Hw7-S05_Solution

# Hw7-S05_Solution - HW 7 SOLUTION ME 304 Stability Steady...

This preview shows pages 1–3. Sign up to view the full content.

HW 7 - SOLUTION ME 304 Stability & Steady State Error CONTROL SYSTEMS (01) May 4, 2005 PROBLEM 1: + - 2 s R(s) C(s) s2 s(s K) + + 2 32 2 2(s 2) C(s) 2(s 2) s( s K ) 2(s 2) R(s) s Ks 2s 4 1 ) + + + == + ++ + + + Characteristic equation, sK s40 + = A stability margin of 1 is required. This can be done by replacing ‘s’ with ‘z-1’ in the original characteristic equation and arranging the parameter K to ensure stability of the new characteristic equation. s4 + = 0 1) K(z 2(z 4 0 => (z +−+− + = s (K 3)s (5 2K)s (K 0 +− +− ++= Hurwitz Test for stability, K-3>0 => K>3 5-2K>0 => K<2.5 K+1>0 => K>-1 This system can not have a stability margin of 1 for any value of K. PROBLEM 2: 0.1 + - 2 1 K K s + C(s) 1 0 + 50 s1 0 + + Plant T (Load torque) Disturbance - R(s) Actuator Controller Tachometer Let T(s)=0, 12 Ks K 15 0 50(K s K ) C(s) s (s 20) (s 10) 0 R(s) s 30s (200 5K )s 5K 10 . 1 s (s 20) (s 10) + + + +++ + + Prof. Dr. Y. Samim Ünlüsoy, Res. Asst. Kemal Çal ı ş kan 1

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

View Full Document
HW 7 - SOLUTION ME 304 Stability & Steady State Error CONTROL SYSTEMS (01) May 4, 2005 Let R(s)=0 32 12 50 C(s) 50s(s 20) (s 10) Ks K 15 0 T(s) s 30s (200 5K )s 5K 10 . 1 s (s 20) (s 10) −+ + == + +++ + + ++ Then, 50(K s K ) 50s(s 20) C(s) R(s) T(s) s
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 5

Hw7-S05_Solution - HW 7 SOLUTION ME 304 Stability Steady...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online