Explanation Given: A block diagram of the given system. Where, R ( s ) = input reference signal. E (s) = error signal. F (s) = control signal. D (s) = disturbance signal. C (s) = output signal. M (s) = actuating signal. Calculation: For the given system, the output equation is given by C ( s ) = 6 15 s + 2 M ( s ) − D ( s ) ........ (1) The equation for actuating signal is given by M ( s ) = 4 K 3 s + 1 E ( s ) ........ (2) The error equation is given by E ( s ) = R ( s ) − C ( s ) ......... (3) Substituting for E ( s ) = R ( s ) − C ( s ) in equation (2) M ( s ) = 4 K 3 s + 1 R ( s ) − C ( s ) ........ (4) Substituting the above value of M ( s ) in equation (1), we get C ( s ) = 6 15 s + 2 4 K 3 s + 1 R ( s ) − C ( s ) − D ( s ) C ( s ) = 6 15 s + 2 4 K 3 s + 1 R ( s ) − 4 K 3 s + 1 C ( s ) − D ( s ) C ( s ) = 6 15 s + 2 4 K 3 s + 1 R ( s ) − 6 15 s + 2 4 K 3 s + 1 C ( s ) − 6 15 s + 2 D ( s ) C ( s ) + 6 15 s + 2 4 K 3 s + 1 C ( s ) = 6 15 s + 2 4 K 3 s + 1 R ( s ) − 6 15 s + 2 D ( s ) 1 + 6 15 s + 2 4 K 3 s + 1 C ( s ) = 24 K 15 s + 2 3 s + 1 R ( s ) − 6 15 s + 2 D ( s ) 1 + 24 K 15 s + 2 3 s + 1 C ( s ) = 24 K 15 s + 2 3 s + 1 R ( s ) − 6 15 s + 2 D ( s ) ....... (5) If disturbance D ( s ) = 0, we get 1 + 24 K 15 s + 2 3 s + 1 C ( s ) = 24 K 15 s + 2 3 s + 1 R ( s ) 15 s + 2 3 s + 1 + 24 K 15 s + 2 3 s + 1 C ( s ) = 24 K 15 s + 2 3 s + 1 R ( s ) C ( s ) R ( s ) = 24 K 15 s + 2 3 s + 1 + 24 K Hence, the expression for Transfer function is given by C ( s ) R ( s ) = 24 K 15 s + 2 3 s + 1 + 24 K ........ (6) When input is 0, that is R ( s ) = 0 in equation (5) we get, 1 + 24 K 15 s + 2 3 s + 1 C ( s ) = 24 K 15 s + 2 3 s + 1 0 − 6 15 s + 2 D ( s ) 1 + 24 K 15 s + 2 3 s + 1 C ( s ) = − 6 15 s + 2 D ( s ) C ( s ) D ( s ) = − 6 15 s + 2 15 s + 2 3 s + 1 15 s + 2 3 s + 1 + 24 K C ( s ) D ( s ) = − − 6 3 s + 1 15 s + 2 3 s + 1 + 24 K C ( s ) D ( s ) = − − 6 3 s + 1 45 s 2 + 21 s + 2 + 24 K Thus, the value of C ( s ) D ( s ) = − − 6 3 s + 1 45 s 2 + 21 s + 2 + 24 K ........ (7) We have the expression of C ( s ) in terms of E ( s ) C ( s ) = 6 15 s + 2 4 K 3 s + 1 E ( s ) − D ( s ) ........ (8) Substituting C ( s ) = 6 15 s + 2 4 K 3 s + 1 E ( s ) − D ( s ) in E ( s ) = R ( s ) − C ( s ) we get, E ( s ) = R ( s ) − 6 15 s + 2 4 K 3 s + 1 E ( s ) − D ( s ) E ( s ) + 6 15 s + 2 4 K 3 s + 1 E ( s ) = R ( s ) + 6 15 s + 2 D ( s ) 1 + 24 K 15 s + 2 3 s + 1 E ( s ) = R ( s ) + 6 15 s + 2 D ( s ) 15 s + 2 3 s + 1 + 24 K 15 s + 2 3 s + 1 E ( s ) = R ( s ) + 6 15 s + 2 D ( s ) If disturbance D ( s ) = 0, we get 15 s + 2 3 s + 1 + 24 K 15 s + 2 3 s + 1 E ( s ) = R ( s ) E ( s ) R ( s ) = 15 s + 2 3 s + 1 15 s + 2 3 s + 1 + 24 K E ( s ) R ( s ) = 45 s 2 + 21 s + 2 45 s 2 + 21 s + 2 + 24 K

3rd Edition

ISBN: 8220100254963

Author: Palm

Publisher: MCG

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Question

Chapter 10, Problem 10.4P

To determine

The output C (s), error E (s) and actuator M (s) equations and also obtain the characteristic polynomial.

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Ch. 10 - Prob. 10.1P Ch. 10 - 10.2 Draw the block diagram of a system using...Ch. 10 - Prob. 10.3P Ch. 10 - Prob. 10.4P Ch. 10 - Prob. 10.5P Ch. 10 - Prob. 10.8P Ch. 10 - Prob. 10.9P Ch. 10 - Prob. 10.12P Ch. 10 - Prob. 10.13P Ch. 10 - Prob. 10.14P

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