MECH 466 - Lecture21-LeadLagDesigninFrequencyDomain-2009W

5 db gm 0 50 100 10 2 10 1 10 0 10 1 10 2 100 lead

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: 00 10 -2 10 -1 10 0 10 1 10 2 -100 Lead These values are too small for good transient response! -150 -200 -250 10 2008/09 MECH466 : Automatic Control 7 2008/09 -2 10 -1 MECH466 : Automatic Control 10 0 10 1 10 2 8 2 Gain compensation Bode plot for C(s)=0.286 50 PM is specified to be 50 deg. PM In this example, to increase PM by gain In compensation, we need to lower the gain curve. Low freq. gain 0 decreases. -50 -100 10 -2 10 -1 10 0 10 1 10 2 Uncompensated (C(s)=1) Uncompensated (C(s)=1) -100 Gain compensated -150 -200 -250 10 2008/09 MECH466 : Automatic Control 9 Step responses -2 10 2008/09 -1 10 0 10 1 10 2 MECH466 : Automatic Control 10 Phase-lag compensator (review) 1.4 dB 1.2 0 1 -20 0.8 -5 -10 -15 0.6 K=0.455 (PM=35deg) 0.4 K=0.286 (PM=50deg) 0.2 -20 -2 10 deg K=0.158 (PM=65deg) 0 0 0 10 1 10 2 3 10 10 0 -20 -40 5 10 15 -45 -60 -2 10 2008/09 -1 10 MECH466 : Automatic Control 11 2008/09 -1 10 MECH466 : Automatic Control 0 10 1 10 2 3 10 10 12 3 Phase-lag C(s) design After Step 1 OK We try to design phase-lag C(s) which gives phaseC(s) • PM 50deg • Low frequency gain same as the original plant. 50 0 -50 1. To satisfy low frequency requirement, adjust DC gain of OL system by a constant gain K. -100 10 C(s) C(s) Analys...
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