8/31/2015 EE 481 Lab #1
Section 001
Chad Ryan Weiss & Mireille Mballa
THE PENNSYLVANIA STATE UNIVERSITY
1 | P a g e
Table of Contents
Table of figures...................................................................................................................................1
Introduction.......................................................................................................................................2
OBJECTIVES ....................................................................................................................................2
METHODS.......................................................................................................................................2
Mechanical Spring Mass Damper System......................................................................................3
First Differential Equation............................................................................................................5
Second Differential Equation........................................................................................................7
Results...............................................................................................................................................9
Conclusion .......................................................................................................................................10
References.......................................................................................................................................11
Table of figures
Figure 1 Mechanical system to be analyzed..........................................................................................2
Figure 2 Simulink graphfor the mechanical system...............................................................................3
Figure 3 MATLAB graph for the mechanical system...............................................................................4
Figure 4 Simulink graphfor the first differential equation .....................................................................5
Figure 5 MATLAB graph for the first differential equation .....................................................................6
Figure 6 Simulink for the second differential equation..........................................................................7
Figure 7 MATLAB graph for the second differential equation.................................................................8
2 | P a g e
Figure 1 Mechanicalsystemto be analyzed
Introduction
OBJECTIVES
In this lab we will be using MATLAB and Simulink to model a pair of differential equations as
well as one mechanical spring-mass damper system.
METHODS
For the followingdifferential equationsandspring-massdampersystem, performthe following:
A. Write the frequency domain(Laplace)equationsthatmodel the system.
B. Findthe transferfunctionthatrelatesthe displacementvariableX(s) tothe inputvariable F(s).
C. Calculate the time domainexpressionx(t)byhand,usinginverseLaplace transformtechniques.
D. Write a simple scriptinMATLAB that will generate asolutionforandplotx(t).
E. Create a simple blockdiagraminSimulinkthatcontainsasource,transferfunctionandsink;run
a Simulinksolutiontoplotx(t).
F. Verifythe plotsobtainedinMATLABand Simulinkforx(t) asfollows:
1. Doesx(t) meetthe initial conditions?
2. Doesx(t) converge tothe correct final value?
3. Doesx(t) settle afterapproximately4time constants,usingthe slowestexponential
termin the system?
4. Doesx(t) exhibitthe expectedtransientresponse (overdamped,underdamped,or
criticallydamped) foritssecondordercomponents?
MECHANICAL SYSTEM
𝑘 = 50 N/m
𝑚 = 2 kg
𝑏 = 12 N-s/m
g = 9.8 m/s2
𝑥(0) = 0 m
𝑥̇(0) = 0 m/s
DIFFERENTIAL EQUATIONS
a) 𝒙⃛( 𝒕) + 𝟏𝟔𝒙̈ ( 𝒕) + 𝟔𝟓𝒙̇ ( 𝒕) + 𝟓𝟎𝒙( 𝒕) = 𝟓𝒖( 𝒕); 𝒙( 𝟎) = 𝒙̇ ( 𝟎) = 𝒙̈ ( 𝟎) = 𝟎
b) 𝒙̈ ( 𝒕) + 𝟒𝒙̇ ( 𝒕) + 𝟒𝒙( 𝒕) = 𝒖̇ ( 𝒕) + 𝟑𝒖( 𝒕); 𝒙( 𝟎) = 𝒙̇ ( 𝟎) = 𝟎
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Figure 2 Simulink graph forthemechanicalsystem
Mechanical Spring Mass Damper System
𝒎𝒙̈ ( 𝒕) + 𝒃𝒙̇ ( 𝒕) + 𝒌𝒙( 𝒕) = 𝑭( 𝒕)
𝒙( 𝟎) = 𝒙̇ ( 𝟎) = 𝟎
2𝑥̈( 𝑡) + 12𝑥̇( 𝑡) + 50𝑥( 𝑡) = 𝐹( 𝑡)
2[s2
X(s)-sx(0)-x’(0)]+12[sX(s)-x(0)]+50X(s)=F(s)
[2s2
+12s+50]X(s)=F(s)
𝑋(𝑠)
𝐹(𝑠)
=
1
2𝑠2 + 12s + 50
assume F(s) = unitstep(1/s)
𝑿( 𝒔) = (
𝟏
𝒔
)(
𝟏
𝟐𝒔 𝟐+𝟏𝟐𝒔+𝟓𝟎
) (TransferFunction)
Partial Fraction Expansion
𝑋( 𝑠) =
𝐴
𝑠
+
𝐵𝑠 + 𝐶
2𝑠2 + 12𝑠 + 50
Thisrepresentsaunitstepsummedwitha generic oscillatory decay function.
