{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# Chapter6 - CHAPTER 6 6.1 THE PERFORMANCE OF FEEDBACK...

This preview shows pages 1–4. Sign up to view the full content.

CHAPTER 6 THE PERFORMANCE OF FEEDBACK SYSTEMS 6.1 (a) Refer Section 6.3 for the derivation (b) The characteristic equation is s 2 + 10 s + 100 = s 2 + 2 zw n s + w n 2 = 0 (i) w n = 10 rad/sec; z = 0.5; w d = w n 1 2 - z = 8.66 rad/sec (ii) t r = p z w - - cos 1 d = 0.242 sec; t p = p w d = 0.363 sec M p = e - - pz z / 1 2 = 16.32%; t s = 4 zw n = 0.8 sec (iii) K p = lim s fi 0 G ( s ) = ¥ ; K v = lim s fi 0 sG ( s ) = 10; K a = lim s fi 0 s 2 G ( s ) = 0 (iv) e ss = 1 1 + K p = 0; e ss = 1 K v = 0.1; e ss = 1 K a = ¥ 6.2 Characteristic equation is s 2 + 10 s + 10 K A = s 2 + 2 zw n s + w n 2 = 0 (a) K A = 2.5 gives z = 1 which meets the response requirements. (b) K p = ¥ , K v = 2.5 e ss = 5 1 1 + + K K p v = 0.4 6.3 (a) Using Routh criterion, it can be checked that the close-loop system is stable. (i) e ss = 10 K v = 10 ¥ = 0 ; (ii) e ss = 10 0 2 K K a v + . = 10 0 0 1 ¥ + .2 . = 2 (b) The closed-loop system is unstable. 6.4 The closed-loop system is stable. This can easily be verified by Routh criterion. G ( s ) = 5 0 1 1 1 0 1 ( . )( )( .2 ) s s s + + +

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
SOLUTION MANUAL 43 e ss | r = 10 = 10 1 + K p = 10 1 5 + Y s W s ( ) ( ) = 0 1 1 1 0 1 0 1 1 5 . ( )( .2 )( . ) s s s s + + + + + For W ( s ) = 4/ s , y ss = 4/6 e ss | total = 7/3 6.5 (a) The characteristic equation of the system is 900 s 3 + 420 s 2 + 43 s + 1 + 0.8 K A = 0 Using Routh criterion, we find that the system is stable for K A < 23.84. (b) Rearrangement of the block diagram of Fig. P6.5 is shown below. + q r q 50 1 0.016 30 1 s + 10 1 s + 3 1 s + K A G ( s ) = 0 016 50 3 1 30 1 . ( )( ) · + + K s s A H ( s ) = 1 10 1 s + ; the dc gain of H ( s ) is unity. K p = lim s fi 0 G ( s ) = 0.8 K A ; e ss = 10 1 + K p = 1 This gives K A = 11.25 6.6 (a) The system is stable for K c < 9. K p = lim s fi 0 D ( s ) G ( s ) = K c ; e ss = 1 1 + K c e ss = 0.1 (10%) is the minimum possible value for steady-state error. Therefore e ss less than 2% is not possible with propor- tional compensator.
44 CONTROL SYSTEMS: PRINCIPLES AND DESIGN (b) Replace K c by D ( s ) = 3 + K s I . The closed-loop system is stable for 0 < K 1 < 3. Any value in this range satisfies the static accuracy requirements. 6.7 (a) K v = 1000 10 K c = 1000 z = 10 1000 2 100 + · K D = 0.5 These equations give K c = 10 and K D = 0.09. (b) The closed-loop poles have real part = – zw n = – 50. The zero is present at s = – K c / K D = – 111.11. The zero will result in pronounced early peak. z is not an accurate estimate of M p .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern