Chapter 4

Continuous Time SignalsContinuous Time SignalsTime ResponseTime Response

Poles and Zeros

Poles: (1) the values of the Laplace transform variable, s, that cause

the transfer function to become infinite(2) Any roots of the denominator of the transfer function that

are common to roots of the numerator.Zeros:(1) the values of the Laplace transform variable, s, that cause

the transfer function to become zero(2) Any roots of the numerator of the transfer function that

are common to roots of the denominator.

Figure 4.1

a. System showinginput and output;b. pole-zero plotof the system;c. evolution of asystem response.Follow blue arrowsto see the evolutionof the responsecomponent generatedby the pole or zero.

Poles and Zeros

1) A pole of the input function generates the form of the forced response ( steady state 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 of the form e-at

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

Figure 4.2

Effect of a real-axispole upon transientresponse

Ex: Evaluating response using poles

Problem:Given the system, write the output, c(t), in general terms. Specify the

natural and forced parts of the solution.

By inspection, each system pole generates an exponential as part of the natural response, and the input’s pole generates the s s response, thus

1( ) ( )( 3)

( 2)( 4)( 5)R s C ss s

s s s

31 2 4

2 4 51 2 3 4

( ) ( 2) ( 4) ( 5)

( ) t t t

kk k kC s

s s s s

c t k k e k e k e

Forced Natural Response Response

Figure 4.4

a. First-order system;b. pole plot

Steady State Step Response

The steady state response (provided it exists) for a unit step is given by

where G(s) is the transfer function of the system.

First Order Systems

We define the following indicators:

Steady state value, y: the final value of the step response (this is meaningless if the system has poles in the RHP).

Time Constant, 1/a: The time it takes the step response to rise to 63% of its final value. Or the time for e-at to decay 37% of its initial value.

First Order Systems

Rise time, Tr: The time elapsed for the waveform to go from 0.1 to 0.9 of its final value

Settling time, Ts: the time elapsed until the step response enters (without leaving it afterwards) a specified deviation band, ±, around the final value. This deviation , is usually defined as a percentage of y, say 2% to 5%.

Overshoot, Mp: The maximum instantaneous amount by which the step response exceeds its final value. It is usually expressed as a percentage of y

Figure 4.3: Step response indicators

Figure 4.5

First-order systemresponse to a unitstep

Figure 4.6

Laboratory resultsof a system stepresponse test

Figure 4.7

Second-ordersystems, pole plots,and stepresponses

Figure 4.8

Second-orderstep response componentsgenerated bycomplex poles

Figure 4.9

System forExample 4.2

Factoring the denominator we get s1 = -5+j13.23 & s2 = -5-j13.23

The response is damped sinusoid

Figure 4.10

Step responsesfor second-ordersystemdamping cases

The General Second Order system

Natural FrequencyIs the frequency of oscillation of the system without damping Damping Ratio

General System

n

Exponential decay Frequency 1 Natural period (seconds)Natural Frequency 2 Exponentail timec onstant

2

2 2 2( )2

n

n n

bG ss as b s s

2n

n

ba

Figure 4.11

Second-orderresponse as a function of damping ratio

21,2s 1n n

Relating and the pole locations we haven to

Figure 4.12 Systems for Example 4.4

2n

n

ba

Since then

2ab

Using values of a & b from each system we find for (a) overdamped for (b) critically damped for (c) underdamped

1.155

1 0.894

Figure 4.13

Second-orderunderdampedresponses fordamping ratio values

Figure 4.14

Second-orderunderdampedresponsespecifications

Figure 4.15

Percentovershoot vs.damping ratio

Figure 4.16

Normalized risetime vs. dampingratio for asecond-orderunderdampedresponse

Figure 4.17

Pole plot for anunderdamped second-ordersystem

Figure 4.18

Lines of constantpeak time, Tp , settlingtime,Ts , and percentovershoot, %OSNote: Ts

2 < Ts

1 ;

Tp2

< Tp1; %OS1 <

%OS2

Figure 4.19

Step responsesof second-orderunderdamped systemsas poles move:a. with constant real part;b. with constant imaginary part;c. with constant damping ratio

Figure 4.20 Pole plot for Example 4.6

Find , , ,% n p sT OS and T

Solution:1cos cos[tan (7 / 3)] 0.394

2

2 2

( / (1 ) ) 100

7 3 7.616

0.449 sec7

% OS = e 26%and approximate settling time is

4 4T 1.333 seconds3

n

pd

X

sd

T ond

Figure 4.22

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Figure 4.23

Component responses of a three-pole system:a. pole plot;b. componentresponses: nondominant pole is neardominant second-order pair (Case I), far from the pair (Case II), andat infinity (Case III)

Figure 4.24 Step responses with additional poles

1 2

2 2

3 2

21

10 22

3 22

24.542( )4 24.542

245.42( )( 10)( 4 24.542)

73.626( )( 3)( 4 24.54)

( ) 1 1.09 cos(4.532 23.8 )

( ) 1 0.29 1.189 cos(4.532 53.34 )

( ) 1 1.14 0.707 cos(4.532

t

t t

t t

T ss s

T ss s s

T ss s s

c t e t

c t e e t

c t e e t

78.63 )

Note that c2(t) is better approximate of c1(t) than c3(t) since the 3rd pole is farthest from dominant poles

Figure 4.25 Effect of adding a zero to a two-pole system

The system has 2 poles at -1+j2.828 and -1-j2.828

Note: the farther the location of the zero from the dominant poles the lesser its effect on the response

Transfer Functions for Continuous Time State Space ModelsTaking Laplace transform in the state space model equations yields

and hence

G(s) is the system transfer function.

