Control System Laboratory Report(1)

7/30/2019 Control System Laboratory Report(1)

1/55

55

DEPARTMENT OF ELECTRICAL ENGINEERINGKALYANI GOVT. ENGINEERING COLLEGE, KALYANI

CONTROL SYSTEM LABORATORYPaper Code: EE-593 5 th Semester

INDEXSl.No.

Name of the Experiment Page No. Date of Submission

TeachersSignature

1. Familiarization with MATLABControl System Toolbox,

MATLAB Simulink Toolbox.

2 - 16 19 th November,2012

2. Simulation of Step Response &Impulse Response for Type 0,Type 1 & Type 2 Systems

with Unity Feedback usingMATLAB.

17 - 27 19 th November,2012

3. Determination of Root Locus,Bode Plot, Nyquist Plot using

MATLAB Control SystemToolbox for a Second Order System & Determination of Different Control Systems

Specifications from the Plot.

28 - 45 19 th November,2012

4. Determination of PI, PD & PIDControl Action of a First Order

Simulated Process usingMATLAB.

46 - 55 19 th November,2012


2/55

55



EXPERIMENT NO.: 01

TITLE: FAMILIARIZATION WITH MATLAB CONTROL SYSTEM TOOLBOX,MATLAB SIMULINK TOOLBOX.

OBJECTIVE: To get acquaintance with the MATLAB control system toolbox & SIMULINK toolbox.

ASSIGNMENTS:

1. Enter the system transfer function given below by separately specifying the numerator& denominator as vectors containing co- efficient of S in descending powers of S.

G1(S) = & G2(S) =

a) Display G1(S) & G2(S).

Soln: n1 = [10];d1 = [1 1 0];s1 = tf(n1,d1)

n2 = [1 1];d2 = [1 0 0];

s2 = tf(n2,d2)

Result:

s1 =

10-------s^2 + s

Continuous - time transfer function.

s2 =

s + 1-----s^2



3/55

55



b) Find the transfer function of the resultant system when G2(S) is connected in series withG1(S) & display the result.

Soln: n1 = [10];d1 = [1 1 0];s1 = tf(n1,d1)

n2 = [1 1];d2 = [1 0 0];s2 = tf(n2,d2)

sys = s1*s2

Result:

s1 =

10-------s^2 + s


s2 =

s + 1-----s^2


sys =

10 s + 10---------s^4 + s^3



4/55

55



c) Find the transfer function of the resultant system when G2(S) is connected in parallelwith G1(S) & display the result.

Soln: n1 = [10];d1 = [1 1 0];s1 = tf(n1,d1)

n2 = [1 1];d2 = [1 0 0];s2 = tf(n2,d2)

sys = s1+s2

Result:

s1 =

10-------s^2 + s


s2 =

s + 1-----s^2


sys =

s^3 + 12 s^2 + s----------------

s^4 + s^3 Continuous - time transfer function.


5/55

55



d) Find the transfer function of the resultant system when G1(S) is connected in forwardpath with G2(S) in the feedback path with

i.) positive feedback & ii.) negative feedback & display the result in each case.

Soln: i.)n1 = [10];d1 = [1 1 0];s1 = tf(n1,d1)

n2 = [1 1];d2 = [1 0 0];s2 = tf(n2,d2)

sys = feedback(s2,s1,1)

Result:

s1 =

10-------s^2 + s


s2 =

s + 1-----s^2


sys =

s^3 + 2 s^2 + s---------------------s^4 + s^3 - 10 s - 10



6/55

55



ii.)n1 = [10];

d1 = [1 1 0];s1 = tf(n1,d1)

n2 = [1 1];d2 = [1 0 0];s2 = tf(n2,d2)

sys = feedback(s2,s1,-1)

Result:

s1 =

10

-------s^2 + s


s2 =

s + 1-----s^2


sys =

s^3 + 2 s^2 + s---------------------s^4 + s^3 + 10 s + 10



7/55

55



2. Obtain the pole zero map of the following system:

G(S) =

Soln: num = [1 2];den = conv(conv([1 0],[1 1]),[1 3]);sys = tf(num,den)

pzmap(sys)

Result:


8/55

55



3. See the step response of an open loop transfer function using Simulink toolbox.

Soln:

Result:


9/55

55



4. Change the input block to Sine Wave Generator & set the amplitude to 1, frequency to10 rad/sec. & phase to 0 rad. See the output response using Simulink toolbox.

