The output response of a system is the sum of two responses: the forced response (steady-state response) and the natural response (zero input response)

The poles of a transfer function are (1) the values of the Laplace transform variable, s , that cause the transfer function to become infinite, or (2) any roots of the denominator of the transfer function that are common to roots of the numerator.

The zeros of a transfer function are (1) the values of the Laplace transform variable, s , that cause the transfer function to become zero, or (2) any roots of the numerator of the transfer function that are common to roots of the denominator.

Out response, Poles, and Zeros

Figure 4.1a. System showing

input and output;b. pole-zero plot

of the system;c. evolution of

a system response.

1. A pole of the input function generates the form of the forced response.

2. A pole of the transfer function generates the form of the natural response.

3. A pole on the real axis generates an exponential response.

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

)5)(4)(2(

3

sss

sssR

1)( )(sC

542)( 4321

s

k

s

k

s

k

s

ksC

Natural response

Forced response

ttt ekekekktc 54

43

221)(

First-Order Systems

Figure 4.4a. First-order system

b. pole plot

Figure 4.5First-order system

response to a unit step

The time constant can be described as the time for to decay to 37% of its initial value. Alternately, the time is the time it takes for the step response to rise to of its final value.

The reciprocal of the time constant has the units (1/seconds), or frequency. Thus, we call the parameter the exponential frequency.

)()()()(

ass

asGsRsC

ate

atnf etctctc 1)()()(

37.01

1

eeat

at

63.037.011)(1

1

at

at

atetx

a

Rise time: Rise time is defined as the time for the waveform to go from 0.1 to 0.9 of its final value.

Settling time: Settling time is defined as the time for the response to reach, and stay within, 2% (or 5%) of its final value.

Second-Order Systems

Figure 4.7Second-order

systems, pole plots,and step responses

1. Overdamped response:

Poles: Two real at

Natural response: Two exponentials with time constants equal to the reciprocal of the pole location

2. Underdamped responses:

Poles: Two complex at

Natural response: Damped sinusoid with and exponential envelope whose time constant is equal to the reciprocal of the pole’s real part. The radian frequency of the sinusoid, the damped frequency of oscillation, is equal to the imaginary part of the poles

21,

tt ekektc 2121)(

dd j

)cos()( tAetc dtd

3. Undamped response:

Poles: Two imaginary at

Natural response: Undamped sinusoid with radian frequency equal to the imaginary part of the poles

4. Critically damped responses:

Poles: Two real at

Natural response: One term is an exponential whose time constant is equal to the reciprocal of the pole location. Another term is the product of time and an exponential with time constant equal to the reciprocal of the pole location

1j

)cos()( 1 tAtc

1

tt tekektc 1121)(

Figure 4.10Step responses for second-order

system damping cases

Figure 4.8Second-order step response components

generated by complex poles

Natural Frequency: The natural frequency of a second-order system is the frequency of oscillation of the system without damping.

Damping Ratio: The damping ratio is defined as the ratio of exponential decay frequency to natural frequency.

Consider the general system:

Without damping, bass

bsG

2)(

bbs

bsG n

2)(

b

a

radfrequencyNatural

frequencydecaylExponentia

n

2/||

sec)/(

2

4

2,

2

4

2

2

2

2

1

abj

as

abj

as

na 2 22

2

2)(

nn

n

sssG

Figure 4.11Second-order response as a

function of damping ratio

Underdamped Second-Order Systems

)1()(

1)(

)2()(

222

2321

22

2

nn

nn

nn

n

s

KsK

s

K

ssssC

)1()(

11

)(11

222

2

2

nn

nn

s

s

s

ttetc nn

tn 2

2

2 1sin1

1cos1)(

Step response

Taking the inverse Laplace transform

te ntn 2

21cos

1

11

where

2

1

1tan

Figure 4.13Second-order underdamped

responses for damping ratio values

Peak time: The time required to reach the first, or maximum, peak. Percent overshoot: The amount that the waveform overshoots the

steady-state, or final, 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 (or ) of the steady-state value.

Rise time: The time required for the waveform to go from 0.1 of the final value to 0.9 of the final value.

Evaluation of peak time:

%2 %5

22

2

2)()]([

nn

n

ssssCtcL

)1()(

11

222

2

2

nn

nn

s

tetc ntn n 2

21sin

1)(

Setting the derivative equal to zero yields

Peak time: 2

2

11

n

n

ntornt

21

n

pt

Figure 4.14Second-order underdamped

response specifications

Evaluation of percent overshoot ( ):

Evaluation of settling time:

The settling time is the time it takes for the amplitude of the decaying

sinusoid to reach o.o2, or

,

where is the imaginary part of the pole and is called the damped

frequency of oscillation, and is the magnitude of the real part of the

pole and is the exponential damping frequency.

100% max

final

final

c

ccOS

)1/(max

2

1)( etcc p 100% )1/( 2

eOS

02.01

12

tne

nnst

4102.0ln( 2

dn

pt

21

dnst

44

d

d

Figure 4.17Pole plot for an underdamped

second-order system

)1)(1(2)(

22

2

22

2

nnnn

n

nn

n

jsjssssG

Figure 4.18Lines of constant peak time,Tp , settlingtime,Ts , and percent overshoot, %OS

Note: Ts2 < Ts1 ;Tp2 < Tp1; %OS1 <%OS2

Figure 4.19Step responses of

second-orderunderdamped systems

as poles move:a. with constant

real part;b. with constant imaginary part;

c. with constant damping ratio

dn

pt

21 100% )1/( 2

eOS

2173 nndd jjj

dnst

44

Find peak time, percent overshoot, and settling time from pole location.

, ,

Figure 4.21Rotational mechanical system

Design: Given the rotational mechanical system, find J and D to yield 20% overshoot and a settling time of 2 seconds for a step input of torque T(t).

Under certain conditions, a system with more than two poles or with zeros can be approximated as a second-order system tat has just two complex dominant poles. Once we justify this approximation, the formulae for percent overshoot, settling time, and peak time can be applied to these higher-order systems using the location of the dominant poles.

rdn

dn

s

D

s

CsB

s

AsC

22)(

)()(

Figure 4.23Component responses of a three-pole system:

a. pole plot;b. component responses: nondominant pole is near dominant second-order pair (Case I),

far from the pair (Case II), and at infinity (Case III)

Remark:If the real pole is five timesfarther to the left than the dominant poles, we assumethat the system is represented by its dominant second-order pair of poles.

cs

bcac

bs

cbab

cs

B

bs

A

csbs

assT

)/()()/()(

))(()(

System response with zeros

If the zero is far from the poles, then is large comared to and . ca b

))((

)/(1)/(1)(

csbs

a

cs

bc

bs

cbasT

Figure 4.25Effect of adding

a zero to a two-pole system

)828.21( j

Figure 4.26Step response of a nonminimum-phase system

)()()()( saCssCsCas

Time Domain Solution of State Equations

)()()( tButAxtx

t tAAt dBuexetx0

)( )()0()(

t

dButxt0

)()()0()(

where is called the state-transition matrix.Atet )(

)0()()]0()([)]([ 1xAsIxttx LL

)(])[( 1 tAsI -1L