# ch07 - SEVEN Forced Response Errors SOLUTIONS TO CASE...

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S E V E N Forced Response Errors SOLUTIONS TO CASE STUDIES CHALLENGES Antenna Control: Steady-State Error Design via Gain a. () ( ) 76.39 150 1.32 K Gs ss s = ++ . System is Type 1. Step input: ( ) 0 e ∞ = ; Ramp input: 1 1 2.59 76.39 150 1.32 v e K KK ∞= = = × ; Parabolic input: ( ) e ∞ =∞ . b. 12 . 5 9 0.2 v == . Therefore, 12.95 K = . Now test the closed-loop transfer function, 32 989.25 151.32 198 989.25 Ts s = + , 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

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7-2 Chapter 7: Forced Response Errors Video Laser Disc Recorder: Steady-State Error Design via Gain a. The input, 15t 2 , transforms into 30/s 3 . ( ) 30/ 0.005 a eK ∞= = . 3 123 0.2*600 *6 1 0 20000 a KK K K K K K == × . Therefore: () 3 3 30 5 10 61 0 a KKK = = × × . Therefore 6 10 = . b. Using 6 10 = , ( ) 5 24 2 10 600 21 0 s Gs ss ×+ = . Therefore, 5 34 25 8 2 10 600 2 10 2 10 1.2 10 s Ts sss = + × . Making a Routh table, s 3 1 5 0 × s 2 4 0 × 1. 8 0 × 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)
Solutions to Problems 7-3 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 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

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7-4 Chapter 7: Forced Response Errors SOLUTIONS TO PROBLEMS 1. () () ( ) () 00 lim lim 1 ss sR s es E s Gs →→ ∞= = + where ( ) ( ) 2 450 12 8 15 38 2 28 s s s ++ + = + . For step, 0 e . For 37 tu ( t ), 2 37 Rs s = . Thus, ( ) 2 6.075 10 e × . For parabolic input, e ∞=∞ . 2. a . From the figure 5 3 2 ss ss ss er c =−= = b . Since the system is linear, and because the original input was ( )( ) 2.5 rt t ut = , the new steady state error is 2 0.8 2.5 ss e == . 3. ( ) ( ) 3 0 2 lim lim 1 60/ lim 0.9375 20 3 4 8 1 21 5 s sR s E s sss s = + +++ + 4. Reduce the system to an equivalent unity feedback system. ( ) 2 5 11 55 53 11 5 3 16 15 1 11 s s s s + = + + + +
Solutions to Problems 7-5 Hence, () () 0 1 lim 3 p s eG s K ∞= = = ± .

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## This note was uploaded on 11/07/2011 for the course MEM 355 taught by Professor Ani during the Fall '08 term at UPenn.

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ch07 - SEVEN Forced Response Errors SOLUTIONS TO CASE...

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