Soln:

Result:


10/55

55



5. Change the input block to Pulse Generator & observe the result.

Soln:

Result:


11/55

55



6. Reduce the following blocks using MATLAB commands:

a)

Soln: n1 = [1 0];d1 = [1 0 2];s1 = tf(n1,d1)

n2 = [1];d2 = [1 1];s2 = tf(n2,d2)

n3 = [5 0];d3 = [1 3];s3 = tf(n3,d3)

sys1 = feedback(s3,s2,-1)sys2 = feedback(sys1,s1,-1)

step(sys2)

Result:

sys2 =

5 s^4 + 5 s^3 + 10 s^2 + 10 s--------------------------------------s^4 + 14 s^3 + 10 s^2 + 18 s + 6


R(S) C(S)


12/55

55




13/55

55



b)

Soln: n1 = [1 0];d1 = [1 5];s1 = tf(n1,d1)

n2 = [3 0];d2 = [1 0 6];s2 = tf(n2,d2)

n3 = [1 1 0];d3 = [1 5 6];

s3 = tf(n3,d3)

n4 = [7];d4 = [1];s4 = tf(n4,d4)

n5 = [1];d5 = [1 0 1];s5 = tf(n5,d5)

sys1 = feedback(s2,1,-1)sys2 = series(sys1,s3)sys3 = feedback(sys2,s4,-1)sys4 = series(s1,sys3)sys5 = feedback(sys4,s5,1)

step(sys5)

Result:

sys5 =

3 s^6 + 3 s^5 + 3 s^4 + 3 s^3----------------------------------------------------------------------------------s^7 + 34 s^6 + 194 s^5 + 319 s^4 + 466 s^3 + 468 s^2 + 276 s + 180

R(S) C(S)


14/55


15/55

55



c)

Soln: n1 = [1 0];d1 = [1 0 2];s1 = tf(n1,d1)

n2 = [1];d2 = [1 1];s2 = tf(n2,d2)

n3 = [1 0];d3 = [3 5];s3 = tf(n3,d3)

n4 = [1];d4 = [1 3];s4 = tf(n4,d4)

n5 = [1];d5 = [1 2];s5 = tf(n5,d5)

sys1 = series(s2,s5)sys2 = feedback(s1,sys1,1)sys3 = series(s2,s3)sys4 = feedback(sys3,s4,-1)sys5 = series(sys2,sys4)sys6 = feedback(sys5,1,-1)

step(sys6)

Result:

sys6 =

s^5 + 6 s^4 + 11 s^3 + 6 s^2---------------------------------------------------------------------------------3 s^7 + 26 s^6 + 94 s^5 + 194 s^4 + 273 s^3 + 284 s^2 + 195 s + 60

R(S) C(S)


16/55

55




DISCUSSIONS:


17/55

55



EXPERIMENT NO.: 02

TITLE: SIMULATION OF STEP RESPONSE & IMPULSE RESPONSE FOR TYPE 0,TYPE 1 & TYPE 2 SYSTEMS WITH UNITY FEEDBACK USING MATLAB .

OBJECTIVE: To simulate the step response & impulse response for Type 0, Type 1 & Type 2systems with unity feedback using MATLAB.

ASSIGNMENTS:

1. Consider the open loop transfer function of the following unity feedback systems.Obtain the output response curves of each system with unity step input. Also find out

the steady state errors.

G(S) = Type 0 System

G(S) = .. Type 1 System

G(S) = Type 2 System


18/55

55



Soln:

Type- 0 System

num = [1];den = [1 1];s = tf(num,den)sys = feedback(s,1,-1)

step(sys)

Result:

Steady state error is 0.5.


