55040110 Control System Lab Manual by AMIT KUMAR

PRESTIGE INSTITUTE OF ENGINEERING & SCIENCE, INDORE

ELECTRICAL & ELECTRONICS DEPARTMENT

BATCH : 2008-2012

PRESTIGE INSTITUTE OF ENGINEERING & SCIENCE PRESTIGE INSTITUTE OF ENGINEERING & SCIENCE INDORE (M.P.)ELECTRICAL AND ELECTRONICS DEPARTMENT

CONTROL SYSTEMS LABORATORY

Roll No: 0863ex081003 Date: 09/05/2011

Remarks if any, Signature of lecturer:

EXPERIMENT No. --1

Objective:

Study of Error Detectors. a) Potentiometer Error Detectors.b) Synchro Error Detectors.

Theory:

a) Potentiometer Error Detector:

All feedback control systems operate from the error signal which is generated by a comparison of the reference and the output. Error detectors perform the crucial task of comparing the reference and output signals. In a purely electrical system where the reference and output are voltages, the error detector is a simple comparator.

In some other systems with non-electrical outputs, the output signal is converted into electrical form through a measurement or transducer block, and then error detection is performed on the electrical signals. A position control system, with both input and output variables as mechanical positions (linear or angular), may however consist of two potentiometers - reference and output, which function as an error detector. Other devices which could be used in similar applications include synchro sets (for a.c. systems), sine-cosine potentiometers, hall effect-potentiometers etc.

The Potentiometer Error Detector consists of two identical potentiometers electrically connected in parallel and supplied by a voltage source as shown in figure 1a.

Fig.1a Basic Circuit Diagram

The symbolic circuit representation of a potentiometer error detector is

shown in figure 1b below.

Fig.1b Circuit Diagram of Potentiometer Error Detector

The input shaft is coupled to the potentiometer marked A, and is held fixed at the desired angular position say θr while the output shaft is coupled to the potentiometer marked B and the position is indicated asθo. The potential difference between the variable points of potentiometers A and B is proportional to the angular difference (θr−θo ). Therefore, the error is given by, e∝(θ ¿¿ r−θo)¿ e ¿K e(θ¿¿ r−θo)¿ ------- (1) Or e ¿K eθeθe=(θ¿¿ r−θo)¿Where K e known as the gain of error detector, and expressed in V/rad.The Laplace transform of the above equation is,

E ( s) ¿K e [θ¿¿ r (s)−θo (s)]¿ E ( s) ¿K eθe(s) --------- (2)

The block diagram relating the output E(s) and the input θr (s ) drawn and shown in figure 2.

Block Diagram of Potentiometer error detector

And the transfer function is given as,

E ( s)θe (s )

=K e

Or E ( s)

θr(s)−θo(s)=K e

b) Synchro Error Detector :-

Note:- Students have to derive & write the Transfer Function of the

Synchro Error Detector.



Roll No: 0863ex081003 Date: 09/05/201


EXPERIMENT No. -- 2

Objective:

Determination of Transfer Function of DC Motor.

c) Armature Controlled DC Motor.

d) Field Controlled DC Motor.

Theory:

Write DC motor description covering following points

1. Construction and principle of working

2. Types and their characteristics, applications.

Derivation of Transfer Function:-

(a) Armature Controlled DC Motor.

The following assumptions are made while deriving the transfer function.

(1)Air gap flux is proportional to the field current.(2)Armature reaction is negligibly small.(3)The back emf is proportional to motor speed.(4)Field current is constant.

Figure shows armature controlled motor,

Figure: Armature Controlled DC Motor

Let, Ra= armature resistance La= armature inductanceJ= total moment if inertia of the rotor.Θ=Shaft position Ia=armature current of viscous firction.

The voltage equation of motor is given by,

Va=Eb+ IaRa+La dIadt

Where, Eb= back emf.

Taking Laplace Transform of both sides,

Va(s) = Eb(s) +Ia(s) [Ra+sLa]

Ia ( s)=Va ( s)−Eb (s )Ra+sLa

- - - - - - - (1)

Now, torque developed by motor is proportional to product of air gap flux and armature current,

Tm α φ Ia

Tm = Km Ia (φ constant)

Where, Km is constant having unit N-ω/Amp of armature current.

