linear systems control and vibrations

8/21/2019 Linear Systems Control and Vibrations

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Linear Systems Control and Vibrationsfor Mechanical Engineers

Dr. Robert L. Williams II

Mechanical Engineering, Ohio University

NotesBook Supplement forME 3012 System Vibrations, Analysis, and Control

2015 Dr. Bob Productions

[email protected]/people/williar4

These notes supplement the ME 3012 NotesBook by Dr. Bob

This document presents supplemental notes to accompany the ME 3012 NotesBook. The outline

given in the Table of Contents on the next page dovetails with and augments the ME 3012 NotesBook

outline and hence is incomplete here.


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ME 3012 Supplement Table of Contents

1. INTRODUCTION ...............................................................................................................................................3

1.1 CONTROL SYSTEMS AND MECHANICAL VIBRATIONS DEFINITIONS ................................................................3

1.4 MATLABINTRODUCTION ...............................................................................................................................6

2. LINEAR SYSTEM SIMULATION ...................................................................................................................7

2.1 SOLUTION OF ORDINARY DIFFERENTIAL EQUATIONS .....................................................................................7

2.1.1 First-Order ODEs ......................................................................................................................................9

2.1.2 Second-Order ODEs ................................................................................................................................13

2.1.3 MATLAB Functions for ODEs .................................................................................................................19

2.2 THE LAPLACE TRANSFORM ............................................................................................................................20

2.2.2 ODE Solution via Laplace Transforms ...................................................................................................22

3. TRANSFER FUNCTIONS AND BLOCK DIAGRAMS ...............................................................................23

3.1 TRANSFER FUNCTIONS ...................................................................................................................................23

3.2 BLOCK DIAGRAMS .........................................................................................................................................24

3.4 FEEDBACK ......................................................................................................................................................27

4. TRANSIENT RESPONSE ................................................................................................................................28

4.1 SECOND-ORDER SYSTEM DAMPING CONDITIONS .........................................................................................28

4.2 SECOND-ORDER SYSTEM PERFORMANCE SPECIFICATIONS ...........................................................................36

4.3 OPEN-LOOP AND CLOSED-LOOP SYSTEM EXAMPLE...................................................................................... 42

4.4 FIRST-AND SECOND-ORDER TRANSIENT RESPONSE CHARACTERISTICS ......................................................46

4.5 SYSTEM TYPE .................................................................................................................................................51

4.7 TERM EXAMPLE OPEN-LOOP TRANSIENT RESPONSE ....................................................................................60

5. CONTROLLER DESIGN ................................................................................................................................64

5.1 CONTROLLER DESIGN INTRODUCTION ...........................................................................................................64

5.2 ROOT-LOCUS METHOD ..................................................................................................................................69

5.8 CONTROLLER DESIGN EXAMPLE 2UNSTABLE G(S) .................................................................................... 725.9 CONTROLLER DESIGN EXAMPLE 3G(S)WITH A ZERO ................................................................................76

5.10 HIGHER-ORDER SYSTEM CONTROLLER DESIGN ..........................................................................................80

5.11 CLOSED-LOOP CONTROLLER INPUT EFFORT ...............................................................................................84

5.12 DISTURBANCE EVALUATION AFTER CONTROLLER DESIGN .........................................................................91

5.13 WHOLE T(S)MATCHING CONTROLLER DESIGNTHE J-METHOD..............................................................96

5.14 TERM EXAMPLE CONTROLLER DESIGN..................................................................................................... 111

5.15 TERM EXAMPLE DISTURBANCES AND STEADY-STATE ERROR.................................................................. 123

6. VIBRATIONAL RESPONSES ...................................................................................................................... 128

6.1 FREE VIBRATIONAL RESPONSES .................................................................................................................. 128

6.1.1 Undamped Second-Order System Free Responses: Simple Harmonic Motion ..................................... 128

6.1.2 Damped Second-Order System Free Responses ................................................................................... 1516.2 FORCED VIBRATIONAL RESPONSES ............................................................................................................. 171

6.2.1 Undamped Second-Order System Harmonically-Forced Responses .................................................... 171

6.2.2 Damped Second-Order System Harmonically-Forced Responses ........................................................ 197

6.2.3 General Forcing Functions ................................................................................................................... 245

7. HARDWARE CONTROL IN THE OHIO UNIVERSITY ROBOTICS LAB .......................................... 254

7.1 QUANSER/SIMULINK CONTROLLER ARCHITECTURE................................................................................... 254

7.2 PROGRAMMABLE LOGIC CONTROLLER (PLC)ARCHITECTURE................................................................... 260


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1. Introduction

1.1 Control Systems and Mechanical Vibrations Defini tions

This section presents some important definitions for linear control systems and mechanical

vibrations concepts, used throughout the ME 3012 NotesBook. You may be familiar with some these

terms if some are unfamiliar, dont panic, we will discuss them later.

linear system Satisfies the principles of linear superpositionand homogeneity(see p.6).

control To make an engineering dynamic system behave as desired using anactuator, a sensor, and a computer software controller.

actuator The physical component in a control system that provides the open-loopsystem input (such as a servomotor) whose values can be changed to affect

the output.

sensor The physical component in a control system that measures the outputvalue continuously to provide the feedback.

controller The software component in a feedback control system that automaticallycalculates the actuator input every time cycle to improve the output.

PID Proportional, Integral, Derivative; popular industry-standard controller.

PLC Programmable Logic Controller; popular industry-standard controller

hardware, programmed by Ladder Logic. Not part of ME 3012.

SISO Single-Input, Single-Output

MIMO Multiple-Input, Multiple-Output

stability A system is said to be stable if its output is bounded for all bounded

inputs.

open-loop A control system that functions without sensor feedback

closed-loop A control system that functions with sensor feedback and a controller

disturbances Unknown, unwanted effects that open-loop and closed-loop control

systems are subjected to in the real world, modeled at the input level.

model The set of ordinary differential equations and algebraic equations that

represents the simplified dynamics of a real-world system.

simulation Solving the model equations and plotting to predict and display the

dynamic motion of the system.


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digital A data system that uses discrete values, such as computers.

discrete Discontinuous values.

analog A data system that uses continuous values, such as voltage-based sensors.

continuous Non-discontinuous values.

Laplace transform A mathematical transformation from the time domain to the frequency

domain.

input The external forcing function that drives a dynamic system.

output The variable of interest in motion of a dynamic system.

initial conditions The values of the output variable and its derivatives at t= 0. There must

be ninitial conditions given where nis the system order.

transfer function The Laplace transform of the output of a block divided by the Laplacetransform of the block input, with zero initial conditions.

characteristic polynomial The denominator polynomial of a transfer function, whose order is the

system order. Also appears in ODE solutions.

poles The roots of the transfer function denominator. The roots of the

characteristic polynomial.

zeroes The roots of the transfer function numerator

root locus A plot in theRe-Implane demonstrating how the closed-loop poles moveas a controller parameter Kis varied from zero to infinity.

frequency response The plots of system amplitude and phase angle vs. independent variable

input frequency.

degrees of freedom dof the number of independent parameters required to fully specify the

location of a device. The number of different ways something can move.

dynamics The study of motion with regard to forces/torques.

free-body diagram FBD, a diagram drawn out of context for each separate mass or inertiawith all external and internal forces and moments shown to give context.

cycle A repeating unit of motion.

period T, time for one cycle, sec.

cyclical frequency f, number of cycles per second, the inverse of time period, 1 / T, Hz.


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circular frequency 2 f , scalar measure of rotational rate, i.e. the magnitude of theangular velocity vector, rad/sec.

natural frequency The rate at which an unforced, undamped dynamic system tends to

vibrate, either cyclical frequencyfn, or circular frequency n.

spring Idealized massless mechanical element that provides the oscillatorymotion in a dynamic system.

dashpot Idealized massless mechanical element that provides the energy

dissipation in a dynamic system.

mass Idealized mechanical element that provides the inertia in a dynamic

system.

mass moment of inertia Idealized mechanical element that provides the rotational inertia in a

dynamic system.

damping A linear model for energy loss due to friction and other energy dissipaters,

for either translational or rotational motion. Damping is provided by a

virtual viscous dashpot (an automotive shock system is a real example).

amplitude The maximum magnitude of a sinusoidal oscillation.

differential equation An equation in which the unknown appears with its time derivatives.

initial conditions Given values at time t= 0 for the output and its time derivatives. Theremust be as many initial conditions as the order of the differential equation.

transient response Dynamic motion that goes to zero given enough time.

homogeneous solution Transient response due to initial conditions and the external input

forcing function, yielding zero in the ODE.

steady-state response Dynamic motion that does not disappear as time increases.

particular solution Steady-state responsedue to the external input forcing function. Thereis also a transient solution due to the external input forcing function.

forcing function The input actuation provided by an external source (actuator).

damped frequency The rate at which an unforced, damped dynamic system tends to vibrate,

either cyclicalfd, or circular d.


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Linear System Definition

A linear system is one in which all governing equations (differential, algebraic) are linear.

Linear systems satisfy the principles of linear superpositionand homogeneity. Let u(t) be the input

andy(t) be the output. Further, ( ) ( )u t y t indicates a system yielding outputy(t) given input u(t).

1) linear superposition

if 1 1( ) ( )u t y t and 2 2( ) ( )u t y t then 1 2 1 2( ) ( ) ( ) ( )u t u t y t y t

2) homogeneity

if ( ) ( )u t y t then ( ) ( )au t ay t

where ais any constant.

