Using block reducion technique, obtain the transfer function.

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written 5.9 years ago by

teamques10 ★ 68k

• modified 5.8 years ago

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II) Shifting of take off point after a block

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III) Blocks in series

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IV) Eliminate feedback loop. enter image description here

V) Shifitng take of point after a block

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VI) Blocks in series

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VII) Eliminate feedback loop.

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Now simplify,

$$\frac{C(s)}{K(s)} = \frac{\frac{G2.G3.G4}{1+ G3.G4.H2}}{1+ [(\frac{G2.G3.G4}{1+G3.G4.H2})*(\frac{H1}{G4})]}$$

$$\frac{C(s)}{K(s)} = \frac{G2.G3.G4}{1+G3.G4.H2+G2.G3.H1}$$

=> VII) Becomes,

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VIII) Blocks in series:

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IX) Eliminate feedback loop and simplify,

$$ \frac{C(s)}{K(s)} = \frac{\frac{G1.G2.G3.G4}{1+G3.G4.H2+G2.G3.H1}}{1+ [\frac{G1.G2.G3.G4}{1+G3.G4.H2+G2.G3.H1}][\frac{H3+G3.G4.H2.H3}{G3.G4}]}$$

$$ \frac{C(s)}{K(s)} = \frac{G1.G2.G3.G4}{1+G3.G4.H2+G2.G3.H1 + (G1.G2)(H3+G3.G4.H2.H3)}$$

$$ \frac{C(s)}{K(s)} = \frac{G1.G2.G3.G4}{1+G3.G4.H2+G2.G3.H1 + G1.G2.H3 + G1.G2.G3.G4.H2.H3)}$$

$$ \therefore \frac{C(s)}{K(s)} = \frac{G1.G2.G3.G4}{1+G3.G4.H2+G2.G3.H1+G2.G3.H3+G1.G2.G3.G4.H2H}$$

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