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7. Reduction of Multiple Subsystem

8. System Response – Poles/ Zeros, Second Order

System, Steady State Error, Stability Analysis

Muhammad Faizal bin Ismail

Dept. of Electrical Engineering

PPD, UTHM

ifaizal@uthm.edu.my

013-7143106

7. Reduction of Multiple Subsystem

• Subsystem is represented as a block with an input, output and transfer function.

• Many systems are composed of multiple subsystems.

• When multiple subsystem are interconnected, new element must be added like summing junction and pickoff points.

• The main purpose of reduction multiple subsystem is to reduce more complicated systems to a single block.

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Component of a block diagram for a

linear, time-invariant system

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Example 1Cascade form

7. Reduction of Multiple Subsystem

Example 2Parallel form

7. Reduction of Multiple Subsystem

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Example 3

Cascade

Eliminate feedback

Cascade

Example 4

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8. System response

• Time is an independent variable mostly used in a control system evaluation, also referred as time response.

• In order to find a system response, 2 steps are commonly used:

1. differential equation

2. inverse Laplace transformation

8. System response (cont.)

• In general, system response contains 2 parts:

1. Transient response (natural response)

• Part of the time response that goes to zero as time becomes very large.

2. Steady state response (forced response)

• Part of the time response that remains after the transient has died out

0000(t)(t)(t)(t)yyyytttt limlimlimlim tttt ====∞∞∞∞→→→→

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Example 1

Find the response of the system for a step input .

Solution :

Poles, Zeros and System

Response

• Poles is the values of the Laplace transform variable(s) that cause the transfer function to become infinite or any roots of the denominator of the transfer function .

• Zeros is the values of the Laplace transform variable(s) that cause the transfer function to become zero, or any roots of the numerator of the transfer function .

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Example 2

Given the transfer function G(s) in Figure below, a pole exists

at s = -5, and a zero exists at -2. These values are plotted on

the complex s-plane, using x for the pole and О for the zero.

Example 2 (cont.)

Solution:

To show the properties of the poles and zeros, lets applied

unit step response of the system. Multiplying the transfer

function by a step function yields :

( )

( )5ss

2sC(s)

+

+=

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Second Order System

• A system where the closed loop transfer function possesses

two poles is called a second-order system

Second Order System (cont.)

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• Step

input

Second Order System (cont.)

Second Order System (cont.)

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1. Rise Time, Tr, Time required for response to to rise

from 10% to 90% of its final value.

2. Peak Time, Tp, Time required for response the reach

the first peak of overshoot.

3. Max overshoot, Mp, Maximum peak value of the

response curve measured from unity.

4. Settling time, Ts, Time required for the response

curve to reach and stay within the range about the

final value of sized specified by percentage of the

final value( 2% or 5%)

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Second Order System (cont.)

Example 3

The transfer function of a position control system is given by:

Determine:

Rise time, tr

Peak time, tp

Percent overshoot, %Mp

Settling time ts for 2% criterion

7505.27

750

)(

)(2

++=

sss

s

i

o

θ

θ

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Solution:

27.39 rad/s

2 x 27.39

Example 3

Example 3

Solution:

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Steady State Error

• Steady-state error is the difference between the

input and the output for prescribed test input as t �

∞.

• Test inputs used for steady-state error analysis are:

– Step Input

– Ramp Input

– Parabolic Input

• Figure has a step input and two possible output. Output 1 has

zero steady-state error, and Output 2 has a finite steady-state

error e2(∞).

Steady State Error

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• The formula to obtain steady-state error is as

follow:

Steady State Error (cont.)

Step input

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Ramp Input

Parabolic Input

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Stability Analysis

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

• A system is unstable if any bounded input yields an unbounded output.

• A system is marginally stable/neutral if the system is stable for some bounded input and unstable for the others (as undamped) but remains constant or oscillates.

Stability Analysis in the Complex

s-Plane

There have 3 condition of poles location that indicates transient response, which is:

1. Stable systems have closed-loop transfer function with poles only in the left half-plane(LHP).

2. Unstable systems have closed loop transfer function with at least one pole in the right half-plane (RHP).

3. Marginally stable systems have closed loop transfer function with only imaginary axis poles.

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Stability Analysis in the Complex

s-Plane

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Example 4

Determine whether the unity feedback system below is stable

or unstable.

Solution:

Let equation = 0 to find the poles,

)2)(1(

31

)2)(1(

3

)()(1

)()(

+++

++=

+=

sss

sss

sHsG

sGsT

3)2)(1(

3

+++=

sss323

323

+++=

sss

032323

=+++ sss

Thus, the roots of characteristic equation are

-2.672

-0.164 + j1.047

-0.164 - j1.047

Plot pole-zero on the s-plane and sketch the response of the system >

all of the roots are locate at left half-plane, therefore the system is

stable.

Example 4 (cont.)

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Determine the stability of the system shown below.

Solution:

)2)(1(

71

)2)(1(

7

)()(1

)()(

+++

++=

+=

sss

sss

sHsG

sGsT

7)2)(1(

7

+++=

sss 723

723

+++=

sss

Example 5

Let equation = 0 to find the poles,

Thus, the roots of characteristic equation are,

-3.087

0.0434 + j1.505

0.164 - j1.505

Plot pole-zero on the s-plane and sketch the response of the system >

only one of the roots are locate at left half-plane and the others 2

roots locate at right half-plane, therefore the system is unstable

072323

=+++ sss

Example 5 (cont.)