MECH 466 - Lecture21-LeadLagDesigninFrequencyDomain-2009W

1 cs0286 cs0286 pm50deg g05 70 csclags

Info iconThis preview shows page 1. Sign up to view the full content.

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

Unformatted text preview: =1 Kv=2) C(s)=CLag(s) (Kv=2) C(s)=C (s) (Kv=2) (Kv=0.572) (Kv=0.572) 0.5 Small overshoot is due to larger PM. Slower response is due to smaller wg. 0 0 2008/09 5 10 MECH466 : Automatic Control 65 64 67 15 67.5 68 68.5 69 69.5 70 Smaller steady-state error is due to larger Kv. steadyKv. 19 2008/09 MECH466 : Automatic Control 20 5 Phase-lead compensator (review) An example: Revisited Consider a system Consider C(s) C(s) 15 Increasing ωg 10 5 0 -2 10 Plant Specs Specs -1 0 10 10 1 2 10 10 3 10 Stable Stable PM at least 50 deg PM Settling time < 4s Settling 60 40 Note large settling time by previous designs Note (next two slides) 20 0 -2 10 G(s) G(s) Controller 20 -1 0 10 10 1 2 10 10 3 10 Max phase lead at Stabilizing effect 2008/09 MECH466 : Automatic Control 21 Lag C(s) design (review) 2008/09 MECH466 : Automatic Control 22 Step responses (previous designs) 50 2 C(s)=1 C(s)=1 -50 (PM=12deg, ωg=1.1) C(s)=0.286 C(s)=0.286 0 (PM=50deg, ωg=0.5) C(s)=CLag(s) (PM=52.3deg, ωg=0.4) C(s)=C (s) 1.5 -100 -2 10 -1 10 10 0 10 1 2 10 1 -100 PM=50 -150 0.5 -200 -250 -2 10 2008/09 -1 10 0 10 MECH466 : Automatic Control 10 1 0 0 2 10 23 2008/09 5 10 MECH466 : Automatic Control 15 24 6 Phase-lead...
View Full Document

This document was uploaded on 03/05/2014 for the course MECH 466 at UBC.

Ask a homework question - tutors are online