Ta Proj 1 Reportl

7/27/2019 Ta Proj 1 Reportl

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ECSE 4440 Control System Engineering

Project 1

Controller Design of a Second Order System

TA

Contents

1. Abstract

2. Introduction3. Controller Design for a Single Pendulum

4. Conclusion


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1. Abstract

The purpose of this project is to design a controller for a second order system(themost common prototype control problem). A Proportional Integral derivative(PID)

will be adopted for the give system. The study will be approached analytically and

experimentally.

2. IntroductionPractically, there are few systems that dont have any control system inside. More

complicated become systems, more sophisticated controllers are needed. To design acontroller, several issues should be considered e.g., modeling, system performance,tuning gain and stability. For a simple second order system, those issues will be

studied in this project.The given single pendulum system is a basic second order system to be controlled.

The system will be modeled as a linear system. The poles and zeros of the systemwill be found to test the stability of the given system. The controller will be

developed in the order of proportional(P) controller, proportional-derivative(PD)controller-proportional and proportional- integral-derivative(PID) controller. To meetthe given specifications and stability, the gains will be tuned.

The controllers will be implemented in continuous time domain(S plane) and indiscrete time domain(Z plane), respectively. The analytical controller will be verified

by simulation with simulink of Matlab.

3. Controller Design for A Single Pendulum3.1 Linearization(Task 1)

The analysis and control design are far easier for the linear than for nonlinear models.

Linearization is the process of finding a linear mode that approximates a nonlinearone. Linearization process depends on the expanding the nonlinear state equation in

to a Taylor series. In the given dynamic equation (1), two nonlinear functions exist,

e.g., sin),sgn(.

.

( ) KvmglFFmlII gcvgm ==+++++ sin)sgn(....

22(1)

Parameter Name Value

m Mass .048Kg

I Link Inertia .000187 Kg m2

mI Motor Inertia 2.2e-7 Kg m

2

gl Distance .051 m

N Gear Ratio 70.35

cF Coulomb Friction Coefficient .014 N-m

vF Viscous Friction Coefficient .0034 N-m-sec

K Torque Constant .01447


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)sgn(.

can be set to be zero since at the equilibrium state, .0.

= For sin , the firsttwo terms(linear terms) of the Taylor series expansion(2) are used.

( ) ( ) ++= 2sin5.cossinsin (2)

The the linearized system becomes

( ) desgdesgvgm mglmglFmlINI sincos...

22 =++++ (3)where ( )des =

The Laplace transform of the equation is

( ) ( ) ( )ssmglsFs desgv =++ cos2

(4)

where22

gm mlINI ++= and I(s) is the Laplace transform of desgmgl sin .

Then, the transfer function of the system is

desgv mglsFssIssH

cos1

)()()(

2 ++==(5)

From the transfer function, it has poles at

2

cos4

2

desgvvmglFF

s

= (6)

and no zeros. For the system to be stable, every poles should be in left hand side of s-

plane. In the equation (6), 0cos >des is the condition for the system to be stable. In

other words,

2

2

3

2

0


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Figure 1

Figure 2

With only the single pendulum system, the step response goes infinity in figure 3.

Figure 3

3.3 P Feed Back Controller(task 3)


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As a basic controller operation, the controller is simply an amplifier with a constant

gain pK and a feedback loop, ( )despK = . Hence the output of P controller isrelated with the input of the controller by a proportional constant. Adding a Pcontroller to the given system results in changing the transfer function of overall

system such as

pdesgv

des

pdesgv KmglsFasd

KmglsFss

++++

+++=

coscos 22(7)

where disturbance desgmgld sin= .

From (7), the location of the poles depends on the given parameters and pK . The

poles are

2

cos4

2

pdesgvvKmglFF

s

+

= (8)

The locations of the poles are changing by varyingp

K . Ifdesgp

mglK cos

>, the system

becomes stable. With given parameters and pK = 5, the poles are located at

i1194274.2 for =des and2

=des such that the system is stable. The analytical

result is verified by the simulation( pK =5, iK =0 and iK =0). In figure 4, the system is

stable for both values.

Figure 4

3.4 PD Feed Back Controller(task 4)

Even though the system with P controller is stable, the output has relatively high peakovershoot and is oscillating. The oscillation results from the excessive amount of


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torque and the lack of damping. Adding the derivative of the input makes the systemcritical damped. In the equation (9), the locations of poles are

2

cos4

2

pdesgdvdvKmglKFKF

s

+

+

+

= (9)

As it is shown in (9), tuning dK and pK make it possible for the system to meet the

given specifications e.g., rise time (90%), settling time (2%) small overshoot(less 5%)

and steady state error(less then 2%). Once the inside of the square root is negative,the damping depends on its magnitude.

For fixed 1.=dK , the step responses are shown for various pK values in the figure 5.

Figure 5

The plots show increasing pK results in decreasing rise time( pK =1,10) but

increasing overshoot( pK =10).

