Explanation Given: The block diagram for the given system is as shown below: Figure 1. Concept Used: The transfer functions for the block diagram as shown in figure 1 are: C ( s ) R ( s ) = s K P + K I I s 2 + s ( c + K P ) + K I C ( s ) D ( s ) = − s I s 2 + s ( c + K P ) + K I The transfer functions for the block diagram as shown in figure 2 are: C ( s ) R ( s ) = s K P + K I I s 2 + s ( c + K P K f ) + K I K f C ( s ) D ( s ) = − s I s 2 + s ( c + K P K f ) + K I K f The steady-state error of a system using final value theorem is: e s s = lim s → 0 s E ( s ) Calculation: From the block diagram as shown in figure 1, the transfer functions are as: C ( s ) R ( s ) = s K P + K I I s 2 + s ( c + K P ) + K I C ( s ) D ( s ) = − s I s 2 + s ( c + K P ) + K I Therefore, the response C ( s ) for the system is: C ( s ) = s K P + K I I s 2 + s ( c + K P ) + K I R ( s ) − s I s 2 + s ( c + K P ) + K I D ( s ) And from the block diagram shown in figure 1, we have E ( s ) = ( R ( s ) − C ( s ) ) ⇒ E ( s ) = ( 1 − s K P + K I I s 2 + s ( c + K P ) + K I ) R ( s ) + s I s 2 + s ( c + K P ) + K I D ( s ) ∵ C ( s ) = s K P + K I I s 2 + s ( c + K P ) + K I R ( s ) − s I s 2 + s ( c + K P ) + K I D ( s ) In order to modify this system, add a feedforward compensation of constant value K f such that the block diagram appears as shown in figure 2. Figure 2 From the block diagram as shown in Figure 2, the transfer functions are as: C ( s ) R ( s ) = s K P + K I I s 2 + s ( c + K P K f ) + K I K f C ( s ) D ( s ) = − s I s 2 + s ( c + K P K f ) + K I K f Therefore, the response C ( s ) for the system is: C ( s ) = s K P + K I I s 2 + s ( c + K P K f ) + K I K f R ( s ) − s I s 2 + s ( c + K P K f ) + K I K f D ( s ) And from the block diagram shown in figure 2, we have E ( s ) = ( R ( s ) − C ( s ) ) ⇒ E ( s ) = ( 1 − s K P + K I I s 2 + s ( c + K P K f ) + K I K f ) R ( s ) + s I s 2 + s ( c + K P K f ) + K I K f D ( s ) ∵ C ( s ) = s K P + K I I s 2 + s ( c + K P K f ) + K I K f R ( s ) − s I s 2 + s ( c + K P K f ) + K I K f D ( s ) Case 1. For the given system: Since, E ( s ) = ( 1 − s K P + K I I s 2 + s ( c + K P ) + K I ) R ( s ) + s I s 2 + s ( c + K P ) + K I D ( s ) ⇒ E ( s ) = ( s ( I s + c ) I s 2 + s ( c + K P ) + K I ) R ( s ) + s I s 2 + s ( c + K P ) + K I D ( s ) Therefore, for unit-step command response R ( s ) and zero disturbance response D ( s ) , we have E ( s ) = ( s ( I s + c ) I s 2 + s ( c + K P ) + K I ) R ( s ) + s I s 2 + s ( c + K P ) + K I D ( s ) ⇒ E ( s ) = ( s ( I s + c ) I s 2 + s ( c + K P ) + K I ⋅ 1 s + s I s 2 + s ( c + K P ) + K I ⋅ 0 ) ⇒ E ( s ) = ( I s + c ) I s 2 + s ( c + K P ) + K I Thus, the steady-state error for this design is: e s s = lim s → 0 s E ( s ) ⇒ e s s = lim s → 0 s ( I s + c ) I s 2 + s ( c + K P ) + K I ∵ E ( s ) = ( I s + c ) I s 2 + s ( c + K P ) + K I ⇒ e s s

3rd Edition

ISBN: 9780073398068

Author: III William J. Palm

Publisher: MCG

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Chapter 10, Problem 10.30P

To determine

The modified block diagram by adding a feedforward command compensation of constant value Kf. Also, check whether the compensation can remove the steady-state error in the system due to step and ramp commands.

Expert Solution & Answer

Chapter 10 Solutions

System Dynamics

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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