙𝑙 and 𝜔𝑛𝑜−𝑙𝑜𝑎𝑑 and hence calculate the electrical constants. Going from the previous equation:

(𝑅𝑎 + 𝐿𝑎𝑠)𝑇𝑚(𝑠)𝐾𝑡

+ 𝐾𝑏𝑠𝜃𝑚(𝑠) = 𝐸𝑎(𝑠)

And letting 𝐿𝑎 = 0, then we can rearrange for 𝑇𝑚, the constant torque which results when a DC voltage 𝑒𝑎 is applied causing a constant angular velocity 𝜔𝑚.

𝑅𝑎𝐾𝑡𝑇𝑚(𝑠) + 𝐾𝑏𝑠𝜃𝑚 = 𝐸𝑎(𝑠)

Taking the inverse Laplace transform:

From the dynamometer test, we see a linear relationship:

𝑇𝑠𝑡𝑎𝑙𝑙 =𝐾𝑡𝑅𝑎

𝑒𝑎

𝜔𝑛𝑜−𝑙𝑜𝑎𝑑 =𝑒𝑎𝐾𝑏

Upon rearranging we find that:

𝜃𝑚(𝑠)𝐸𝑎(𝑠) =

𝐾𝑡𝑅𝑎𝐽𝑚

𝑠�𝑠 + 1𝐽𝑚 �𝐷𝑚 + 𝐾𝑡𝐾𝑏

𝑅𝑎��

𝑇𝑚 = −𝐾𝑏𝐾𝑡𝑅𝑎

𝜔𝑚 +𝐾𝑡𝑅𝑎

𝑒𝑎

𝐾𝑡𝑅𝑎

=𝑇𝑠𝑡𝑎𝑙𝑙𝑒𝑎

𝐾𝑏 =𝑒𝑎

𝜔𝑛𝑜−𝑙𝑜𝑎𝑑

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Given the system and torque-speed curve shown, find the transfer function 𝜃𝐿(𝑠)/𝐸𝑎(𝑠)

Example 2.21

The general equation for the motor’s transfer function is given by:

𝜃𝑚(𝑠)𝐸𝑎(𝑠) =

� 𝐾𝑡𝑅𝑎𝐽𝑚

�

𝑠 �𝑠 + 1𝐽𝑚 �𝐷𝑚 + 𝐾𝑡𝐾𝑏

𝑅𝑎��

We find that:

𝐽𝑚 = 𝐽𝑎 + 𝐽𝐿 �𝑁1𝑁2�2

= 5 + 700 �1

10�2

= 12

𝐷𝑚 = 𝐷𝑎 + 𝐷𝐿 �𝑁1𝑁2�2

= 2 + 800 �1

10�2

= 10

𝐾𝑡𝑅𝑎

=𝑇𝑠𝑡𝑎𝑙𝑙𝑒𝑎

=500100

= 5

𝐾𝐵 =𝑒𝑎

𝜔𝑛𝑜−𝑙𝑜𝑎𝑑=

10050

= 2

Thus:

𝜃𝑚(𝑠)𝐸𝑎(𝑠) =

� 512�

𝑠 �𝑠 + 112 �10 + 5(2)��

=0.417

𝑠(𝑠 + 1.667)

Now since we have:

𝜃𝑚(𝑠)𝜃𝐿(𝑠) =

𝑁2𝑁1

→ 𝜃𝑀(𝑠) =𝑁2𝑁1

𝜃𝐿(𝑠)

∴𝜃𝐿(𝑠)𝐸𝑎(𝑠) =

𝑁1𝑁2

𝜃𝑚(𝑠)𝐸𝑎(𝑠) =

0.0417𝑠(𝑠 + 1.667)

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LINEARISATION A linear system possess two characteristics:

• Superposition • Homogeneity

Which can be summarized as:

𝐻{𝑎𝑥1 + 𝑏𝑥2} = 𝑎𝐻{𝑥1} + 𝑏𝐻{𝑥2}

Where H is the transformation (the system’s transfer function). If a system does not possess these qualities, then the system is said to be nonlinear.

An electronic amplifier is linear over a specific range, but exhibits the nonlinearity called saturation at high input voltage. A motor that does not respond at very low input voltages due to frictional forces exhibits a nonlinearity called dead zone. Gears that do not fit tightly exhibit a nonlinearity called backlash.

In many situations, it is possible to make a linear approximation to a nonlinear system with small variations around the point of interest. When we linearise we:

• Recognise the nonlinear component and write the non linear differential equation • We then linearise it for small signal inputs about the steady state solution when the small signal input

is equal to zero • We linearise the nonlinear differential equation and take the Laplace transform, assuming zero initial

conditions.

Consider a point A on the curve which we want to linearise. Suppose that the differential at point A yields 𝑚𝑎, then using the point gradient formula of a straight line:

𝑓(𝑥) − 𝑓(𝑥0) ≈ 𝑚𝑎(𝑥 − 𝑥0)

Since: 𝛿𝑓(𝑥) ≈ 𝑚𝑎𝛿𝑥

Thus:

𝑓(𝑥) ≈ 𝑓(𝑥0) + 𝑚𝑎𝛿𝑥

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Linearise:

𝑑2𝑥𝑑𝑡2

+ 2𝑑𝑥𝑑𝑡

+ 𝑐𝑜𝑠𝑥 = 0

Example 2.22

About 𝑥 = 𝜋4

If we let 𝑥 = 𝛿𝑥 + 𝜋4 then:

𝑑2 �𝛿𝑥 + 𝜋4�

𝑑𝑡2=𝑑2𝛿𝑥𝑑𝑡2

𝑑 �𝛿𝑥 + 𝜋4�

𝑑𝑡=𝑑𝛿𝑥𝑑𝑡

cos 𝑥 = cos (𝛿𝑥 +𝜋4

)

Since:

𝑓(𝑥) − 𝑓(𝑥0) ≈𝑑𝑓𝑑𝑥| 𝑥=𝑥0

𝛿𝑥

cos �𝛿𝑥 +𝜋4� − cos �

𝜋4� =

𝑑𝑑𝑥|𝑥=𝜋4

cos 𝑥 𝛿𝑥

= − sin �𝜋4� 𝛿𝑥

cos �𝛿𝑥 +𝜋4� = cos �

𝜋4� − sin �

𝜋4� 𝛿𝑥 =

√22−√22𝛿𝑥

∴𝑑2𝛿𝑥𝑑𝑡2

+ 2𝑑𝛿𝑥𝑑𝑡

−√22

𝛿𝑥 = −√22

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3.1 STATE SPACE The state space approach is a unified method for modelling, analysing and designing a wide range of systems. It provides a more power tool for analysis because it can also be used to represent non linear systems, systems with nonzero initial conditions, time varying systems and multiple input – multiple output systems. We first enumerate the steps involved and then present an example to consolidate the process.

1. We select a particular subset of all possible system variables and call the variables in this subset state variables

2. For an nth order system, we write n first order differential equations in terms of the state variables. These simultaneous differential equations are known as state equations

3. If we know the initial condition of all of the state variables at 𝑡0 as well as the system input for 𝑡 ≥ 𝑡𝑜 we can solve the simultaneous differential equations for the state variables for 𝑡 ≥ 𝑡0

4. We algebraically combine the state variables with the system’s input and find all of the other system variables for 𝑡 ≥ 𝑡0. We call this algebraic equation the output equation

5. The representation of the system using the system and output equations is known as the state space representation.

The minimum number of state variables required to describe a system is equal to the order of the differential equation (or independent energy storage elements in the system). If we can define more state variables than the minimum, we must eliminate the state variables which do not form part of a linearly independent set i.e no state variable can be written as a linear combination of the other state variables:

𝑎1𝒗𝟏 + 𝑎2𝒗𝟐 + ⋯+ 𝑎𝑛𝒗𝒏 = 0

Where 𝑣𝑖 is a set of vectors and 𝑎𝑖 are constants. Linear independence only occurs if the only solution to the above equation is trivially 𝑎𝑖 = 0

Find the state space representation of the system.

Example 3.1

Typically, we select the state variables to be the variables with the differentials – in this case the inductor current and the capacitor voltage. A KVL loop yields:

−𝑣(𝑡) + 𝑅𝑖(𝑡) + 𝐿𝑑𝑖(𝑡)𝑑𝑡

+ 𝑣𝑐(𝑡) = 0

Rearranging yields:

𝑑𝑖(𝑡)𝑑𝑡

=1𝐿�−𝑅𝑖(𝑡) − 𝑣𝑐(𝑡) + 𝑣(𝑡)�

Note that this first order differential equation is in terms of the state variables, inductor current and capacitor voltage and has the input of the voltage source.

Since: 𝐶 𝑑𝑣𝑐(𝑡)𝑑𝑡

= 𝑖(𝑡) → 𝑑𝑣𝑐(𝑡)𝑑𝑡

= 1𝐶

𝑖(𝑡)

These two differentials are the state space representation of the system.

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3.2 GENERAL STATE SPACE REPRESENTATION Let us formally define:

• State variable: any variable that responds to an input or initial condition in a system • State variables: the smallest set of linear independent system variables such that the values of the

members of the set at time 𝑡0 along with known forcing functions completely determine the value of all system variables for all 𝑡 ≥ 𝑡0.

• State vector: a vector whose elements are the state variables • State space: the n dimensional space whose axes are the state variables • State equations: A set of n simultaneous, first order differential equations with n variables, where n

variables to be solved are the state variables • Output equation: the algebraic equation that expresses the output variables of a system as a linear

combination of the state variables and the inputs

A state space representation of a system (state equation, output equation) can be written as:

Where:

𝒙 = 𝑠𝑡𝑎𝑡𝑒 𝑣𝑒𝑐𝑡𝑜𝑟, �̇� = 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑡𝑎𝑡𝑒 𝑣𝑒𝑐𝑡𝑜𝑟

𝒚 = 𝑜𝑢𝑡𝑝𝑢𝑡 𝑣𝑒𝑐𝑡𝑜𝑟, 𝒖 = 𝑜𝑢𝑡𝑝𝑢𝑡 𝑣𝑒𝑐𝑡𝑜𝑟

𝑨 = 𝑠𝑦𝑠𝑡𝑒𝑚 𝑚𝑎𝑡𝑟𝑖𝑥, 𝑩 = 𝑖𝑛𝑝𝑢𝑡 𝑚𝑎𝑡𝑟𝑖𝑥

𝑪 = 𝑜𝑢𝑡𝑝𝑢𝑡 𝑚𝑎𝑡𝑟𝑖𝑥, 𝑫 = 𝑓𝑒𝑒𝑑𝑓𝑜𝑟𝑤𝑎𝑟𝑑 𝑚𝑎𝑡𝑟𝑖𝑥

Given the electrical network shown, find the state space representation of the output is the current through the resistor.

Example 3.2

1. Select state variables – check the number of independent storage elements. We select the inductor current and capacitor voltage as the state variables:

𝐶𝑑𝑣𝑐𝑑𝑡

= 𝑖𝑐 , 𝐿𝑑𝑖𝐿𝑑𝑡

= 𝑣𝐿

�̇� = 𝑨𝒙 + 𝑩𝒖

𝒚 = 𝑪𝒙 + 𝑫𝒖

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2. Write 𝑖𝑐, 𝑣𝐿 as a linear combination of the state variables – 𝑣𝑐 , 𝑖𝐿 and the input 𝑣(𝑡).

𝑖𝑐 = 𝑖𝐿 − 𝑖𝑅 But since R is parallel to C:

𝑖𝑅 =𝑣𝑐𝑅

∴𝑑𝑣𝑐𝑑𝑡

=1𝐶�𝑖𝐿 −

𝑣𝑐𝑅�

Now, the KVL loop around the voltage source-inductor and capacitor yields: −𝑣(𝑡) + 𝑣𝐿 + 𝑣𝑐 = 0

∴𝑑𝑖𝐿𝑑𝑡

=1𝐿

(𝑣(𝑡) − 𝑣𝑐)

3. The output equation is:

𝑖𝑅 =𝑣𝑐𝑅

4. The final state space representation is:

�𝑣�̇�𝚤�̇�� = �

−1𝑅𝐶

1𝐶

−1𝐿

0� �𝑣𝑐𝑖𝐿 � + �

01𝐿� 𝑣(𝑡)

𝑖𝑅 = �1𝑅

0� �𝑣𝑐𝑖𝐿 �

When we have a dependent source, the approach is much the same – we just find the relationship to describe the dependent sources either in a KVL loop or a KCL node.

