Kontrol Ödev

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Prof. Dr. Metin Uymaz Salamc

ADEM AKYZ

101155004

MM403 CONTROL SYSTEMS

HOMEWORK 1

Figure 1

Figure 2

Answer 1:

A Main differences is closed loop and feepback lop. Figure 1 is open-loop control systems

and Figure 2 is closed(feedback) control system.

The number of components used in a closed-loop control system is more than that for a

corresponding open-loop control system. The closed-loop control system is generally

higher in cost and power. A proper combination of open loop and closed loop controls are

usually less expensive and will give satisfactory overall system performance closed loop

control system the reference input is modified by the actual output before entering the

controller.


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Solution 2:

Figure 3

*Block Diagram Reduction Tecnique

Step 1: Move summing point ahead of G1

-

U(s) + + + Y(s)

- +

Step 2: Move branch point G4 right and Combine G3 & G4

-

U(s) + + + Y(s)

- +

H2/G1

G1 G2 G3 G4

H1

H3

G1 G2 G3G4

H1

H3

H2/G1.G4


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Step 3: Combine G1 & G2 and Reduce feedback form on left, internal loop

-

U(s) + + Y(s)

-

Step 4: Combine G1G2 & G3G4/ 1+ G3G4H1

-

U(s) + + Y(s)

-

Step 5: Reduce feedback form on left

U(s) Y(s)

-

G1G2G3G4

1- G3G4H1

H3

H2/G1.G4

G1G

2G

3G

4

1- G3G4H1

H3

H2/G1.G4

G1G2G3G4

1- G3G4H1 + G2G3H2

H3


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Step 6: Reduce feedback form on left

U(s) Y(s)

* Verify by Algebric Manuplations

Y= A.G4 B=C+D C=E.G2 D=Y.H1

A=B.G3 A=E.G2 + Y.H1

Y=G3G4 (E.G2 + Y.H.1) M-F=E F=A.H2 M= G1.N

Y= G3G4 ((G1.N + A.H2).G2 + Y.H1)

Y= G3G4(G1 G2.N+ A.H2.G2 + Y.H1)

Y= G1 G2.G3G4N + G2.G3G4 A.H2 + Y. G3G4H1

U-I=N I=Y.H3 N=U + Y.H3 A=Y/ G4

Y= G1 G2.G3G4U- YG1 G2.G3G4 H3 - YG2.G3 H2 + YG3G4H1

Y(1+ G1 G2.G3G4 H3 + G2.G3 H2 - G3G4H1 )= U.G1 G2.G3G4

G1G2G3G4

1- G3G4H1 + G2G3H2+ G1G2G3G4H3


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Figure 4

Step 1: Move summing point in front of G2

+

R(s) + + + C(s)

- -

Step 2: Reduce feedbacback form on right and sum G1 & G1

R(s) + C(s)

-

G2

H2G2

G1

G3

H1

G1 + G2G3

1+ G3H1

H2G2


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Step 3: Reduce feedbacback form on right interloop

R(s) + C(s)

R(s) C(s)

* Verify by Algebric Manuplations

C=A.G3 B=C.H1 D=RG1 E=F.G2 I=H2.C

A=E+D-B F=R-I E=(R-H2.C).G2

A=(R-H2.C).G2 + RG1 - C.H1

C=((R-H2.C).G2 + RG1 - C.H1)G3

C=RG2G3 - H2G2G3.C + RG1G3 - C.H1G3

C + H2G2G3.C + C.H1G3 = RG2G3 + RG1G3

G1 + G2

G3

1+ G3H1 + G2G3H2

G3G1 +G2G3

1+ G3H1 + G2G3H2


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Solution 3:

i) V.G2=C V=U+D

C=UG2 + D.G2

ii) C=G2V=G2(U+D)=G2[G1(R-B)+D)

=G1G2(R-HC) + G2D

=G1G2R G1G2HC + G2D

[1 + G1G2H ] = G1G2R + G2D

iii) V=U+D U=G1E

E=R-B HC=B >> H.V.G2=B

E=R-HVG2

V=G1R G1G2HV + D

[1+ G1G2H] V= G1R+ D

iv) C=G2V=G2(U+D)=G2[G1(R-B)+D)

=G1G2(R-HC) + G2D

=G1G2R G1G2HC + G2D

[1 + G1G2H ] = G1G2R + G2D

v)


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Solution 4:

Taking Laplace Transforms of these two equations;


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Solution 5:

a) The equation of motion for the system shown in (a)

Taking the Laplace transform of this last equation,

Hence transfer function is given by

b) The equation of motion for the system shown in (b)

Taking the Laplace transform of this last equation;

Hence transfer function is given by;


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c) The equation of motion for the system shown in (c)

Taking the Laplace transform of this last equation;

Hence transfer function is given by;

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