Tm(s) = Km Ia(s) - - - - - - - (2)

The load torque is given by,

Tl=Jd 2θdt 2

+F dθdt

Tl=(Js2 +Fs)θ(s) - - - - - - - - - (3)

Equating (2) and (3) we get,

Km Ia(s) = (Js2 +Fs)θ(s)

Km﴾ Va ( s )−Eb(s)Ra+sLa

) = (Js2 +Fs)θ(s)

Km[ Va(s)- Eb(s)]= (Js2 +Fs)(Ra+sLa)θ(s) - - - - - (4)

But, the back emf is proportional to motor speed

i.e. Eb α dθdt

Eb = Kbdθdt

where, Kb = constant having v/rad/sec.

Eb= s Kb θ(s)

Substituting this value of Eb(s)

Km[ Va(s) - sKbθ(s)]= (Js2 +Fs)(Ra+sLa)θ(s)

Km Va(s) = [(Js2 +Fs)(Ra+sLa) + sKb]θ(s)

θ(s)Va(s)

= Km[(Js 2+Fs)(Ra+sLa)+sKb]

is the required transfer function.

The unknown quantities in above equation are Kb or Km (it can be shown Km=Kb) Ra, La, J and F, out of these Ra and La can be found by bridge methods J is found by retardation test. The only unknowns now are Kb and Kf.

(b)Field Controlled DC Motor:-

Note:- The procedure of deriving Transfer function is similar to Armature Controlled DC Motor.

Students have to derive & write the transfer function of Field Controlled DC Motor.



Roll No: 0863ex081003 Date: 09/05/2011


EXPERIMENT No.--3

Objective:

To determine the Time Response of 1st and 2nd order system using MATLAB Simulink.

Theory:-

Whenever any standard known test signal is applied to the systems, it responds according to design and nature of the system. This response is the function of time and may be divided into two main parts:

(i) Transient Response: - This occurs just after the switching or after any abnormal condition (due to fault or sudden change of load). This is the function of system Transfer Function. This response is mainly due poles of the system.

(ii) Steady State Response: - When the system becomes settled and starts its normal working, that response is known as steady state response and is the function of input. The portion of the response is mainly due to the poles of the input or Forcing function.

Time Response of First Order Systems for Unit Step Input:-

A general expression of a 1st order system is given as,

C( s )R( s )

= 1sT +1

(3.1)

C (s )=R (s ) 1sT +1

(3.2)

For unit step input, R( s )=1

s Figure 1 Block Diagram representation

of a 1st order system

Put R(s) in equation (3.2),

C (s )=1s

1sT +1

Take partial fraction,

C (s )=1s− TsT +1

=1s− 1s+1/T

Take inverse Laplace Transform,

c ( t )=1−e−t /T (3.3)

Figure 2 Time Response of 1st order system for Unit Step input

Error, e ( t )=r ( t )−c ( t )=1−(1−e−t /T )

=e−t /T

(3.4)

Steady State error, ess=Lim

t→∞e−t /T=0

Time Response of Second Order Systems for Unit Step Input:-

A general expression of a second order system is given as,

C( s )R( s )

=ωn

2

s2+2ξωn s+ωn2

(3.5)

Figure 3 Block Diagram of 2nd order Control System

For unit step function R(s) =1,

C (s )=1s

ωn2

s2+2ξωns+ωn2

Take Partial fraction,

C (s )=1s−

s+2ξωn

s2+2ξωn s+ωn2

Or

C (s )=1s−

s+2ξωn

( s+ξω )2+ωn2 (1−ξ2 )

(3.6)

Putωd=ωn√(1−ξ2 )

,

C (s )=1s−

s+2ξωn

( s+ξω )2+ωd2

(3.7)

Rearranging above equation,

C (s )=1s−

s+ξωn

( s+ξω )2+ωd2−ξωn

ωd

ωd

(s+ξω )2+ωd2

Take inverse Laplace transform

c ( t )=1−e−ξωn t cosωd t−

ξωn

ωd

e−ξωn t sin ωd t

(3.8)

Since, ωd=ωn√(1−ξ2 )

c ( t )=1−e−ξωn t cosωd t−

ξωn

ωn√(1−ξ2)e−ξωn t sinωd t

=1− e−ξωn t

√(1−ξ2)[√(1−ξ2 )cosωd t+ξ sinωd t ]

(3.9)