19/55

55



Type 1 System

num = [1];den = conv([1 0],[1 1]);s = tf(num,den)sys = feedback(s,1,-1)

step(sys)

Result:

Steady state error is 0.


20/55


21/55

55



2. Obtain the output response curve of the above systems with unit impulse input.Soln:

Type 0 System

num = [1];den = [1 1];s = tf(num,den)sys = feedback(s,1,-1)

impulse(sys)

Result:


22/55

55



Type 1 System

num = [1];den = conv([1 0],[1 1]);s = tf(num,den)sys = feedback(s,1,-1)

impulse(sys)

Result:


23/55

55



Type 2 System

num = conv([2 1],[4 1]);den = conv([1 0 0],[1 2 8]);s = tf(num,den)sys = feedback(s,1,-1)

impulse(sys)

Result:


24/55

55



3. Consider the following system:

Obtain the i.) Rise Time (t r ), ii.) Peak Time (t p), iii.) Percentage of Peak Overshoot (M p),from the output response curve with unit step input. Also compare these with thecalculated values.

Soln: n = [1];d = conv([1 0],[1 1]);s = tf(n,d)sys = feedback(s,1,-1)

step(sys)

Result:

R(S) C(S)


25/55

55



From the output response curve,

i.) Rise Time (t r ) = 2.44 seconds

ii.) Peak Time (t p) = 3.67 seconds

iii.) Percentage of Peak Overshoot (M p) = 16.3%

From calculations,

i.) Rise Time (t r ) = 2.42 seconds

ii.) Peak Time (t p) = 3.63 seconds

iii.) Percentage of Peak Overshoot (M p) = 16.3%

Hence, the both the results tally.


26/55

55



4. Obtain the step response of the second order system for different values of . Transfer

function of the system is G(S) = , using for loop concept.

Soln: for E = 0.0 : 0.1 : 1.1num = [10];den = [1 20*E 10];sys = tf(num,den)

step(sys)hold on

end

Result:


27/55

55


CONTROL SYSTEM LABORATORYPaper Code: EE-593 5 th Semester DISCUSSIONS:


28/55


29/55


30/55

55



See also: linspace, colon.

Overloaded methods:

distributed/logspace

codistributor2dbc/logspace

codistributor1d/logspace

codistributed/logspace

Reference page in Help browser

doc logspace

>> help margin

margin Gain & phase margins & crossover frequencies.

[Gm,Pm,Wcg,Wcp] = margin(SYS) computes the gain margin Gm, the phase

margin Pm, & the associated frequencies Wcg & Wcp, for the SISO

open-loop model SYS (continuous or discrete). The gain margin Gm is

defined as 1/G where G is the gain at the -180 phase crossing. The

phase margin Pm is in degrees. The frequencies Wcg & Wcp are in

radians/TimeUnit (relative to the time units specified in SYS.TimeUnit,

the default being seconds).

The gain margin in dB is derived by

Gm_dB = 20*log10(Gm)

The loop gain at Wcg can increase or decrease by this many dBs before


31/55

55



losing stability, & Gm_dB


32/55

55



i.) G(S) =

Soln: n = [10];

d = [0.01 0.25 1 0];s = tf(n,d)

bode(s)

Result:

From the output response curve,

Gain Margin = 7.96 dBGain Cross Over Frequency = 10 rad/secondsPhase Margin = 22.5

Phase Cross Over Frequency = 6.08 rad/seconds

ii.) G(S) =


33/55


34/55

55


CONTROL SYSTEM LABORATORYPaper Code: EE-593 5 th Semester Soln: >> help nyquist

nyquist Nyquist frequency response of dynamic systems.

nyquist(SYS) draws the Nyquist plot of the dynamic system SYS. Thefrequency range & number of points are chosen automatically. See BODEfor details on the notion of frequency in discrete-time.

