control systems ii eee notes

8/3/2019 Control Systems II Eee Notes

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UNIT-1

System Variables

The system’s boundary depends upon the defined objective function of the system.• The system’s function is expressed in terms of measured output variables.• The system’s operation is manipulated through the control input variables.

• The system’s operation is also affected in an uncontrolled manner through the disturbanceinput variables.Open-loop control system

A system that utilizes a device to control the process without feedback. Thus the output hasno effect upon the signal to the processClosed-loop feedback control system

A system that uses a measurement of the output and compares it with the desired outputTypes of feedback control systems

– Negative feedback - the output signal is fed back sothat it subtracts (-) from input signal * – Positive feedback

Example 1: Car and Driver

Objective function: to control the direction and speed of the car • Outputs: actual direction and speed of the car • Control inputs: road markings and speed signs

• Disturbances: road surface and grade, wind, obstacles• Possible subsystems: the car alone, power steering system, braking system, . . .Control System Components

System, plant or process (to be controlled)• Actuators (converts the control signal to a power signal)• Sensors (provides measurement of the system output)• Reference input (represents the desired output)• Error detection (forms the control error)• Controller (operates on the control error to form the



control signal, sometimes called compensators)

Control System TerminologyDesired Response – the idealised instantaneous behaviour that we

would like from the system.Transient Response – the gradual change in the system as itapproaches its approximation of the desired response.Steady-State Response – the response of the system once it hasfinished changing and is now approximating the desired response.Error – the difference between the input and the output of the system.Steady-State Error – the difference between the steady-stateresponse and the desired response.Stability – the ability of the system to settle into a steady-stateresponse.Controller – the part of the system that generates the input to the

plant or process being controlled.Open-Loop – a system that does not monitor its output. Open-loopsystems can not correct for disturbances.Closed-Loop – a system that monitors its output and makescorrections to reduce error. By monitoring the output the system cancorrect for disturbances.Disturbance – a signal that is not modelled or calibrated in thesystem leading to corruption of the expected behaviour.Compensator – a system inserted into the controller to improveperformance.Feedback – a path that allows signals from the output of some sub-

system to flow back and affect the input of some sub-system earlier inthe system signal path.Robust – a system that will still work as expected with changes to thesystem parameters, as might be caused by wear of components, or achange in behaviour with temperature.



UNIT-2



• In the s-domain a function is plotted on the imaginary axis (y) and the real axis (x)

• Roots of q(s) give poles• Roots of p(s) give zeros

-6 -5 -4 -3 -2 -1 0 1-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1Pole-ZeroMap

Real Axis

I m a g i n a r y A x i s

Steady-state error is defined as the difference between the input and output of a systemin the limit as time goes to infinity (i.e. when the response has reached the steady state).The steady-state error will depend on the type of input (step, ramp, etc) as well as thesystem type (0, I, or II).

Note: Steady-state error analysis is only useful for stable systems. It is your

responsibility to check the system for stability before performing a steady-state

error analysis. Many of the techniques that we present will give an answer even if

the system is unstable; obviously this answer is meaningless for an unstable system.

Calculating steady-state errors

Before talking about the relationships between steady-state error and system type, we willshow how to calculate error regardless of system type or input. Then, we will startderiving formulas we will apply when we perform a steady state-error analysis. Steady-state error can be calculated from the open or closed-loop transfer function for unityfeedback systems. For example, let's say that we have the following system:

( ))(

)(

)5)(3(

210)(

sq

s p

s s s

s sY =

+++

=



which is equivalent to the following system:

We can calculate the steady state error for this system from either the open or closed-looptransfer function using the final value theorem (remember that this theorem can only beapplied if the denominator has no poles in the right-half plane):

Now, let's plug in the Laplace transforms for different inputs and find equations tocalculate steady-state errors from open-loop transfer functions given different inputs:

• Step Input (R(s) = 1/s):

• Ramp Input (R(s) = 1/s^2):

• Parabolic Input (R(s) = 1/s^3):

When we design a controller, we usually want to compensate for disturbances to asystem. Let's say that we have the following system with a disturbance:



we can find the steady-state error for a step disturbance input with the followingequation:

Lastly, we can calculate steady-state error for non-unity feedback systems:

By manipulating the blocks, we can model the system as follows:

Now, simply apply the equations we talked about above.

System type and steady-state error

If you refer back to the equations for calculating steady-state errors for unity feedback systems, you will find that we have defined certain constants ( known as the static error constants). These constants are the position constant (Kp), the velocity constant (Kv), andthe acceleration constant (Ka). Knowing the value of these constants as well as thesystem type, we can predict if our system is going to have a finite steady-state error.