Put sinφ=√(1−ξ2 )

and cos φ=ξ

c ( t )=1− e−ξωn t

√(1−ξ2)[sinφ cosωd t+cos φ sinωd t ]

c ( t )=1− e−ξωn t

√(1−ξ2)[sin(ωd t+φ )]

(3.10)

Since ωd=ωn√(1−ξ2 )

c ( t )=1− e−ξωn t

√(1−ξ2) [sin (ωn√(1−ξ2 )t+tan−1(√(1−ξ2)ξ )]

(3.11)

Now, e(t)=r(t)-c(t) where, r(t)=1,

e ( t )= e−ξωn t

√(1−ξ2) [sin(ωn√(1−ξ2) t+ tan−1(√(1−ξ2)ξ )]

(3.12)

Steady state error,

ess=Limt→∞

{ e−ξωn t

√(1−ξ2 ) [sin(ωn√(1−ξ2) t+ tan−1(√(1−ξ2 )ξ )]}=0

(3.13)

Case-I: ξ<1

; (Underdamped Response)

Case-II: ξ=0

; (Undamped Response)

From equation (3.11),

c ( t )=1− e−o×ωn t

√(1−02) [sin(ωn√(1−02) t+ tan−1(√(1−02)0 )]

=1−sin(ωn t+ tan−1∞)

=1−sin(ωn t+π /2 )

c ( t )=1−cos (ωn t )

(3.14)

Case-III: ξ=1

; (Critically damped Response)

From equation (3.10),

c ( t )=Limξ→1 [1− e

−ξωn t

√(1−ξ2)[sin(ωn√(1−ξ2) t )+φ )] ]

c ( t )=Limξ→1 [1− e

−ξωn t

√(1−ξ2)[sin(ωn√(1−ξ2) t )cos φ+cos (ωn√(1−ξ2)t )sinφ ) ]]

(3.15)

Now, cos φ=ξ

and sinφ=√(1−ξ2 )

c ( t )=Limξ→1 [1− e

−ξωn t

√(1−ξ2)[sin(ωn√(1−ξ2) t )ξ+cos (ωn√(1−ξ2)t )√(1−ξ2 )) ]]

(3.16)

Now,

Limξ→1

[sin(ωn√(1−ξ2) t )ξ ]→ωn√(1−ξ2 )t

Limξ→1

[cos(ωn√(1−ξ2 )t ) ]→1

Put, in equation (3.16)

c ( t )=Limξ→1 [1− e

−ξωn t

√(1−ξ2)[ωn√(1−ξ2)t+1×√(1−ξ2)) ]]

=Limξ→1 [1− e

−ξωn t

√(1−ξ2)√(1−ξ2 )(ωn t+1 ) ]]

c ( t )=1−e−ωn t ((ωnt+1)

Case-IV: ξ>1

; (Overdamped Response)

MATLAB PROBLEMS:-

1) 1st Order Systems

(i) Plot the Unit Step Response of the systems having Transfer Function G(s) =

a) C b)

1s+2

c)

1s+5

(ii) Plot the Unit step response of previous problem in 1 graph.

(iii) Plot the Unit step response of

1s+1

,

2s+2

,

5s+5

2) 2nd order systems

Plot the Time response of the system-

(i)

1

s2+s+1 ; ξ=0 . 5

(ii)

1

s2+2 s+1 ; ξ=1

(iii)

1

s2+1 ; ξ=0

(iv)

1

s2+1,

4

s2+4; ξ=0

, ωn=1,2

(v) State the difference between time response of,

1

s2+2 s+1 ; ξ=1

&

1

s2+3 s+1; ξ=1 .5



Roll No: 0863ex081003 Date: 09/05/2011


EXPERIMENT No. 4

Objective:

Draw the Root Locus of a system using MATLAB.

Theory:- Root locus analysis is a graphical method for examining how the roots of a system change with variation of a certain system parameter, commonly the gain of a feedback system. The stability of a control system is determined from the roots of the characteristic equation 1+G(s)H(s)=0. For a system to be stable the roots of the characteristic equation should be located in the left half of the s-plane.

A closed loop control system having forward path gain K is shown in Figure 1.The characteristic equation of the system is given by,

1+G(s)H(s)=0 ………..(1)Any root of equation satisfies following two conditions,

|1+G( s )H ( s )=0| ……. (2)∠G( s )H ( s )=(2k+1)180∘

………… (3) Where k=0, 1, 2 …The root locus method of analysis is a process of determining the points in the s plane satisfying equation (2) and (3). Usually forward path gain K is considered as an independent variable and the roots of the equation 1+G(s)H(s)=0 as dependent variables, the roots are plotted in s-plane with K as variable parameter. Form the location of the roots in s-plane the nature of the time response and system stability can be ascertained.