A=0.02 B=-0.02 C=-0.12 a=3 ωd=4
𝑋( 𝑠) =
0.02
𝑠
−
0.02𝑠+0.12
2( 𝑠+3)2+16
𝑥( 𝑡) = [0.02 + 𝑒−3𝑡(−0.02 ∗ 𝑐𝑜𝑠(4𝑡) − 0.015 ∗ 𝑠𝑖𝑛(4𝑡))] ∗ 𝑢(𝑡)
SimulinkSolution
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MATLAB Script Solution
>> X(s) = (1/(2*s^2+12*s+50));
>> U(s) = 1/s;
>> x_of_t= ilaplace(X(s)*U(s));
>> x_of_t
1/50 - (exp(-3*t)*(cos(4*t) +(3*sin(4*t))/4))/50
Figure 3 MATLAB graph for the mechanical system
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First Differential Equation
𝒙⃛( 𝒕) + 𝟏𝟔𝒙̈ ( 𝒕) + 𝟔𝟓𝒙̇ ( 𝒕) + 𝟓𝟎𝒙( 𝒕) = 𝟓𝒖( 𝒕);
𝒙( 𝟎) = 𝒙̇ ( 𝟎) = 𝒙̈ ( 𝟎) = 𝟎
[s3
X(s)+16s2
X(s)+65sX(s)+50X(s)]=5/s
Transfer Function
𝑋(𝑠)/𝑈(𝑠) = (
1
𝑠3 +16𝑠2+65𝑠+50
)
𝑋( 𝑠) = (
5
𝑠
)(
1
𝑠3 + 16𝑠2 + 65𝑠 + 50
)
Partial Fraction Expansion
𝑋( 𝑠) =
0.1
𝑠
+
−0.13
𝑠 + 1
+
0.05
𝑠 + 5
+
−0.01
𝑠 + 10
𝑥( 𝑡) = 0.1 − 0.13𝑒−𝑡 + 0.05𝑒−5𝑡 − 0.01𝑒−10𝑡
SimulinkSolution
Figure 4 Simulink graph for the first differential equation
6 | P a g e
MATLAB Script Solution
>> symss;
>> X(s)=(1/(s^3+16*s^2+65*s+50));
>> U(s) = 5/s;
>> t = linspace(0,10,1000);
>> x_of_t= ilaplace(X(s)*U(s));
>> x_of_t
exp(-5*t)/20- (5*exp(-t))/36- exp(-10*t)/90+ 1/10
Figure 5 MATLAB graph for the first differential equation
7 | P a g e
Second Differential Equation
𝒙̈ ( 𝒕) + 𝟒𝒙̇ ( 𝒕) + 𝟒𝒙( 𝒕) = 𝒖̇ ( 𝒕) + 𝟑𝒖( 𝒕)
𝒙( 𝟎) = 𝒙̇ ( 𝟎) = 𝟎
s2
X(s) + 4sX(s) + 4X(s) = sU(s) + 3U(s) [F.Domain]
(s2
+4s+4)X(s) = (s+3)U(s)
TransferFunction
𝑋(𝑠)
𝑈(𝑠)
=
(𝑠 + 3)
(𝑠2 + 4𝑠 + 4)
Partial Fraction Expansion
𝑋( 𝑠) =
0.75
𝑠
−
0.75
𝑠 + 2
−
0.5
(𝑠 + 2)^2
𝑥( 𝑡) = 0.75 − 0.75𝑒−2𝑡 − 0.5𝑡𝑒−2𝑡
SimulinkSolution
Figure 6 Simulink for the second differential equation
8 | P a g e
MATLAB Script Solution
>> symss;
>> X(s)=(s+3)/(s+2)^2;
>> U(s)=1/s;
>> x_of_t=ilaplace(X(s)*U(s));
>> x_of_t
3/4 - (t*exp(-2*t))/2- (3*exp(-2*t))/4
Figure 7 MATLAB graph for the second differential equation
9 | P a g e
Figure 8 Simulink graph forthemechanicalsystem
≈4τ
Final value
Initial value
≈ 4τ = 2s
≈ 4τ = 4s
Initial value Initial value
Final value
Final value
Results
For our mechanical spring-massdampersystem,we obtainedthe positionof the massasa functionof
time foreveryunitintime,i.e. 𝑥( 𝑡) = [0.02 + 𝑒−3𝑡(−0.02 ∗ 𝑐𝑜𝑠(4𝑡) − 0.015 ∗ 𝑠𝑖𝑛(4𝑡))] ∗ 𝑢(𝑡).This
functioncanbe evaluatedatzeroandinfinitytodetermine ourinitialandfinal values.Bysettingt= 0,
x(0) = 0; by settingt= ∞, x(∞) =0.02 andif we take anotherlookat figure 2,we can see thatour initial
and final valuesare consistentwithoursimulation.Anotherwaytoverifyourdatais basedonthe
transientresponse orthe time constantof thisparticularsystem.The time constantforthisfunctionis
ascertainedviathe lowestdecayingexponential inthe system, (i.e.e-3t
).Time constant(τ) happenstobe
1/3 secondsforthisparticularsystem,meaningthatafterabout4τ or 1.333 seconds,the systemshould
be approximately98percentof the way to approachingthe steady-statevalue. Figure 2will verifythis.
