ch07 - S E V E N Steady-State Errors SOLUTIONS TO CASE...

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S E V E N Steady-State Errors SOLUTIONS TO CASE STUDIES CHALLENGES Antenna Control: Steady-State Error Design via Gain a. G(s) = 76.39K s(s+150)(s+1.32) . System is Type 1. Step input: e( ) = 0; Ramp input: e( ) = 1 K v = 1 76.39K 150 x 1.32 = 2.59 K ; Parabolic input: e( ) = . b. 1 K v = 2.59 K = 0.2. Therefore, K = 12.95. Now test the closed-loop transfer function, T(s) = 989.25 s 3 +151.32s 2 +198s+989.25 , for stability. Using Routh-Hurwitz, the system is stable. s 3 1 198 s 2 151.32 989.25 s 1 191.46253 0 s 0 989.25 0 Video Laser Disc Recorder: Steady-State Error Design via Gain a. The input, 15t 2 , transforms into 30/s 3 . e( ) = 30/K a = 0.005. K a = 0.2*600 20000 * K 1 K 2 K 3 = 6x10 -3 K 1 K 2 K 3 . Therefore: e( ) = 30/K a = 30 6x10 3 K 1 K 2 K 3 = 5x10 -3 . Therefore K 1 K 2 K 3 = 10 6 . b. Using K 1 K 2 K 3 = 10 6 , G(s) = 2x10 5 (s + 600) s 2 (s + 2x10 4 ) . Therefore, T(s) = 2x10 5 (s + 600) s 3 + 2x10 4 s 2 + 2x10 5 s + 1.2x10 8 . Making a Routh table,
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Answers to Review Questions 231 s 3 1 2x10 5 s 2 2x10 4 1.2x10 8 s 1 194000 0 s 0 120000000 0 we see that the system is stable. c. Program: numg=200000*[1 600]; deng=poly([0 0 -20000]); G=tf(numg,deng); 'T(s)' T=feedback(G,1) poles=pole(T) Computer response : ans = T(s) Transfer function: 200000 s + 1.2e008 ------------------------------------ s^3 + 20000 s^2 + 200000 s + 1.2e008 poles = 1.0e+004 * -1.9990 -0.0005 + 0.0077i -0.0005 - 0.0077I ANSWERS TO REVIEW QUESTIONS 1. Nonlinear, system configuration 2. Infinite 3. Step(position), ramp(velocity), parabola(acceleration) 4. Step(position)-1, ramp(velocity)-2, parabola(acceleration)-3 5. Decreases the steady-state error 6. Static error coefficient is much greater than unity. 7. They are exact reciprocals. 8. A test input of a step is used; the system has no integrations in the forward path; the error for a step input is 1/10001. 9. The number of pure integrations in the forward path 10. Type 0 since there are no poles at the origin
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232 Chapter 7: Steady-State Errors 11. Minimizes their effect 12 . If each transfer function has no pure integrations, then the disturbance is minimized by decreasing the plant gain and increasing the controller gain. If any function has an integration then there is no control over its effect through gain adjustment. 13. No 14. A unity feedback is created by subtracting one from H(s). G(s) with H(s)-1 as feedback form an equivalent forward path transfer function with unity feedback. 15. The fractional change in a function caused by a fractional change in a parameter 16. Final value theorem and input substitution methods SOLUTIONS TO PROBLEMS 1. e ( ) = lim s E(s ) s 0 = lim s 0 s R(s ) 1+G(s ) where G(s) = 450(s + 12)(s + 8)(s + 15) s(s + 38)(s 2 + 2s + 28) . For step, e ( ) = 0. For 37tu(t) , R(s) = 37 s 2 . Thus, e ( ) = 6.075x10 -2 . For parabolic input, e( ) = 2. ) = s 0 = s 0 s R(s ) 1+G(s ) =lim s 0 s(60/s 3 ) 1 + 20(s + 3)(s + 4)(s + 8) s 2 (s + 2)(s + 15) = 0.9375 3. Reduce the system to an equivalent unity feedback system by first moving 1/s to the left past the summing junction. This move creates a forward path consisting of a parallel pair, 1 s + 1 in cascade with a feedback loop consisting of G ( s ) = 2 s + 3 and H ( s ) = 7. Thus, G e ( s ) = ( s + 1 s 2/( s + 3) 1 + 14 /( s + 3) ⎟ = 2( s + 1) s ( s + 17) Hence, the system is Type 1 and the steady-state errors are as follows: Steady-state error for 15 u ( t ) = 0.
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This note was uploaded on 04/04/2008 for the course MECH COntrol Sy taught by Professor Khurshid during the Spring '08 term at Michigan State University.

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ch07 - S E V E N Steady-State Errors SOLUTIONS TO CASE...

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