The Procedure for plotting Root Locus:-

The stepwise procedure for plotting the root locus for a given open loop transfer function is given below.

1. Starting Points:- The root locus starts (K=0) from the open loop poles.

2. Ending Points:- The root locus terminates (K=) either on the open loop zero or infinity.

3. Number of branches:- The number of branches of the root locus are,

N=P, if P>Z

=Z, if Z>P

4. Existence on real axis:- The existence of root locus on a section of real axis is confirmed if the sum of open loop poles and zeros to the right of the section is odd.

5. Break away points:- On the root locus between two open loop poles the roots move towards each other as the gain factor K is increased till they are coincident. At this point, the value of K is maximum as far as the portion of root locus between the two open loop poles is concerned. Any further increase in K, breaks the root locus in two parts. The breakaway point can be determined by rewriting the characteristic equation and therefrom solving for the value of s from the equation below,

dKds

=0

6. The angle of asymptotes:- For higher values of K the root locus branches are approximated by asymptotic lines making an angle with the real axis given by,

(2k+1 )×180∘

P−Z Where k=0, 1, 2 …upto (P-Z)-1

7. Intersection of asymptotes on real axis:- The asymptotes intersects at point x on the real axis is given by,

x=∑ Poles−∑ Zeros

P−Z

8. Intersection points on the Imaginary axis:- The value of K and the point at which the root locus branch crosses the imaginary axis is determined by applying Routh criterion of the characteristic equation. The roots at the intersection points are imaginary.

9. The angle of departure from the complex pole:- The angle of departure from the complex pole is given by,

Φd=180∘−(Φ p−Φz )

Where Φ p

is the sum of all angles suspended by remaining poles.

and Φz

is the sum of all zeros suspended by zeros.

The angle of departure is tangent to the root locus at complex pole.

10. The angle of arrival at complex zeros:- The angle of arrival at the complex zero is given by,

Φa=180∘−(Φz−Φ p )

Where Φz

is the sum of all angles suspended by remaining zeros.

and Φ p

is the sum of all zeros suspended by poles.

The angle of arrival is tangent to the root locus at complex zero.

MATLAB PROBLEMS:-

1. Sketch the root locus for the open loop transfer function of a unity

feedback control system,

G (s )H (s)= Ks ( s+1 )(s+3)

Determine (i) The value of K at s=-4

(ii) The value of K for =0.5

(iii) Obtain the closed loop transfer function for K=1.66.

2. Sketch the root locus plot for the system when the open loop transfer

function is given by

G (s )= K

s (s+4 )(s2+4 s+13)

Determine (i) The breakaway points.

(ii) The angle of departure from the complex pole.

(iii) The stability condition.

Note:-

First solve the problem on the paper and attach it with the Practical.

Using MATLAB draw the root locus & compare it with your result.

Take print out of the result and attach it with the practical.



Roll No: 0863ex081003 Date: 09/05/2011


EXPERIMENT No. 5

Objective:

Draw the Nyquist Plot of a system using MATLAB.

Theory:

Nyquist Criterion: - Nyquist criterion is used to identify the presence of roots of characteristic equation of a control system in a specified region of s plane. From the stability view point the specified region being the entire right hand side beyond the imaginary axis of complex s-plane.

Procedure for mapping from s-plane to G(s)H(s)-plane:-

The first step to understand the application of Nyquist criterion in relation to determination of stability of control systems is mapping from s-plane to G(s)H(s)-plane. “s” is considered as independent complex variable and the corresponding value of G(s)H(s) being the dependent variable plotted in another complex plane called G(s)H(s)-plane.

Thus for every point in s-plane there exists a corresponding point in G(s)H(s)-plane. During the process of mapping the independent variable “s” is varied along a specified path in “s” plane and the corresponding points in G(s)H(s)-plane are joined. This compltes the process of making from s-plane to G(s)H(s)-plane.

