Chapter 2 – Analog Control System (cont.)

Reduction of Multiple SubsystemSystem Response – Poles/ Zeros, Second Order System, Steady State Error, Stability Analysis

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.

Component of a block diagram for a linear, time-invariant system

Example 1

Cascade form

Example 2

Parallel form

Example 3

Example 4

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 equation2. inverse Laplace transformation

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

0(t)yt

limt

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 .

Example 1

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.

cont.

Solution:Solution:

To show the properties of the poles and zeros, lets applied unit step 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 response of the system. Multiplying the transfer function by a step function yields :yields :

Thus :Thus :

5s

3/5s

2/55ss2s

C(s)

5te53

52

c(t)

cont.

From the development summarize, the following From the development summarize, the following conclusions can be drawn:conclusions can be drawn:

Second Order System

A system where the closed loop transfer function possesses two poles is called a second-order system

cont.

cont.

Step input

cont.

cont.

cont.

Example 1

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

Determine:

Rise time, tr

Peak time, tpPercent overshoot, %Mp

Settling time ts for 2% criterion

7505.27

750

)(

)(2

sss

s

i

o

cont.

Solution:

27.39 rad/s

2 x 27.39

cont.

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

cont.

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().

cont.

The formula to obtain steady-state error is as follow:

Step input

Ramp Input

Parabolic Input

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.

cont.

Example 1

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

ssssHsG

sGsT

3)2)(1(

3

sss 323

323

sss

0323 23 sss

cont.

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 2

Determine the stability of the system shown below.

Solution:

)2)(1(

71

)2)(1(

7

)()(1

)()(

sss

ssssHsG

sGsT

7)2)(1(

7

sss

723

723

sss

cont.

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

0723 23 sss