Find the state and output equations for the electrical network shown if the output vector is 𝒚 = �𝑣𝑅2𝑖𝑅2 �

Example 3.3

We choose out state variables to be the inductor current and capacitor voltage so:

𝐿𝑑𝑖𝐿𝑑𝑡

= 𝑣𝐿 , 𝐶𝑑𝑣𝑐𝑑𝑡

= 𝑖𝐶

Now let us find expressions for the right hand side of those differentials in terms of 𝑖𝐿 . 𝑣𝑐 and the input 𝑖(𝑡). KVL loop 1 shows:

𝑣𝐿 = 𝑣𝑐 + 𝑖𝑅2𝑅 (1)

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But at node 2 we have:

𝑖𝑐(𝑡) + 4𝑣𝐿 = 𝑖𝑅2

∴ 𝑖𝑛 (1) 𝑣𝐿 =1

1 − 4𝑅2 (𝑣𝑐 + 𝑖𝑐𝑅2)

We now wish to find 𝑖𝑐 in terms of the state variables at node 1:

𝑖𝑐 = 𝑖(𝑡) − 𝑖𝑅1 − 𝑖𝐿

= 𝑖(𝑡) −𝑣𝐿𝑅1

− 𝑖𝐿 (2)

We now have a simultaneous equation:

𝑣𝐿 =1

1 − 4𝑅2 (𝑣𝑐 + 𝑖𝑐𝑅2) (1)

𝑖𝑐 = 𝑖(𝑡) −𝑣𝐿𝑅1

− 𝑖𝐿 (2)

Thus:

(1 − 4𝑅2)𝑣𝐿 − 𝑖𝑐𝑅2 = 𝑣𝑐

−𝑣𝐿𝑅1

− 𝑖𝑐 = 𝑖𝐿 − 𝑖(𝑡)

By the Cramer Rule:

𝑣𝐿 =�

𝑣𝑐 −𝑅2𝑖𝐿 − 𝑖(𝑡) −1 �

Δ=�−𝑣𝑐 + 𝑅2𝑖𝐿 − 𝑅2𝑖(𝑡)�

Δ

𝑖𝑐 =

�1 − 4𝑅2 𝑣𝑐− 1𝑅1

𝑖𝐿 − 𝑖(𝑡)�

Δ=�(1 − 4𝑅2)𝑖𝐿 − (1 − 4𝑅2)𝑖(𝑡) + 𝑣𝑐

𝑅1�

𝛥

Where:

Δ = −(1 − 4𝑅2) −𝑅2𝑅1

Thus:

�𝚤�̇�𝑣�̇�� =

⎣⎢⎢⎡

𝑅2Δ𝐿

−1Δ𝐿

1 − 4𝑅2Δ𝐶

1𝑅1ΔC⎦

⎥⎥⎤�𝑖𝐿𝑣𝑐

� + �−𝑅2𝐿Δ

−1 − 4𝑅2Δ𝐶

� 𝑖(𝑡)

𝒚 = �𝑣𝑅2𝑖𝑅2 � =

⎣⎢⎢⎡𝑅2Δ

−�1 +1Δ�

1Δ

1 − 4𝑅1Δ𝑅1 ⎦

⎥⎥⎤�𝑖𝐿𝑣𝑐

� + �−𝑅2Δ

−1Δ

� 𝑖(𝑡)

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3.3 STATE EQUATIONS FOR MECHANICAL SYSTEMS It is convenient when working with mechanical systems to obtain the state equations directly from the equations of motion rather than from the energy storage elements. Our state variables in mechanical systems will always be the position and velocity of each point of linearly independent motion.

Find the state space representation of the translational mechanical system

Example 3.4

1. Find the Laplace transformed equations of motion for the two linearly independent motions: (𝑀1𝑠2 + 𝐷𝑠 + 𝐾)𝑋1 − 𝐾𝑋2 = 0 −𝐾𝑋1 + (𝑀2𝑠2 + 𝐾)𝑋2 = 𝐹

2. Take the inverse Laplace transform:

𝑀1𝑑2𝑥1𝑑𝑡2

+ 𝐷𝑑𝑥1𝑑𝑡

+ 𝐾𝑥1 − 𝐾𝑥2 = 0

−𝐾𝑥1 + 𝑀2𝑑2𝑥2𝑑𝑡2

+ 𝐾𝑥2 = 𝑓(𝑡)

3. Assign our state variables to be 𝑥1, 𝑣1, 𝑥2𝑣2 and rewrite the above equations:

𝑣1̇ =1𝑀1

(−𝐷𝑣1 − 𝐾𝑥1 + 𝐾𝑥2)

𝑣2̇ =1𝑀2

(𝑓(𝑡) + 𝐾𝑥1 − 𝐾𝑥2)

4. Write the matrix:

�

𝑥1̇𝑣1̇𝑥2̇𝑣2̇

� =

⎣⎢⎢⎢⎢⎡

0 1 0 0

−𝐾𝑀1

−𝐷𝑀1

𝐾𝑀1

0

0 0 0 1𝐾𝑀2

0 −𝐾𝑀2

0⎦⎥⎥⎥⎥⎤

�

𝑥1𝑣1𝑥2𝑣2

� +

⎣⎢⎢⎢⎡

0001𝑀2⎦⎥⎥⎥⎤

𝑓(𝑡)

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3.4 CONVERTING A TRANSFER FUNCTION TO STATE SPACE One advantage of the state space representation over the use of transfer functions is that physical systems can be simulated on the computer.

To perform a conversion from a transfer function to a state space, we select phase variables, where each subsequent state variable is defined to be the derivate of the previous state variable. Consider the differential equation:

𝑑𝑛𝑦𝑑𝑡𝑛

+ 𝑎𝑛−1𝑑𝑛−1𝑦𝑑𝑡𝑛−1

+ ⋯+ 𝑎1𝑑𝑦𝑑𝑡

+ 𝑎0𝑦 = 𝑏0𝑢

We choose the phase variables to be:

𝑥1 = 𝑦

𝑥2 =𝑑𝑦𝑑𝑡

⋮

𝑥𝑛 =𝑑𝑛−1𝑦𝑑𝑡𝑛−1

On differentiating both sides, we get:

𝑥1̇ = 𝑥2

𝑥2̇ = 𝑥3

⋮

𝑥�̇� = −𝑎0𝑥1 − 𝑎1𝑥2 …− 𝑎𝑛−1𝑥𝑛 + 𝑏0𝑢

In Vector matrix form:

⎣⎢⎢⎢⎢⎡𝑥1̇𝑥2̇𝑥3̇⋮

𝑥𝑛−1𝑥𝑛 ̇̇⎦⎥⎥⎥⎥⎤

=

⎣⎢⎢⎢⎢⎡

0 1 0 0 0 … 00 0 1 0 0 … 00 0 0 1 0 … 0…0 0 0 0 0 … 1−𝑎0 −𝑎2 −𝑎3 −𝑎4 −𝑎5 … −𝑎𝑛−1 ⎦

⎥⎥⎥⎥⎤

⎣⎢⎢⎢⎢⎡𝑥1𝑥2𝑥3⋮

𝑥𝑛−1𝑥𝑛 ⎦

⎥⎥⎥⎥⎤

+

⎣⎢⎢⎢⎢⎡

000⋮0𝑏0⎦⎥⎥⎥⎥⎤

𝑢

Yielding an output equation of:

𝑦 = [1 0 0 0 0 0]

⎣⎢⎢⎢⎢⎡𝑥1𝑥2𝑥3⋮

𝑥𝑛−1𝑥𝑛 ⎦

⎥⎥⎥⎥⎤

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Find the state space representation and draw the equivalent block diagram.

Example 3.5

We first transform the transfer function to a differential equation:

𝑑3𝑐𝑑𝑡

+ 9𝑑2𝑐𝑑𝑡

+ 26𝑑𝑐𝑑𝑡

+ 24𝑐 = 24𝑟

Let

𝑥1 = 𝑐

𝑥2 = �̇�

𝑥3 = �̈�

Thus:

𝑥1̇ = 𝑥2, 𝑥2̇ = 𝑥3, 𝑥3̇ = −24𝑥1 − 26𝑥2 − 9𝑥3 + 24𝑟

In Matrix-vector form:

�𝑥1̇𝑥2̇𝑥3̇� = �

0 1 00 0 1

−24 −26 −9� �𝑥1𝑥2𝑥3� + �

00

24� 𝑟

𝑦 = [1 0 0] �𝑥1𝑥2𝑥3�

To draw the functional block diagram, we note that it is an order 3 system, so we need 3 integrators in series with the input being 𝑥3̇ which is the output from the summer as shown.

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In the above example, the numerator was a constant term. If the transfer function had been a polynomial in s, we would handle the numerator and denominator separately by splitting the system up into two subsystems. Consider the system shown and how it is split up into two systems.

The first subsystem is the same as the previous example. The second transfer function would yield:

𝑦(𝑡) = 𝑏2𝑑2𝑥1𝑑𝑡2

+ 𝑏1𝑑𝑥1𝑑𝑡

+ 𝑏0𝑥1

After taking the inverse Laplace transform with zero initial conditions. In the second subsystem, the output is an output equation, so in conformance to what is shown above, the result would be writing the terms in reverse order:

𝑦(𝑡) = 𝑏0𝑥1 + 𝑏1𝑥2 + 𝑏2𝑥3

Find the state space representation for the system shown.

Example 3.6

1. We split the system up as shown:

2. With subsystem 1 we have:

𝑑3𝑥1𝑑𝑡3

+ 9𝑑2𝑥1𝑑𝑡2

+ 26𝑑𝑥1𝑑𝑡

+ 24𝑥1 = 𝑟

Let 𝑥1 = 𝑥1, 𝑥2 = 𝑥1̇. 𝑥3 = 𝑥1̈, then: 𝑥3̇ = −24𝑥1 − 26𝑥2 − 9𝑥3 + 𝑟

The resulting matrix-vector representation is:

�𝑥1̇𝑥2̇𝑥3̇� = �

0 1 00 0 1

−24 −26 −9� �𝑥1𝑥2𝑥3� + �

001� 𝑟

3. For subsystem 2:

𝑐 =𝑑2𝑥1𝑑𝑡2

+ 7𝑑𝑥1𝑑𝑡

+ 2𝑥1

Let 𝑥1 = 𝑥1, 𝑥2 = 𝑥1̇, 𝑥3 = 𝑥1̈

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∴ 𝑐(𝑡) = 𝑥3 + 7𝑥2 + 2𝑥1 Which yields:

𝑦 = [2 7 1] �𝑥1𝑥2𝑥3�

4. The block diagram for subsystem 1 is similar to that of the previous example. The second subsystem collects the derivates from subsystem 1 and sums them as shown.

3.5 CONVERTING FROM STATE SPACE TO A TRANSFER FUNCTION Given the state and output equations:

�̇� = 𝑨𝒙 + 𝑩𝒖

𝒚 = 𝑪𝒙 + 𝑫𝒖

We take the Laplace transforms assuming zero initial conditions:

𝒔𝑿(𝒔) = 𝑨𝑿(𝒔) + 𝑩𝑼(𝒔) (1)

𝒀(𝒔) = 𝑪𝑿(𝒔) + 𝑫𝑼(𝒔) (2)

Now in (1):

𝑿(𝒔) = (𝑠𝑰 − 𝑨)−1𝑩𝑼(𝒔)

And in (2):

𝒀(𝒔) = 𝑪(𝑠𝑰 − 𝑨)−1𝑩𝑼(𝒔) + 𝑫𝑼(𝒔)

Which yields the transfer function:

𝒀(𝒔)𝑼(𝒔) = 𝑪(𝑠𝑰 − 𝑨)−1𝑩 + 𝑫

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Find the transfer function given:

�̇� = �0 1 00 0 1−1 −2 −3

� 𝒙 + �1000� 𝒖

𝒚 = [1 0 0]𝒙

Example 3.7

This means:

𝐴 = �0 1 00 0 1−1 −2 −3

� 𝐵 = �1000� 𝐶 = [1 0 0] 𝐷 = 0

Now:

𝑠𝐼 − 𝐴 = �𝑠 −1 00 𝑠 −11 2 𝑠 + 3

�

So:

(𝑠𝐼 − 𝐴)−1 =𝑎𝑑𝑗(𝑠𝐼 − 𝐴)1

det(𝑠𝐼 − 𝐴)

= ⎣⎢⎢⎢⎢⎡ �𝑠 −12 𝑠 + 3� �0 −1

1 𝑠 + 3� �0 𝑠1 2�

�−1 02 𝑠 + 3� �𝑠 0

1 𝑠 + 3� �𝑠 −11 2 �

�−1 0𝑠 −1� �𝑠 0

0 −1� �𝑠 −10 𝑠 �⎦

⎥⎥⎥⎥⎤𝑇

𝑠3 + 3𝑠2 + 2𝑠 + 1

=

�(𝑠2 + 3𝑠 − 2) −1 −𝑠

(𝑠 + 3) 𝑠(𝑠 + 3) −(2𝑠 + 1)1 𝑠 𝑠2

�

𝑇

𝑠3 + 3𝑠2 + 2𝑠 + 1

=

�(𝑠2 + 3𝑠 + 2) 𝑠 + 3 1

−1 𝑠(𝑠 + 3) 𝑠−𝑠 −(2𝑠 + 1) 𝑠2

�

𝑠3 + 3𝑠2 + 2𝑠 + 1

Now:

𝑇(𝑠) = 𝐶(𝑠𝐼 − 𝐴)−1𝐵 + 𝐷

1 This is an adjoint matrix.

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= [1 0 0]

⎝

⎜⎜⎜⎛�

(𝑠2 + 3𝑠 + 2) 𝑠 + 3 1−1 𝑠(𝑠 + 3) 𝑠−𝑠 −(2𝑠 + 1) 𝑠2

�

𝑠3 + 3𝑠2 + 2𝑠 + 1

⎠

⎟⎟⎟⎞�1000�

=

[(𝑠2 + 3𝑠 + 2) 𝑠 + 3 1] �1000�

𝑠3 + 3𝑠2 + 2𝑠 + 1

=10(𝑠2 + 3𝑠 + 2)𝑠3 + 3𝑠2 + 2𝑠 + 1

LINEARISATION State space representation can also be used to represent systems with nonlinearities. For small perturbations about an equilibrium point, state space representations can be linearised.

Represent the system in stage space. Assume the mass is evenly distributed with the centre of mass at L/2. Then linearise the state equations about the pendulum’s equilibrium point –the vertical position with zero angular velocity.