Examples

linear ODEs( ) ( ) ( )ay t by t u t 2( ) 2 ( ) ( ) ( )

n ny t y t y t u t

non-linear ODEs2 ( ) sin ( ) ( )ay t b y t u t 3 2( ) 2 ( ) cos ( ) ln ( )n ny t y t y t u t

Almost all real-world systems are non-linear due to Coulomb friction, hysteresis, unmodeled dynamics,

non-linear stiffness, large-angle dynamics, etc.

However, many engineering systems have

linear operating ranges, or equations may be piecewise linearized in various operating regions

Model vs. actual system we use a mathematical model to design the controller, but in the end the

controller will run with an actual system. The controller is a software program, a MATLAB/Simulink

block or blocks, taking sensor feedback from the real world and automatically generating actuatorcommands to continuously improve the dynamic performance of the system.

1.4 MATLAB Introduction

MATLAB is a general engineering analysis and simulation software. MATLAB stands forMATrix LABoratory. It was originally developed specifically for control systems simulation and

design engineering, but it has grown over the years to cover many engineering and scientific fields.MATLAB is based on the C language, and its programming is vaguely C-like, but simpler. MATLAB is

sold by Mathworks Inc. (mathworks.com) and Ohio University has a site license. For an extensive

introduction to the MATLAB software, see Dr. Bobs MATLAB Primer:

www.ohio.edu/people/williar4/html/PDF/MATLABPrimer.pdf


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2. Linear System Simulation

2.1 Solution of Ordinary Differential Equations

Simulation means solving linear constant-coefficient ordinary differential equations, subject to

the given initial conditions (ICs). We must solve and plot the system response (output vs. time) given

the input forcing function and the initial conditions. We will consider first- and second-order ODEs.The number of initial conditions must equal the ODE order.

e.g. Solve 2 1 0( ) ( ) ( ) ( )a x t a x t a x t r t forx(t)

Given the input forcing function r(t) and subject to the initial conditions0

0

(0)

(0)

x x

x v

.

General Solution Methods

Slow ME Way

Laplace Transforms

Slow ME Way

total solution = homogeneous solution + particular solution

transient response to ICs steady-state response to input

1. Homogeneous Solution transient response to initial conditions (0), (0)x x

This method works for ODEs of any order it is demonstrated for the second-order ODE above.

2 1 0( ) ( ) ( ) 0H H Ha x t a x t a x t

Assume: ( ) stH

x t Ae

Substitute

2

( )

( )

( )

st

H

st

H

st

H

x t Ae

x t sAe

x t s Ae

into 2 1 0( ) ( ) ( ) 0H H Ha x t a x t a x t

Calculate si the roots of the nth

-order polynomial in s(called the system characteristic polynomial)

The general homogeneous solution form must have nterms. For the second-order case:

1 2

1 2( ) s t s t

Hx t A e A e

Substitute the specific siroots at this stage, but leave theAicoefficients unknown for now (there are nof

each of these, n = 2 in this example).


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2. Particular Solution steady-state response to input forcing function r(t)

2 1 0( ) ( ) ( ) ( )P P Pa x t a x t a x t r t

AssumexP(t) according to the same form as the input forcing function r(t).

Table of Particular Solution Forms

r(t) xP(t)

constant different constant

t B1t+B2

t2 B1t

2+B2t + B3

cosD t 1 2cos sinB t B t

sinD t 1 2cos sinB t B t etc. etc.

Method of Undetermined Coefficients

Determine the unknown coefficientsBi:

Substitute ( ), ( ), ( )P P Px t x t x t into 2 1 0( ) ( ) ( ) ( )P P Pa x t a x t a x t r t

and solve for the unknown coefficientsBiright away.

3. Total Solution

( ) ( ) ( ) ( )T H P

x t x t x t x t

Now determine the unknown homogeneous coefficientsAiusing the initial conditions onxT(t).

# of characteristic polynomial roots = # of initial conditions = ODE order = n


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2.1.1 First-Order ODEs

First-Order System Example 2

Same cksystem and initial condition, but ( ) 5sin2f t t Solve ( ) 50 ( ) 5sin2x t x t t subject tox(0) = 0

Solution

505

( ) 25sin 2 cos 21252

tx t e t t

Equivalent alternate solution form

505

( ) 25.02sin 2 0.041252

tx t e t using sin( ) sin cos cos sina b a b a b

The responsex(t) starts at zero as specified by the initial condition.

The transient solution goes to zero by about t= 0.10 sec. Steady-state solution is ( ) 25.02 sin 2 0.04ssx t t m.

= 2 rad/s = 2f f= 1/= 1/T T= sec

0 1 2 3 4 5 6 7 8-0.1

-0.08

-0.06

-0.04

-0.02

0

0.02

0.04

0.06

0.08

0.1

t (sec)

-5/1252 cos(2 t)+125/1252 sin(2 t)+5/1252 exp(-50 t)

x(t)(m)


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Alternate form for particular solution (Example 2)

Sum-of-angles formulaecos( ) cos cos sin sin

sin( ) sin cos cos sin

a b a b a b

a b a b c a b

25sin 2 cos 2 sin(2 )

sin 2 cos cos 2 sin

t t C t

C t t

sin 2 25 cos

cos 2 1 sin

t C

t C

2 2 2 2 2cos sin 25 1C 626 25.02C

sin 1

cos 25

1

1tan 0.04

25

rad

25sin 2 cos 2 25.02 sin 2 0.04t t t

In general: 2 21 2C B B 1 2

1

tan B

B

when

1 2

( ) sin 2 cos2

sin 2

Px t B t B t

C t


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A Final First-Order ODE exampleR, C series electrical circuit (voltage v(t) input, current i(t) output)

Model (from KVL and electrical circuit element table)1

( ) ( ) ( )Ri t i t dt v tC

substitute charge q(t)( ) ( )

( ) ( )

i t dt q t

i t q t

1

( ) ( ) ( )Rq t q t v tC

Given R= 50 and C= 0.2 mF, solve 50 ( ) 5000 ( ) 1q t q t subject to (0) 0q and a unit step voltageinput v(t). Solution

1001

( ) 15000

tq t e

1001

( ) ( )50

ti t q t e

As expected from the circuit dynamics, the charge q(t) in the capacitor builds up to a constantgiven a constant voltage input.

Also as expected, the capacitor current i(t) goes to zero at steady-state. The steady state charge value is qSS= 1/5000. The time constant is = RC= 0.01, so at 3 time constants (t = 0.03 sec) both the q(t) and i(t)

values have approached 95% of their respective final values.

v(t)

R

+

-i(t) C

0 0.01 0.02 0.03 0.04 0.05 0.060

0.5

1

1.5

2x 10

-4

q(t)

0 0.01 0.02 0.03 0.04 0.05 0.060

0.005

0.01

0.015

0.02

i(t)

t (sec)


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First-Order System Examplesc, k massless translational mechanical system

force inputf(t) and displacement outputx(t)

( ) ( ) ( )cx t kx t f t

m, c springless translational mechanical system

force inputf(t) and velocity output v(t)

( ) ( ) ( ) ( ) ( )mx t cx t mv t cv t f t

cR, kR massless rotational mechanical system

torque input (t) and angular displacement output (t)

( ) ( ) ( )R R

c t k t t

J, cR springless rotational mechanical system

DC servomotor, torque input (t) and angular velocity output (t)

( ) ( ) ( ) ( ) ( )R RJ t c t J t c t t

L,R series electrical circuit

voltage input v(t) and current output i(t)

( )( ) ( )

di tL Ri t v t

dt

R, C series electrical circuit

voltage input v(t) and current output i(t), or charge ( ) ( )q t i t dt

1 1( ) ( ) ( ) ( ) ( )Ri t i t dt Rq t q t v t

C C

The time constants for these 6 examples are1

R

R R

cc m J L RC

k c k c R

All first-order ODEs are solved in the same way as Example 1 (assuming a unit step input), all

have the same type of time response graph, all share the same time constant behavior (after three time

constants 3the output response is within 95% of its final value).Some first-order system models are given in:

www.ohio.edu/people/williar4/html/PDF/ModelTFAtlas.pdf.

f(t)

x(t)

c

k

f(t)

x(t)

mc

k

(t)

R

(t)

cR

J

(t) (t)

cR

v(t) L

R

+

-i(t)

v(t)

R

+

-i(t) C


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2.1.2 Second-Order ODEs

Derivation of underdamped homogeneous solution form

In this case we have complex conjugate characteristic polynomial roots1,2s a bi .

For this example assume 1,2 1 3s i as in the Section 2.1.2 ME 3012 NotesBook example.

Eulers identity must be usedcos sin

cos( ) sin( ) cos( ) sin( )

i

i

e i

e i i

1 2

1 2

( 1 3 ) ( 1 3 )

1 2

3 3

1 2

1 2

( )

cos3 sin 3 cos3 sin 3

s t s t

H

i t i t

t it it

t

x t Ae A e

Ae A e

e A e A e

e A t i t A t i t

where we used = 3tin Eulers identify. Simplifying by collecting (factoring) terms.

1 2 1 2( ) cos 3 sin 3t

Hx t e A A t A A i t

Let

1 1 2

2 1 2

B A A

B A A i

We can only have real solutions when starting with real ODE coefficients, so *1 2

A A (these constants

must be complex conjugates of each other).