First, let's talk about system type. The system type is defined as the number of pureintegrators in a system. That is, the system type is equal to the value of n when the systemis represented as in the following figure:



Therefore, a system can be type 0, type 1, etc. Now, let's see how steady state error relates to system types:

Type 0 systemsStep Input Ramp Input Parabolic Input

Steady State Error Formula 1/(1+Kp) 1/Kv 1/Ka

Static Error Constant Kp = constant Kv = 0 Ka = 0

Error 1/(1+Kp) infinity infinity

Type 1 systems Step Input Ramp Input Parabolic Input

Steady State Error Formula 1/(1+Kp) 1/Kv 1/KaStatic Error Constant Kp = infinity Kv = constant Ka = 0

Error 0 1/Kv infinity

Type 2 systems Step Input Ramp Input Parabolic Input

Steady State Error Formula 1/(1+Kp) 1/Kv 1/Ka

Static Error Constant Kp = infinity Kv = infinity Ka = constant

Error 0 0 1/Ka

http://www.krellinst.org/UCES/archive/classes/control/extras/ess/ess1.html








UNIT-3• The steady-state response of a linear system to sinusoidal excitation may be

determined from the system transfer function T(s) by replacing s by jω

• The resulting frequency response function T(jω ) has the polar form :

• For any value of ω , T(jω ) reduces to a complex number whose amplitude,A(jω ) gives , and φ (ω ) gives .

\

Steady-state vs. general response• In control work. the response of the system to general input functions, e.g. step

input, impulse input, are of most interest (rather than steady state sinusoidalreponse)

• However, with Fourier analysis, any time function can be represented as a seriessum of sinusoids :

Therefore it is possible to predict the nature of a response, y(t) to any given x(t)from a knowledge of how the gain, A, and phase shift, φ , vary with angular frequency

• In particular, the closed-loop behaviour of a feedback system may be usefully predicted from the open-loop gain and phase-shift characteristics

Open loop response : Bode plots

T(s)

X(s ) Y(s)

)()()( ω φ ω ω j

e A jT =

)(

)(

s X

sY

)(

)(arg

s X

sY

22

2)(2

++

=

s s sT

......)2()()( 22110 +−+−+= φ ω α φ ω α α t Sint Sint x

10



• Amplitude is plotted in dB• Phase is plotted in degrees• Used to identify system order, corner frequencies, etc

Open loop response : Polar plots

• Polar plots are more useful in control system design work – open-loop polar plot indicates whether the closed-loop system is stable– also gives a measure of how stable the system is

• Represent T(jw) on an Argand diagram :

• Polar plot is traced out by the trajectory of T(jw) as w increases progressivelyfrom zero

ReIm

T(jw)

A ( j w ) ) fai(jw)(j)



• A polar plot may be constructed from experimental data or from a system transfer function

• If values of w are marked along the contour, a polar plot has the same information

as a Bode plot• Usually, the shape of a polar plot is of most interest ….

Problem:Construct t0 j1) j(T:0 +→ω→ω

o

2

180 j

1

e0) j(T.e.i

) j(T:

−ω

→ω

−→ω∞→ω

2 j) j(T:1 −=ω=ω

2)1s(

1)s(T

+=

2) j1(

1) j(T

ω+=ω

w 0 0.25 0.5 0.75 1.0 1.5 2.0 3.0 5.0

T(jw) 1+j0

0.83-j0.44

0.48-j0.64

0.18-j0.61

0-j0.5

-0.12-j0.28

-0.12-j0.16

-0.08-j0.06

-0.04-j0.01



a s

a sT

+=)(

s sT

1)( =



Nyquist Stability Criterion• This is used to determine stability of a closed loop system• It is an alternative method to the Routh-Hurwitz criterion

– It is applicable to control systems involving time-delays; e.g. open looptransfer functions of the form :

– In addition to determining whether or not a system is stable, the method

also provides a measure of relative stability– It requires only a polar plot of the open-loop frequency response

The Nyquist Contour • As , the contour encloses the entire RHP• Therefore, if a function has poles or zeros in the (finite) RHP, they will be

enclosed by the Nyquist contour • The Nyquist method may be interpreted as determining the region of values taken

by a transfer function when s lies on, or within the Nyquist contour

• For a closed-loop system to be stable, the transfer functionmust have no poles in the RHP :

• This can be determined by ensuring that the open-loop transfer function,for any value of s in the RHP

• I.e. for any value of s enclosed by the Nyquist contour

22

2

2)(

nn

n

s s

sT

ω ξ ω

ω

++=

∞→ R

)()(1

)(

)(

)(

s H sG

sG

s R

sC

+

=

1)()( −≠ s H sG

1)()( −≠ s H sG



• The region of valu

for :

)1s)(10s2s(

20)s(H)s(G2

+++=

T

) ps)....( ps)( ps(

)zs)....(zs)(zs(K )s(H)s(G

n21

m21

−−−

−−−=

∞→R

0)s(H)s(G →

• The plot of the N

the Nyquist stabilit



Th∞→R

Section de: s =

G(

∞→R )1s)(10s2s(

20)s(H)s(G

2 +++

=

)1 j)(102 j(

20) j(H) j(G

2+ω+ω+ω−

=ωω

Problem :Construct tthe following open-lo



E

The complete Nyqu

function :



UNIT-4

control systems ii eee notes

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