Example 3.8

The sum of torques is given by:

𝐽𝑑2𝜃𝑑𝑡2

+𝑀𝑔𝐿

2𝑠𝑖𝑛𝜃 = 𝑇

Recall that the torque experience due to gravity is given by: 𝜏 = 𝐹𝑑 = 𝑀𝑔 �𝐿2� 𝑠𝑖𝑛𝜃

We select the state variables 𝑥1, 𝑥2 as phase variables and let 𝑥1 = 𝜃. 𝑥2 = 𝑑𝜃𝑑𝑡

𝑥1̇ = 𝑥2, 𝑥2̇ = −𝑀𝑔𝐿

2𝐽𝑠𝑖𝑛𝑥1 +

𝑇𝐽

In order to be able to convert these state equations to transfer functions, we must linearise them. We let 𝑥0 = 0, 𝑥1 = 0, 𝑥2 = 0 be perturbed:

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𝑥1 = 𝛿𝑥1, 𝑥2 = 𝛿𝑥2

Now we wish to linearise 𝑠𝑖𝑛𝑥1:

𝑓(𝑥1) − 𝑓(𝑥0) =𝑑𝑑𝑥

𝑓(𝑥)|𝑥=𝑥1𝛿𝑥1

𝑠𝑖𝑛𝑥1 − 𝑠𝑖𝑛0 = 𝑑𝑑𝑥1|𝑥1=0

𝑠𝑖𝑛𝑥1 𝛿𝑥1

𝑠𝑖𝑛𝑥1 = 𝛿𝑥1

𝛿𝑥1̇ = 𝛿𝑥2

𝛿𝑥2̇ = −𝑀𝑔𝐿

2𝐽𝛿𝑥1 +

𝑇𝐽

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4.1 POLES, ZEROS AND SYSTEM RESPONSE Recall that the output response of a system is the sum of two responses:

So far we have neglected to focus on the poles and zeros of a transfer functions even though they serve of

great importance in analysing a system’s response. What we find is that:

The poles cause the transfer function to become infinite, and forms part of the natural response

The poles of the input function generates the form of the forced response

The zeros and the poles generate the amplitudes for both the forced and natural responses.

Consider a unit step response into the system as shown:

4.2 FIRST ORDER SYSTEMS A first order system without zeros can be described as:

If we find the output and take the inverse transform we get:

The most significant (and only) parameter in this system is a, the exponential frequency, which determines the

transient response. When

:

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TIME CONSTANT

The time constant is defined as:

The time it takes for the step response to rise to 63% of the final value.

RISE TIME

For a first order system this would be:

SETTLING TIME

The rise time is defined as the time for the wave to go from 0.1 to 0.9 of its final value.

The time it takes for the response to reach and stay within 2% of its final value:

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Example 4.1

The curve shown shows the response of a system subjected to a step input. Find the transfer function given

that it is a first order system:

Note that the output equation is given by:

The final value is 0.72, which means:

When

, the output has fallen 63%:

From the curve, it takes a=0.13s to reach 0.45, thus:

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4.3 SECOND ORDER SYSTEMS Whereas varying a first order system’s parameter simply changes the speed of the response, changes in the

parameters of a second order system can change the form of the response. The forms of the response vary

widely, and can be summarised as either overdamped, underdamped, undamped or critically damped.

System Pole-zero plot Criteria Response

Two real poles

Two complex (conjugate) poles

Two completely imaginary poles

Double real roots

Thus:

1. Overdamped response: Two real poles at –

2. Underdamped response: Two complex poles –

3. Undamped response: Two imaginary poles

4. Critically Damped: Two real poles –

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4.4 GENERAL SECOND ORDER SYSTEM The natural frequency, is the frequency of oscillation of the system without damping. The damping ratio

is defined to be:

We can represent a second order transfer function as:

Solving for the poles of the transfer function yields:

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Example 4.2

Find the value of and the type of response expected

a)

b)

c)

4.5 UNDERDAMPED SECOND ORDER SYSTEM In underdamped systems, it is possible to relate transient specifications with the pole locations. Recall that a

general second order system with step input can be written as:

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The inverse Laplace transform is:

Note that the lower the value of , the more oscillatory the response. From our definition of and , we can

define other parameters:

Rise time the time required for the waveform t go from 0.1 to 0.9 of the final value

Peak time : the time to reach the first maximum or peak

Percent overshoot: %OS: the amount that the waveform overshoots the steady state value at the

peak time, expressed as a percentage of the steady-state value

Settling time, : the time required for the transient’s damped oscillations to reach and stay within

of the steady state value.

EVALUATION OF

To evaluate the peak time, we

differentiate and find the

stationary point:

Completing the squares yields:

Which gives us:

Thus, the stationary point is at:

Where

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EVALUATING %OS

We evaluate and :

For a unit step input.

But:

Notice that the percent overshoot is a function only of the damping ratio .

EVALUATING

In order to evaluate the settling time, must stay within 2% of the steady state value. Thus since

And the cosine term will tend towards 1 when the transient response dies down, then we are left with:

Which gives us:

RISE TIME

The rise time cannot be calculated analytically.

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4.6 RELATIONSHIP BETWEEN POLES AND THE SECOND ORDER

UNDERDAMPED SYSTEM PARAMETERS Consider the pole plot shown for an underdamped second order system. What we find is that:

The angle subtended by the pole’s location and the x-axis gives us , the damping ratio.

The distance between the pole and the origin is the natural frequency

The imaginary component, the damped frequency oscillation is given by

The real part, the exponential damping frequency is given by

We thus find:

Example 4.3

Determine what parameters are changing and which are constant, for each of the pole-zero plots.

a) Since is inversely proportional to the imaginary part of the poles, lines of constant imaginary value

(horizontal lines) imply that each pair of poles produce a constant peak time.

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b) Lines of constant real value (vertical lines) show that the settling time is constant since

.

c) Since , radial lines are lines of constant . And since percent overshoot is only a function of

, radial lines are thus lines of constant percent overshoot.

d) We find that the line % yields a

smaller angle than , this means

that the damping factor is greater for

since which gets its

maximum when .

The line also indicates that the real part

of is greater, indicating that the

settling time is shorter.

Since the imaginary part is greater for

, this indicates that the peak time

is shorter.

In summary:

To decrease the settling time, we move the pole to the left.

To decrease the peak time, we move the pole higher.

To decrease the percent overshoot, we make the angle shallower

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5.1 BLOCK DIAGRAMS In control system block diagrams, the main elements are signals, summing junctions, pickoff points and

systems. Using these elements we can represent a system with different topologies. As we shall see, there are

several common topologies.

5.1.1 CASCADE FORM

In cascade form, each subsystem is lined up one after the other, and the resulting output of each stage, is the

product of the input and the transfer function e.g . For the system shown then, the

equivalent transfer function of the system is . This assumption holds as long as the

interconnected subsystems do not load adjacent subsystems.

5.1.2 PARALLEL FORM

Parallel subsystems have a common input and output formed by the algebraic sum of the outputs from all the

subsystems. The equivalent transfer function is thus:

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5.1.3 FEEDBACK FORM

The feedback system forms the basis for control systems engineering given that it forms a close loop system.

We can easily derive the transfer function for negative and positive feedback. Now:

But:

Note that represents the negative feedback system while represents the positive

feedback system. The open loop gain of this system then is given by .

5.2 MOVING BLOCKS TO CREATE FAMILIAR FORMS In some situations, we may need to move blocks to create familiar forms. This may involve moving the blocks

to the left or right of summing junctions and pickoff points. The figures show the equivalent block diagrams

after the movement of the blocks.

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Example 5.1

Reduce the system shown to a single transfer function.

We begin by noting the feedback loop between and . Thus the transfer function block will be:

We also shift the pickup point in the last summer to

The feedforward path consists of and

. This simplifies to

. We also shift to the right of

the second summer so that we can remove one of the summers. We thus replace with and

with

.

Note that parallel blocks, and

. Thus, we replace this block with one block labelled

.

We also simplify the cascaded block with one block labelled

.

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We finally note the feedback loop which has a transfer function of:

This then yields a cascaded system, which simplifies down to:

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5.3 ANALYSIS AND DESIGN OF FEEDBACK SYSTEMS In some situations, systems can reduce down to second

order systems. Consider a situation in which we have a

feedback control system with a open loop gain of

.

The closed loop transfer function is thus:

Where models the amplifier gain, the ratio of the output voltage to the input voltage. As K varies, the poles

move through the three ranges of operations of second order systems – overdamped, critically damped and

underdamped. Note that when the poles are given by:

The poles are real, and the system is overdamped. When

, then the system a double pole, which yields

a critically damped system. Whereas if increases beyond

then the poles are complex:

And the system is underdamped. We find that the peak time decreases and the percent overshoot increases,

while the settling time, which is given by

remains constants.

Example 5.2

Find the peak time, percent overshoot and settling

time for the system shown.

The closed loop transfer function is:

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5.4 SIGNAL FLOW GRAPHS Signal flow graphs consist only of branches, which represent systems and

nodes, which represent signals. This offers an alternative to block diagrams. The

conversion from a block diagram to a signal flow diagram is simple:

1. Draw a node for each signal

2. Then draw branches to represent each transfer function that is associated

with the particular signal

3. Simplify

Example 5.3

Convert the block diagram into a signal flow graph.

1. We first identify all the signals and draw nodes for them:

And the transfer functions:

2. Start at and move through the signals.

This is a unity transfer function multiplied by the input, . Thus we connect a line from to

, labelled 1. also subtracts off , so there is a line labelled -1.

Then is produced by . We thus connect the signals and

and label them appropriately.

We continue for all the signals.

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3. To simplify, we combine branches with unity gain with other transfer functions. For example we

combine and to be – . This reduced the signal flow graph to:

5.4 SIGNAL FLOW GRAPHS OF STATE EQUATIONS In this section we draw signal flow graphs from state equations.

1. Identify the nodes to be the variables and their derivatives

2. Identify the input and output nodes.

3. Interconnect the state variables and their derivatives with the defining of integration,

.

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Example 5.4

Drawn the signal diagrams based on the following state and output equations:

1. We select and their derivatives to be the nodes. The input is r and the output is y.

2. Join to with a line labelled

. Do the same .

3. Starting from which is represented by in the signal flow diagram, we interconnect the signals.

We then proceed with and .

In the next section, the signal flow model will help us visualise the process of determining alternative

representations in state space of the same system. We will see that even though a system can be the same

with respect to its input and output terminals, the state space representations can be many and varied.

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5.5 ALTERNATIVE REPRESENTATIONS IN STATE SPACE So far, we have represented a system in state space with the phase variable form. Here, we see that there can

be many representations that yields the same output for a given input. Such variations in representations

allow us to select a model which allows us to determine a solution easily.

5.5.1 CASCADE FORM

We can represent a system:

As a cascaded system of first order blocks as shown. From this, we can determine the signal flow graph, and

consequently the state equation in cascade form.

For each block, in the form of:

That yields a inverse Laplace transform:

We can represent the block as shown with an integrator and a

closed loop. We thus cascade the system as shown.

We can now easily determine the state equations from the signal flow graph by looking at each node that

represents a differential.

The output equation is written by inspection:

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The state space representation in vector-matrix form is thus:

5.5.2 PARALLEL FORM

By considering the same transfer function, we can perform a partial fraction expansion:

This shows the sum of three terms, with each term

being the first order subsystem with as the

input. Each term is thus parallel with the other 2

terms, as shown. The state equations are then:

And in Vector Matrix form:

This representation always yields a diagonal system matrix, A. The result is that each equation is a first order

differential equation in only one variable such that the equations are decoupled. Note that with repeated real

roots, we will not obtain a diagonal matrix but a Jordan canonical form.

Example 5.5

Determine the parallel representation of the state space equations for the transfer function:

Since there are 3 terms, there are 3 parallel branches. The contains 2 feedback loops as shown.

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The state and output equations are thus:

And in vector-matrix form this is:

5.5.3 CONTROLLER CANONICAL FORM

This form is obtained from the phase variable form simply by ordering the phase variables in the reverse order

i.e becomes .

Example 5.6

Determine the controller canonical form for the transfer function:

Which has a phase variable form given by:

Then rearranging in ascending order:

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5.5.4 OBSERVER CANONICAL FORM

The observer canonical form yields left companion system matrix. The process of obtaining the state equations

in observer canonical form is done over a few processes:

1. Divide by the highest power of

2. Cross multiplying, obtain terms of life power of integration

3. Identify the state variables as the outputs of the integrators

The result is a form:

Where

Example 5.7

Determine the observer canonical form:

Now, we let the innermost bracket represent the input translate to r and for each translate to

Thus in vector matrix form:

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We can now draw the signal flow diagram either using the vector matrix form, or the factorised expression for

. For the factorised expression for , we start with the outermost bracket and find that the output is

connected to an integral with the integral being fed the input and a feedback of . Then we do the

same for the other state variables.

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6.1 STABILITY An unstable system cannot be designed for a specific transient response or steady state error requirement.

Recall that the total response of the system is given by:

Using this, we define a:

Linear, time invariant stable system: A system which has a decaying natural response

Linear, time invariant unstable system: A system which has a natural response that grows without

bound

Linear, time invariant marginally stable system: A system which has a natural response that does not

decay nor grow but remains constant or oscillates.

An alternative definition of stability is the BIBO stability:

We also find that stable systems have closed loop transfer functions with poles only in the left half plane, since

this yields an exponential decay or damped sinusoidal natural response as time tends to infinity. Poles in the

right half plane yield either pure exponentially increasing or exponentially increasing sinusoidal natural

responses, leading to an unstable system. Moreover, poles of multiplicity greater than 1 on the imaginary axis

lead to the sum of responses of the form where , which also approaches infinity.

A marginally stable system is obtained when the closed loop transfer functions have only poles on the

imaginary axis of multiplicity 1 and all other poles in the left half plane.

A system is stable if and only if every bounded input yields a bounded output.

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6.2 ROUTH-HURWITZ CRITERION Using this criteria, we can determine how many closed loop system poles are in the left half plane and in the

right half plane. The process required :

1. Generate a Routh Table

2. Determine how many times the numbers change sign to determine how many right half plane poles

The power of the Routh-Hurwitz criterion allows one to determine what values a parameter can be to yield a

stable design.