1

2

RE IM

RE IM

A A A i

A A A i

1

2

2

2

RE

IM

B A

B A

Therefore 1 2( ) cos 3 sin 3t

Hx t e B t B t

Biare the real constant unknown homogeneous solution coefficients

And the general underdamped homogeneous solution form given1,2s a bi is:

1 2( ) cos sinat

Hx t e B bt B bt

That is, the real part of the poles is placed in the exponential and the imaginary part of the poles is

placed as the circular frequency of the cosand sinfunctions.


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Derivation of alternate solution form for Second-Order ODE Example 3

The solution

( ) 0.20 cos 3 0.05 sin 3 0.30tx t e t t

from the ME 3012 NotesBook may be expressed in an equivalent form, using a single sinusoid (e.g. sin

here; cos is also possible) with phase angle. Clearly the steady-state response 0.30 will not change sonow we will work only on the transient response.

Though this underdamped x(t) solution is unique, there are alternate solution forms that are allmathematically-equivalent. A second-order system has two-dof, expressed in this solution by a cosine

term and a sine term (two modes), both with the same damped natural frequency d= 3, but of differentamplitudes. An alternate solution form which is mathematically equivalent is found by applying the

following trigonometric identity for sine (the sum-of-angles formula):

sin( ) sin cos cos sina b a b a b

Here the two-dof are expressed as an overall amplitudeAand a phase angle .

( ) (0.20cos3 0.05sin 3 ) sin(3 ) [sin 3 cos cos3 sin ]t t tH

x t e t t e A t e A t t

Matching terms, we havecos 0.05

sin 0.20

A

A

Solving for the alternate solution unknownsAand :

2 20.05 0.20 0.2062A atan2(0.20,0.05) 1.326

where we further used the trigonometric identity 2 21 1

cos sin 1 . atan2 is the MATLAB

quadrant-specific inverse tangent function and the result is given in radians. The input to atan2must

be (num,den)rather than a single ratio as in the atancase.

So the alternate solution form is:

( ) 0.206 sin 3 1.326 0.30tx t e t

Note the choice of sine in the above alternate solution was arbitrary; we could have equallychosen cosine.

Recall that the cosine function leads the sine function by /2 rad, or 90 deg.


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Second-Order System Example 1

m= 1 kg, c= 7Ns/m, k= 12N/m

Solve ( ) 7 ( ) 12 ( ) ( ) 3 ( )x t x t x t f t u t subject to(0) 0.10

(0) 0.05 /

x m

x m s

This system is overdampedReal distinct roots, relatively slower response, no overshoot

This solution is left to the interested reader.


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Second-Order System Example 2

m= 1 kg, c= 6Ns/m, k= 9N/m

Solve ( ) 6 ( ) 9 ( ) ( ) 3 ( )x t x t x t f t u t

Subject to(0) 0.10

(0) 0.05 /

x m

x m s

1. Homogeneous Solution ( ) 6 ( ) 9 ( ) 0H H Hx t x t x t

Assume ( ) stH

x t Ae 2( 6 9) 0sts s Ae

Characteristic polynomial 2 26 9 ( 3) 0s s s

1,2 3, 3s Real, repeated roots

Homogeneous solution form 3 31 2( )

t t

Hx t A e A te

2. Particular Solution ( ) 6 ( ) 9 ( ) 3P P Px t x t x t

( )Px t B 0 6(0) 9 3B so ( ) 1/ 3Px t B

3. Total Solution3 3

1 2

3 3 3

1 2 2

( ) ( ) ( ) 1/ 3

( ) 3 3

t t

H P

t t t

x t x t x t Ae A te

x t Ae A e A te

Now apply initial conditions

1 2

1 2

(0) 0.10 (0) 0.33

(0) 0.05 3

x A A

x A A

1

2

0.233

0.65

A

A

3( ) (0.233 0.65 ) 0.33tx t t e

This system is critically-dampedReal, repeated roots fastest response without overshoot

Check solution

Plug answerx(t) plus its two derivatives into the original ODE. Also check the initial conditions.

Plot check transient and steady state solutions, plus total solution.


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Second-Order System Example 2 Forcedm-c-kSystem

Model ( ) ( ) ( ) ( )mx t cx t kx t f t

( ) 6 ( ) 9 ( ) 3 ( )x t x t x t u t (0) 0.10

(0) 0.05 /

x m

x m s

2 26 9 ( 3) 0s s s

Solution3 1( ) (0.233 0.65 )

3

tx t t e

Plot of x(t) vs. t

The total solutionx(t)starts at 0.1 m, ( )x t is non-zero, as specified by the initial conditions. Transient approaches zero after t = 2.5 sec Critically damped; 3 root goes to zero slightly faster alonethan witht Steady-state valueisxSS= 1/3 m.

0 1 2 3 4

-0.2

-0.1

0

0.1

0.2

0.3

0.4

t (sec)

x(t)


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Second-Order System Example 4: Forcedm-kSystem

Model ( ) ( ) ( )mx t kx t f t (no damping)

( ) 9 ( ) 3 ( )x t x t u t (0) 0.10

(0) 0.05 /

x m

x m s

2 9 ( 3 )( 3 ) 0s s i s i

Solution ( ) (7 30)cos3 (1 60)sin3 1 3x t t t

Plot of x(t) vs. t

x(t) starts at 0.1 m, ( )x t is non-zero, as specified by the initial conditions. Simple harmonic motion Undamped; zero viscous damping coefficient

Transient solution oscillates forever about the particular solution 1/3 = 3 rad/s = 2f f= 3/2= 1/T T= 2/3 = 2.09 sec

This system is undampedComplex-conjugate roots with 0 real part, simple harmonic motion, no damping, theoretically

never stops vibrating.


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2.1.3 MATLAB Functions for ODEsIn solving ODEs the slow ME way, we have to do the work manually, so MATLAB doesnt help

all that much (with the exception of symbolicMATLAB, see dsolvebelow). We can find the roots of

the characteristic equation; the remaining code below plots the results that were obtained manually.

cl c; cl ear ;char = [ 1 2 10] ; % Char act er i st i c pol ynomi al and r oot ss = r oot s( char ) ;

% Pl ot second- or der syst em sol ut i on r esul t sTee = i nput ( ' Ent er [ i ni t i al , del t a, and f i nal t i mes (sec) ] : ' ) ;t 0 = Tee( 1) ; dt = Tee( 2) ; t f = Tee( 3) ;t=[ t 0: dt : t f ] ;x = - exp( - t ) . *( 0. 20*cos( 3*t ) +0. 05*si n( 3*t ) ) + 0. 30;ss = 0. 3*ones( si ze( t ) ) ; % St eady- st at e sol ut i onf i gure;pl ot ( t , x, t , ss, ' : ' ) ; gr i d;set ( gca, ' Font Si ze' , 18) ;

axi s ( [ t 0 t f 0 0. 4] ) ;xl abel ( ' t i me' ) ; yl abel ( ' X( t ) ' ) ;

Other MATLAB ODE-related functions

dsolve symbolic ODE solution (see below)

ode45 numerical ODE solution using 4th

/5th

order Runge-Kutta method

After we get into Laplace Transforms, we will learn other useful MATLAB functions for ODE solution.

ODE solution via symbolic MATLAB function dsolve

The MATLAB code below shows how to analytically solve first- and second-order ODEs (first-

order ODE Example 1 and second-order ODE Example 3), using MATLAB function dsolve, display

the analytical solution results to the screen, and then plot results using MATLAB function ezplot

(which reversesthe input data compared to most MATLAB plot commands).%- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -% Anal yt i cal sol ut i on f or ODEs usi ng symbol i c MATLAB% Dr . Bob, ME 3012%- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -cl c; cl ear ;x1 = dsol ve( ' Dx+50*x=5' , ' x(0) =0' ) ;pretty(x1)f i gure;

t 1 = [ 0: 0. 001: 0. 16] ;ezpl ot ( x1, t 1) ; gr i d; axi s( [ 0 0. 16 0 0. 105] ) ;x2 = dsol ve( ' D2x+2*Dx+10*x=3' , ' Dx( 0) =0. 05' , ' x(0) =0. 10' ) ;pretty(x2)f i gure;t 2 = [ 0: 0. 01: 6] ;ezpl ot ( x2, t 2) ; gr i d; axi s([ 0 6 0 0. 4] ) ;


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2.2 The Laplace Transform

Partial Laplace Transform Table

( )f t ( )F s

1 Dirac delta (t) 1

2 unit step u(t)1

s

3 unit ramp r(t) = t2

1

s

4 tn

n

sn

!1

5 e at 1( )s a

6 1 ate ( )

a

s s a

71

( ) ( )

at bt e e

ab a a b b b a

1

( )( )s s a s b a b

8 n att e

1!

( )nn

s a

9 sin t

s2 2

10 cos ts

s2 2

11 e tat sin 2 2( )s a

12 e tat cos 2 2( )

s a

s a


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Partial Laplace Transform Table (continued)

( )f t ( )F s

13 cos sinat b a

e t t

2 2( )

s b

s a

142 2( )

sin( )atb a

e t

1tanb a

2 2( )

s b

s a

15 2sin

1

ntnde t

;

21d n ; 0 1 2

2 22

n

n ns s

16 21 sin( )

1

nt

d

et

; 1cos ; 0 1 2

2 2( 2 )

n

n ns s s

17 2 2 2 21

sin( )ate

ta a

; 1tan

a

2 2

1[( ) ]s s a

18

2 2

2 2 2 2

1 ( )sin( )at

b b ae t

a a

1 1tan tan

b a a

2 2[( ) ]

s b

s s a


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2.2.2 ODE Solut ion via Laplace TransformsThis subsection presents an alternate solution for the first-order ODE problem from the ME 3012

NotesBook. Again, this ODE is solved using the Laplace Transform method but now, if we have aLaplace Transform table of sufficient detail, as on the preceding pages, we can skip the partial fraction

expansion step and use the table directly.