Application of Nyquist Criterion to determine the stability of closed Loop systems:-

The overall transfer function of a closed loop system is given by,

C( s )R( s )

=G( s )

1+G( s )H ( s )

(1)

G( s )H ( s )

is the open loop transfer function and C is the characteristic equation. For a system to be stable the zeros of the characteristic equation should not have positive real part. In order to investigate the presence of any

zero of 1+G( s )H ( s )=0

in the right half of the s-plane, let us chose a contour which completely encloses this right half of the s-plane.

The closed path is ab→bc→ cd→da

starting at a (s=+ j 0 )

through b (s=+ j∞)

, along a semicircular arc bc of infinite radius, c ( s=− j∞)

, d ( s=− j 0 )

and finally from d to a semicircular arc of radius r in

anticlockwise direction in such a manner that r→0

.

On traversing the closed path for independent variable s in

s-plane as above, the corresponding point to point values of G( s )H ( s )

are

plotted in G( s )H ( s )

plane and change in argument of G( s )H ( s )

is noted.

A number indicating the zeros of G( s )H ( s )

which are located in right half of s-plane is calculated by the relation:

Δ AugG (s )H ( s )=2π (P−Z )

Where P is the number of poles of G( s )H ( s )

having positive real part and is a known quantity.

Note that the change in argument of G( s )H ( s )

is measured w.r.t. origin

(0, 0) in G( s )H ( s )

plane. This procedure calculates the number of zeros of G( s )H ( s )=0

and not the zeros of1+G( s )H ( s )=0

.

However, from the plot of G( s )H ( s )

the number of zeros of 1+G( s )H ( s )=0

which have +ve real part can also be determined if the origin

of G( s )H ( s )

plane is shifted to the point (-1+j0) and the change in argument

of G( s )H ( s )

plot is measured w.r.t. the point (-1+j0) in G( s )H ( s )

plane. The number of zeros thus calculated is also the number of roots of characteristic

equation1+G( s )H ( s )=0

.

As the s-plane region wherein only poles and zeros of 1+G( s )H ( s )=0

with +ve real part is under investigation, the expression relating the number of poles and zeros present in the R.H.S. of s-plane is written as,

Δ AugG (s )H ( s )=2π (P+−Z+ )

Or

Δ AugG (s )H (s2π

)=(P+−Z+ )

Or N=(P+−Z+ )

Where ,

N = Number of encirclements of the point (-1+j0) by G( s )H ( s )

plot. The positive direction of encirclement being anticlockwise.

P+ = Number of poles of G( s )H ( s )

with +ve real part.

Z+ = Number of zeros of 1+G( s )H ( s )=0

with +ve real part.

For a stable system Z+ =0, therefore, the condition for a control system to be stable is

N=(P+−0 )

N=P+

For most of the control systemP+=0

, therefore for such cases the condition of stability be, N=0

The closing of Nyquist plot from s=-j0 to s=+j0 is given the following table

according to type of G( s )H ( s )

:

nType of G(s)H(s)

Angle through which Nyquist plot

is to be closed from =-0 to =+0

Magnitude of

G( jω )H ( jω)

0 0 0The points =-0 &

=+0 are coincident

1 1 -

2 2 -2

3 3 -3

“ ‘ ‘ ‘

“ ‘ ‘ ‘

n N -n

Note:-

-n= clockwise rotation;

n= type of transfer function;

MATLAB PROBLEMS:-

1. Draw the Nyquist plot for the open loop transfer function given below and comment on the closed loop stability.

G (s )H ( s )= 2.2

s(s2+2 s+2) (s+1 )

2. The open loop transfer function of a unity feedback control system is given below

G (s )= s+0.25

s2 (s+1 )(s+0.5)

Determine the closed loop stability by applying Nyquist Criterion.

NOTE:-


Using MATLAB draw the Nyquist plot & compare it with your result.




Roll No: 0863ex081003 Date: 09/05/2011


EXPERIMENT No. 6

Objective:

Draw the Bode Plot of a system using MATLAB.

Theory:

Bode Plot:

The Bode plot method gives a graphical procedure for determining the stability of a control system based on sinusoidal frequency response. The transfer function of a system for sinusoidal input response can be obtained by substituting j in place of Laplace operator “s”. Therefore, if the open loop

transfer function of a system is G( s )H ( s )

, the corresponding sinusoidal open-

loop transfer function is G( jω )H ( jω)

which can be expressed in the form of magnitude and phase angle.

