Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 21: Stability Margins and Closing the Loop
Overview In this Lecture, you will learn: Closing the Loop Effect on Bode Plot Effect on Stability Stability Effects Gain Margin Phase Margin Bandwidth Estimating Closed-Loop Performance using Open-Loop Data Damping Ratio Settling Time Rise Time M. Peet Lecture 21: Control Systems 2 / 31
Review Recall: Frequency Response Input: u(t) = M sin(ωt + φ) Output: Magnitude and Phase Shift 2.5 2 1.5 1 Linear Simulation Results y(t) = G(ıω) M sin(ωt + φ + G(ıω)) Amplitude 0.5 0 0.5 1 1.5 2 0 2 4 6 8 10 12 14 16 18 20 Time (sec) Frequency Response to sin ωt is given by G(ıω) M. Peet Lecture 21: Control Systems 3 / 31
Review Recall: Bode Plot Definition 1. The Bode Plot is a pair of log-log and semi-log plots: 1. Magnitude Plot: 20 log 10 G(ıω) vs. log 10 ω 2. Phase Plot: G(ıω) vs. log 10 ω Bite-Size Chucks: G(ıω) = i G i (ıω) M. Peet Lecture 21: Control Systems 4 / 31
Complex Poles and Zeros We left off with Complex Poles: 1 G(s) = ( ( ) ) 2 s ω n + 2ζ s ω n + 1 M. Peet Lecture 21: Control Systems 5 / 31
Closing The Loop Now we examine the effect of closing the loop on the Frequency Response. Use simple Unity Feedback (K = 1). Closed-Loop Transfer Function: G cl (ıω) = G(ıω) 1 + G(ıω) We are most concerned with magnitude: G cl (ıω) = G(ıω) 1 + G(ıω) - u(s) + k G(s) Figure : Unity Feedback y(s) M. Peet Lecture 21: Control Systems 6 / 31
Closing The Loop On the Bode Plot 20 log G cl (ıω) = 20 log G(ıω) 20 log 1 + G(ıω) Which is the combination of The original bode plot The new factor log 1 + G(ıω) 20 10 Bode Diagram We are most concerned with the effect of the new term 20 log 1 + G(ıω) Specifically, as 1 + G(ıω) 0 lim 20 log 1 + G(ıω) = 1+G(ıω) 0 An unstable mode! Magnitude (db) Phase (deg) 0 10 20 30 40 50 60 70 80 0 45 90 135 180 10 1 10 0 10 1 10 2 Frequency (rad/sec) Figure : Open Loop: Blue, CL: Green M. Peet Lecture 21: Control Systems 7 / 31
Closing The Loop Stability Margin Instability occurs when 1 + G(ıω) = 0 For this to happen, we need: G(ıω) = 1 G(ıω) = 180 Stability Margins measure how far we are from the point ( G = 1, G = 180 ). Definition 2. The Gain Crossover Frequency, ω gc is the frequency at which G(ıω c ) = 1. This is the danger point: If G(ıω c ) = 180, we are unstable M. Peet Lecture 21: Control Systems 8 / 31
Closing The Loop Phase Margin Definition 3. The Phase Margin, Φ M is the phase relative to 180 when G = 1. Φ M = 180 G(iω gc ) ω gc is also known as the phase-margin frequency, ω ΦM M. Peet Lecture 21: Control Systems 9 / 31
Closing The Loop Gain Margin Definition 4. The Phase Crossover Frequency, ω pc is the frequency (frequencies) at which G(ıω pc ) = 180. Definition 5. The Gain Margin, G M is the gain relative to 0dB when G = 180. G M = 20 log G(ıω pc ) G M is the gain (in db) which will destabilize the system in closed loop. ω pc is also known as the gain-margin frequency, ω GM M. Peet Lecture 21: Control Systems 10 / 31
Closing The Loop Stability Margins Gain and Phase Margin tell how stable the system would be in Closed Loop. These quantities can be read from the Open-Loop Data. M. Peet Lecture 21: Control Systems 11 / 31
Closing The Loop Stability Margins: Suspension System 20 10 0 10 Bode Diagram Gm = Inf db (at Inf rad/sec), Pm = 42.1 deg (at 0.891 rad/sec) Magnitude (db) 20 30 40 50 60 70 80 0 45 Phase (deg) 90 135 180 10 1 10 0 10 1 10 2 Frequency (rad/sec) M. Peet Lecture 21: Control Systems 12 / 31
Closing The Loop Stability Margins Φ M = 35 G M = 10dB M. Peet Lecture 21: Control Systems 13 / 31
Closing The Loop Stability Margins Note that sometimes the margins are undefined When there is no crossover at 0dB When there is no crossover at 180 M. Peet Lecture 21: Control Systems 14 / 31
Transient Response Closing the Loop Question: What happens when we Close the Loop? We want Performance Specs! We only have open-loop data. Φ M and G M can help us. Unity Feedback: Step Response 1.4 - u(s) + k G(s) y(s) 1.2 1 We want: Damping Ratio Settling Time 0 Peak Time Amplitude 0.8 0.6 0.4 0.2 0 1 2 3 4 5 6 7 8 9 10 Time (sec) M. Peet Lecture 21: Control Systems 15 / 31
Transient Response Quadratic Approximation Assume the closed loop system is the quadratic Then the open-loop system must be ω 2 n G cl = s 2 + 2ζω n s + ωn 2 G ol = Assume that our open-loop system is G ol Use Φ M to solve for ζ ω 2 n s 2 + 2ζω n s Set 20 log G cl = 3dB to solve for ω n M. Peet Lecture 21: Control Systems 16 / 31