ODE Solution via Laplace Transforms Examples

First-Orderckmechanical system Example 1

Solve ( ) 50 ( ) 5x t x t forx(t), subject tox(0) = 0 and a step input of magnitude 5.

We did this ODE solution via the slow ME way and the Laplace Transform method with partial fraction

expansion.

Take the Laplace transform of both sides (dont forget the initial condition but it is given as zero).

5

( ) (0) 50 ( )

5( ) 50 ( )

sX s x X s

s

sX s X ss

Solve for the variable of interest,X(s), which is the Laplace transform of the answerx(t).

5( 50) ( )

5( )

( 50)

s X ss

X ss s

Up to this point the solution is identical to that in the ME 3012 NotesBook. But now we can skip thepartial fraction expansion if we use the following Laplace Transform table entry.

( )( )

aF s

s s a

( ) 1 atf t e

We must algebraically modify X(s) so that the same constant a appears in the numerator and

denominator, by multiplying by 1 (10/10).

10 5 1 50( )

10 ( 50) 10 ( 50)X s

s s s s

Taking the inverse Laplace transform ofX(s) yields the solutionx(t).

1 1 1

50

1 50 1 50( ) ( )

10 ( 50) 10 ( 50)

1( ) (1 )

10

t

x t X ss s s s

x t e

This is the same solution obtained twice previously.


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3. Transfer Functions and Block Diagrams

3.1 Transfer Functions

Addi tional Transfer Funct ion Example

4. R-L-C Parallel Electrical Circuit

Model:( ) 1 1

( ) ( ) ( )dv t

C v t v t dt i t dt R L

input: current i(t) output: voltage v(t)

Open-loop transfer function :

2

( ) 1 1( ) ( ) ( )

1 1 ( )( ) ( ) ( )

1 1( ) ( )

( ) 1( ) 1 1 1 1( )

dv tC v t v t dt i t

dt R L

V sCsV s V s I s

R L s

Cs V s I sR Ls

V s sG s I s

Cs Cs sR Ls R L

Open-loop block diagram

i(t) R L C v(t)

+

-

i (t)L


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3.2 Block Diagrams

Method to define transfer functions in MATLAB

Define polynomials via an array using square brackets with the numerical polynomial

coefficients given in descending power of s. See denGbelow.

Define the numerator and denominator polynomials num and den for your transfer function;

then use:

SysName = tf(num,den); % define transfer function in MATLAB

to define the transfer function in MATLAB. You then use the name SysName in various MATLAB

functions that require a transfer function as one of its inputs.

Examples2

1( )

2 8G s

s s

1( )

2C

sG s

s

1( )

3H s

s

numG = [1];denG = [1 2 8];sysG = tf(numG,denG);

numGc = [1 1];denGc = [1 2];sysGc = tf(numGc,denGc);

numH = [1];denH = [1 3];

sysH = tf(numH,denH);


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MATLAB to reduce block diagrams to one transfer function

[num,den] = zp2tf(Z,P,K); % convert zero-pole to transfer function

[Z,P,K] = tf2zp(num,den); % convert transfer function to zero-pole

Z zeroes

P poles

K scalar gain

num,den transfer function numerator and denominator

[num,den] = series(num1,den1,num2,den2); % series connection of two systems

or [sys] = series(sys1,sys2);

sysi = tf(numi,deni) % transfer function description of block i

[sysT]=feedback(sys1,sys2,sign) %close the feedback loop determine T(s)

sys1 = tf(num1,den1); % transfer function for GC(s)G(s)

sys2 = tf(num2,den2); % transfer functionH(s)

sign: +1 for positive feedback, 1 for negative (the default)

Make sure you can do each of these by hand!

Methods to solve ODEs

Slow ME way, Laplace Transforms MATLAB dsolve analytical MATLAB numerical solutions: impulse, step, lsim, ode45

MATLAB functions to numerically solve ODEs

impulse impulse response of continuous-time linear systems

1. impulse(SysName); % plots to screen

2. impulse(SysName,T); % plots to screen, user controls timeT3. [Y,X] = impulse(SysName); % saves to [ Y, X] for plotting later

4. [Y,T,X] = impulse(SysName,T); % same as 3, user controls timeT

where:

SysName=tf(num,den) name for transfer function-based system

num,den transfer function numerator and denominator

T = [t0:dt:tf]; evenly-spaced, user-supplied time array

Y output

X system state ( ,Y Y stored column-wise)

step step response of continuous-time linear systems

step(SysName); % same options and terms as impulse

lsim simulation of continuous-time linear systems for any general given input

lsim(SysName,U,T);plots to screen

[Y,X] = lsim(SysName,U,T,X0); % saves to [ Y, X] for plotting later

U given input, same length asT

X0 initial conditions


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Simulink model and plots for the Example of Section 3.2

Open-loop Closed-loop

See Dr. Bobs on-line MATLAB Primer for a Simulink tutorial to get you started:


Step Ydes

Step U

StepResponses

1

s+3

H

s+1

s+2

Gc

1

s +2s+82

G Open

1

s +2s+82

G

Ydes E U Closed Yact

Yact

Yact

Ysens

U Y Open


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3.4 Feedback

Four reasons for using feedback

1. To modify the transient response of the system. We can change open-loop poles to closed-looppoles with more desirable behavior to ensure stability and modify the system transientperformance.

2. To reduce steady-state error in the system.

3. To decrease the sensitivity of the closed-loop system to variations in the open-loop plant transfer

function. Sensitivity is like a derivative( )

( )

T s

G s

.

4. To reduce the effects of disturbances, unmodeled dynamics, uncertainties, nonlinearities,parameters changing with time, and noise. To increase system robustness.

Feedback is not free. A closed-loop feedback system is more expensive and complex, and thus

less reliable, than an open-loop system. Therefore, the engineer must determine if closed-loop feedback

control is justified or if the open-loop system can perform adequately.

There are a host of real-world dynamic systems which demand closed-loop feedback control.


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4. Transient Response

4.1 Second-Order System Damping Condit ions

Here is the MATLAB program to create the figure for the example of Section 4.1, comparing the

overdamped, critically-damped, underdamped, and undamped second-order system responses to a unit

step input.

%- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - % Over - , Cr i t i cal l y, Under - , and Undamped cases usi ng st ep f unct i on% Dr . Bob, ME 3012%- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - cl ear ; cl c;

num = [ 1] ;denOVER = [ 1 6 4] ; % Over dampeddenCRI T = [ 1 4 4] ; % Cr i t i cal l y- dampeddenUNDR = [ 1 2 4] ; % Under dampeddenUN = [ 1 0 4] ; % Undamped

pol esOVER = r oot s( denOVER) ; % Pol es f or each casepol esCRI T = r oot s( denCRI T) ;pol esUNDR = r oot s( denUNDR) ;pol esUN = r oot s( denUN) ;

OVER = t f ( num, denOVER) ;CRI T = t f ( num, denCRI T) ;UNDR = t f ( num, denUNDR) ;UN = t f ( num, denUN) ;

t = [ 0: 0. 01: 8] ;[ yOVER, xOVER] = st ep( OVER, t ) ; % Uni t st ep r esponses[ yCRI T, xCRI T] = st ep( CRI T, t ) ;[ yUNDR, xUNDR] = st ep( UNDR, t ) ;[ yUN, xUN] = st ep( UN, t ) ;

% Pl ot uni t st ep r esponsesf i gure;pl ot ( t , yOVER, ' r ' , t , yCRI T, ' g' , t , yUNDR, ' b' , t , yUN, ' m' ) ;set ( gca, ' Font Si ze' , 18) ;

gr i d; yl abel ( ' \ i t y( t ) ' ) ; xl abel ( ' \ i t t ( \ i t sec) ' ) ;l egend( ' Over ' , ' Cr i t ' , ' Under ' , ' Un' ) ;


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Over-, Critically-, Under-, and Un-damped Examples using impulse

Solve

( ) 6 ( ) 4 ( ) ( )

( ) 4 ( ) 4 ( ) ( )

( ) 2 ( ) 4 ( ) ( )

( ) 0 ( ) 4 ( ) ( )

y t y t y t t

y t y t y t t

y t y t y t t

y t y t y t t

fory(t), all subject to zero initial conditions(0) 0

(0) 0

y

y

and an impulse input.

For the impulse responses, solve the same ODEs as the step examples in the ME 3012

NotesBook, subject to zero initial conditions and impulse input (t). Note that though we specified zero

initial conditions (0) 0y

and (0) 0y , for the impulse responses only the y(0) initial condition can besatisfied, i.e. all (0)y initial velocities take their own value, different from zero as seen below.

Impulse responses plot

The impulse input final value is 0. Why are the slopes non-zero at t = 0? See Section 4.4 in thisSupplement.

The above figure was generated using the same MATLAB program as given above, but

substituting MATLAB function impulsefor step.