The variation of the magnitude of sinusoidal transfer function expressed in decibel and the corresponding phase angle in degrees being plotted w.r.t. frequency on a logarithmic scale (i.e. log10) in rectangular axes. The plot thus obtained is known as Bode Plot. The logarithmic Bode Plot has an advantage of being approximated by asymptotic straight lines.

Relative stability of a closed loop control system can be determined by plotting its open loop transfer function by Bode Plot method. The Gain Margin and Phase Margin is determined directly from Bode Plot.

Bode Plot for Transfer Functions:-

The open loop transfer function of a closed loop transfer function can be expressed as:

G( s )H ( s )=K [ (1+sT 1 )(1+sT 2 ). .. . ]ωn

2

sN [ (1+sT a )(1+sT b ). .. .. .( s2+2ξωn s+ωn2 ) ]

(1)

The sinusoidal form of transfer function (1) is obtained by substitutings= jω

, therefore

G( jω )H ( jω)=K [ (1+ jωT 1)(1+ jωT2 ) .. .. ]ωn

2

( jω)N [ (1+ jωT a )(1+ jωT b ). .. . ..((ωn2−ω2 )+2ωnω ) ]

(2)

In decibel equation (2) can be expressed as:

20 log10|G( jω )H ( jω)|=20 log10K+20 log10|1+ jωT1|+.. . .. .. . ..

. .. ..−20N log10| jω|−20 log10|1+ jωTa|.. . .. .. .. .

. .. ..−20N log10|ωn

2−ω2

ωn2

+( j2ξωnω )

ωn2

|

(3)

And for phase angle,

∠G( jω )H ( jω)=tan−1ωT 1+tan−1ωT 2+. .. .. .−N∠90∘−tan−1ωT a−tan−1ωT b

. .. .− tan−1 [ (2 ξωnω )

(ωn2−ω2 ) ]

(4)

The bode plot is a graph obtained from equation (3) and (4) consisting of two parts as follows:

I. Magnitude of G( jω )H ( jω)

in decibel

II. Phase angle of ∠G( jω )H ( jω)

versus log10

Procedure for drawing Bode Plot:-

1. Identify the corner frequencies.

2. Draw the asymptotic bode plot. The slope will change at each corner frequency by +20db/dec for zero and -20db/dec for pole. For complex conjugate pole and zero the slope will change by db/decade.

3. (i) For type zero system draw a line upto first (lowest) corner frequency having 0db/dec slope.

(ii) For type one system draw a line having slope -20db/dec upto =K. Mark first (lowest) corner frequency.

(iii) For type two system draw a line having slope -40db/dec upto ω=√K

and so on. Make first corner frequency.

4. Draw a line upto second corner frequency by adding the slope of next pole or zero to the previous slope and so on.

5. Calculate phase angles for different values of and join all points.

6. Obtain Gain Margin and Phase Margin from the Bode Plot.

MATLAB PROBLEMS:-

1. Sketch the bode plot for the transfer function given below:

G (s )H (s)=2(s+0.25)

s2 ( s+1 ) (s+0.5 )

From the bode plot determine:

(a) The phase crossover frequency (b) The gain crossover frequency

(c) Gain Margin (d) Phase Margin

2. The open loop transfer function of a system is given below

G (s )H (s)= 4s (1+0.5 s )(1+0.08 s)

Determine

(a) Gain Margin

(b) Phase Margin

(c) Closed loop stability

3. Draw the Nyquist plot for the open loop transfer function given below and comment on the closed loop stability.

G (s )H ( s )= 2.2

s(s2+2 s+2) (s+1 )

NOTE:-


Using MATLAB draw the Bode plot & compare it with your result.




Roll No: 0863ex081003 Date: 09/05/2011


EXPERIMENT No. 7

Objective:

Determination of State Space Model for classical transfer function using MATLAB.

Theory:

State Space Analysis:-

The procedure for determining the state of the system is called state variable analysis. The state of a dynamic system is the smallest set of variables such that the knowledge these variables at t=t0 with the knowledge of input at tt0 completely determines the behavior of the system for any time tt0. This set of variables is called as State variables.

Advantages of State Space system:-

1. This approach can be applied to linear or nonlinear, time variant or time invariant systems.

2. It is easier to apply where the Laplace Transform cannot be applied.

3. nth order differential equations can be expressed as ‘n’ equations of first order whose solutions are easier.

