B EG144 Sec 11 Block Diagram Algebra PDF

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11. Block Diagram Algebra A complex control system is usually analysed with the aid of a block diagram. This provides a graphical rendering of the system with its individul component parts represented by blocks which are interconnected by signal paths and possibly summers and/or take-off points. 11.1Graphical Elements (i) Transfer Function Block

i/p X

F(s) TF Y(s) = F(s)X(s)

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o/p Y

(ii) Summer

X

Z

+

Y Z = X + Y (iii) Subtractor

X

+

Z – Y Z = X – Y

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(iv) Take-Off Point

X

X X

Each o/p signal path carries the same signal as that applied to the i/p. Note: This is not the same as an electrical circuit diagram where the currents at the o/p’s sum to give the current at the i/p. These are signal flow diagrams, they are not electrical circuit diagrams. 11.1Cascade Connection of Blocks

X

F(s)

Y

G(s)

Z

We have, Y = F.X and Z = G.Y Therefore, Z = G.(F.X) = F.G.X The overall TF is then Z/X = F.G Therefore the TF’s of cascaded blocks multiply

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11.2Parallel Connection of Blocks (a) With adder

X

F

+

Z

G Here we have, Z = F.X + G.X = (F + G).X Hence the overall TF is, Z/X = F + G (b) With subtractor

X

F

+

Z –

G Here we have, Z = F.X – G.X = (F – G).X Hence the overall TF is, Z/X = F – G

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11.3Negative Feedback Loops

X +

E

Y

G

–

H

G is called the feed-forward path and H is called the feed-back path

Here we have:

E  X  H .Y and Y  G. E

Eliminating E gives, Y  G.( X  H.Y ) Y  G. X  G.H .Y or  1  G. H  Y G. X  the overall TF is

Y G  X 1  G.H

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E  X  H.Y and Y  G. E

We had:

Eliminating Y gives, E  X  H  G. E  E  X  G.H .E

or

 1  G. H  E  X  the TF for the error E is

1 E  X 1  G.H

The formulae for Y/X and E/X are classical and are used regularly without proof. All control engineers know them by heart. Note: In the case of a positive feedback then H in the above formulae is replaced by – H

X +

E

G

Y

+

H

Thus

Y G E 1 and   X 1  G .H X 1  G.H 11- 6

11.4Analysing Complex Systems e.g. X

E

+ –

1 s 1

A

+ –

F

1 s

+

Y

+

2 1 s 2

The innermost parallel combination simplifies as, Y 1 2s  1  2  F s s Then the inner feedback loop has 2 s 1 1 and H  s s 2 Then the inner feedback loop can be reduced to,

G

 2s  1   2s  1   s   s  Y     2s  1    1  s  s 1.5 A  2s  1  1  s 1 1    s    2  s  2 

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Now the outer feedback loop has in the feedforward path

1 2s  1 A Y   in cascade with E s 1 A s  s 1.5  2 s  1 2 s  1   1 G     s  1 s  s  1.5  s  s  1  s  1.5  The feedback path is just a simple connection,

 H 1 and the overall closed loop TF for the whole system is,   2 s  1      s s s 1 1.5     Y G G      X 1  G. H 1  G    2 s 1  1     s s s 1 1.5       Y 2 s 1  Simplifying, X s  s  1  s  1.5   2s  1 

Y 2s 1  3 X s  2.5s 2  3.5s  1

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