0 1 2 3 4 5 6 7 8-0.5

-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

y(t)

t (sec)

Over

Crit

Under

Un


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30

MATLAB Simulink

MATLAB has a powerful graphical user interface (GUI) called Simulink. From the MATLAB

command window type simulinkand you can create and simulate any control system described by

block diagrams and transfer functions. Look around to find the transfer function, step input, and scope(plot) output blocks. Simply drag and drop the desired block to the Simulink workspace. Double-click

on any block to change its parameters. Connect blocks with lines using the mouse/cursor. Simply press

the play button to run your model. Its fast, fun, and addicting! Make sure you can do the steps byhand, but you are free to use this tool for homework.

For example, the Simulink diagram below recreates the unit step responses for the overdamped,critically-damped, underdamped, and undamped second-order system example of Section 4.1. All

transfer functions are fed by the same Stepinput. The mux, or multiplexer, block is used to compare 4

results in one plot, in this case combining 4 scalar signals to one vector signal, to send to the Plots

scope. The resulting plot is identical in form to that shown in the ME 3012 NotesBook, though the

Simulink graphics do not look as nice.

A Simulink tutorial is given in Dr. Bobs on-line MATLAB Primer:



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Impulse Input in Simulink

One can simulate the impulse responses in Simulink also. Since there is no Impulse input

block, we must create our own impulse input. There are three possible methods.

1. The Dirac Delta impulse input (t) should have infinite magnitude but infinitesimal duration. It is

normalized in the sense that ( ) 1t , i.e. the area under the curve is 1. One way to approximate theimpulse input (t) is to turn on a step input at t= 0 with a large magnitudeMand then subtract from thisstep input another of equal magnitudeM, starting at time t = 1/Msec, as shown in the figure below. Ihave had success withM= 1000 and 10000, among others. Be sure to scope the resulting impulse input

to ensure it is what you intended.

2. We know that( )

( ) du t

tdt

, where u(t) is the unit step input and (t) is the impulse input. This

concept is implemented in the figure below. It appears that this approach may not work when starting at

t= 0 (it does not generate (t) but gives zero instead). If you start the Simulink simulation at t= 1, itshould work fine. Again, be sure to scope the resulting impulse input to ensure it is what you intended.

Step dt

Step

Scope

10

s +s+102

G(s)

impulse input

du/dt

derivativeStep @ t=1

Scope

10

s +s+102

G(s)

impulse input


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32

3. Either of the previous two methods work fine. Here is a third method instead of using a discrete

step input as in Method 1 on the previous page, use a steep ramp for a very short time, followed by a

symmetric steep ramp with negative slope to send the input signal back to zero. Again the area underthis steep triangular input must be 1.

Let the final input value be a, the time to rise from 0 to abe called t1, and the final time when theinput signal reaches back to zero be called t2(by symmetry t2= 2t1. This type of approximated Dirac

delta input is constructed in Simulink as follows. Add three ramp inputs, the first starting at t= 0 with a

steep positive slope of c= a/ t1, the second starting at t= t1with a negative slope of 2c, and the third

starting at t= t2with a positive slope of c(see the Simulink model shown below). For example, use a=1000, t1= 0.001 sec, c= 1,000,000, and t2= 0.002 sec. Again, be sure to scope the resulting ramped-

unramped Dirac delta approximation to ensure it is what you expected. The general equations for this

case are summarized below:

12 1

2

1

1

a ct

at

ta

tc


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Other Inputs in Simulink

Simulink provides step, ramp, sine inputs among many other built-in possibilities. Thissubsection discusses and presents some other inputs that can be built in Simulink.

Ramped-step input

The unit step input is very popular and common in the controls field. Actually it is a poor input

signal in the real world because no dynamic system can change its value instantaneously to match what

the unit step input is demanding. Instead, a ramped-step inputmay be better for real-world systems tofollow. The ramped-step input signal starts at zero (not instantaneously starting at a constant value) and

linearly changes, with constant slope, to a final constant value, which is held as long as the simulation

proceeds. Let the final input value be a, and the time to rise from 0 to a be called t1. This generalramped-step input may be constructed in Simulink as follows. Add two ramp inputs, the first starting at

t= 0 with a positive slope of c= a/ t1, and the second starting at t= t1with a negative slope of c(or,

equivalently, subtract the second ramp with a positive slope as shown in the Simulink model below).The example shown assumes a= 3 and t1= 2 sec. be sure to scope the resulting ramped-step to ensure it

is what you expected.

Scope

Ramp @ t1

Ramp

ramped-step input


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Step and ramped-step inputs work well for controlling position variables (either translational or

rotational). They also work for controlling velocity when that is the output of interest. However, many

projects may need to move positions based on velocity control. In these cases consider using atrapezoidal step input, which is similar to the ramped-step, but turns off rather than holding a constant

value indefinitely.

Trapezoidal-step input

The trapezoidal-step input signal starts at zero, linearly changes with constant slope to a final

constant value, which is held for a specified time. Then the input signal linearly decreases in asymmetric manner back to zero, which is held as long as the simulation proceeds. Let the final input

value be a, the time to rise from 0 to abe called t1, the time to start the linear decrease back to 0 be

called t2, and the final time when the input signal reaches back to zero be called t3(by symmetry t3 t2=t1. This general trapezoidal-step input is constructed in Simulink as follows. Add four ramp inputs, the

first starting at t= 0 with a positive slope of c= a/ t1, the second starting at t= t1with a negative slope

of c, the third starting at t= t2with a negative slope of c, and finally the fourth starting at t= t3with apositive slope of c(see the Simulink model shown below). The example shown assumes a= 1, t1= 2, t2

= 6, and t3= 8 sec. Again, be sure to scope the resulting ramped-step to ensure it is what you expected.


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Trapezoidal Veloci ty Input, Associated Position Command

The example trapezoidal-step velocity input is shown in magenta in the plot above. Integrating

this desired signal yields the associated position command. Since the controlled velocity returns to zero,

the associated position can be controlled to move from one displacement to another. This appliesequally to translational or rotational velocity/position.

Assuming the example trapezoid-step represents angular velocity (t), the associated final

angular displacement Fis:

2 2

0 1 2 1 3 2

2

0 1 1

1 1( ) ( )

2 2

2

F

F

ct a t t c t t

ct at

where in the example above, initial angle 0 = 0 and by symmetry t1 = t3 t2. Further we chose

2 1 12t t t (this can be changed by the user). Since the slope is c= a/ t1, the above equation simplifiesto:

01

3

Fta

Using the above equation, one can specify the initial and final angles (or translational positions), plus theconstant velocity term a, and solve for t1and construct any desired trapezoidal-step velocity input like

the example above.


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4.2 Second-Order System Performance Specifications

This section presents plots for the four performance specifications presented in the 3012

NotesBook, based on the analytical formulas for the underdamped case.

Rise Time vs. for various nvalues

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 15

10

15

20

25

30

(unitless)

tR(sec)

n= 0.1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.5

1

1.5

2

2.5

3

(unitless)

tR(sec)

n= 1

n= 10

n= 100


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Rise Time vs. nfor various values

Peak Time vs. d


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38

Percent Overshoot vs.

Settling Time vs. n


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39

MATLAB Function stepinfo

The ME 3012 NotesBook presented a fast and easy method for MATLAB to display the rise time

tR, peak time tP, percent overshoot PO, and settling time tS for any given step response. This right-

clicking in a stepwindow is accurate for any order and type of system since MATLAB applies the

rules to calculate these results numerically, rather than applying the generic-second-order-system-only

equations.

This section presents an alternative MATLAB function to obtain the same information. The

drawback is there is no graphical output in this case, but then more information is given. We now

demonstrate this approach for the same example from the NotesBook.

num = [10];den = [1 0.5 10];Sys = tf(num,den);stepinfo(Sys);

The result of this piece of m-code is a structure with the following data:

RiseTime: 0.3505SettlingTime: 15.1295SettlingMin: 0.3924SettlingMax: 1.7795Overshoot: 77.9467Undershoot: 0

Peak: 1.7795PeakTime: 0.9966

These answers agree well with the results presented earlier in the ME 3012 NotesBook. The

familiar rise time tR, peak time tP, percent overshoot PO, and settling time tSare given on lines 1, 8, 5,and 2 above, respectively.

The additional information given is as follows (from help stepinfoin MATLAB):

SettlingMin: min value of Y once the response has risenSettlingMax: max value of Y once the response has risenUndershoot: percentage undershoot

Peak: peak absolute value of Y

This extra information is rather redundant in this example; the SettlingMaxvalue is identical to the

Peakvalue, both of which were already known from percent overshoot PO.

Two items are new and potentially interesting: SettlingMin gives the value of the lowest

valley, which is directly following the maximum peak, and the Undershootvalue gives the percent

undershoot (negative from zero) relative to the final value. Refer to the ME 3012 NotesBook to see that

these values of 0.3924and 0, respectively, are correct for this example.

See also: lsiminfo.


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Specifying Desired Poles for Critically-Damped or Overdamped Performance

In the ME 3012 NotesBook, to specify good behavior for the closed-loop controller, all we haveare underdamped performance specification equations. Many real-world controllers can benefit from

critically-damped or overdamped performance specifications. This original work is presented below.

