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Unformatted text preview: Page: 1 of 5 SCHOOL OF ENGINEERING EXAMINATION MODEL ANSWER SHEET Paper Title: Paper No : EG-243 MAY/JUNE 2009 Examiner(s) : JSM Question No Marks JSM/EG-243_09 1. (a) ( 29 K c s b s a s K s R C + + + + = ) )( )( ( [2 marks] (b) ( 29 K abc K R C + = [2 marks] (c) Type 0 system so: abc K K p = . Thus sse for: unit step = 1/(1+K p ) = abc/(abc+K) and ramp [5 marks] (d) Nyquist diagram [5 marks] (e)Bode diagrams [6 marks] (f) Root locus [5 marks] arg(0) K abc-1 Re(GH(jw) Im(GH(jw)) a b c ( ) arg(0) K GH j abc ( ) arg( 270) GH j - jw -c -b -a Product of these 3 vectors give value of K at stability limit Page: 2 of 5 SCHOOL OF ENGINEERING EXAMINATION MODEL ANSWER SHEET Paper Title: Paper No : EG-243 MAY/JUNE 2009 Examiner(s) : JSM Question No Marks JSM/EG-243_09 2. (a) ( 29 ) 1 2 ( ) arg( , 1 , 1 + = = = + n GH GH s GH , ie imaginary terms are zero magnitude is unity, phase is odd multiple of ....
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This note was uploaded on 05/17/2010 for the course ENGINEERIN EG 243 taught by Professor Murphy during the Spring '10 term at Swansea UK.
- Spring '10