📖 Explanation
GHq(s)Necessary:Kp=(ssKp+KI)(s3+4s2+5s+22)=s4+4s3+5s2+s(2+2kp)+2kI>−1;KI>0s4s3s2s1s014418−2Kp(9−Kp)(1+Kp)−8KI2KI52+2Kp2KI2KI
Sufficient:
418−2Kp⇒Kp∴−1(18−2Kp)(2+2Kp)−32KI32KI∴0>0<9<Kp<9>0<36+32Kp−4Kp2<KI<3236+32Kp−4Kp2
If KpIf Kp=−1⇒kI=0=9⇒kI=0
,
∴dKpdKI⇒32−8Kp=0=0=0⇒Kp=4
∴ For Kp=4,KI is maximum, which is KI=3236+32×4−64=3.125 For Kp=4,KI<3.125 for stability ∴KImax=3.125