Laplace transform solution: Example

Problem: Given the system represented in state space by the following equations: a) solve the state equation and find output b) find eigenvalues and poles

Solution:

First we find (sI-A)-1 =

0 1 0 00 0 1 024 26 9 1

1 1 0

1(0) 0

2

tX X e

y X

X

2

2

2

3 2

( 9 26) ( 9) 124 ( 9 )

24 (26 24)

9 26 24

s s ss s s

s s s

s s s

Laplace transform solution: Example Cont.Using U(s) = 1/(s+1), laplace transform of e-t, and B and x(0) from given equations, we

sustitue in X(s) = (sI-A)-1[x(0) + BU(s)], we get

The output is

3 2

1

2

2

2

3

( 10 37 29)( 1)( 2)( 3)( 4)

(2 21 24)( 1)( 2)( 3)( 4)

(2 21 24)( 1)( 2)( 3)( 4)

s s sXs s s s

s sXs s s s

s s sXs s s s

1

2 1 2

3

3 2

( )( ) 1 1 0 ( ) ( ) ( )

( )

( 12 16 5)( )( 1)( 2)( 3)( 4)6.5 19 11.5 =

2 3 4

X sY s X s X s X s

X s

s s sY ss s s s

s s s

Laplace transform solution: Example Cont.

Taking Laplace transform of Y(s) we get

b)We set det(sI-A) = 0 to find poles and eigenvalues which are -2, -3, and -4

2 3 4( ) 6.5 19 11.5t t ty t e e e

Solution of Continuous Time State Space Models

A key quantity in determining solutions to state equations is the matrix exponential defined as

The explicit solution to the linear state equation is then given by

Time domain solution, Example:

Problem: for the state equation and initial state vector shown, where u(t) is a unit step. Find the state-transition matrix and then solve for x(t).

Solution: Find eigenvalues using det(sI-A) = 0. Hense s2+6s+8 = 0 and s1= -2 s2

= -4. Now we assume state transition matrix as

Solve for the constants using

0 1 0( ) ( )

8 6 1

1(0)

0

x x t u t

x

2 4 2 41 2 3 4

2 4 2 45 6 7 8

( ) ( )( )

( ) ( )

t t t t

t t t t

K e K e K e K et

K e K e K e K e

(0) and (0)=AI

2 4 2 4

2 4 2 4

(2 ) (1 / 2 1 / 2 )( )

( 4 4 ) ( 1 2 )

t t t t

t t t t

e e e et

e e e e

Time domain solution, Example: Cont.

Also,

Then,

2( ) 4( )

2( ) 4( )

2 4

4

(1 / 2 1 / 2 )( )

( 2 )and

(2 )( ) (0)

( 4 4 )

t t

t t

t t

t

e et B

e e

e et x

e e

2 2 4 4

0 0

0 2 2 4 4

0 0

2 4

2 4

1 / 2 1 / 2 )( ) ( )

2 )

1 / 8 1 / 4 1 / 8 =

1 / 2 1 / 2

t tt t

t

t tt t

t t

t t

e e d e e dt Bu d

e e d e e d

e e

e e

Time domain solution, Example Cont.

0

2 4

2 4

( ) ( ) (0) ( ) ( )

1 / 8 7 / 4 7 / 8

-7/2 7 / 2

t

t t

t t

x t t x t Bu d

e e

e e

Final result is,

State Space via Laplace transform Example:

Problem: Find the state transition matrix using laplace for the following system

Solution: we find

First find

For which

0 1 0( ) ( )

8 6 1

1(0)

0

x x t u t

x

1( ) as inverse Laplace transform of ( - )t sI A

1( )

8 ( 6)s

sI As

2 21

2

2 2

6 1 6 18 6 8 6 8( )

86 86 8 6 8

s ss s s s ssI A

ss ss s s s

State Space via Laplace transform Example Cont.

Expanding each term in the matrix using partial fractions

Finally, taking the inverse Laplace transform, we obtain

1

2 1 1 / 2 1 / 2( ) ( )2 4 2 4( )

4 4 1 2( ) ( )2 4 2 4

s s s ssI A

s s s s

2 4 2 4

2 4 2 4

(2 ) (1 / 2 1 / 2 )( )

( 4 4 ) ( 2 )

t t t t

t t t t

e e e et

e e e e

Step Response for Canonical Second Order Transfer Function

On applying the inverse Laplace transform we finally obtain

Figure 4.5: Pole location and unit step response of a canonical second order system.

Summary

There are two key approaches to linear dynamic models: the, so-called, time domain, and the so-called, frequency domain

Although these two approaches are largely equivalent, they each have their own particular advantages and it is therefore important to have a good grasp of each.

Time domain In the time domain,

systems are modeled by differential equations systems are characterized by the evolution of their

variables (output etc.) in time the evolution of variables in time is computed by

solving differential equations

Frequency domain In the frequency domain,

modeling exploits the key linear system property that the steady state response to a sinusoid is again a sinusoid of the same frequency; the system only changes amplitude and phase of the input in a fashion uniquely determined by the system at that frequency,

systems are modeled by transfer functions, which capture this impact as a function of frequency.