Lastly,we knowthat the systemis underdamped due tothe complex rootsandthe simulationshowsus
justthat.
For our setof differential equations,the followingsolutionswere obtained
𝑥( 𝑡) = [0.1 − 0.13𝑒−𝑡 + 0.05𝑒−5𝑡 − 0.01𝑒−10𝑡] 𝑢 (𝑡) & 𝑥( 𝑡) = [0.75 − 0.75𝑒−2𝑡 − 0.5𝑡𝑒−2𝑡] 𝑢(𝑡).
For the two differential equation
solutionsetsshownbelow,youcan
see that the exactsame methodfor
verifyingthe datainourmechanical
spring-massdampersystemverifies
the solutionstoourdifferential
equations.Also,if we gobackand
take a look at the roots,we will see
that theyare appropriatelymatched
to theircorrespondingoutputcurves.
The leftD.E. isoverdampedandthe
rightD.E. is criticallydamped.
10 | P a g e
Conclusion
In conclusion,thislabservedas anecessary introductiontomodelingsimpleandcomplex systems
(mechanical,electrical orboth). The lessonbehindthislabwasthatthere are basicallytwotypesof
systemconfigurations:open-loopandclosed-loop(feedback) systemsnotcountingcomputercontrolled
systemswhichcanapplyto eithercase (openorclosed).We alsolearnedaboutthe mainanalysisand
designobjectiveswhenitcomestocontrol systems, “analysisbeingthe processbywhichasystem’s
performance isdetermined.”[1]
Withour simulations,we wereable todetermine how the systemwouldbehave givenacertaininput.
Doingso, we were able toevaluate the transientaswell asthe steady-state response of the system.
Although there wasnodesigncomponenttothislab,we have learnedthatstabilityisthe mainobjective
and that “designisthe processbywhicha system’sperformance iscreatedorchanged”, i.e. tomaintain
or achieve stability.[1]
The resultsof thislab were thatof verysimple systems(one mechanical spring-massdampersystemand
twodifferential equations). Usingpartof the designprocess,we were able tolookata simple system
and determine the mathematical model byintegratingthe physical attributesof the systemintoan
input-outputfunctionof time calleda linear, time-invariantdifferentialequation.Withthisequationin
hand,our analysiscommenced.
Finally, we modeledandanalyzedoursystems.Knowingthatcontrol systemsare designedprimarilyfor
the purpose of stabilization,we came tothe conclusionthatnodesigneffortsneedtake place because
all of oursystem’snatural responseseventuallydecayedtozero (i.e.the systemsettledonasteady-
state value).
AfterhavingusedMATLAB& Simulink tomodel these simplesystems,we now know the basic
fundamentalsof constructingblockdiagramsandanalyzingsimple systems.Thisanalysisanddesign
practice that we have learnedinthislab will eventually allowustomodel,analyze andoptimize real-
world,future applications.
11 | P a g e
References
[1] N.Nise,Control SystemsEngineering,6th
edition.New Jersey,Wiley,2011, pp. 33-116.
[2] S. vanTonningen,“EE481 – Laboratory1: UsingMATLAB andSimulinkforModelingand
Simulation,”Course Handout,2015.