nyquist(SYS,{WMIN,WMAX}) draws the Nyquist plot for frequencies betweenWMIN & WMAX in radians/TimeUnit (relative to the time units specified inSYS.TimeUnit, the default being seconds).

nyquist(SYS,W) uses the vector W of frequencies (in radians/TimeUnit) toevaluate the frequency response. See LOGSPACE to generate logarithmicallyspaced frequency vectors.

nyquist(SYS1,SYS2,...,W) plots the Nyquist response of several systemsSYS1, SYS2,... on a single plot. The frequency vector W is optional.You can specify a color, line style, & marker for each model,For example:

nyquist(sys1,'r',sys2,'y--',sys3,'gx') .

[RE,IM] = nyquist(SYS,W) & [RE,IM,W] = nyquist(SYS) return the realparts RE & imaginary parts IM of the frequency response (along withthe frequency vector W if unspecified). No plot is drawn on the screen.If SYS has NY outputs & NU inputs, RE & IM are arrays of size[NY NU LENGTH(W)] & the response at the frequency W(k) is given byRE(:,:,k)+j*IM(:,:,k) . The frequencies W are in rad/TimeUnit.

Response uncertainty computation:[RE,IM,W,SDRE,SDIM] = nyquist (SYS) also returns the st&ard deviationsof RE & IM for the identified system SYS.

See also : nyquistplot, bode, nichols, sigma, freqresp, ltiview, DynamicSystem.

Overloaded methods :DynamicSystem/nyquistresppack.ltisource/nyquist

Reference page in Help browserdoc nyquist


35/55

55



4. Obtain the nyquist plot of the system given by G(S) = . Find the gain

margin from the graph & compare with that obtained by using margin.

Soln: n = [11 22];d = [1 3 0 10];s = tf(n,d)

nyquist(s)

Result:

Gain Margin = -0.823 dB


36/55

55



Using margin ,

n = [11 22];

d = [1 3 0 10];s = tf(n,d)

margin(s)

Result:

Gain Margin = -0.823 dB

Hence, results from both the curves tally.


37/55

55



5. Note the function rlocus & rlocfind by using help.

Soln: >> help rlocus

rlocus Evans root locus.

rlocus(SYS) computes & plots the root locus of the single-input,single-output LTI model SYS. The root locus plot is used to analyzethe negative feedback loop

& shows the trajectories of the closed-loop poles when the feedback gain K varies from 0 to Inf. rlocus automatically generates a set of

positive gain values that produce a smooth plot.

rlocus(SYS,K) uses a user-specified vector K of gain values.

rlocus(SYS1, SYS2,...) draws the root loci of several models SYS1,SYS2,...on a single plot. You can specify a color, line style, & marker for each model, for example:

rlocus(sys1,'r',sys2,'y:',sys3,'gx') .

[R,K] = rlocus(SYS) or R = rlocus(SYS,K) returns the matrix R of complex root locations for the gains K. R has LENGTH(K) columns& its j-th column lists the closed-loop roots for the gain K(j).

See RLOCUSPLOT for additional graphical options for root locus plots.

See also : rlocusplot, sisotool, pole, ISSISO, lti.

Overloaded methods :DynamicSystem/rlocusresppack.ltisource/rlocus

Reference page in Help browserdoc rlocus


38/55

55



>> help rlocfind

rlocfind Find root locus gains for a given set of roots.

[K,POLES] = rlocfind(SYS) is used for interactive gain

selection from the root locus plot of the SISO system SYS

generated by RLOCUS. rlocfind puts up a crosshair cursor

in the graphics window which is used to select a pole location

on an existing root locus. The root locus gain associated

with this point is returned in K & all the system poles for

this gain are returned in POLES.

[K,POLES] = rlocfind(SYS,P) takes a vector P of desired root

locations & computes a root locus gain for each of these

locations (i.e., a gain for which one of the closed-loop roots

is near the desired location). The j-th entry of the vector K

gives the computed gain for the location P(j), & the j-th

column of the matrix POLES lists the resulting closed-loop poles.