4. It is a time domain approach.

5. This method is suitable for digital computer computation because this is a time domain approach.

6. The system can be designed for optimal conditions w.r.t. given performance indices.

State Space representation of Systems:-

The nth order differential equation is given as,

dn ydt n

+a1dn−1 ydtn−1

+. .. ..+an y=u ( t ) (1)

It is noted that the input u(t) and the initial conditions

y (0 ) ,dy (0 )dt

, .. . ..dn−1 y (0 )dt n−1

at t=0 are given.

The dynamic behavior of differential equation (1) van be determined

completely from the knowledge of u(t), y.

( t ), y(t).

The terms y(t), y.

( t ),………

yn−1( t )can be considered as a set of n state

variables.

Let, y=x1

y.

=x2(2)

yn−1=xn

The set of equations (2) can be written as,

x1

.

=x2

x2

.

=x3(3)

x.n−1=xn

xn.

=−an x1−an−1 x2−.. .. . .. .. . ..−a1 xn−1+u

A general representation of set of equations (3) can be expressed in the form of state equation,

x.

=Ax+Bu

Where,

x=[x1

x2

::::xn];

A=[0 1 .. 0 00 0 .. 1 00 00 0 0 1: : : :: : : :

−an −an−1 0 a1

];

B=[000:::1]

Where x=state vector (n1) u=control (scalar)

A=Matrix (nn) B=Matrix (n1)

The output equation is given as,

y= [1 0 0 . . 0 ] [x1

x2

::xn] (4)

Or y=Cx

WhereC=[1 0 0 .. 0 ]

, y= output (scalar), C=matrix (1n)

MATLAB Problems:-

Q.1) Obtain the state model for the transfer function given below:

C(s)R(s)

=(s+2)

(s+1 )(s+3)

Q.2) The transfer function of a system is given below,

Y (s)U (s)

= s2+s+2s3+9 s2+26 s+24

Determine the state model

Q.3) Write MATLAB script to determine the state transition matrix for

A=[ 1 4−2 −5]

Note:




Roll No: 0863ex081003 Date: 09/05/2011


EXPERIMENT No. 8

Objective:

Study of Lead and Lag Compensator using MATLAB Simulink.

Theory:-

If the performance of the control system is not upto expectations as per desired specifications, then it is required that some change in the system is needed to obtain the desired performance. The change can be in the form of adjustment of the forward path gain or inserting a compensating device in control systems.

The gain adjustment improves the steady state accuracy of the system at the cost of driving the system towards instability. However, improvement in steady state performance of a control system by gain adjustment is possible in cases where the system is found to be within the desirable stability limit even after altering forward path gain. In such a cases a compensation network is introduced in the system.

The compensation network can be introduced in the system as shown below,

Figure 1 Arrangements for introducing compensation network in a control system

The compensation network introduces additional poles/zeros in the original system thereby altering its original transfer function. The effect of adding zero to an unstable system results in making the system stable as shown in root locus plots of Figure 2.

Figure 2 Unstable system made stable by adding a zero

Figure 2 shows the root locus plot of an unstable system for higher-path gain. Addition of a zero to the system transfer function changes the shape of root locus as shown below, which indicates the stable system for all values of forward path gain.

The networks used for cascade compensation are described below.

Phase Lead Compensation:-

Figure 3 shows a phase lead network wherein the phase of output voltage leads the phase of input voltage for sinusoidal input.

Figure 3 Phase-Lead Network

The Transfer function of a phase lead network shown above is given by,

Eo (s )Ei( s )

=α (1+sT )(1+sαT )

(8.1)

Where, <1

α=R2

R1+R2 and

T=R1C

The transfer function given by equation (8.1) can be expressed in sinusoidal form as

Eo ( jω)Ei( jω)

=α (1+ jωT )(1+ jωα T )

(8.2)

The pole zero configuration and bode plot for transfer function (8.2) is shown in

Figure 4.