E E 481 Lab 1

  • 1.
    8/31/2015 EE 481Lab #1 Section 001 Chad Ryan Weiss & Mireille Mballa THE PENNSYLVANIA STATE UNIVERSITY
  • 2.
    1 | Pa g e Table of Contents Table of figures...................................................................................................................................1 Introduction.......................................................................................................................................2 OBJECTIVES ....................................................................................................................................2 METHODS.......................................................................................................................................2 Mechanical Spring Mass Damper System......................................................................................3 First Differential Equation............................................................................................................5 Second Differential Equation........................................................................................................7 Results...............................................................................................................................................9 Conclusion .......................................................................................................................................10 References.......................................................................................................................................11 Table of figures Figure 1 Mechanical system to be analyzed..........................................................................................2 Figure 2 Simulink graphfor the mechanical system...............................................................................3 Figure 3 MATLAB graph for the mechanical system...............................................................................4 Figure 4 Simulink graphfor the first differential equation .....................................................................5 Figure 5 MATLAB graph for the first differential equation .....................................................................6 Figure 6 Simulink for the second differential equation..........................................................................7 Figure 7 MATLAB graph for the second differential equation.................................................................8
  • 3.
    2 | Pa g e Figure 1 Mechanicalsystemto be analyzed Introduction OBJECTIVES In this lab we will be using MATLAB and Simulink to model a pair of differential equations as well as one mechanical spring-mass damper system. METHODS For the followingdifferential equationsandspring-massdampersystem, performthe following: A. Write the frequency domain(Laplace)equationsthatmodel the system. B. Findthe transferfunctionthatrelatesthe displacementvariableX(s) tothe inputvariable F(s). C. Calculate the time domainexpressionx(t)byhand,usinginverseLaplace transformtechniques. D. Write a simple scriptinMATLAB that will generate asolutionforandplotx(t). E. Create a simple blockdiagraminSimulinkthatcontainsasource,transferfunctionandsink;run a Simulinksolutiontoplotx(t). F. Verifythe plotsobtainedinMATLABand Simulinkforx(t) asfollows: 1. Doesx(t) meetthe initial conditions? 2. Doesx(t) converge tothe correct final value? 3. Doesx(t) settle afterapproximately4time constants,usingthe slowestexponential termin the system? 4. Doesx(t) exhibitthe expectedtransientresponse (overdamped,underdamped,or criticallydamped) foritssecondordercomponents? MECHANICAL SYSTEM 𝑘 = 50 N/m 𝑚 = 2 kg 𝑏 = 12 N-s/m g = 9.8 m/s2 𝑥(0) = 0 m 𝑥̇(0) = 0 m/s DIFFERENTIAL EQUATIONS a) 𝒙⃛( 𝒕) + 𝟏𝟔𝒙̈ ( 𝒕) + 𝟔𝟓𝒙̇ ( 𝒕) + 𝟓𝟎𝒙( 𝒕) = 𝟓𝒖( 𝒕); 𝒙( 𝟎) = 𝒙̇ ( 𝟎) = 𝒙̈ ( 𝟎) = 𝟎 b) 𝒙̈ ( 𝒕) + 𝟒𝒙̇ ( 𝒕) + 𝟒𝒙( 𝒕) = 𝒖̇ ( 𝒕) + 𝟑𝒖( 𝒕); 𝒙( 𝟎) = 𝒙̇ ( 𝟎) = 𝟎
  • 4.
    3 | Pa g e Figure 2 Simulink graph forthemechanicalsystem Mechanical Spring Mass Damper System 𝒎𝒙̈ ( 𝒕) + 𝒃𝒙̇ ( 𝒕) + 𝒌𝒙( 𝒕) = 𝑭( 𝒕) 𝒙( 𝟎) = 𝒙̇ ( 𝟎) = 𝟎 2𝑥̈( 𝑡) + 12𝑥̇( 𝑡) + 50𝑥( 𝑡) = 𝐹( 𝑡) 2[s2 X(s)-sx(0)-x’(0)]+12[sX(s)-x(0)]+50X(s)=F(s) [2s2 +12s+50]X(s)=F(s) 𝑋(𝑠) 𝐹(𝑠) = 1 2𝑠2 + 12s + 50 assume F(s) = unitstep(1/s) 𝑿( 𝒔) = ( 𝟏 𝒔 )( 𝟏 𝟐𝒔 𝟐+𝟏𝟐𝒔+𝟓𝟎 ) (TransferFunction) Partial Fraction Expansion 𝑋( 𝑠) = 𝐴 𝑠 + 𝐵𝑠 + 𝐶 2𝑠2 + 12𝑠 + 50 Thisrepresentsaunitstepsummedwitha generic oscillatory decay function. A=0.02 B=-0.02 C=-0.12 a=3 ωd=4 𝑋( 𝑠) = 0.02 𝑠 − 0.02𝑠+0.12 2( 𝑠+3)2+16 𝑥( 𝑡) = [0.02 + 𝑒−3𝑡(−0.02 ∗ 𝑐𝑜𝑠(4𝑡) − 0.015 ∗ 𝑠𝑖𝑛(4𝑡))] ∗ 𝑢(𝑡) SimulinkSolution
  • 5.