For fixed pK =1 in the figure 6, the step responses are shown for various dK values.


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Figure 6In the figure, decreasing dK values leads to decrease the rise time. For the given

specification, dK =.1 and pK =2 make the system meet the specifications very well in

Figure 7.

Figure 7From the given transfer function for the system with PD controller (10), steady state

error can be calculated by setting .0=s

( ) ( ) pdesgdvdes

pdesgdv KmglsKFas

d

KmglsKFs

s

+++++

++++=

coscos 22

where desgmgld sin= (10)

Therefore the steady state error is the function of des in (11).

pdesg

desg

ssKmgl

mgle

+

=

cos

sin(11)


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With simulation with the parameters2

=des , pK =1, the error in Figure 8 is almost

same with .024 calculated by (11).

Figure 8

3.5 Washout Filter Design(task 5)

In practice,.

is not measured. In stead of that, one high pass filter which has onezero at the origin in feedback loop.

t1

t i m e

s

s+3

T r a n s f e r F c n

t h e t a

T o W o r k s p a c e

T e r m i n a t o r

S u b S y s t e m

S t e p

In1

t h e t a

t het adot

S i n g l e

P e n d u l u m

S i m u l a t o r

S c o p e

K p . s + K i

s

P I c o n t r o l

K d

G a i n

C l o c k

d ou ble d ou bl e

d o u b l e

d o u b l e

d o u b l e

d o u b l e

d o u b l e

d o u b l e

d o u b l e

Figure 8-1

The step response of the washout filter controller is oscillated at transient part and hasdamping. But the filter reduces the steady state error in Figure 8-2.


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Figure 8-2

3.6 PD controller in sample data implementation(task 6)So far, the controller has been implemented in continuous time domain. The discrete

implementation, however, becomes more popular by appearing computers and smalldigital microprocessors. In continuous domain, a differential equation can be

approximated with a difference equation e.g., ( ) ( ) ( )( ) stkkk /1

.

With the approximation, the PD controller system is implemented in figure 9.

Figure 9

The simulation result is well approximated in sampled data implementation in figure10.


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Figure 10

3.7 PID controller in continuous and sampled data implementation (task 7,8)To compensate the steady state error, the integral controller should be added. One

obvious effect of the integral control is that it increases the type of the system by one;that is, if the steady-state error to a given input is constant, the integral controlreduces it to zeros. The transfer function of the PID controller system is

( ) ( ) ( ) ( )ddesgdv

des

ipdesgdv

idd

KsKmglsKFas

ds

KsKmglsKFs

KsKsK

++++++

+++++++

=

coscos 323

2

wheredesg

mgld sin

=(12)

then0

sdes

goes to zero as 0S . In other word, the integral controllercompensates steady state error. The continuous and sampled data implemented PIDcontrollers are simulated and compared in figure 11 and figure 12.

t1

t i m e

t h e t a

T o W o r k s p a c e

S u b S y s t e m

S t e p

I n 1

t h e t a

t h e t a d o t

S i n g l e

P e n d u l u m

S i m u l a t o r

S c o p e

K p . s + K i

s

P I c o n t r o l

K dG a i n

C l o c k

d o u b l e d o u b l e d o u b l e d o u b l e

d o u b l e

d o u b l e

d o u b l e

d o u b l e

Figure 11


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The performances of both controllers meet the given specifications in figure 12 andfigure 13.

Figure 13

Figure 14

Parameter Continuous Controller Sampled ZOH controller

=des 2/ =des =des 2/ =des Overshoot(%) 4.42 3.5 4.37 3.45

pt (sec) .27 .354 .269 .368

rt (sec) .079 .079 .079 .08

st (sec) 1.113 1.053 1.113 1.056Steady error (%) 1.4e-4 1.5e-4 1.47e-4 1.57e-4

3.8 Tracking(task 9)Up to now, the input has been a step function. But in real system, the input tends to be a

time varying function such as a sinusoidal function. The given input ( ) ( )ttdes 4sin=


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increases the transfer function by 2 because the Laplace transform of the input is

22

2

16

4

+s. The total system becomes (13).

( ) ( )

( ) ( ) iddesgdv

ipdesgdv

idd

KsKmglsKFas

ds

sKsKmglsKFs

KsKsK

+++++

++

+++++

++=

cos

16

4

cos

3

22

2

23

2

(13)

As shown in (13), the system is a fifth order system. First of all, dominant second order

system needs to be found which has the poles most close to imaginary axis. Then tuningthe poles of the second system to make the system meet the specifications.With the same gains with task 8, the output is shown in Figure 15.

Figure 15

3.9 Experimental

4. ConclusionIn this project, the controller design of a second order system(a single pendulum) hasbeen studied. As a controller, PID controller has been adopted. Mathematical

analysis of the transfer function has been used and simulation justify the analysis.Even though, a PID controller is simple, the robust of the controller has been

experienced and justified.

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