For the critically-damped and overdamped cases, percent overshoot is zero and peak time is

infinite. Generally settling time is more important than rise time so this work will focus only on settling

time. We cannot use the underdamped equation for calculating 2% settling time for the critically-damped and overdamped specifications. However, the definition for settling time is identical (the time

at which the critically-damped or overdamped response crosses the 98% line of the final value for aunit step response. Using MATLAB step and right-clicking on the resulting plot can easily calculate thesettling time numerically this is the approach followed here.

For the critically-damped case, s1= s2 (both poles are negative, real, and identical). There areinfinite overdamped cases; here we use s2= 2s1 (both poles are negative and real). All poles must be

negative for stability, but the plots below show positive s1 for convenience of viewing the data

represents only negative poles.

For various pole values in the critically-damped case the settling time is found numerically using

MATLAB step and right-clicking in the plot (Characteristics Settling Time). The settling time tSvs.pole s1 plot is shown below. The same data was entered in Excel and the best exponential curve fitswere found to be:

Critically-Damped Overdamped

1

1

5.83( )St s

s 2R 1

1

1

4.6( )St s

s 2R 1

It was found that including s1values less than 1 made for worse curve fitting; therefore the curve

fits reported above are only for 11 10s and the table below reports discrete settling time results for

11s , for completeness. Again the overdamped case uses s2= 2s1. The two R

2values indicate that

these are perfect curve fits.

s1 tS(critically-damped) tS(overdamped)

0.25 23.30 18.40

0.50 11.70 9.20

0.75 7.78 6.14

The two plots below present settling time results for the critically-damped and overdampedcases, respectively. To use these results for specifying good behavior in controller design, one can

specify the desired settling time, choose critically-damped or overdamped, and solve for s1 from the

appropriate best curve fit equation above (or try to read the s1value from the appropriate plot). Once s1is known, use s2= s1for the critically-damped case and s2= 2s1for the overdamped case. If you desire a

different type of overdamping, you can use trial-and-error.


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For all design cases, be sure to evaluate the expected dynamic response over time for your

desired behavior to ensure it agrees with what you specified (in terms of settling time and 0%

overshoot).

Settling Time vs. s1

Critically-Damped s2= s1and Overdamped s2= 2s1

1 2 3 4 5 6 7 8 9 100

1

2

3

4

5

6

s1

tS(sec)

critically-damped

overdamped


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4.3 Open-Loop and Closed-Loop System Example

Given the following open-loop system model ( ) 2 ( ) 4 ( ) 4 ( )y t y t y t u t

subject to (0) 0, (0) 0y y and a unit step input.

Open-Loop Behavior

Open-loop transfer function and block diagram:

2

2 2 2

4( )

2 2 4

n

n n

G ss s s s

Open-loop characteristic polynomial OL(s), natural frequency n, dimensionless damping ratio ,damped frequency d, and poles s1,2 (complex conjugate poles with negative real part, therefore this

system is underdamped):

2( ) 2 4OL

s s s 2n 0.5 3d 1,2 1 3s i

Open-Loop System Unit Step Response

Open-loop performance specificationstR 0.84 stP = 1.81 s

PO = 16.30%tS 4.00 s

Open-loopySS use the final value theorem (FVT)

2 20 0 0 0

4 1 4 4lim ( ) lim ( ) lim ( ) ( ) lim lim 1

2 4 2 4 4OLSS

t s s s sy y t sY s sG s U s s

s s s s s

0 1 2 3 4 5 60

0.2

0.4

0.6

0.8

1

1.2

yo(t)

t (sec)


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Closed-Loop Controller Evaluation

Choose a simple controller ( ) 10CG s K Assume a perfect sensor, i.e.H(s) = 1.

Control law ( ) ( ) ( ) ( )CU s G s E s KE s

Closed-loop block diagram and transfer function:

2 24 40( )

2 4 4 2 44KT s

s s K s s

Closed-loop characteristic polynomial CL(s), natural frequency n, dimensionless damping ratio ,damped frequency d, and poles s1,2 (complex conjugate poles with negative real part, therefore this

closed-loop system is also underdamped):

2( ) 2 44CL s s s 6.63n 0.15 6.56d 1,2 1 6.56s i

Open- and Closed-Loop System Unit Step Responses

Closed-loop performance specs Open-loop performance specstR 0.14 s(in error) tR 0.84 stP = 0.48 s tP = 1.81 s

PO = 62.1% PO = 16.3%tS 4.02 s tS 4.00 s

The MATLAB m-file to generate this plot is given later in this section.

Closed-loopySS use the final value theorem (FVT)

2 20 0 0 0

40 1 40 40lim ( ) lim ( ) lim ( ) ( ) lim lim 0.909

2 44 2 44 44CLSS

t s s s sy y t sY s sT s R s s

s s s s s

0 1 2 3 4 5 60

0.5

1

1.5

t (sec)

y(t)

Open-loop

Closed-loop


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Steady-state error

Let us assume that the steady-state valueySSof the open-loop system is the desired output value.

We calculate the steady-state error of the closed-loop system relative to the desired open-loop value.

100%OL CL

OL

SS SS

SS

SS

y ye

y

For the example40

144 100% 9.1%

1SSe

Closed-loop steady-state error Open-loop steady-state error

eSS = 9.1% eSS = 0%

Summary

Choosing a random simple controller made the step response worse than the original open-loop

case! Controller design must be done properly or things can easily get worse.

For ( ) 10CG s K , the closed-loop system is more underdamped than the open-loop

system: the riseand peak timesare much faster (not necessarily good for the dynamics of the system),

the percent overshootis much worse, the settling timeturned out to be about the same, and a steady-

state error was introduced. The steady-state error arose because the controller introduced a virtual

spring, in addition to the existing spring in the open-loop system, making the closed-loop system stiffer,

resulting with a smaller displacement given the same unit step input force.

Simulink Simulation

The MATLAB Simulink diagram below recreates the previous plot comparing the open- and

closed-loop systems for the current example.

Two new blocks are used, the circular summing junction and the triangular gain (representing thesimple proportional controller). More complex controllers can be represented by transfer functions as

we will see later. The feedback loop takes the closed-loop system output back to the summing junction.

Themuxblock (two signals into one above) allows us to compare both results on one scope (Plots).

Both open- and closed-loop simulations use the same Stepinput for a fair comparison (pickoff point).

If we choose a final simulation time of 6 sec (entered via the time box on the simulation windowmenu), the Simulink plots are identical to that shown earlier.


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Here is the MATLAB program to create the figure for the example of this section, comparing the

open- and closed-loop system responses to a unit step input.

%- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - % Open- l oop/ Cl osed- l oop Exampl e% Dr . Bob, ME 3012%- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - cl ear ; cl c;% Open- l oop uni t st ep r esponsenumo = [ 4] ; deno = [ 1 2 4] ;SysO = t f ( numo, deno) ;zeroso = r oot s( numo) ; pol eso = r oot s( deno) ;[ wno, zet ao] = damp(deno) ;t = [ 0: 0. 01: 6] ;yo = st ep( SysO, t ) ; % Open- l oop uni t st ep response

% Cl osed- l oopK = 10; % Si mpl e pr opor t i onal cont r ol l er numc = [ 4*K] ; denc = [ 1 2 4+4*K] ;SysC = t f ( numc, denc) ;zerosc = r oot s( numc) ; pol esc = r oot s( denc) ;

[ wnc, zet ac] = damp(denc) ;yc = st ep( SysC, t ) ; % Cl osed- l oop uni t st ep r esponse

% Check hand- deri vat i on of T( s) =[ numc/ denc] vi a MATLABnumgc = [ K] ;dengc = [ 1] ;[ numa, dena] = ser i es( numgc, dengc, numo, deno) ;[ numc2, denc2] = f eedback( numa, dena, [ 1] , [ 1] ) ;

TCheck = t f ( numc2, denc2) % Di spl ay T( s) check r esul t s

% Pl ot open- and cl osed- l oop st ep r esponsesf i gure;pl ot ( t , yo, ' r ' , t , y c , ' g' ) ;set ( gca, ' Font Si ze' , 18) ; l egend( ' Open- l oop' , ' Cl osed- l oop' ) ;axi s([ 0 6 0 1. 5] ) ;gr i d; yl abel ( ' \ i t y( t ) ' ) ; xl abel ( ' \ i t t ( \ i t sec) ' ) ;

% Per f ormance specs: open- l oopwdo = wno( 1) *sqrt ( 1- zet ao( 1) 2) ; % Damped nat ural f r equencyt r o = ( 2. 16*zet ao(1) + 0. 60) / wno(1) % Ri se t i met po = pi / wdo % Peak t i mepoo = 100*exp( - zet ao(1) *pi / sqr t ( 1- zet ao(1) 2) ) % Per cent overshoott so = 4/ ( zet ao( 1) *wno( 1) ) % Set t l i ng t i me

% Per f ormance specs: cl osed- l oopwdc = wnc( 1) *sqr t ( 1- zetac( 1) 2) ; % Damped nat ural f r equencyt r c = ( 2. 16*zet ac( 1) + 0. 60) / wnc( 1) % Percent overshoot

t pc = pi / wdc( 1) % Peak t i mepoc = 100*exp( - zet ac( 1) *pi / sqr t ( 1- zet ac( 1) 2) ) % Per cent overshoott sc = 4/ ( zet ac( 1) *wnc( 1) ) % Set t l i ng t i me

% Ri ght - cl i ck f or per f ormance specs: open- and cl osed- l oopf i gur e; st ep( SysO) ; gr i d;f i gure; step( SysC) ; gr i d;


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4.4 First- and Second-Order Transient Response Characteristics

The Cartesian representation of the 3x4 subplots (step responses) from the ME 3012 NotesBook

Section 4.4 is for convenience only, since that is the way the MATLAB subplot works. For more

accuracy, the underdamped cases c. should be represented using polar coordinates (for this example

shown to scale below) rather than Cartesian coordinates.