THE BASIC ROUTH TABLE

Consider the transfer function:

To create the Routh Table, we label the rows with

powers of s starting from the highest power. We

then start with the coefficient of the highest power

of s in the denominator and list, horizontally in the

first row, every second coefficient. In the second

row, list horizontally, starting with the 2nd

highest

power of s, and every other coefficient that was

skipped in the first row.

We then fill the remaining entries by noting that each entry is a negative determinant of entries in the

previous two rows divided by the entry in the first column directly about the calculated row. The left hand

column of the determinant is always the first column of the previous two rows, and the right hand column is

the elements of the column above and to the right.

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Example 6.1

Make a Routh Table for the system shown.

The closed loop transfer function is given by:

1 31 0

10 1030 0

0 0

1030 0 0

INTERPRETING THE ROUTE TABLE

Thus, if a closed loop transfer function has all

poles in the left half of the s-plane, the

system is stable. For the previous example,

there were 2 sign changes, which indicates 2

poles in the right half plane and 1 pole in the

left half plane.

The number of roots of the polynomial that are in the right half plane is equal to the

number of sign changes in the first column

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Example 6.2

For the system shown, find the range of values of gain K (which is to be positive), for which the system is

stable.

1 77 0

18 K 0

0 0

K 0 0

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6.3 STABILITY IN STATE SPACE We can determine the stability of a system represented in state space by finding the eigenvalues of the system

matrix, A and determining their locations on the s-plane. We stated that the solution of:

Give us the poles of the transfer function:

Example 6.3

Given the system

Find out how many poles are in the left half plane and the right half plane.

We must first find

1 -7

-6 -26

-1 0

-26 0

Since there is one sign change in the first column, the system has one right half plane pole and two left half

plane poles. Thus, it is unstable.

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6.4 STEADY STATE ERROR Recall that the steady state error is the difference between the input and the output for a prescribed test input

as . Test waveforms such as the step, ramp and parabola allows us to determine the state of the system

under different conditions:

Step input: represent constant

position and are useful in

determining the ability of the

control system to position itself

with respect to a stionary target

Ramp input: represent constant-

velocity inputs to a position control

system by their linearly increasing

amplitude.

Parabolas: represent constant

acceleration inputs to position

control systems and can be used to

represent accelerating targets.

The steady state errors that we look at here arise from

the configuration of the system itself and the type of

applied input.

Consider the system shown which has an error of

In steady state, if equals ,

will be zero. But with a pure gain, K, the error,

cannot be zero if is to be finite and nonzero. Thus,

by virtue of the configuration of the system, an error

must exist:

Thus, the larger the value of K, the smaller the value of would have to be to yield a similar value

of .

On the other hand, if an integrator is placed in the forward path as shown, there will be zero error in steady

state for a step input. This is because as increases, the error decreases until there is zero error. Note that

there will still be a value for since an integrator can have a constant output without any input.

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6.5 STEADY STATE ERROR FOR UNITY FEEDBACK SYSTEMS Consider the figure shown.

But

To find the steady state error, we use the final value theorem:

Let us consider now a feedback control system as

shown. As we have shown previously:

If the system is stable then:

From this, we determine the steady state error for the 3 most common test signals.

6.5.1 STEP INPUT

A step input has a Laplace transform of 1/s so:

is considered the DC gain of the forward transfer function, since s, the frequency variable, is

approaching zero. In order to have zero steady state error then

This occurs if there is at least one pure integration in the forward path resulting in the forward transfer

function having the form:

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6.5.2 RAMP INPUT

A ramp input is given by:

. We thus obtain:

To have zero steady state error then:

Which indicates that must be in the form:

In other words, there must be at least two integrations in the forward path. If only one integration exists in the

forward path then

will be finite and given by

. This leads to a constant steady state error for a

ramp input. If there are no integrations in the forward path result in

, and the steady state error

would be infinite.

6.5.2 PARABOLA INPUT

For a parabolic input: . Hence:

In order to have zero steady state error for a parabolic input, we must have:

Which can only occur if:

If only 2 integrations exist in the forward path, we obtain an finite and constant steady state error, while a

lower number of n will yield an infinite steady state error.

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Example 6.4

Find the steady state errors for the inputs of to the system shown.

We verify that the closed loop system is indeed stable by using the Routh-Hurwitz criterion. We then use the

final value theorem on the step input:

Where

For a ramp input though:

For a parabolic input:

But

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6.6 STATIC ERROR CONSTANTS & SYSTEM TYPES The steady state error performance specifications are called static error constants. Recall that:

The three limit terms are known as static error constants with the position constant defined as:

The velocity given by:

And the acceleration constant given by:

Since the steady state errors are dependent upon the number of

integrations in the forward path, we give a name to this system

attribute. We define the system type to be the value of n in the

denominator, or, equivalently, the number of pure integrations

in the forward path. Thus indicates a type 0 system.

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6.7 STEADY STATE ERROR SPECIFICATIONS Static error constants can be used to specify the steady state error characteristics of control systems. For

example, if a control system has the specification , we can draw from this:

That the system is stable

The system is a Type 1, since only type 1 systems have a finite

A ramp input is the test signal. Since is specified as a finite constant, and the steady state error for

the ramp input is inversely proportional to , we know that the test input is a ramp

The steady state error between the input ramp and the output ramp is per unit of input slope.

Example 6.5

Given the control system in the figure, find the value of K so that there is a 10% error in the steady state.

This is a type 1 system since there is 1 pure integration in the forward path gain. The input signal must also be

a ramp, in order to yield a finite error in a type 1 system. Thus:

Applying the Routh-Hurwitz criterion, we see that the system is stable at this gain.

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6.8 STEADY STATE ERROR FOR DISTURBANCES The advantage of using feedback is that

regardless of disturbances, the system can

be designed to follow the input with small

or zero error. Consider the following

system subject to disturbances

between the controller and the plant. The

transform of the output is given by:

Using

To find the steady state value of the error, we apply the final value theorem to obtain:

Where is the steady state error due to , and is the steady state error due to the

disturbance. If we assume a step disturbance

, then:

Can be decreased by increasing the DC gain of (which lowers the value of that is fed back) or

decreasing the DC gain of , which yields a smaller value of as predicted by the feedback formula.

Example 6.6

Find the steady state error component due to a step disturbance with

This shows that the steady state error produced by the step disturbance is inversely proportional to the DC

gain of .

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6.9 STEADY STATE ERROR FOR NONUNITY FEEDBACK SYSTEMS

The steady state error is given by:

If we now consider step input and disturbances then:

For zero error:

It is possible that the steady state error is zero if:

1. The system is stable

2. is a type 1 system

3. is a type 0 system

4. H(s) is a type 0 system with a DC gain of unity

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Example 6.7

Determine the system type, error constant, and the steady state error for a unit step.

We first create a system with unity feedback as shown.

We then simplify and note that

This then creates a feedback loop. Let

. Thus:

This is a type 0 system since there is no pure integration. The appropriate static error constant is

The negative value for steady state error implies that the output step is larger than the input step.

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6.10 SENSITIVITY The degree to which changes in system parameters affect system transfer functions and hence performance, is

called sensitivity. A system with zero sensitivity is ideal.

Example 6.8

Find the sensitivity of the steady-state error to changes in parameter K and parameter a for the system shown.

This is a type zero system so the steady state error is:

For parameter a:

For parameter k:

Mathematically, sensitivity is the ratio of the fractional change in the function to the fractional change in

the parameter as the fractional change of the parameter approaches zero:

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6.11 STEADY STATE ERROR FOR SYSTEMS IN STATE SPACE There are two methods for calculating the steady state error:

1. Analysis via final value theorem

2. Analysis via input substitution

6.11.1 FINAL VALUE THEOREM APPROACH

Consider a closed loop system represented in state space by:

The Laplace transform of the error is:

But: , thus:

Where is the closed loop transfer function given by . This was proved previously.

6.11.2 INPUT SUBSTITUTION APPROACH

This method avoids taking the inverse of and can be expanded to multiple input and multiple output

systems.

STEP INPUTS Given the state equations as shown initially, and an input unit step where , a steady state solution,

for is:

Where is constant. Also,

We thus get:

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But the steady state error is the difference between the steady state input and the steady state output. The

final result for the steady state error for a unit step input into a system represented in state space is:

RAMP INPUTS

For unit ramp inputs, , a steady state solution for is:

Where and are constants. Hence,

The state equations are then:

From (1):

Substituting into (2) yields:

This yields a steady state error:

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7.1 VECTOR REPRESENTATION OF COMPLEX NUMBERS Any complex number, , can be represented by a vector with a magnitude M and angle : . If the

complex number if substituted into a complex function , another complex number will result.

For example, we can represent in two ways:

1. The traditional radius vector approach

2. The vector drawn from the zero of the function to the point s.

Using the segment approach, we find that the magnitude M of at a point s is given by:

Where is the magnitude of the vector drawn from the zero of F(s) at – to the point s, and the pole

length being the magnitude of the vector from the pole of F(s) at – to the point s. The angle of

at the point s is then given by:

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7.2 ROOT LOCUS DEFINITION The root locus technique can be used to analyse and design the effect of loop gain upon the system’s transient

response and stability. Consider the system shown and the table which represents the nature of the change in

the gain with respect to changes in the poles. The representation of the paths of the closed loop poles as the

gain is varied is called the root locus.

It provides solutions for systems of order higher than

two

Describes qualitatively the performance of a system as

various parameters are changes

Gives a graphical representation of a system’s stability

From the given root locus, we can see that the system is

overdamped for gains less than 25 (since the poles are

completely real). When the gain is 25, the system is critically

damped, and beyond this, the system is underdamped. When

the system is underdamped, the settling time is constant since

the real part of the poles do not change. Also, as we increase

the gain, the damping ratio diminishes, and the percent

overshoot increases.

7.3 PROPERTIES OF THE ROOT LOCUS Consider the general control system which has a transfer function:

From this, we see that a pole exists (and are part of the root locus) when the denominator becomes zero. This

condition can also be expressed as:

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Example 7.1

Given a unity feedback system with a forward transfer function:

a) Calculate the angle of at the point by finding the algebraic sum of the angles of the

vectors drawn from the zeros and poles to the given point.

b) Determine if the point specified is on the root locus

c) If the point specified in part (a) is on the root locus, find the gain, K,

using the lengths of the vectors.

a) We first find the poles and zeros:

From the plot, we can easily find the sum of the angles:

b) The point can be on the root locus since the sum of angles is an odd number of 180 degrees. The only

other criteria is that:

c) We use:

The gain K, at a point for which the angles add up to an odd multiple of 180 degrees is

found by dividing the product of the pole lengths by the product of the zero lengths.

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7.4 SKETCHING THE ROOT LOCUS

1. Number of branches: is equal to the number of closed loop poles.

2. The root locus is symmetrical about the real axis

3. On the real axis, for K>0, the root locus exists to the left of an odd number of real axis, finite open

loop poles and/or finite open loop zeros.

4. The root locus begins at the finite and infinite poles of and ends at the finite and infinite

zeros of .

5. Every function of s has an equal number of poles and zeros if we include infinite poles and zeros as

well as the finite poles and zeros. The root locus approaches straight lines asymptotes as the locus

approaches infinity. The equation of these asymptotes is given by the real axis intercept and the

angle :

Example 7.2

Sketch the root locus for the system shown.

We can easily plot the open loop poles and zeros and determine the root locus on the real axis, as shown.

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We then wish to find the asymptotes which are given by:

Thus, the asymptotes cut the real axis at -4/3. The angle of intersection is given by:

Note that the number of asymptotes

obtained is equal to the difference

between the number of finite poles

and number of finite zeros.

Since there are more open loop finite

poles than zeros, then there must be

zeros at infinity. The asymptotes tell us

how we get to these zeros at infinity.

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7.5 REFINING THE ROOT LOCUS The following rules help us to find specific points on the root locus such that we can better understand what

the root locus looks like.

REAL AXIS BREAKAWAY AND BREAK-IN POINTS

The point where the locus leaves the real axis, called the

break away point occurs where the gain is a maximum,

and breaks into the real axis at the break in point where

the gain is a minimum,

We can find the break away and break in point by using

differential calculus and noting that the points are on

the real axis such that . But on the real axis we get:

The breakaway and break-in points can then be found by solving:

The breakaway and break-in points also satisfy the relationship:

Where and are the negative of the zero and pole values of

We then solve for

which yield the breakaway and break-in points without differentiating.

Example 7.3

Find the breakaway and break-in points for

the root locus shown using differential

calculus, if we are given that the open loop

system has a root locus of:

To find the breakaway and breakin points

on the real axis we note that they lie on the

root locus so:

And

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Thus:

, which are the breakaway and break-in points respectively. We do not need to check that

they are indeed minimums or maximums since we know that there is a maximum between -1 and -2 and a

minimum between 3 and 4.

We verify this using the transition method:

AXIS CROSSINGS

The axis crossing is a point on the root locus that separates the

stable operation of the system from the unstable operation. The

value of at the axis crossing yields the frequency of oscillation.

ROUTH-HURWITZ CRITERION

As already seen, forcing a row of zeros in the Routh table will yield

the gain. Going back one row to the even polynomial equation and

solving for the roots yields the frequency at the imaginary axis

crossing.

Example 7.4

Find the frequency and gain, K, for which the root locus crosses the imaginary axis for the closed loop system

shown with a closed loop transfer function:

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We start off by drawing the Routh table:

1 14 3K

7 8+K

90-K 21K

21K

Somce K>0, then the only line that can be zero is the line.

