advertisement

Definitions - Unity Feedback R(s) + C(s) Num( s) Den(s) K - Stability in Frequency Domain Open-loop transfer function Num( s) = 1 = 0 dB Magnitude criterion: K Den (s) Num ( s ) angle Den ( s ) Phase criterion: • Gain Margin, GM - How much open-loop gain K can we add such that the 0 db crossover occurs at -180° phase? • Phase Margin, Φ M - How much phase angle could we subtract at 0 dB crossover (unity gain) to reach -180° phase? Both gain and phase margin computed from the open-loop transfer function’s frequency response = − 180 o Phase Margin and ζ Estimations • We can estimate some unity feedback, closed-loop time responses from the open loop frequency response: –damping ratio from phase margin –static error constants from low frequency slopes • From p. 587 of text, for an ideal 2nd order system, Φ M = tan − 1 2ς − 2ς 2 + 4 1 + 4ς Eqn. 10.73 A plot of phase margin vs. damping ratio is shown in Fig. 10.48 and on next slide Plot of Phase Margin vs. ζ Example #1 100 90 R(s) + True Phase Margin Phase Margin, degrees 80 Phase~100*Damping Ratio 70 - K C(s) 2 s( s + 12 s + 20 ) 60 50 Find phase margin, ΦM, for K = 30 40 30 "Good" Approximation Use Matlab™ What are the closed loop roots at K = 30? 20 10 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Damping Ratio, z 1 Example #2 • Figure 10.51 of your text shows how to determine the static error constants from the low frequency response of the open-loop system C(s) K 2 s( s + 4 s + 8 ) - Find phase margin, ΦM, for K = 10 Use Matlab™ What are the closed loop roots at K = 10? – Kp (position error constant) for Type 0 systems – Kv (velocity error constant) for Type 1 systems – Ka (acceleration error constant) for Type 2 systems Example #4 90 20 10 60 10 0 30 0 0 -20 -30 Type 0 system, so 20log10(Kp)=14dB -30 -40 -60 -90 -50 -120 -60 -70 -80 Magnitude -90 Phase K p = 10 (+14 20 ) ≈ 5 -100 1 10 100 1000 -150 -180 180 150 Type 1 system, so find Kv -10 Magnitude, dB -10 Phase Angle, degrees Magnitude, dB Example #3 20 -20 120 90 60 -30 30 -40 0 Magnitude = 0 dB at ω=2, so Kv=2 rad/sec -50 -60 -30 -60 -70 -90 -210 -80 Magnitude -120 -240 -90 Phase -150 -270 -100 10000 -180 1 Definition - Bandwidth – “zero” frequency would be approximated by a very small frequency – if you roughly approximated your closed loop system as first order (i.e., a low pass filter), this would be the “break” frequency 10 100 1000 10000 Frequency, rad/sec Frequency, rad/sec • Bandwidth (ω BW), is the frequency at which the closed-loop frequency response is 3 dB below its “zero” frequency value (p. 579 of text) Phase Angle, degrees R(s) + Static Error Constants Example #5 R(s) + - K C(s) 2 s( s + 12 s + 20 ) Find bandwidth, ω MW, for K = 30 Use Matlab™ 2