Using block reducion technique, obtain the transfer function.

33views

written 6.1 years ago by

teamques10 ★ 69k

• modified 6.1 years ago

enter image description here

II) Shifting of take off point after a block

enter image description here

III) Blocks in series

enter image description here

IV) Eliminate feedback loop. enter image description here

V) Shifitng take of point after a block

enter image description here

VI) Blocks in series

enter image description here

VII) Eliminate feedback loop.

enter image description here

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,

enter image description here

VIII) Blocks in series:

enter image description here

IX) Eliminate feedback loop and simplify,

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

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

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

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

ADD COMMENT EDIT