Transient Response Damping Ratio The Quadratic Approximation gives the Closed-Loop Damping Ratio as Φ M = tan 1 2ζ 2ζ 2 + 4ζ 4 + 1 Φ M is from the Open-Loop Data! A Handy approximation is ζ = Φ M 100 Only valid out to ζ =.7. Given Φ M, we find closed-loop ζ. M. Peet Lecture 21: Control Systems 17 / 31
Transient Response Bandwidth and Natural Frequency We find closed-loop ζ from Phase margin. We can find closed-loop Natural Frequency ω n from the closed-loop Bandwidth. Definition 6. The Bandwidth, ω BW is the frequency at gain 20 log G(ıω BW ) = 20 log G(0) 3dB. Closely related to crossover frequency. The Bandwidth measures the range of frequencies in the output. For 2nd order, Bandwidth is related to natural frequency by ω BW = ω n (1 2ζ 2 ) + 4ζ 4 4ζ 2 + 2 M. Peet Lecture 21: Control Systems 18 / 31
Transient Response Finding Closed-Loop Bandwidth from Open-Loop Data Question: How to find closed-loop bandwidth? Finding the closed-loop bandwidth from open-loop data is tricky. Have to find the frequency when the Bode plot intersects this curve. Heuristic: Check the frequency at 6dB and see if phase is = 180. M. Peet Lecture 21: Control Systems 19 / 31
Finding Closed-Loop Bandwidth from Open-Loop Data Example Bode Diagram 20 10 0 10 Magnitude (db) 20 30 40 50 60 70 80 0 45 Phase (deg) 90 135 180 10 1 10 0 10 1 10 2 Frequency (rad/sec) At phase 135, 5dB, we get closed loop ω BW = 1. M. Peet Lecture 21: Control Systems 20 / 31
Transient Response Bandwidth and Settling Time We can use the expression T s = 4 ζω n to get ω BW = 4 (1 2ζ T s ζ 2 ) + 4ζ 4 4ζ 2 + 2 Given closed-loop ζ and ω BW, we can find T s. M. Peet Lecture 21: Control Systems 21 / 31
Transient Response Bandwidth and Peak Time We can use the expression T p = π ω n 1 ζ 2 to get π ω BW = (1 2ζ 2 ) + 4ζ 4 4ζ 2 + 2 T p 1 ζ 2 Given closed-loop ζ and ω BW, we can find T p. M. Peet Lecture 21: Control Systems 22 / 31
Transient Response Bandwidth and Rise Time Using an expression for T r, we get a relationship between ω BW T r and ζ. Given closed-loop ζ and ω BW, we can find T r. M. Peet Lecture 21: Control Systems 23 / 31
Transient Response Example Question: Using Frequency Response Data, find T r, T s, T p after unity feedback. First Step: Find the phase Margin. Frequency at 0dB is ω gc = 2 G(2) = 145 Φ M = 180 145 = 35 M. Peet Lecture 21: Control Systems 24 / 31
Transient Response Example Step 2: Closed-Loop Damping Ratio ζ = Φ M 100 =.35 Step 3: Closed-Loop Bandwidth Intersect at = ( G = 6dB, G = 170 ) Frequency at intersection is ω BW = 3.7 M. Peet Lecture 21: Control Systems 25 / 31
Transient Response Example Step 4: Settling Time and Peak Time ω BW = 3.7, ζ =.35 ω BW T s = 20 implies T s = 5.4s ω BW T p = 4.9 implies T p = 1.32 M. Peet Lecture 21: Control Systems 26 / 31
Transient Response Example Step 5: Rise Time ω BW = 3.7, ζ =.35 ω BW T r = 1.98 implies T r =.535 M. Peet Lecture 21: Control Systems 27 / 31
Transient Response Example Step 6: Experimental Validation. Use the plant G(s) = 50 s(s + 3)(s + 6) 0.7 0.6 We find T p = 1.6s predicted 1.32 T r =.7s predicted.535 T s = 4s Predicted 5.4 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 0.5 0.4 0.3 0.2 0.1 0 Figure : Step Response M. Peet Lecture 21: Control Systems 28 / 31
Steady-State Error Finally, we want steady-state error. Steady-State Step response is lim G(s) = lim G(ıω) s 0 ω 0 Steady-state response is the Low-Frequency Gain, G(0). Close The Loop to get steady-state error e ss = 1 1 + G(0) M. Peet Lecture 21: Control Systems 29 / 31
Steady-State Error Example A Lag Compensator lim 20 log G(ıω) = 20dB ω 0 So lim ω 0 G(ıω) = 10. Steady-State Error: e ss = 1 1 + G(0) = 1 11 =.091 Phase (deg) Magnitude (db) 40 30 20 10 0 10 20 0 45 Bode Diagram 90 10 0 10 1 10 2 10 3 10 4 10 5 Frequency (rad/sec) M. Peet Lecture 21: Control Systems 30 / 31
Summary What have we learned today? Closing the Loop Effect on Bode Plot Effect on Stability Stability Effects Gain Margin Phase Margin Bandwidth Estimating Closed-Loop Performance using Open-Loop Data Damping Ratio Settling Time Rise Time Next Lecture: Compensation in the Frequency Domain M. Peet Lecture 21: Control Systems 31 / 31