See also : rlocus.

Overloaded methods :

DynamicSystem/rlocfind


39/55

55



6. Obtain the root locus of the following OLTF of a unity feedback system. Find the valueof K at points where the locus crosses over the right half of the S plane & at the break away points. Find the dominant poles where the damping factor is 0.5. Also find out thevalue of K at that pole.

i.) G(S) =

Soln: num = [1];den = conv(conv([1 0],[1 1]),[1 2]);sys = tf(num,den)

rlocus(sys)

Result:

The value of K where the locus crosses over the right half of the S plane is 6.The value of K at break away point is 0.385.


40/55

55



For damping factor = 0.5,

The dominant poles are -0.33 + 0.571j & -0.33 0.571j.

The value of K is 1.02.


41/55

55



ii.) G(S) =

Soln: num = conv([1 1],[1 2]);den = conv([1 -1],[1 0.1]);sys = tf(num,den)

rlocus(sys)

Result:

The value of K where the locus crosses over the right half of the S plane is 2.999.The values of K at break away points are 0.0937 & 12.9.


42/55

55



For damping factor = 0.5,

The dominant poles are -0.471 + 0.816j & -0.471 0.816j.

The value of K is 0.894.


43/55


44/55

55



As can be seen from the response curve, the closed loop system is unstable.

ii.) G(S) =

Soln: n = [1 0.25];d = [1 1.5 0.5 0 0];s = tf(n,d)

nyquist(s)

Result:

As can be seen from the response curve, the closed loop system is unstable.


45/55


46/55

55



EXPERIMENT NO.: 04

TITLE: DETERMINATION OF PI, PD & PID CONTROL ACTION OF A FIRST ORDER SIMULATED PROCESS USING MATLAB.

OBJECTIVE: To study the performances of P, PI, PD & PID controller action of a first order simulated process using MATLAB commands & also using SIMULINK toolbox.

ASSIGNMENTS:

1. Obtain the step response & observe the steady state error for an open loop system with

transfer function G(S) = .

Soln:

n = [1];d = [5 1];sys = tf(n,d)

step(sys)

Result:


47/55

55



Steady State Error is 0.

2. Obtain the step response with K p = 100 & find t r , t p, %M p , ess, t s.

Soln:

n = [1];d = [5 1];s = tf(n,d)sys1 = series(100,s)sys2 = feedback(sys1,1,-1)

step(sys2)


48/55


49/55


50/55

55



Result:

Rise Time, t r = 0.0991 seconds

Peak Time, t p = 0.331 seconds

Percentage of Peak Overshoot, %M p = 3.11

Steady State Error, e ss = 0

Settling Time, t s = 0.81 seconds


51/55

55



4. Obtain the step response with K p = 100, K d = 5 & find t r , t p, %M p , ess, t s

Soln:

n1 = [1];d1 = [5 1];s1 = tf(n1,d1)

n2 = [1 0];d2 = [1];s2 = tf(n2,d2)

sys1 = series(5,s2)

sys2 = parallel(100,sys1)sys3 = series(sys2,s1)sys4 = feedback(sys3,1,-1)

step(sys4)


52/55


53/55

55






5. Obtain the step response with K p = 100, K i = 100, K d = 5 & find t r , t p, %M p, ess, t s.

Soln:

n1 = [1];d1 = [5 1];s1 = tf(n1,d1)

n2 = [1];d2 = [1 0];s2 = tf(n2,d2)

n3 = [1 0];d3 = [1];s3 = tf(n3,d3)

sys1 = series(100,s2)sys2 = series(5,s3)


54/55

55



sys3 = 100+sys1+sys2sys4 = series(sys3,s1)sys5 = feedback(sys4,1,-1)

step(sys5)

Result:

Rise Time, t r = 0.185 seconds


55/55



Peak Time, t p = 0.555 seconds




DISCUSSIONS:

Documents

Control System Laboratory Report(1)