Then in the line we substitute to get the polynomial equation:

Solving for this, we find that . Thus, the root locus crosses the axis at at a gain of 9.65.

We conclude that the system is stable for .

SUM OF ANGLES

At the axis crossing, the sum of angles from the finite open loop poles and zeros must add to .

By searching on the axis for when this situation occurs, we can determine the axis crossing.

CHARACTERISTIC EQUATION

By letting , in the characteristic equation, and equating both the real and imaginary part to zero, we can

solve for and K.

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ANGLES OF DEPARTURE AND ARRIVAL

We can calculate the root locus departure angle from the complex poles and the arrival angle to the complex

zeros. Recall that on the root locus:

If we assume a point on the root locus close to a

complex pole, the sum of angles drawn from all the

finite poles and zeros to this point is an odd

multiple of . Except for the pole that is close

to the point, we assume all angles drawn from all

other poles and zeros are drawn directly to the pole

that is near the point. Thus, the only unknown angle

in the sum is the angle drawn from the pole that is

close, which corresponds to the angle of departure.

To determine the angle of arrival at a complex pole,

consider a point on the root locus close to the

complex zero. Again, we sum the angles drawn from

all finite poles and zeros and equate this to an odd

multiple of 180 There is only one unknown angle in

the sum, the angle drawn from the zero that is

close, that corresponds to the angle of arrival:

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Example 7.5

Given the unity feedback system, find the angle of departure from the complex poles and sketch the root

locus.

We note that the open loop poles are found by solving:

, giving us , and there is one open loop zeros at s=-2.

Since the complex poles are complex conjugates, we only need to work with

one of the poles and use symmetry to find the angle of departure.

Now:

PLOTTING AND CALIBRATING THE ROOT LOCUS

Let us assume we want to find the exact point at which the locus crosses the 0.45 dmping ratio line and the

gain at that point. We plot the root locus plot.

If a few test points along the line are selected, we can evaluate which points yield:

And the corresponding gain by:

This is shown in the figure.

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Example 7.6

Given a unity feedback system that has the forward transfer function of:

a) Sketch the root locus

Recall that the number of branches is equal to the

number of closed loop poles. This happens to be 2.

Solving yields . These two finite

poles will be part the start of the root locus, and the two

zeros will be the end of the root locus, as shown.

b) Find the imaginary axis crossing

We can use the Routh Table since the closed loop transfer

function is:

We force a row of zeros for row since it is the highest row which can be all zero. Solving this yields

. Then using the row above we find:

Thus, the axis crossing is at

c) Find the gain K, at the imaginary axis crossing

As shown in the last section, the gain, K=1.

d) Find the break-in point

Since must lie between 2 and 4, then .

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e) Find the point where the locus crosses the 0.5 damping ratio line.

Searching along for the 180 point, we find

f) Find the gain at the point where the locus crosses the 0.5 damping line.

This yields:

g) Find the range of gain, K, for which the system is stable.

Since the root locus starts on the left side and ends on the right, the crossing gain represents the

maximum gain for which the system is still stable. Thus:

7.6 TRANSIENT RESPONSE DESIGN VIA GAIN ADJUSTMENT Recall that our analysis of transient response was limited to second order

systems. However, if we can justify that a second order approximation is

possible, then we may use the same techniques:

1. Higher order poles are much farther to the left of the s plane that

dominant second order pair of poles.

2. Closed loop zeros near the closed loop second order pole pair are

nearly cancelled by the close proximity of higher order closed loop

poles

3. Closed loop zeros not cancelled by the close proximity of higher order

closed loop poles are far removed from the closed loop second order

pole pair.

The design of a high order system can be summarized as:

1. Sketching the root locus

2. Assuming the system is a second order system without any zeros and then find the gain to meet the

transient response specification

3. Justify the second order assumption by finding the location of all higher order poles and evaluating

the fact that they are much farther from the axis than the dominant second order pair –a factor of

5 times approximately. If closed loop zeros are not cancelled by higher order closed loop poles, then

they should be far removed from the dominant second order pole pair.

4. Otherwise simulate the results.

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7.7 GENERALISED ROOT LOCUS In many cases, we wish to know how the closed loop poles change as a function of another parameter.

Consider the open loop forward path transfer function to be:

We now find the closed loop function, but also in the process want to appear as the forward path gain

:

Then divide by so that the denominator

is in the form :

This conceptually implies that:

We can thus sketch the root locus now. The zero is at

s=-2 and the poles are at .

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-5 -4 -3 -2 -1 0 1 2 3-5

-4

-3

-2

-1

0

1

2

3

4

5

Root Locus

Real Axis

Imagin

ary

Axis

-3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1-4

-3

-2

-1

0

1

2

3

4

Root Locus

Real Axis

Imagin

ary

Axis

-3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1-4

-3

-2

-1

0

1

2

3

40.5

0.050.10.150.20.250.30.350.40.450.5

1

2

3

4

1

2

3

4

0.050.10.150.20.250.30.350.40.45

Root Locus

Real Axis

Imagin

ary

Axis

8.8 MATLAB For the unity feedback system with a forward path of:

a) Find the operating point at which the damping ratio is 0.45.

b) The axis

c) The breakaway point

d) The range of K within which the system is stable

Plot the root locus

Zoom in

Sgrid – show damping line and natural frequencies

‘

>> sgrid(z,wn)

>> clf

>> clear all

>> numg=[1 -4 20];

>> deng=poly([-2 -4]);

>> GH=tf(numg,deng)

Transfer function:

s^2 - 4 s + 20

--------------

s^2 + 6 s + 8

>> rlocus(GH)

>> axis([-3 1 -4 4])

%axis([xmin xmax ymin ymax])

>>z=0:0.05:0.5;

%damp ratio from 0-0.5 in steps of 0.5

>> wn=0:1:10;

%natural frequency in steps of 1 from 1-10

>> sgrid(z,wn) %plot on graph

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Select points (3 points on graph)

a)

b)

c)

d) Using the plot of the 3 points, we find that the maximum gain occurs on the imaginary axis, with

an associated gain of 1.5. Thus the system is stable for 0<k<1.5

Smoothing the root locus

-3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1-4

-3

-2

-1

0

1

2

3

40.45

0.45

Root Locus

Real Axis

Imagin

ary

Axis

>> for k=1:3

[K,p]=rlocfind(GH)

end

>> K=0.005

%specify range of gain to smooth

Rlocus(G,K)

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0 1 2 3 4 5 6 7 80

0.2

0.4

0.6

0.8

1

1.2

1.4

Step Response

Time (sec)

Am

plit

ude

Input command

Allows custom keyboard entry of value

Find closed loop transfer function

After selecting the operating point using the

[K,p]=rlocfind(G) command:

Simulate design – step input

>> pos=input(‘type %OS’)

>> T=feedback(K*G,1)

>> step(T)

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8.1 IMPROVING TRANSIENT RESPONSE & STEADY STATE ERROR

TRANSIENT RESPONSE

Our goal here is to get the desired transient response even though the operating point is not on the original

root locus. Rather than change the existing system, we augment, or compensate, the system with additional

poles and zeros, so that the compensated system has a root locus that goes through the desired pole location

for some value of gain.

The compensated system can be realized using a passive or active network (ideal compensators) and presents

little need to interfere with the power requirements of the system or the load for that matter. A disadvantage

is that the compensated system is of a higher order – making it necessary to simulate the system after the

design is complete.

We have seen that the transient response will be improved if we insert a differentiator in the forward path in

parallel with the gain since large changes yield a large derivative signal that drives the plant, while small

changes yield a small derivative signal, and the output from the differentiator becomes negligible compared to

the output from the gain.

STEADY STATE ERROR

If we use dynamic compensators, compensating networks can be designed that will allow us to meet transient

and steady state error specifications simultaneously.

We have seen that the steady state error can be improved by adding an open loop pole at the origin in the

forward path, thus increasing the system type and driving the associated steady state error to zero. This

additional pole at the origin is realized with an integrator.

8.2 IMPROVING STEADY STATE ERROR VIA CASCADE COMPENSATION Compensated systems can be arranged using

either cascade compensation or a feedback

compensation. Here, we will only be looking at

the cascade compensation, whose system

diagram is as shown.

There are two techniques:

Ideal integral compensation: uses a pure integrator to place an open loop, forward path pole at the

origin, thus increasing the system type and reducing the error to zero. This type of system is called the

proportional-plus-integral (PI) controller, since the implementation consists of feeding the error and

the integral of the error forward to the plant.

Lag compensation: does not use a pure integration and places the pole near the origin, but is unable

to reduce the error to zero.

In summary:

Improvements in transient response require differentiators

Improvements in the steady state error require integrators in the forward path

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IDEAL INTEGRAL COMPENSATOR

To show how steady state error is improved without affect the

transient response, consider the root locus shown. In this case,

the root locus goes through point A. Suppose that this is the

desired operating point. To reduce the steady state error, we

then add a pole at the origin; the root locus no longer passes

through A. To solve this problem, we add a zero closed to the

pole at the origin as shown. The total angular contribution of

the compensator zero and pole cancel out, and point A is now

on the root locus.

LAG COMPENSATION TO IMPROVE STEADY STATE ERROR

With lag compensation, the pole and zero are moved to the left of the origin. This does not increase the

system type and improves the static error on an uncompensated system. Similarly, we find that the transient

response is not changed much since the total angular contribution of the compensator pole and zero is

approximately zero degrees. Similarly, the gain remains relatively unchanged since the length of the vectors

drawn from the lag compensator are approximately equal.

A compensator with a pole at the origin and a zero close to the pole is called an ideal integral

compensator.

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Without loss of generality consider a type 1 static error constant:

The new static error constant after lag compensation is:

To minimize the error, the static error constant must be larger, and this is achieved by placing both the pole

and zero close to the origin, with the zero left of the pole.

Example 8.1

Compensate the system shown, whose root locus is shown, so that the steady state error is improved by a

factor of 10 if the system is operating with a damping ratio of 0.174.

As seen in the root locus, the desired operating point is at , with a gain of K=164.6. This

yields:

The new steady state error must be:

Lag compensation offers an improvement of the static error constant by a factor of .

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The improvement in is the required ratio of the compensator zero and pole:

We choose

We then must check:

Where the dominant second order poles are and the associated gain:

Where the third and fourth closed loop poles are: -11.55, -0.101

Whether the fourth pole of the compensated system cancels its zero.

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8.3 IMPROVING TRANSIENT RESPONSE VIA CASCADE COMPENSATION There are two ways to improve the transient response of a system via compensation:

1 Ideal derivative compensation: uses a pure differentiator and is referred to as a proportional-plus-

derivative (PD) controller. This technique is prone to high frequency noise.

2 Lead compensation: This does not use a pure differentiation but rather a passive network which has a

zero and pole closed to the origin, with the pole being more distant.

PD CONTROLLER

By adding an additional zero to the forward

path, we speed up the original system. The

PD controller does just that and has a transfer

function of:

The basic process of designing a PD controller is to:

Evaluate the sum of angles from the open loop poles and zeros to a design point that is the closed

loop pole that yields the desired transient response.

The difference between 180 and the calculated angle must be the angular contribution of the

compensator zero.

Trigonometry is then used to locate the position of the zero to yield the required difference in angle

Example 8.2

Given the system, design an ideal derivative compensator with 16% overshoot with a threefold reduction in

settling time. The operating point with the associated overshoot is . Assume a

second order approximation.

The settling time is:

This is the real part of the root locus (which must be in the left plane for stability) that intersects the damping

line . The imaginary part is thus given by:

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Now, the additional zero must have an

angular contribution equal to that of the sum

of the other open loop zeros and poles:

LEAD COMPENSATION

With lead compensation:

A pole and zero pair must be introduced, with the pole farther from the imaginary axis than the zero,

resulting in the angular contribution of the compensator to be still positive, and thus approximating

an equivalent single zero.

Implementation is done using passive components

There is less sensitivity to noise

We arbitrarily select a lead compensator pole or zero during design

There are an infinite number of lead compensators that can meet the transient response

requirements, but the design usually is constrained by the static error constant, gain and second

order approximation

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Example 8.3

Design a lead compensator that will reduce the settling time by a factor of 2 while maintaining 20% overshoot.

Assume a second order approximation.

The new operating point is thus .

Let us choose . We then find the angle contribution

of the current zeros and poles to be:

Using trigonometry, we can determine the pole location, as shown in the figure:

The pole contribution angle must be

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8.4 IMPROVING STEADY STATE ERROR & TRANSIENT RESPONSE The process of improving both the steady state error and transient response is a two step process:

We first improve the transient response by using a ideal derivative or lead compensator

Then we improve the steady state error by using an ideal integrator or lag compensator

A problem with this approach is that improvement in the transient response in some cases yields deterioration

in the improvement of the steady state error, but this is not always the case.

Such a design process yields

A PD controller followed by a PI controller, or alternatively a proportional-plus-integral-plus-

derivative (PID) controller.

If we first design a passive lead compensator and then design a passive lag compensator, the resulting

compensator is called a lag-lead compensator.

PID CONTROLLER DESIGN

The general transfer function is:

Which has two zeros plus a pole at the origin – one zero,

pole pair used as an ideal integral compensator and the

other zero to be used as an ideal derivative compensator.

The process of designing such a PID controller involves:

1. Evaluate the performance of the uncompensated system to determine how much improvement in

transient response is required.

2. Design the PD controller (lead compensator) to meet the transient response specifications – zero and

loop gain or for a lead compensator, also the pole location.

3. Verify the system has met the requirement. Redesign if not.

4. Design the PI controller (lag compensator) to yield the steady state error

5. Determine the gains

6. Simulate the system to be sure all requirements have been met.

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Example 8.4

Given the system shown, design a PID controller so that the system can operate with a peak time, two thirds of

that of the uncompensated system at 20% overshoot and with zero steady state error.

