Review last lectures. procedures for drawing block diagram 1. Write the equations that describe the...

Review last lectures

procedures for drawing block diagram

1. Write the equations that describe the dynamic behavior for each component.

2. Take Laplace transform of these equations, assuming zero initial conditions.

3. Represent each Laplace-transformed equation individually in block form.

4. Assembly the elements into a complete block diagram.

block diagram: example

Let consider the RC circuit:

R

Cei eoi

The equations for this circuit are: C

idte

R

eei o

oi

block diagram: exampletake Laplace transform:

R

sEsE

R

tetesI oioi )()()()()(

L

Cs

sI

C

idtsEo

L

block diagram: example

block representations for Laplace transforms:

R

sEsEsI oi )()()(

R

1_+

)(sI

sEo

)(sEi

Cs

sIsEo

Cs

1)(sI sEo

block diagram: exampleAssembly the elements into a complete block diagram.

R

1_+

)(sI

sEo

)(sEi

Cs

1)(sI sEo

block diagram reductionRules for reduction of the block diagram: Any number of cascaded blocks can be reduced by a

single block representing transfer function being a product of transfer functions of all cascaded blocks.

The product of the transfer functions in the feedforward direction must remain the same.

The product of the transfer functions around the loop mast remain the same.

block diagram: reduction example

R_+

_+1G 2G 3G

1H

2H

++

C


R_+

_+

1G 2G 3G

1H

1

2

G

H

++

C


R_+

_+

21GG 3G

1H

1

2

G

H

++

C


R_+

_+ 21GG 3G

1H

1

2

G

H

++

C


R_+

_+

121

21

1 HGG

GG

3G

1

2

G

H

C


R_+

_+

121

321

1 HGG

GGG

1

2

G

H

C


R_+

232121

321

1 HGGHGG

GGG

C


R

321232121

321

1 GGGHGGHGG

GGG

C

block diagram: reduction exampleNOTICE: Numerator of the closed-loop transfer function

C(s)/R(s) is the product of the transfer functions of the feedforward path.

1. The denominator of the closed-loop transfer function C(s)/R(s) is equal to:

1-Σ( product of the transfer functions around each loop)

2. The positive feedback loop yields a negative term in the denominator.

signal flow graph

input node (source)

b1x a

2x c

4x

d

1

3x

3x

output node (sink)

mixed node

input node (source)

mixed node

forward path

path

loop

branch

node

transmittance

flow graphs of control systems

)(sR)(sG

)(sC )(sG

)(sR )(sC

block diagram: signal flow graph:

flow graphs of control systems)(sR

_+

)(sH

)(sG)(sC)(sE

block diagram:

signal flow graph:)(sG

)(sR)(sC

1

)(sE )(sH

flow graphs of control systems

)(sR

_+

)(sH

)(sG)(sC)(sE

block diagram:

signal flow graph:)(sG

)(sR )(sC

1

)(sE )(sH

++

)(sN

)(sN1

1)(sC

signal flow graph algebra

1xa

2x12 axx

b

1x

a

2x 3x 1x

ab

2x


b

a

2x

1x c

4x3xbc

ac

2x

1x

4x

1x

ba 2x

1x 2x

a

b


1xbc

ab

13x

1x 3xa bc

2x

ab1x

bc

3x

bc

ab

x

x

11

3

312

23

cxaxx

bxx

313 bcxabxx

flow graphs of linear systems113132121111 ubxaxaxax

223232221212 ubxaxaxax

3332321313 xaxaxax

3x

1x2x1

11u1b

12a

11a

13a

1x 3x2x

2u 2b23a

22a

21a

33a31a

32a

Transient and steady state response analyses

Input signal not known ahead of time but is random in nature!

In analyzing and designing we need basic comparison of performance so we need input signal by specifying particular test input signals.

Typical Test Signals

1. Step functions

2. Ramp functions

3. Acceleration (Parabolic) functions

4. Impulse functions

5. Sinusoidal functions

6. Polynomial functions

Note: which input signals we must use? Depend on the system normal operation inputs

ttf

ttf

function impulseUnit ,0 As

otherwise ,022

- ,/1

Test Input Signals

The impulse input is useful when we consider the convolution integral for the output y(t) in terms of an input r(t):

This relationship is shown in the block diagram:

If the input is a unit impulse function then

t

sRsGdttgty 1L

tgty

Transient response and Steady- State response

Time response parts Steady state

Output which behaves as t approach infinity

Transient

Which Goes from initial state to the final state

Or output minus steady state output!

First order systems response

Final value Theorem for steady state response

Second order systems response(steady state part)

Second order systems types

Over damped real poles

Critically damped Real equal poles

Under damped Imaginary poles

Over damped response

Critically damped

Under damped

Under damped 2

Performance of a Second-Order System

Consider the system:

The closed loop output is:

22

2

22

2

2

2

input, stepunit aWith

2

Or

1

nn

n

nn

n

ssssY

sRss

sY

sRKpss

KsR

sG

sGsY

10 ,cos ,1

sin1

1

:isoutput transientThe

12

n

nt tety

UnstableStable

Performance of a Second-Order System The transient response of this second-order system is shown below. As ζ decreases, the closed-loop roots approach the imaginary axis,

and the response becomes oscillatory.

tety ntn 2

21sin

1

11:on Based

Performance of a Second-Order System For the unit impulse (R(s)=1) the output is:

The transient response is:

The impulse response of the

second order system is shown here

22

2

2 nn

n

sssY

tety ntn n 2

21sin

1

Performance of a Second-Order System Standard performances are usually defined in terms of the step

response of a system as shown below:

The swiftness of theresponse is measuredby the rise time (Tr),and the peak time (Tp).

Performance of a Second-Order System The percent overshoot for the unit step input is defined as:

Mpt: is the peak value of the time response fv: is the final value of the response

Normally fv is the magnitude of the input, but many systems have a final value that is different from the desired input magnitude.

%100P.O.

fv

fvMtp


Settling time (Ts): the time required for the system to settle within a certain percentage, δ, of the input amplitude.

The settling time is four time constants (τ=1/ζωn) of the dominant roots of the characteristic equation.

The steady-state error of the system may be measured on the step response of the system as shown in the previous figure.

Performance of a Second-Order System Consider the second order system with closed-loop damping

constant ζωn and a response described by

We seek to determine the time, Ts, for which the response remains within 2% of the final value.

ns

sn

T

T

T

e sn

44

4

02.0

tety ntn 2

21sin

1

11


The transient response of the system may be described in terms of two factors: The swiftness of the response, as represented by

the rise time and the peak time The closeness of the response to the desired

response, as represented by the overshoot and settling time

Performance of a Second-Order System The peak time relationship for this second-order system is:

The peak response is:

The percent overshoot is:

21

n

pT

211

eMtp

21100O..P

e

Performance of a Second-Order System The swiftness of step response can be measured as the time it takes to

rise from 10% to 90% of the magnitude of the step input. This is the rise time (Tr).

The swiftness of a response to a step input is dependent of ζ and ωn.

For a given ζ, the response is faster for larger ωn.

The overshoot is independent of ωn.

For a give ωn, the response is faster for lower ζ. The swiftness of the response will be limited by the overshoot that can

be accepted

rT 10 ,cos ,1 12 n

Quiz and Home work

Don’t forget quiz in next timeHome works of chapter 1 No: 1, 2, 3, 4, 5, 6, 7, 8, 13.

Review last lectures. procedures for drawing block diagram 1. Write the equations that describe the...

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