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Jigsaw Puzzle 3 rd Team Angela Purkey Jason Powell Mena Aziz Engr 328 4-10-2007.

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Presentation on theme: "Jigsaw Puzzle 3 rd Team Angela Purkey Jason Powell Mena Aziz Engr 328 4-10-2007."— Presentation transcript:

1 Jigsaw Puzzle 3 rd Team Angela Purkey Jason Powell Mena Aziz Engr 328 4-10-2007

2 Outline Controller Gain Ultimate Controller Gain Block Diagram Conclusion

3 Controller Gain Controller gain,Kc, is the change in output divided by the change in input (units) % You have control to set what the system Gain would be but you can’t manipulate the ultimate controller Gain, Kcu

4 Ultimate Controller Gain Kcu = 1/ AR Kcu obtains a phase shift of (-180 °) When the (amplitude ratio intersects with utltimate frequency) At that instance Kcu reaches a phase angle shift of (-180°) Amplitude ratio = 1/kcu for ◦ Ex: (0.08 %/(lb/min)) = (1/12 %/(lb/min))

5 Block Diagram M(t) = input= A sin ( ωt), A= Amplitude C(t)= (AR) * A sin(ωt+u), AR= Amplitude Ratio e(t) = (r(t)- c(t)) Error = setpoint – system output.

6 Phase Angle = (-180 °) C(t)= (AR) * A sin ( ωt-180°) sin(wt-180 ° )= ◦ =sin( ωt)*cos(-180) + cos(ωt)* sin(-180) ◦ =sin( ωt)*(-1) + cos(ωt)*(0) ◦ =- sin( ωt) C(t)= -(AR) * A sin ( ωt-180°) Evaluated using the dougle angle formula using trig Identity.

7 Block Diagram Algebra e(t) = (r(t)- c(t))=0-[-(AR)*Asin(ωt)] e(t)=(AR)* Asin(ωt) KcSystem e(t) r(t) m(t) c(t)

8 e(t)*Kc=m(t) m(t)=(AR)* Asin(ωt)*kc m(t)=Kc*(AR)*Asin( ωt) But earlier: m(t)=Asin(ωt) Block Diagram Algebra e(t) m(t) Kc

9 Asin(ωt)= Kc*(AR)*Asin( ωt) Asin( ωt)/ Asin( ωt)=( Kc*(AR)*Asin( ωt))/ Asin( ωt) 1=Kc*(AR) AR=1/Kc Block Diagram Algebra m(t) c(t) System

10 Conclusion When the phase angle =-180, AR=1/Kc An advantage of knowing this is that we know where AR=1/Kc occurs.


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