Recall for underdamped poles the polar representation is r= nand1sin ; thus the two symmetric

angles in the examples above are 5.7 and 30 .

Also, remember all complex conjugates occur in pairs, as shown above (two poles generate onetime response plot in all second-order cases, including the non-complex-conjugate cases).

c1

n

n

Im

Re

x

x

x

xx

x

x

x

c1

c2

c2

c3

c3

c4

c4

n

n


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We can also perform the example of Section 4.4 using the impulse function in place of the

stepfunction; this is shown below.

Transient Response Characteristics vs.Re-Im-plane pole locations

(impulseresponses, generic second-order system, zero initial conditions)

Figure key

5.0 n 2 s i1 2 1 173, .

c1. underdamped

01. n 2 s i1 2 02 199, . .

c2. underdamped

0 n 2 s i1 2 2,

d1. undamped

01. n 2 s i1 2 0 2 199, . .

e1.

0 5. n 1s i1 2 05 0866, . .

c3. underdamped

01. n 1s i1 2 01 0995, . .

c4. underdamped

0 n 1 s i1 2,

d2. undamped

01. n 1 s i1 2 01 0 995, . .

e2.

2 n 1s1 2 373 0 27, . , .

a. overdamped

1 n 1s1 2 1,

b. critically-

damped

0 n 0 s1 2 0,

d3. special

undamped

1(-1.005)

n 1( 0.99n )s1 2 1, (0.9,1.1)

f. (g.)

Note While cases fand gmay look identical between the step (ME 3012 NotesBook) and impulse (this

page) responses, they are different the impulse responses go to infinity faster than the step cases.

0 5 10 15 20-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

0 5 10 15 20-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

0 5 10 15 20-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

0 5 10 15 20-100

-50

0

50

100

0 5 10 15 20-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

0 5 10 15 20-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

0 5 10 15 20-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

0 5 10 15 20-10

-5

0

5

10

0 5 10 15 20-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

0 5 10 15 20-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

0 5 10 15 200

5

10

15

20

0 5 10 15 200

2

4

6

8

10

12

14

16x 10

9

f

g


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Here is the MATLAB program to create the figure for the example of Section 4.4, comparing the

various second-order system transient responses to a unit step input, based on pole locations. To find

the impulse responses of the previous page, simply substitute MATLAB function impulsefor step.

%- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -% Gener at e pl ot s t o di spl ay t r ansi ent uni t st ep r esponse char act er i st i cs% gi ven di f f er ent di mensi onl ess dampi ng r ati os and natural f r equenci es% f or t he generi c second- order system Dr. Bob, ME 3012%- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -cl c; cl ear ; f i gure;

dt = 0. 01; t f = 20; t = [ 0: dt : t f ] ;Xmi n = 0; Xmax = t f ; Ymi n = 0; Ymax = 2;

subpl ot ( 3, 4, 1) ;zet a = 0. 5; wn = 2; num = [ wn 2] ; den = [ 1 2*zet a*wn wn 2] ; y = st ep( num, den, t ) ;pl ot( t , y) ; axi s( [ Xmi n Xmax Ymi n Ymax] ) ; r 1 = r oots ( den) ;


subpl ot ( 3, 4, 3) ;zet a = 0; wn = 2; num = [ wn 2] ; den = [ 1 2*zet a*wn wn 2] ; y = st ep(num, den, t ) ;pl ot( t , y) ; axi s( [ Xmi n Xmax Ymi n Ymax] ) ; r 3 = r oots ( den) ;

subpl ot ( 3, 4, 4) ;zet a = - 0. 1; wn = 2; num = [ wn 2] ; den = [ 1 2*zet a*wn wn 2] ; y = st ep( num, den, t ) ;pl ot ( t , y); axi s( [ Xmi n Xmax - 50 50] ) ; r 4 = r oot s( den) ;




subpl ot ( 3, 4, 8) ;zet a = - 0. 1; wn = 1; num = [ wn 2] ; den = [ 1 2*zet a*wn wn 2] ; y = st ep( num, den, t ) ;pl ot ( t , y); axi s( [ Xmi n Xmax - 10 10] ) ; r 8 = r oot s( den) ;


subpl ot ( 3, 4, 10) ;zet a = 1; wn = 1; num = [ wn 2] ; den = [ 1 2*zet a*wn wn 2] ; y = st ep(num, den, t ) ;

pl ot ( t , y) ; axi s( [ Xmi n Xmax Ymi n Ymax] ) ; r 10 = r oot s( den) ;

subpl ot ( 3, 4, 11) ;zet a = 0; wn = 0; num = [ 1] ; den = [ 1 0 0] ; y = st ep( num, den, t ) ;pl ot( t , y) ; axi s( [ Xmi n Xmax 0 200] ) ; r 11 = r oot s( den) ;

subpl ot ( 3, 4, 12) ;zet a = - 1; wn = 1; num = [ wn 2] ; den = [ 1 2*zet a*wn wn 2] ; yf = st ep( num, den, t ) ;zet a = - 1. 005; wn = sqr t ( 0. 99) ; num = [ wn 2] ;den = [ 1 2*zet a*wn wn 2] ; yg = st ep( num, den, t ) ;pl ot ( t , yf , ' b' , t , yg, ' g' ) ; l egend( ' f ' , ' g' ) ; axi s( [ Xmi n Xmax 0 16e09] ) ;


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For the impulse response transient characteristics plots presented two pages back, no initial

conditions were specified in the MATLAB program, which means that MATLAB assumes zero initial

conditions. We can see zero initial condition on position are achieved in all of the plots. However, none

of these impulse response plots has a zero slope at t = 0, which means that the impulse response cannot

allow zero velocity initial condition. Why? Let us address this issue via an example.

Example analytical impulse response solution for the generic second-order system

Solve 2 2( ) 2 ( ) ( ) ( )n n n

y t y t y t t fory(t) given , n, and zero initial conditions:

0

0

(0) 0

(0) 0

y y

y v

Here the Laplace Transform method will be used. The first step is to take the Laplace Transformof both sides of the ODE, including the zero initial conditions, and solving algebraically for the

unknown in the Laplace domain, ( ) { ( )}Y s y t .

2 2

2 2 2

2 2 2

2

2 2

( ) 2 ( ) ( ) ( )

( ( ) (0) (0)) 2 ( ( ) (0)) ( ) (1)

( 2 ) ( )

( )2

n n n

n n n

n n n

n

n n

y t y t y t t

s Y s sy y sY s y Y s

s s Y s

Y ss s

Note that the system characteristic polynomial 2 22n n

s s appears when using the Laplace

Transform method.

For simplicity in solution, let us assume this system is overdamped, i.e. there are two negative,real, distinct poles (our impulse response conclusion will apply to all damping cases).

2 2

2 2

1 2

( )2 ( )( )

n n

n n

Y ss s s s s s

The Heaviside Partial Fraction expansion of Y(s) is:2

1 2

1 2 1 2

( )( )( ) ( ) ( )

n C CY ss s s s s s s s

Solving for the residues C1and C2:

2 2

1 2 1 2 2 1

1 2 1 2 1 2 1 2

0( ) ( )

( ) ( ) ( )( ) ( )( ) ( )( )

n nsC C C s s C s s

s s s s s s s s s s s s s s s s


50/262

50

1

1 2

0 2

1 2 2 1

0

n

s C C

s C s C s

2

1

2 1

2

2

1 2

n

n

Cs s

Cs s

This solution is always valid since due to the overdamped assumption, 1 2s s .

To find the solutiony(t) in the time domain, we must take the inverse Laplace Transform of Y(s).

1 2

1

21

1 2

1 1 2

1 2

1 2

( ) ( )

( )( )( )

( )( ) ( )

( )

n

s t s t

y t Y s

y ts s s s

C Cy t

s s s s

y t C e C e

Checking the zero initial conditions from this solution, it is seen that the initial position is zero,

but the initial velocity is NOT zero:

1 2(0) (0)

1 2 1 2

2 2 2 2

2 1 1 2 2 1 2 1

(0) (1) (1)

0

s s

n n n n

y C e C e C C

s s s s s s s s

1 2

1 2

1 1 2 2

(0) (0)

1 1 2 2 1 1 2 2

2 2 2 2

1 2 1 2

2 1 1 2 2 1 2 1

2

2 1

2 1

2

( )

(0) (1) (1)

( )

0

s t s t

s s

n n n n

n

n

y t s C e s C e

y s C e s C e s C s C

s s s ss s s s s s s s

s s

s s

The impulse responses for all other damping cases similarly result in non-zero initial velocities(slopes ofy(t) at t= 0).


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4.5 System Type

For a system transfer function expressed in the following zero-pole form (mzeroeszjand nnon-

zero polespi):

1 2

1 2

( )( ) ( ) ( )( )

( )( ) ( ) ( )

m

r

n

K s z s z s z Y sG s

s s p s p s p U s

ris the System Type, i.e. ris the number of poles at s= 0. The System Type is the number of system

integrators. This section presents transient response characteristics for example Type 0, I, and IIsystems.