Step 1

We thus find the dominant poles to be the intersection of the root

locus and the damping line of 0.456:

Step 2

We determine the original peak time and find the new peak time:

The new peak time is to be 2/3 of 0.297=0.198. The imaginary part of the new dominant poles is thus:

The real part must be:

Step 3

We then find the total angular contribution of the poles and zeros:

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The zero must then contribute:

Then, using trigonometry:

The PD controller is described by:

We then simulate the result to check the reduction in the peak time.

Step 4

We now design the ideal integral compensator

for zero steady state error. Any ideal integral

compensator will work as long as the zero is

placed close to the origin:

We then check the root locus characteristics:

Step 5

We now determine the 3 values for the gain:

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Step 6

We verify with simulations that we have indeed improved both the transient and steady state error. The

complete PID controller further improved the steady state error with appreciably changing the transient

response designed with the PD controller – it is slightly longer than the PD controller.

If we want a faster response, we must move the zero farther from the origin for the PI controller.

Example 8.5

Design a lag lead compensator with 20% overshoot and a twofold reduction in settling time and a 10 fold

improvement in steady state error for a ramp input.

Step 1: Evaluate performance:

corresponds to a 20% overshoot. We find the dominant poles at , with a gain of

192.1.

Step 2: Evaluate settling time:

The new real part of the new dominant pole in the negative direction is:

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The imaginary real part of the dominant pole is:

Step 3: Design lead compensator

A lead compensator has both a zero and pole. We select the compensator zero to be -6. Now, the sum of the

angles to the design point is:

The angular contribution of the pole must be

By using Matlab, we find the gain at the design point is 1977. We also verify results using simulation.

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Step 4: Steady state error using lag compensator

The steady state error of the uncompensated system is:

The new steady state error must be:

Note that the current lag compensated system has a steady state error of

Step 5: Select values for lag compensator

We select a pole to be close to the origin: 0.01. Thus:

Step 6: Combine lag-lead system

The final transfer function of the open loop is:

The new dominant, closed loop poles are at with a gain of 1971. Notice that the lag-lead

compensation has indeed increased the speed of the system, as witnessed by the settling time or the peak

time. The steady state error for a ramp input has also decreased by about 10 times.

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8.5 PHYSICAL REALISATION With active circuit realizations, the transfer function can be formed by

the use of an inverting operational amplifier. By judicious choice of

, this circuit can be used as a building block to implement

compensators and controllers.

By using the basic building blocks, we can cascade compensators to build

larger compensators such as the lag lead compensator.

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We may also choose to do passive circuit realization. Note that cascading these networks require buffers since

one network can load the other – thus changing the overall transfer function.

8.6 SUMMARY Function Compensator Transfer Function Characteristics

Improve steady state error

PI

Increases system type

Error is zero

Zero at – is small and negative

Active circuit implementation

Improve steady state error

Lag

Error improved

Pole is small and negative

Zero is left of pole

Passive implementation

Improve transient response

PD Zero selected to put design point on root locus

Can cause noise and saturation

Active circuit implementation

Improve transient response

Lead

Zero and pole selected to put design point on root locus

Pole is to the left of zero

Passive Implementation

Improve steady state error and transient

response

PDI

Lag zero left of pole at origin improves steady state error

Lead zero improves transient response

Active circuit implementation Can cause noise and saturation

Improve steady state error and transient

response

Lead-lag

Lag zero left of pole at origin improves steady state error

Lead zero improves transient response

Passive Implementation

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9.1 CONCEPT OF FREQUENCY RESPONSE In this chapter, we present the design of feedback control systems through gain adjustment and compensation

networks from the perspective of frequency response. This method has distinct advantages:

Transfer functions can be modeled from physical data

We can design compensators that meet both steady state error and transient response requirements

We can find stability in nonlinear systems

Settles ambiguities when sketching a root locus

Before doing so we define the concept of frequency response of a

system consisting of a relationship that describes the changes in the

output magnitude and phase component of a signal when given a

particular input.

We consider a mechanical system with a block diagram as shown.

We find that that the system can be describes as:

We thus show that the magnitude response is:

And the phase response is:

More generally, we can express it as:

The analytical expression for the frequency response given a transfer function is thus:

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Example 9.1

Find the analytical expressions for the magnitude and phase responses of

Rationalising:

Then:

The resulting bode plot is shown. We shall consider in the next section how we can get approximate sketches

of the bode plot by hand.

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9.2 BODE PLOTS Bode plots are widely used in the engineering discipline because they characterize both the magnitude and

phase response of a system. In order to obtain bode plots, we must first look at the transfer function, and if we

are doing so by hand, we make generalizations about the transfer function.

As we have just seen, any transfer function can be written in the form:

Where denotes the DC gain, i.e . The frequencies represent the zeroes and the

poles of the transfer function respectively.

BODE’S RULES

The task of plotting the magnitude and phase

response becomes tedious, and for this reason, we

make the following approximations for first order

systems:

As passes each pole frequency, the slope

of the magnitude plot decreases by

20 dB/dec – a tenfold change in H for a

tenfold increase in frequency.

As passes each zero frequency, the slope

of increases by 20 dB/dec.

We determine the phase response at low

frequencies, the break frequencies and at

high frequencies.

The table shows a summary of bode straight-line

magnitude and phase plots.

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BODE PLOTS FOR

We first normalize the expression:

We then let . Note that as , . This equates to a magnitude response of:

This magnitude is approximate in the range of to .

When

We note that this is a 6 dB increase for the doubling of the frequency. Finally for high frequencies:

Which is equivalent to:

If we plot against , we get a straight line with a slope of 20, that is a slope of 20 dB per

decade or a slope of 6dB/octave, where an octave is a doubling of the frequency.

To get the phase response, we note that at low frequencies, the phase is completely real, that is , and at

high frequencies we have

, which is imaginary so . The slope is 45 per decade upwards

for the zero.

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BODE PLOTS FOR

When

Which in dB is a constant:

When :

In dB is a 20dB decade slope downwards:

We can normalize the bode plot by removing the 1/a term

and plotting

. This gives the following magnitude

response.

From the above analysis, we also find that at low frequencies, the response is real, while at high frequencies

the phase is . Again we normalize the plot so that the corner frequency is at .

BODE PLOTS FOR G(S)=S AND G(S)=1/S

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Example 9.2

Construct the bode plot for the transfer function

We first put the transfer function in standard form:

This indicates that there are 2 corner/break frequencies at and a zero at 0. Hence, the plot should

initially have a 20dB/dec increase, then countered by a -20dB/dec decrease to bring it linear, and then finally

another 20dB/dec decrease.

We can double check by converting the transfer function into decibels:

The dotted lines show where each term is. We then add the lines together to get our magnitude plot. To get

our phase plot

We then consider the cases in which:

Now, we draw the dotted lines for the terms of inverse tan. We note that at

, the phase is

approximately zero, while at , the phase is approximately . We then add the lines together.

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Example 9.3

Sketch the bode plots for:

We first rearrange so:

The break frequencies are at 1,2 and 3. Since s=0 is a pole, then the magnitude plot begins a decade below the

lowest break frequencies and extend a decade above the highest frequencies.

The magnitude K determines whether the curve should be moved up or down the magnitude plot, but has

not effect on the phase plot. We first take k=1 and denormalise later. We make a table as shown:

Frequency (rad/s)

0.1 (start at 0) 1 (start at -1) 2 (start at -2) 3 (start at -3)

Pole at 0 -20 -20 -20 -20

Pole at -1 0 -20 -20 -20

Pole at -2 0 0 -20 -20

Zero at -3 0 0 0 20

Total Slope (dB/dec) -20 -40 -60 -40

Thus using the table, we draw the required magnitude plot:

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We take a similar approach in the phase response but the breaks exist at a decade above and below the break

frequency. Thus, our table will consist of 0.1, 0.2, 0.3, 0, 20, 30. The zero at 0 is shifted a decade below to

approximately 0.1 but is not shifted to 10 since 10 is significantly far away from 0.

Frequency (rad/s)

0.1 0.2 0.3 0 20 30

Pole at -1 -45 -45 -45 0

Pole at -2 -45 -45 -45 0

Pole at -3 45 45 45 0

Total slope (deg/dec) -45 -90 -45 0 45 0

We show that the contributions of the breaks in the decade below the break frequency and we show that the

contribution is negligible at a decade above the breakfrequency.

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9.3 SECOND ORDER BODE PLOTS The second order polynomial is of the form:

Without correction, the asymptotic approximation and the actual frequency response can be great for some

values of . We will thus introduce a correction technique.

For a general second order system, at low frequencies:

As such, the magnitude response is:

At high frequencies though,

So, the magnitude response becomes:

This indicates that the slope is at 40 dB/decade or 12 dB/octave. We also note that the break frequency is ,

and at that point:

And this indicates that the phase is and that the phase plot increases by 90 degrees per decade.

NORMALISATION

For convenience, we normalize and scale

our findings by dividing the magnitude by

, and the scale frequency by . As such

we plot:

This results in having a low frequency

asymptote at 0 dB and a break frequency of

1 rad/s.

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CORRECTION TO SECOND ORDER BODE PLOTS

At the natural frequency, the correction on a normalized log magnitude:

On an unscaled magnitude plot, the correction at the natural frequency gives a magnitude of is:

SECOND ORDER PLOTS

In this case, the plots are reverse in polarity:

The magnitude curve breaks at the natural frequency and decreases at a rate of -40 dB/decade.

The phase is 0 at low frequencies and decreases at at until , where it

levels off at .

The correction is on the normalized log magnitude plot or a magnitude of on

an unscaled plot.

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9.4 SUMMARY OF BODE’S RULES In summary we state that:

As passes each pole frequency, the slope

of the magnitude plot decreases by

20 dB/dec

As passes each zero frequency, the slope

of increases by 20 dB/dec.

We determine the phase response at low

frequencies, the break frequencies and at

high frequencies.

For second order systems, the correction to

the normalised magnitude plot is or

for the unscaled magnitude

plot.

For second order system, the phase plot

breaks at a decade below the break

frequencies and has negligible impact on the

phase at a decade above the break

frequency, as shown in the diagram.

Example 9.4

Draw the bode plots for

We draw a magnitude plot after normalizing the transfer function:

Note the 3/50 DC gain, we thus start the plot at

Start plot Start at pole -2 Start at zero -3 Start at

Pole at -2 0 -20 -20 -20

Zero at -3 0 0 20 20

0 0 0 -40

Total slope (dB/dec) 0 -20 0 -40

The corrected magnitude at the natural frequency is thus or the correction to

the magnitude is , as shown.

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We now turn to the phase plots:

0.2 (start pole

at -2)

0.3 (start zero at

-3)

0.5 (start at -5)

20 (end pole at

-2)

30 (end zero at

-3)

50 (end )

Pole at -2 -45 -45 -45 0

Zero at -3 45 45 45 0

-90 -90 -90 0

Total slope (dB/dec)

-45 0 -90 -45 -90 0

Since the second order system is dominant, the result is that the plot ends at as shown.

9.5 NYQUIST CRITERION The Nyquist criterion relates the stability of a closed loop

system to the open loop frequency response and open pole

location. We can thus make statements about the stability of

the closed loop system knowing the open loop system, just

like we did with the root locus.

By Letting:

We show that:

From this:

1. The poles of are the same as the poles of , the open loop system

2. The zeros of are the same as the poles of , the closed loop system.

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We also make a note that contours are a set of points drawn out in a mapping where we take points drawn out

by contour A and substitute it into a function:

Assuming a clockwise direction for mapping the points on contour A to contour B:

From the last point about rotation cancelling we extend this idea to show that:

1. The contour B maps in a clockwise direction if contour A encircles only zeros or only does

not encircle poles

2. The contour B maps in a anti-clockwise direction if contour A encircles a pole

3. The contour B encircles the origin if contour A encircles either a pole or zero.

4. The contour B does not encircle the origin if contour A encircles a pole and zero due to

rotation cancelling.

Where N is the number of counterclockwise rotations of contour B about the origin

P is the number of poles of inside contour A, or the number of poles of

Z is the number of zeros of inside contour A, or the poles of the closed loop system

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The contour mapping is the same for and , except that the previous is translated one

unit to the left – and this calls for us to count the rotations about -1 instead of about the origin.

Finally, we state that the Nyquist stability criterion is:

Example 9.5

Determine the number of right half plane poles in the following mapping.

Since contour A does not encircle any poles or zeros, the contour B rotates clockwise. Also since contour A

does not encircle any poles or zeros, the mapping does not encircle -1. As such:

Which means that there are no right half plane poles and the system is stable.

Example 9.6

Determine the number of right half plane poles in the following mapping.

Since there are two unknown zeros of , or alternatively the poles of the closed loop system,

there are two clockwise encirclements about -1 in the Nyquist plot. P=0, N=-2 so, Z=-2. The system is then

unstable.

If a contour A, that encircles the entire right half plane is mapped through , then the number of

closed loop poles, Z, in the right half plane equals the number of open loop poles, P, that are in the right

half plane minus the number of counterclockwise revolutions, N, around -1:

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9.6 SKETCHING NYQUIST PLOTS A sketch can be obtained by looking at the vectors of and their motion along the contour.

If we detour to the right around the pole, the pole vectors rotate through . If we detour to the left of

open loop poles, the pole vectors rotate through an angle of as we detour around it.

Example 9.7

Given the following system, sketching the Nyquist diagram for the system.

Conceptually, we substitute points in the right half plane into:

The resultant vector R is the product of the zero vectors divided by the product of the pole vectors. As we

move in a clockwise direction around the contour from A to point C, the resultant angle goes from to

, since the poles gain in a counterclockwise direction, which explains the decrease in angle of the

function .