    4 | Pa g e MATLAB Script Solution >> X(s) = (1/(2*s^2+12*s+50)); >> U(s) = 1/s; >> x_of_t= ilaplace(X(s)*U(s)); >> x_of_t 1/50 - (exp(-3*t)*(cos(4*t) +(3*sin(4*t))/4))/50 Figure 3 MATLAB graph for the mechanical system
  • 6.
    5 | Pa g e First Differential Equation 𝒙⃛( 𝒕) + 𝟏𝟔𝒙̈ ( 𝒕) + 𝟔𝟓𝒙̇ ( 𝒕) + 𝟓𝟎𝒙( 𝒕) = 𝟓𝒖( 𝒕); 𝒙( 𝟎) = 𝒙̇ ( 𝟎) = 𝒙̈ ( 𝟎) = 𝟎 [s3 X(s)+16s2 X(s)+65sX(s)+50X(s)]=5/s Transfer Function 𝑋(𝑠)/𝑈(𝑠) = ( 1 𝑠3 +16𝑠2+65𝑠+50 ) 𝑋( 𝑠) = ( 5 𝑠 )( 1 𝑠3 + 16𝑠2 + 65𝑠 + 50 ) Partial Fraction Expansion 𝑋( 𝑠) = 0.1 𝑠 + −0.13 𝑠 + 1 + 0.05 𝑠 + 5 + −0.01 𝑠 + 10 𝑥( 𝑡) = 0.1 − 0.13𝑒−𝑡 + 0.05𝑒−5𝑡 − 0.01𝑒−10𝑡 SimulinkSolution Figure 4 Simulink graph for the first differential equation
  • 7.
    6 | Pa g e MATLAB Script Solution >> symss; >> X(s)=(1/(s^3+16*s^2+65*s+50)); >> U(s) = 5/s; >> t = linspace(0,10,1000); >> x_of_t= ilaplace(X(s)*U(s)); >> x_of_t exp(-5*t)/20- (5*exp(-t))/36- exp(-10*t)/90+ 1/10 Figure 5 MATLAB graph for the first differential equation
  • 8.
    7 | Pa g e Second Differential Equation 𝒙̈ ( 𝒕) + 𝟒𝒙̇ ( 𝒕) + 𝟒𝒙( 𝒕) = 𝒖̇ ( 𝒕) + 𝟑𝒖( 𝒕) 𝒙( 𝟎) = 𝒙̇ ( 𝟎) = 𝟎 s2 X(s) + 4sX(s) + 4X(s) = sU(s) + 3U(s) [F.Domain] (s2 +4s+4)X(s) = (s+3)U(s) TransferFunction 𝑋(𝑠) 𝑈(𝑠) = (𝑠 + 3) (𝑠2 + 4𝑠 + 4) Partial Fraction Expansion 𝑋( 𝑠) = 0.75 𝑠 − 0.75 𝑠 + 2 − 0.5 (𝑠 + 2)^2 𝑥( 𝑡) = 0.75 − 0.75𝑒−2𝑡 − 0.5𝑡𝑒−2𝑡 SimulinkSolution Figure 6 Simulink for the second differential equation
  • 9.
    8 | Pa g e MATLAB Script Solution >> symss; >> X(s)=(s+3)/(s+2)^2; >> U(s)=1/s; >> x_of_t=ilaplace(X(s)*U(s)); >> x_of_t 3/4 - (t*exp(-2*t))/2- (3*exp(-2*t))/4 Figure 7 MATLAB graph for the second differential equation
  • 10.