Figure 4 Pole-zero configuration and Bode Plot for phase-lead network

The two corner frequencies are

ω= 1T

; Lower corner frequency

ω= 1αT

; Upper corner frequency

The maximum phase-lead m occurs at mid frequency m between upper and lower corner frequencies.

log10ωm=12 [ log10( 1

T )+log10( 1αT )]

ωm=1

√αT (8.3)

The phase angle ∠Eo( jω)/Ei( jω)

can be calculated as,

∠Eo ( jω)Ei( jω)

= tan−1 (ωT )−tan−1 (ωαT )

At

ω=ωm=1

√αT the phase angle is m,

φm=tan−1( 1

√αTT )−tan−1(ωα T )

φm= tan−1( 1

√α )− tan−1(√α )

φm=tan−1(1

√α−√α

1+ 1√α √α )

Or

tanφm=1−α2√α

(8.4)

and

sinφm=1−α1+α

(8.5)

From pole-zero configuration of phase lead network, it is observed that the zero is nearer to origin as compared to pole, hence the effect of zero is dominant, therefore, the phase lead network, when introduced in cascade with forward path transfer function, the phase shift is increased.

The bode plot of phase lead network reveals that the lead network allows to pass higher frequencies and low frequencies are attenuated.

Phase-Lag Network:-

Figure 5 shows a phase lag network wherein the phase of output voltage lags the phase of input voltage for sinusoidal inputs.

Figure 5 Phase-Lag Network

The Transfer function of a phase lead network shown above is given by,

Eo (s )Ei( s )

= 1+sT1+sβT

(8.6)

Where, >1 and

β=R1+R2

R2, T=R2C

The transfer function given by equation (8.6) can be expressed in sinusoidal form as

Eo ( jω)Ei( jω)

= 1+ jωT1+ j ωβ T

(8.7)

The pole zero configuration and bode plot for transfer function (8.7) is shown in Figure 6. The two corner frequencies are

ω= 1T

; Upper corner frequency

ω= 1βT

; Lower corner frequency

Figure 6 Pole-zero configuration and Bode Plot for phase-lag network

The maximum phase-lag m occurs at mid frequency m between upper and lower corner frequencies.

log10ωm=12 [ log10( 1

βT )+log10( 1T )]

ωm=1

√αT (8.8)

The phase angle ∠Eo( jω)/Ei( jω)

can be calculated as,

∠Eo ( jω)Ei( jω)

= tan−1 (ωT )− tan−1 (ωβT )

At

ω=ωm=1

√βT the phase angle is m,

tanφm=1−β2√β

(8.9)

and

sinφm=1−β1+ β

(8.10)

Note. As β>1

, (1−β ) /(1+ β )

is negative indicating that m is negative which means that the phase angle is lagging.

Phase Lead-Lag Compensation:-

Introduction to phase-lead network in a control system shifts the gain crossover point to a higher value and, therefore, the bandwidth is increased thus improving the speed of response and overshoot is reduced but the steady state error does not show much improvement.

The use of phase lag compensation network in a control system shifts the gain crossover frequency point to a lower value thus decreasing the bandwidth and improvement in steady state error is noted but speed of response is reduced.

The speed of response and, steady state error can be simultaneously improved, if both phase-lag and phase-lead compensation is used. However instead of using two separate lag and lead network, a single network known as phase-lag-lead network shown in Figure is used which combines the effect of both lag and lead networks.

Figure 7 Phase-Lag-Lead Network

The Transfer function of phase-lag-lead network is given as,

Eo (s )Ei( s )

=(1+sT 1 )(1+sαT1 )

(1+sT 2 )(1+sβT 2 )

(8.11)

Lead Lag

Where, T 1=R1C1

, T 2=R2C2

, <1, >1 and =1.

The Transfer function given by (8.11) can be expressed in sinusoidal form as

Eo ( jω)Ei( jω)

=(1+ jωT 1 )(1+ jωα T1 )

(1+ jωT 2)(1+ jωβ T 2 )

(8.12)

The pole zero plot and bode plot for the transfer function (8.12) is shown in Figure 8.

Figure 8 Pole-zero configuration and Bode Plot for phase-lag-lead network

MATLAB PROBLEMS:-

1. The open loop transfer function of a unity feedback control system is given by,

G (s )= Ks (1+0.2 s )

Design a suitable compensator such that the system will have Kv=10 and P.M.=50.

2. The open loop transfer function of a unity feedback control system is given by,

G (s )= Ks (1+0.5 s)(1+0.2 s )

It is desired that,

a) For a unit ramp input the steady state error of the output position be less than 0.125 degrees/(degree/second).

b) The Phase margin P.M. .

c) The Gain margin G.M. .

NOTE:-


Using MATLAB draw the Bode plot & compare it with your result.


Documents

55040110 Control System Lab Manual by AMIT KUMAR