The resultant goes from a finite value at zero frequency to zero magnitude at infinite frequency since the

length of the 3 infinite poles lengths is infinite.

Since:

When an open loop pole is situated along the contour enclosing the right half plane, we detour around the

poles on the contour.

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Then:

The Nyquist plot starts at 50/3 with an angle of . Now as increases the real part remains positive and the

imaginary part remains negative until

. When the imaginary part goes to zero, the Nyquist diagram

crosses the negative real axis:

The real part of this is:

At infinite frequency,

Which is approximately zero at .

Around the infinite semicircle from point C to

point D, the vectors rotate clockwise each by

, and thus the resultant undergoes

counterclockwise rotation of ,

starting at point C’ and ending at point D’.

At point C, the angles are all 90 , and hence

the resultant is given by:

Which is, , and for point D, the

angles are all , so the resultant is

. By selecting intermediate points, we

can verify the spiral.

The mapping of the imaginary axis is a mirror

image of the mapping of the positive

imaginary axis. Thus the mapping of the

section of the contour from points D to A is the

mirror of the mappings from points A to C.

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Example 9.8

Sketch the Nyquist plot of

At point A, the two open loop poles contribute and the zero contributes nothing. This gives a

total angle at point A of , since the angle contribution is the sum of zero angles minus the sum of poles

angles. The zero is negligible. Close to the origin, the fuction is infinite, thus point A maps into point A’, located

at infinity at an angle of -180 .

At point B, the zero makes an additional contribution of , and the total contribution is at B’. The

magnitude is also zero. Alternatively consider the low and high frequencies:

Which yield:

Moving along BCD, the magnitude stays at 0, but the vectors change; at point C, the total angle contribution is

, but at point D, the total angle contribution is . Thus the total change in the

vector from to is .

The mapping from D to E is the mirror of the mapping of A to B.

Finally over the detour, the resultant magnitude approaches infinity, since substituting the pole would yield an

infinite magnitude. At point E:

Since the zero has negligible contributions, but each of the poles contribute . And at F:

And finally at A again:

Drawing a test radius shows one counterclockwise revolution and one clockwise revolution, yielding zero

encirclements.

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9.7 STABILITY VIA THE NYQUIST DIAGRAM We previously used the Routh-Hurwitz criterion and the root locus to find system stability. If we manipulate

the gain, we manipulate the resultant by a constant factor in the Nyquist diagram.

1. Set K=1 and sketch the Nyquist diagram

2. Consider the critical point to be at -1/K rather than at -1

3. Adjust the value of K to yield stability

Example 9.9

Sketch the Nyquist diagram and determine what values of K yield stability for:

We first set K=1, and then sketch the Nyquist plot. The right half contour encircles 2 open loop poles, and we

thus require the number of counterclockwise encirclements to be N=2 about -1/K.

Starting at the origin we have

, then moving to B, the contribution of the angles is zero, and the

magnitude is small. To find the negative axis crossing:

We take the imaginary part and solve it for the roots:

Thus the negative axis is cut at -1.33. Finally, we note that since poles are encircled, the number of

counterclockwise revolutions is 2.

If the Nyquist diagram intersects the real axis at -1, then the system is marginally stable.

We use for stability, the number of right half plane poles, Z must be zero:

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So long as the critical point at , which yields -1.33 in encircles

, the system is stable. Thus, K

can be decreased until:

For a stable system, and if , the system is marginally stable.

STABILITY VIA MAPPING ONLY THE POSITIVE AXIS

Consider the diagram shown, which represents a system that is stable at low values of gain and unstable at

high values of gain. Since the contour does not encircle open loop poles, the Nyquist criterion tells us that we

must have no encirclements of -1 for the system to be stable. Thus, if the gain is small, the mapping will pass

to the right of -1 for a stable system.

For such a system then:

The loop gain K, must be so that the open loop magnitude is less than unity at that frequency

where the phase angle is i.e the negative real axis.

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For a system with stability achieved at high gains, the contour looks like that shown. In this case, we require

two counterclockwise encirclements of the critical point for stability. Thus:

Example 9.10

Find the range of gain for stability and the gain for marginal stability for the unity feedback system with:

Since the open loop poles are all in the left half plane, the Nyquist criterion tells us we want no encirclements

about -1 for stability. Thus a gain of less than unity is required on the negative real axis.

Let K=1. The point A maps to ¼, and the curve moves clockwise since there are no encirclements required.

Now at B, the total angle contribution yields , but the magnitude is zero so the Nyquist diagram is as

shown.

The system is stable if the open loop magnitude is greater than unity at that frequency where

the phase angle is i.e the negative real axis.

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To find the negative axis crossing:

We set the imaginary part to zero and get:

We substitute this back into the open loop transfer function and this yields -1/20. Thus, the gain can be

increased to:

The system is marginally stable for K=20 (frequency of oscillation is ) and unstable for K>20.

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9.8 GAIN MARGIN AND PHASE MARGIN VIA THE NYQUIST DIAGRAM We define:

For the given diagram, since no encirclements are required, the gain margin expressed in dB is the log of the

reciprocal of the real axis crossing:

The angle drawn out by the negative real axis and a line drawn from the origin to the point Q’, the intersection

of the Nyquist plot and the unit circle is the phase margin. Unfortunately, this is hard to find by hand, but can

be done using Bode plots very easily or by computational tools

Gain margin : The gain margin is the change in open loop gain, expressed in dB, required at

180 of phase shift to make the closed loop system unstable.

Phase Margin : The phase margin is the change in open loop phase shift required at unity gain

to make the closed loop system unstable.

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Example 9.11

Determine the gain margin for the unity feedback system with the forward transfer function:

We thus find the negative axis crossing:

Thus,

Which yields a real part of:

Thus, the gain margin can be increased by

before the real part becomes -1. Hence the gain margin

is:

9.9 STABILITY, GAIN MARGIN, PHASE MARGIN USING BODE PLOTS Stability can be determined using bode plots.

If we require no encirclements, then the

gain should be less than unity, when the

bode phase plot is

This is the case for

If we take K=40, the bode plots look like that

shown. At , the magnitude plot is -20 dB. As

such a gain of +20 dB is possible before the system

is unstable. Since the plot is scaled for a gain of 40,

the gain for instability is .

Thus, stability is achieved when 0<k<200.

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Example 9.12

If:

Find the gain margin and the phase margin.

We use the bode plots for K=40. Thus, now that K=200, the magnitude plot would start at .

Since the plots have taken 20log40 as 0dB, the plot required should be 13.98 dB higher.

Finding the gain margin is done by look at when the phase reaches . This occurs at . And

the magnitude plot shows a gain of . The correction needed is then . Thus:

The phase margin is found when the gain is 0dB. However, a correction of must be made, so we are now

seeking the phase margin at the point -13.98 dB on the plot. The phase is at . Thus the

gain margin is .

Gain margin : Is how far below unity the magnitude plot is at a frequency where the

phase plot is at

Phase Margin : The phase margin is how far above the phase plot is at the frequency

, where the gain is unity.

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9.10 RELATION BETWEEN CLOSED LOOP TRANSIENT AND CLOSED LOOP

FREQUENCY RESPONSES Consider the second order feedback control

system shown. The closed loop log

magnitude plot is shown.

By finding the magnitude of the system and finding the maximum through differentiation, we can find the

peak value of the closed loop magnitude response by just knowing the damping ratio:

This also indicates that is also related to the

percent overshoot. We also note that is not

the natural frequency, but for low values of the

damping ratio, we can assume the peak occurs

at the natural frequency.

Note that no peak will occur at frequencies

above zero if .

At a frequency, of:

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Since we know the relationships: and

, then we find the bandwidth is related to the

settling and peak time of a second order system.

Example 9.13

Find the closed loop bandwidth required for 20% overshoot and a 2 second settling time.

The bandwidth of the closed loop frequency response is defined as the frequency, , at which the

magnitude response curve is 3 dB lower than its initial value at DC:

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9.11 RELATION BETWEEN CLOSED LOOP TRANSIENT AND OPEN LOOP

FREQUENCY RESPONSE In this section, we consider situations in which we

are given the open loop frequency response and wish

to find information about:

The system bandwidth

The peak time

The settling time

The rise time

To do so, we must find the phase margin in terms of

the damping ratio and also a relationship between

the closed loop bandwidth and open loop frequency

response.

We simply state without proof that the phase margin can be determined as:

The closed loop bandwidth can be approximated as the frequency at which the open loop

magnitude response is between -6 and -7.5 dB if the open loop phase response is between

and -223

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Example 10.14

Given the system shown and the bode diagrams (open loop transfer

function always), find the settling time and peak time.

We attempt to find by first selecting the region

between -6 and -7.5 dB and 135 and 225 . This is

approximately 3.7 rad/s. The phase margin is about .

We then use the following graph or

to determine

Then:

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9.12 STEADY STATE ERROR CHARACTERISTICS FROM FREQUENCY RESPONSE We show how to find values of the static error constants for equivalent unity feedback systems from

unnormalised and unscaled bode plots.

POSITIVE ERROR CONSTANT

A type 0 system has the form:

Which will always yield a constant term at DC when we

take the logarithm. Given that:

We find that this value can be obtained from the low

frequency axis on an unscaled magnitude plot yielding:

VELOCITY ERROR CONSTANT

Since a type 1 system has an open loop transfer

function of the form:

Which always yields a 20 dB decrease in the

magnitude from 0 frequency. If we extend this, the

intersection of the -20 dB/dec drop with the

frequency axis yields:

Which is the same as the velocity error constant.

The position error constant is found at the intersection of the magnitude at DC. A Type

0 system always starts with a horizontal magnitude plot.

A Type 1 system always starts with a -20 dB/dec drop. The extension of a -20 dB/dec slope to the

frequency axis yields the velocity error constant,

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ACCELERATION CONSTANT

For a type 2 system:

There is an initial -40 dB/dec drop due to the

term part of the function:

And the intersection with the frequency axis yields:

Example 10.15

For each unnormalised and unscaled bode plot, find the system type and the appropriate static error constant.

This is a type 0 system due to the initial horizontal magnitude. Thus:

This is a type 1 system due to the -20 dB initial drop. The zero dB crossing is approximately at 0.55 rad/s, and

as such .

A Type 1 system always starts with a -40 dB/dec drop. The extension of a -40 dB/dec slope to the

frequency axis yields the square root of the acceleration error constant,

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This is a type 2 system with initial 40 dB drop. The zero crossing is at 3 rad/s and this corresponds to so,

.

MATLAB COMMANDS There are two ways to create a transfer function: zpk() and tf(). The first is used when we are given the transfer

function is factorised form, while the second method is used when we know the coefficients of the transfer

function numerator and denominator.

To get the bode plot/Nyquist for

To plot

To find information about margins:

G=zpk([-20], [-1 -7 -50], 1)

Bode(G); grid on

Nyquist(G)

Numg[ 1 3];

Deng=conv([1 2], [1 2 25]);

G=tf(numg, deng)

Bode(G)

[mag, phase, w]=bode(G);

Points=[20*log10(mag(:,:)’, phase(:,:)’, w]

[Gm, Pm, Wcg, Wcp]=margin(G);

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10.1 TRANSIENT RESPONSE VIA GAIN ADJUSTMENT As we have already seen, we can change the transient

response by adjusting the gain of a system. This was

previously done by mathematical analysis and by the

root locus method. We now use the frequency

response tools to design systems.

The procedure involves:

1. Drawing the bode plots for a convenient value

of gain

2. We determine the required phase margin

from the percent overshoot:

3. We find the frequency on the phase diagram that yields the desired phase margin given by the

segment CD.

4. We must then change the gain by an amount AB to force the magnitude curve to go through 0 dB at

.

Example 10.1

For the position control system shown, find the value of preamplifier gain, K, to yield a 9.5% overshoot in the

transient response for a step input.

1. For convenience, we normalize the bode plot with K=3.6 such that 0 dB is at

2. An overshoot of 9.5% corresponds to .

3. We locate the point on the phase plot which is above i.e . This occurs at 14.8

rad/s.

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4. At a frequency of 14.8 rad/s, the gain is -44.2 dB. This magnitude has to be raised to 0 dB to yield the

required phase margin. Since the log magnitude plot was drawn for K=3.6, a 44.2 dB increase or

will yield the required phase margin.

The gain adjusted open loop transfer function is then:

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10.2 LAG COMPENSATION The function of the lag compensator is to:

1. Improve the static error constant by increasing

only the low frequency gain without any

resulting instability

2. Increase the phase margin of the system to

yield the desired transient response

Consider the bode plot show with an unstable system

since the gain is higher than unity at . The gain

compensator, while not changing the low frequency

gain, will reduce the high frequency gain. Thus the low

frequency gain can be made high to yield a large

without creating instability. The process of designing

such a lag compensator is based on knowning that the

transfer function of the lag compensator is:

Where

The general procedure is:

1. Set the gain, K, to the value that satisfies the steady state error specification and plot the bode plots

2. Find the frequency where the phase margin is to greater than the phase margin that yields the

desired transient response. This step compensates for the fact that the phase of the lag compensator

may still contributed to of phase at the phase margin frequency.

3. Select a lag compensator whose magnitude response yields a composite bode magnitude diagram

that goes through 0 dB at the frequency found at step 2.

a. Draw the compensator’s high frequency asymptote to yield 0 dB for the compensated

system at the frequency found in step 2.

b. Select the upper break frequency to be 1 decade below the frequency found in step 2

c. Select the low frequency asymptote to be at 0 dB

d. Connect the compensator’s high and low frequency asymptotes with a -20 dB/decade line to

locate the lower break frequency

4. Reset the system gain, K, to compensate for any attenuation in the lag network in order to keep the

static error constant the same as that found in step 1.