    9 | Pa g e Figure 8 Simulink graph forthemechanicalsystem ≈4τ Final value Initial value ≈ 4τ = 2s ≈ 4τ = 4s Initial value Initial value Final value Final value Results For our mechanical spring-massdampersystem,we obtainedthe positionof the massasa functionof time foreveryunitintime,i.e. 𝑥( 𝑡) = [0.02 + 𝑒−3𝑡(−0.02 ∗ 𝑐𝑜𝑠(4𝑡) − 0.015 ∗ 𝑠𝑖𝑛(4𝑡))] ∗ 𝑢(𝑡).This functioncanbe evaluatedatzeroandinfinitytodetermine ourinitialandfinal values.Bysettingt= 0, x(0) = 0; by settingt= ∞, x(∞) =0.02 andif we take anotherlookat figure 2,we can see thatour initial and final valuesare consistentwithoursimulation.Anotherwaytoverifyourdatais basedonthe transientresponse orthe time constantof thisparticularsystem.The time constantforthisfunctionis ascertainedviathe lowestdecayingexponential inthe system, (i.e.e-3t ).Time constant(τ) happenstobe 1/3 secondsforthisparticularsystem,meaningthatafterabout4τ or 1.333 seconds,the systemshould be approximately98percentof the way to approachingthe steady-statevalue. Figure 2will verifythis. Lastly,we knowthat the systemis underdamped due tothe complex rootsandthe simulationshowsus justthat. For our setof differential equations,the followingsolutionswere obtained 𝑥( 𝑡) = [0.1 − 0.13𝑒−𝑡 + 0.05𝑒−5𝑡 − 0.01𝑒−10𝑡] 𝑢 (𝑡) & 𝑥( 𝑡) = [0.75 − 0.75𝑒−2𝑡 − 0.5𝑡𝑒−2𝑡] 𝑢(𝑡). For the two differential equation solutionsetsshownbelow,youcan see that the exactsame methodfor verifyingthe datainourmechanical spring-massdampersystemverifies the solutionstoourdifferential equations.Also,if we gobackand take a look at the roots,we will see that theyare appropriatelymatched to theircorrespondingoutputcurves. The leftD.E. isoverdampedandthe rightD.E. is criticallydamped.
  • 11.
    10 | Pa g e Conclusion In conclusion,thislabservedas anecessary introductiontomodelingsimpleandcomplex systems (mechanical,electrical orboth). The lessonbehindthislabwasthatthere are basicallytwotypesof systemconfigurations:open-loopandclosed-loop(feedback) systemsnotcountingcomputercontrolled systemswhichcanapplyto eithercase (openorclosed).We alsolearnedaboutthe mainanalysisand designobjectiveswhenitcomestocontrol systems, “analysisbeingthe processbywhichasystem’s performance isdetermined.”[1] Withour simulations,we wereable todetermine how the systemwouldbehave givenacertaininput. Doingso, we were able toevaluate the transientaswell asthe steady-state response of the system. Although there wasnodesigncomponenttothislab,we have learnedthatstabilityisthe mainobjective and that “designisthe processbywhicha system’sperformance iscreatedorchanged”, i.e. tomaintain or achieve stability.[1] The resultsof thislab were thatof verysimple systems(one mechanical spring-massdampersystemand twodifferential equations). Usingpartof the designprocess,we were able tolookata simple system and determine the mathematical model byintegratingthe physical attributesof the systemintoan input-outputfunctionof time calleda linear, time-invariantdifferentialequation.Withthisequationin hand,our analysiscommenced. Finally, we modeledandanalyzedoursystems.Knowingthatcontrol systemsare designedprimarilyfor the purpose of stabilization,we came tothe conclusionthatnodesigneffortsneedtake place because all of oursystem’snatural responseseventuallydecayedtozero (i.e.the systemsettledonasteady- state value). AfterhavingusedMATLAB& Simulink tomodel these simplesystems,we now know the basic fundamentalsof constructingblockdiagramsandanalyzingsimple systems.Thisanalysisanddesign practice that we have learnedinthislab will eventually allowustomodel,analyze andoptimize real- world,future applications.
  • 12.
    11 | Pa g e References [1] N.Nise,Control SystemsEngineering,6th edition.New Jersey,Wiley,2011, pp. 33-116. [2] S. vanTonningen,“EE481 – Laboratory1: UsingMATLAB andSimulinkforModelingand Simulation,”Course Handout,2015.