Example Transfer Functions G(s) for different System Types

Type 0 Type I Type II

1( )G s

s

2

1( )G s

s

1

( ) ( 1)G s s 1

( ) ( 1)G s s s 21

( ) ( 1)G s s s

2

10( )

( 10)G s

s s

2

10( )

( 10)G s

s s s

2 2

10( )

( 10)G s

s s s

Example Unit Step Responses (outputy(t) vs. time, 0 ICs) for different System Types



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Type I, II, and higher systems have no effective spring, so their transient responses go to infinity.

The steady-state results are constant, linear, and parabolic for Types 0, I, II, respectively.

Analytical Unit Step Response Solutions for the System Types examples above


( )y t t 2

( )2

ty t

( ) 1 ty t e ( ) 1ty t e t 2

( ) 12

t ty t e t

0.5( ) 1 (cos3.12 0.16sin 3.12 )t

y t e t t

0.51 1( ) (cos3.12 3.04sin 3.12 )

10 10

ty t t e t t

20.59 9( ) (cos3.12 0.52sin 3.12 )

2 10 100 100

tt ty t e t t

For they(t) solutions summarized in the above table, what is the transient solution in each case?

The steady-state solutions?

System Order, Poles, and Stability Conditions for the System Types examples above

Type 0 Type I Type IIfirst

10s

marginally-stable

second

1 20s s

marginally-stable

first

1 1s stable

second

1 0s 2 1s marginally-stable

third

1 2 0s s 3 1s marginally-stable

second

1,2 0.5 3.12s i

stable

third

1 0s 2,3 0.5 3.12s i

marginally-stable

fourth

1 2 0s s 3,4 0.5 3.12s i

marginally-stable

Note that the natural frequency nand the dimensionless damping ratio , defined for the generic secondorder system, have no meaning for Type I or II systems (other than as modes in the third line above).

The first Type 0 example is typical of a stable first-order system step response. The second Type0 example is typical of a stable underdamped second-order system step response. The top Type II

example was previously identified in the ME 3012 NotesBook as case d3, the special undamped case

with 1 2 0s s .


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The middle Type I example is typical of an electric motor model with input torque (t), output

angle (t), and transfer function1

( )( )R

G ss Js c

, where J is the motor shaft/load mass moment of

inertia and cR is the rotational viscous damping coefficient. A distributed block diagram explicitlyshows the angular-velocity-to-angle integrator in this system:

Analytical ODE Solutions

Since only two of these ODEs are familiar in our previous analytical ODE solutions in ME 3012

(the stable first- and stable underdamped second-order systems), let us perform an analytical solution for

one example: the second-order Type I system from above. Solve

( ) ( ) ( )y t y t u t

fory(t), given that input u(t) is a unit step function and subject to given zero initial conditions (0) 0y and (0) 0y .

For funlet us solve this problem in three ways, two alternate methods using the slow ME Way

and also using the Laplace Transform approach. Since this is a linear ODE, all methods must yield the

one correct answer, stated earlier in the Analytical Unit Step Response Solutionstable.

1a. Slow ME Way second-order solution

1. Homogeneous Solution

( ) ( ) 0H Hy t y t

Assume: ( ) stH

y t Ae so that2

( )

( )

st

H

st

H

y t sAe

y t s Ae

2 2( ) ( ) ( ) 0st st st H H

y t y t s Ae sAe A s s e

The system characteristic polynomial is2

s s , leading to a system characteristic equation2 ( 1) 0s s s s , whose two roots are 1 0s and 2 1s . The general homogeneous solution form isthen:

1 2 (0 ) ( 1)

1 2 1 2 1 2( ) s t s t t t t

Hy t A e A e A e A e A A e

We must leave the homogeneous coefficientsAiunknown for now.


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2. Particular Solution

( ) ( ) ( ) 1P Py t y t u t

Our standard particular solution assumption of ( )Py t B will not work in this case so we immediatelytry:

( )Py t B

whereBis an unknown constant, which must be determined in this step. Clearly ( ) 0P

y t and so:

( ) ( ) 0 1P Py t y t B

And so the particular solution for velocity is ( ) 1P

y t B . However, we need the particular solution fory(t), not its first derivative, and so we must integrate:

0 0( ) ( )

t t

P Py t y t dt dt t c

where cis a constant of integration, in this case left unknown until Step 3.

3. Total Solution

1 2( ) ( ) ( ) t

H Py t y t y t A A e t c

An alert student may ask, how can we solve for three constantsA1,A2, and cwith only two initial

conditions? The solution to this little dilemma is quite easy, recognizing that 1A c is simply a single

constant, call it 3 1A A c . Then the total solution is expressed by:

3 2( ) ty t A A e t

to which we apply the two given initial conditions (0) 0y and (0) 0y in order to solve for theunknown constantsA1andA3.

(0)3 2

2 3

(0) 0

0 (0)

0

y

A A e

A A

2( ) 1ty t A e (0)

2

2

2

(0) 0

0 1

1 0

1

y

A e

A

A

And since2 3 0A A , we then have

3 2 1A A


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and the total solution is:

( ) 1t

y t e t

The transient part of this solution ist

e

and the steady-state part of this solution is 1t .

To demonstrate the transient and steady-state components of this solution better, it is instructive

to plot this response, both its components and the total solution (which was shown previously, the centerType I case).

Detailed Plot for Center Type I System Example Unit Step Response

Transient, Steady-State, and Total Responses


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In order to check this solution, we can first check the resulting initial conditions. From the

solution ( ) 1t

y t e t we clearly see that 0(0) 0 1 1 0 1 0y e as required. Also, for the

first derivative of the solution ( ) 1ty t e , we have 0(0) 1 1 1 0y e as required.

A more complete solution check involves substituting the solution y(t) and its two time

derivatives into the original ODE and prove than an equality results.

( ) 1

( ) 1

( )

t

t

t

y t e t

y t e

y t e

?( ) ( ) ( ) 1

1 1t t

y t y t u t

e e

Since an equality results, 1 = 1, this proves that the solution is the correct one.


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1b. Slow ME Way firs t-order solut ion

A clever alternate solution method arises (either the Slow ME Way or using Laplace Transforms)

when we use the equivalence ( ) ( )v t y t in the original ODE to be solved. Solve

( ) ( ) ( )v t v t u t

for v(t), given that input u(t) is a unit step function and subject to given zero initial condition

(0) (0) 0v y . The findy(t), by integration, using the second given initial condition (0) 0y .

1. Homogeneous Solution

( ) ( ) 0H H

v t v t

Assume: ( ) stH

v t Ae so that ( ) stHv t sAe

( ) ( ) ( 1) 0st st st H Hv t v t sAe Ae A s e

The system characteristic polynomial is 1s , leading to a system characteristic equation 1 0s , whoseone root is 1 1s . The homogeneous solution form is then:

1

1 1( ) s t t

Hv t A e A e

We must leave the homogeneous coefficientA1unknown for now.

2. Particular Solution

( ) ( ) ( ) 1P Pv t v t u t

Assume ( )P

v t B , where B is an unknown constant which must be determined in this step. Clearly

( ) 0Pv t and so:

( ) ( ) 0 1P P

v t v t B

And so the particular solution for velocity is ( ) 1Pv t B .

3. Total Solution

1( ) ( ) ( ) 1t

H Pv t v t v t A e

Applying the given initial condition (0) (0) 0v y in order to solve for the unknown constantA1:


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(0)

1

1

(0) 0

0 1

1

v

Ae

A

the total solution for v(t) is then:

( ) 1 tv t e

However, we need the total solution fory(t), not its first derivative v(t), and so we must integrate:

0 0( ) ( ) (1 )

t tt ty t v t dt e dt t e c

where cis a constant of integration, that is found by applying the remaining initial condition (0) 0y .

(0)(0) 0 (0) 1

1

y e c c

c

and so the total solution is:

( ) 1t

y t e t

which agrees with the Slow ME Way second-order solution.


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2. Laplace Transforms solution

Take the Laplace transform of both sides, with initial conditions.

2

( ) ( ) ( )

1

( ) (0) (0) ( ) (0)

y t y t u t

s Y s s sY s s

Combine Y(s) terms on the left.

2 1( ) ( ) ( 1) ( )s s Y s s s Y ss

Solve for the variable of interest, Y(s) the Laplace transform of the answer,y(t).

2

1( )

( 1)Y s

s s

The partial fraction expansion for this Y(s) is given below, where the constants C1,2,3are the residues.

31 2

2 2

1( )

( 1) 1

CC s CY s

s s s s

Express the partial fraction form over a common denominator.2 2

3 1 2 3 1 3 1 2 21 2

2 2 2 2

( )( 1) ( ) ( )1( )

( 1) 1 ( 1) ( 1)

C C s C s C s C C s C C s C C s CY s

s s s s s s s s

Equate like powers of sin the numerator to get 3 equations to solve for the residues C1,2,3.

21 3

1

1 2

0

2

0

0

1

s C C

s C C

s C

1

2

3

1

1

1

C

C

C

Solution by inverse Laplace transform

1 1 1 1 131 22 21 1 1

( ) ( )1 1

CC s Cy t Y s

s s s s s

( ) 1 ty t t e

which agrees with the solutions we previously derived, but as expected is a more-direct one-stageprocess and faster to obtain.


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4.7 Term Example Open-Loop Transient Response

linear systems control and vibrations

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