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Example 10.2

Given the system shown, use bode diagrams to design a lag compensator to yield a ten fold improvement in

steady state error over the gain compensated system while keeping the percent overshoot at 9.5%.

1. Find value of k which satisfies steady state

We first find a value of K which satisfies the transient percent over shoot.

Recall that we found that and the phase margin was . This required K to be adjusted to:

Thus the open loop system is:

Which is a type 1 system, with a velocity error constant of:

A ten fold increase in error means that the new velocity error constant must be:

Thus, the new value of K which satisfies the steady state is:

And the new open loop transfer function is:

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2. Now fix the transient response by finding desired phase margin

Since for 9.5% overshoot, the required phase margin is . We add for compensation so

the phase margin is , and the phase now must be .

This occurs at 9.8 rad/s. At this frequency, the magnitude plot must go through 0 dB, so we must introduce a -

24 dB lag compensator at that particular frequency.

3. We select the high frequency asymptote as -24 dB starting from a decade below the phase margin

frequency,

. Then starting at 0.98 rad/s, draw a 20 dB/decade line until 0dB is

reached. The lower break frequency is the intersection of this line with the frequency axis, and is

found to be 0.062 rad/s.

4. To retain the value of , the compensator must have a DC gain of unity, so:

Thus:

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11.3 LEAD COMPENSATION The lead compensator:

1. Increases the bandwidth by increasing the gain crossover frequency

2. Increases the phase margin and phase margin frequency to reduce the percent overshoot (larger

phase margin) with smaller peak times (higher phase margin frequencies)

The uncompensated system has a small phase

margin (B) and a low phase margin frequency (A).

Using a phase lead compensator, the phase angle

plot is raised for higher frequency. At the same

time, the gain crossover frequency in the

magnitude plot is increased from A rad/s to C rad/s.

These effects yield a larger phase margin (D), a

higher phase margin frequency (C), and a larger

bandwidth.

, with

setting the compensator to unity.

Some useful analytical expressions include finding

the frequency , at which the maximum phase

angle occurs:

The maximum phase shift of the compensator, is given by:

With the compensator’s magnitude at being:

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DESIGN PROCEDURE

1. Find the closed loop bandwidth required to meet the settling time, peak time, or rise time

requirements using:

2. Since the lead compensator has negligible effect at low frequencies, set the gain, K, of the

uncompensated system to the value that satisfies the steady state error requirement.

3. Plot the bode magnitude and phase diagrams for this value of gain and determine the

uncompensated system’s phase margin

4. Find the phase margin to meet the damping ratio or percent overshoot requirement. Then evaluate

the additional phase contribution required from the compensator

5. Determine the value of by using the additional phase contribution of the compensator.

6. Determine the compensator’s magnitude at the peak of the phase curve

7. Determine the new phase margin frequency by finding where the uncompensated system’s

magnitude curve is the negative of the lead compensator’s magnitude at the peak of the

compensator’s phase curve.

8. Design the lead compensator’s break frequencies to find T and the break frequencies

9. Reset the system gain to compensate for the lead compensator’s gain

10. Check the bandwidth requirement is met

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Example 10.3

Given a unity feedback system with a forward transfer function of

, design a lead

compensator to yield a 20% overshoot, with a peak time of 0.1 seconds.

1. We first find the closed loop bandwidth requirement given and

2. Since

And then:

3. We make the bode plots.

4. Next the required phase margin is:

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Since the lead compensator also increases the phase margin frequency, we add a correction factor

such that:

But currently, the system’s phase margin is and . Thus, the total phase margin

contribution of the lead compensator is:

This will start to yield a system with a phase margin of with a bandwidth of 46.6rad/s.

5.

This yields

6. Then the compensator’s magnitude at is

7. Next, we find where the uncompensated system has a magnitude of -3.76 dB. This occurs at 39 rad/s,

so our new phase margin frequency is

8. Then we find T:

We now require to find the components 1/T,

to derive the equation of the compensator

Hence:

9. The final compensated system is thus:

10. We now check the lead compensated open loop magnitude response.

Recall that the connection between the open loop bode plots to the closed loop bandwidth is that the

open loop response must be between -6 and -7.5 dB and 135 and 225 degrees.

As such we estimate that , which exceeds our bandwidth requirement.

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10.4 LAG-LEAD COMPENSATION Our approach to such design problems

is to first perform the preliminary work

to find the desired bandwidth, gain K to

yield the static error constant, the

required phase margin, selection of a

new phase margin frequency around

and to work out the phase

contribution of a lead compensator

given that the phase contribution of a

lag compensator is about .

The next step is to find the lag

compensator using the new phase

margin frequency. Then using the

above graph, we find a value of , to

help us find:

Such that:

Where is the higher break frequency determined to be 1 decade below the new phase margin

frequency and

it determined to yield 0 dB (unity gain) when s=0.

Our final step is to design the lead compensator given that we know to be the selected new phase

margin frequency, and we also know . Thus:

Where is the higher break frequency found by using:

And:

Is the lower break frequency.

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Example 10.4

Given a unity feedback system where

, design a passive lag-lead compensator using

bode diagrams to yield a 13.25% overshoot, a peak time of 2 seconds and

Preliminary Work

1. We first find the required bandwidth given

2. Given

3. We make the bode plots

4. The required phase margin is:

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5. We select a new phase margin frequency around , let’s say .

6. The uncompensated phase is . The lag compensator will add about contribution so this

gives a total of . We require a phase given by and as such, require the

lead compensator to provide .

Lag Compensator

The lag compensator is not critical and only provides stabilization, not phase margin design.

7. Since we chose , our higher break frequency will be at 0.18 rad/s. We find using the graph

shown:

We can now find our lower break frequency since:

Must be unity at DC so,

Lead Compensator

8. At , the lag compensated system has a phase angle of . Sinev we know ,

, then:

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11.1 DESIGN VIA STATE SPACE In chapter 8, we used the root locus method to design a system that met both steady state and transient

response. In the previous chapter, we used the frequency method to design our system. To finish off, we use

the state space method to design systems which are:

Non linear

MIMO systems

Specified for all higher poles

The disadvantage though is that we cannot specify the closed loop zero locations which frequency domain

methods do allow through placement of the lead compensator zero.

CONTROLLER DESIGN

Consider an nth order feedback control system with an nth order closed loop characteristic equation:

Let us consider a plant:

If we introduce feedback of each state variable to the control, u:

The resulting signal flow diagram changes from the phase variable representation:

To the state variable feedback representation:

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We shall see that the phase variable form with its typical lower companion system matrix, or the controller

canonical form, with its typical upper companion system matrix, yields the simplest evaluation of the feedback

gains.

The main design principles in this chapter consists of:

Using the phase variable of controller canonical form

Equating the characteristic equation of a closed loop system to a desired characteristic equating and

then finding the values of the feedback gains

11.2 POLE PLACEMENT IN PHASE VARIABLE FORM (MATCHING COEFFICIENTS) The procedure is to:

1. Represent the plant in phase variable form

2. Feed back each phase variable to the input of the plant through gain

3. Find the characteristic equation for the closed loop system represented

4. Decide upon all closed loop pole locations and determine an equivalent characteristic equation

5. Equate like coefficients of the characteristic equations and solve for

Recall that the phase variable representation of the plant is given by:

The characteristic equation of the plant is thus:

With the closed loop state variable feedback being:

Where

This gives:

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Which gives a characteristic equation of the closed loop system of:

Given that we have a desired characteristic equation for proper pole placement,

, we can compare the coefficients to find the feedback gains since .

Example 11.1

Design a phase variable feedback gains to yield 9.5% overshoot and a settling time of 0.74 second for the given

plant:

1. We first use the specifications to find the poles:

Thus:

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Now the original system is a third order, so we must select another closed loop pole. Given a zero at -5, it

would be wise to choose -5 as a pole so that it will cancel. We however choose -5.1 to show the effect of such

a selection. The desired characteristic equation is thus:

2. We first draw a signal diagram of the system and deduce the phase variable form

From the transfer function:

Using the phase variable form of the matrix (or the linear equations of them), we can draw our signal diagram

with the feedback of the gains.

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The resulting closed loop system’s state equation will be the same except for the system matrix which is now

:

The characteristic equation is thus:

Comparing coefficients:

We thus obtain:

Since the zero is unchanged, the transfer function numerator is unchanged, only our denominator has changed

which is given by

When we take a simulation of the closed loop

system, there is a large steady state error given

that the steady state response is 0.24 and not

unity. A better system would be achieved if we

had actually chosen -5 as our pole.

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The following example shows the same approach when the system is not in phase variable form.

Example 11.2

Given a plant with a transfer function:

Design a state feedback for the plant represented in cascade form to yield a 15% overshoot with a settling time

of 0.5 seconds.

In cascade form:

We then break this down in the signal diagram:

We can derive the matrix representation of the cascade system by noting:

When adding feedback, our system matrix will then become:

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And our characteristic equation is:

Our desired characteristic equation given:

Thus:

Matching coefficients:

11.3 CONTROLLABILITY

Conversely, if an input to a system can be found that takes every state variable from a desired initial state to a

desired final state, the system is said to be controllable.

Consider the parallel system shown. If could not be controlled by u, and also exhibited an unstable

response due to a nonzero initial condition, there would be no way to effect a state feedback design to

stabilize . In such a case, a state feedback design is not possible.

If any one of the state variables cannot be controlled by the control u, then we cannot

place the poles of the system where we desire.

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The state equation for the controllable system is:

While the state equation for the uncontrollable system shows that is decoupled from u.

Recall that the rank of a matrix is determined by the number of linear independent rows or columns, or put

simply in row echelon form, the number of non-zero rows. Using this, and the controllability matrix of an

nth order plant whose state equation is

The controllability matrix must be of rank n. Another method of determining whether the rank of is n is if

the determinant of is non-zero.

Example 11.3

Given the system shown, determine its controllability

A system with distinct eigenvalues and a diagonal system matrix is controllable if the

input coupling matrix B does not have any rows that are zero.

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Thus, the state equation is:

Thus:

The determinant of is not zero, so the system is controllable.

11.4 OBSERVER DESIGN Controller design relies upon access to the state variables for feedback through adjustable gains – but this may

not be available or viable. An observer sometimes called an estimator is used to calculate state variables that

are no accessible from the plant. Here the observer is a model of the plant.

Assuming a plant with state equation

And an observer with the state equation

We take the difference:

Note that the speed of convergence between the actual and estimated state is the same as the transient

response of the plant since the characteristic equation is the same. From the figure shown, we can increase

the speed of convergence between the actual and estimated states using feedback.

In designing an observer, the observer canonical form yields the easy solution for the observer gains. Thus,

the state equations for the plant and observer should be in observer canonical form.

Similar to the design of the controller vector, K, the design of the observer consists of evaluation the constant

vector, L, so that the transient response of the observer is faster than the response of the controlled loop in

order to yield a rapidly updated estimate of the state vector.

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With the feedback in place in the closed loop observer we find that:

Then:

The design then consists of solving for the value of to yield a desired characteristic equation:

We then select the eigenvalues of the observer to yield stability and a desired transient response that is 10

times faster than the controlled closed loop response.

For an nth order plant, the observer canonical form is:

The characteristic equation for A-LC is then:

We then compare this characteristic equation with the desired characteristic equation to find the elements of

.

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Example 11.4

Design an observer for the plant represented in observer canonical form. The observer will respond 10 times

faster than the closed loop control system with poles at and -10.

1. Draw the estimated plant in observer canonical form:

In phase variable

In controller form we reverse the order of the phase variables and rearrange:

Using the relationship of we derive the observer canonical form:

2. Take the difference between the plant’s actual output y, and the observer’s estimated output and

add the feedback paths from this difference to the derivate of each state variable. Note the only

difference is in the system matrix which is now A-LC

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3. From step 2, we take the system matrix and find the observer error:

The observer characteristic polynomial is given by:

4. Our desired characteristic polynomial needs to be 10 times faster. Thus the poles are 10 times greater

than the original poles at and -10.

5. Comparing the coefficients:

Again, for plants not in observer canonical form, we can do exactly the same thing although the calculations

may be more difficult.

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Example 11.5

Design an observer for the phase variables with a transient response described by and given

the plant is described by:

In phase variable form:

Then:

Thus:

From the problem, we find that the desired characteristic polynomial is:

Matching the coefficients then:

Our final solution is then:

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11.5 OBSERVABILITY

Given a plant in the form (note can be in phase variable, observer canonical etc):

We define the observability matrix as:

If the system is observable then, the rank of the observability matrix must be of rank n.

Example 11.6

Determine whether the system show is observable

If any state variable has no effect upon the output, then we cannot evaluate this state

variable by observing the output.

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Then:

Thus:

Since the determinant of the observability matrix is 0, the system is not observable.

11.6 STEADY STATE DESIGN VIA INTEGRAL CONTROL We now extend our design of the PI controller by introducing a feedback path from the output to form the

error, e, which is fed forward to the controlled plant via an integrator. The integrator increases the system

type and reduces the previous finite error to zero.

The error is given by:

Writing the state equations we have:

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We then produce an augmented matrix:

But, from the diagram:

We thus obtain:

As the system type has been increased, we now have an additional pole to place. By using the characteristic

equation given by the determinant of

, we can design K and .

Example 11.7

Design a controller with integral control to yield a 10% overshoot and settling time of 0.5 second given a plant:

Then:

The characteristic equation of the observer is then:

The desired characteristic polynomial has poles at :

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Since this is a third order system, we must introduce a third pole. We select the third pole to be -100 i.e at

least 5 times greater than the real parts of the second order poles.

Thus:

By matching the coefficients:

Then,

We can find the integral controller transfer function to be:

And steady state error for unit step input to be:

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