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    10 elec3114 10 elec3114 Presentation Transcript

    • 1 Frequency Response Techniques• The definition of frequency response• How to plot frequency response• How to use frequency response to analyze stability• How to use frequency response to analyze a systems transient and steady state error performance• How to use frequency response to design the gain to meet stability specifications Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
    • 2 IntroductionFrequency response technique has distinct advantages:1. when modeling transfer functions from physical data2. when designing compensators to meet a steady-state error requirement and a transient response requirement3. when finding the stability of nonlinear systems4. in settling ambiguities when sketching a root locus Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
    • 3The Concept of Frequency Response t The steady-state output sinusoid is: … magnitude frequency response … phase frequency response Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
    • 4Analytical expression for the frequency response can be obtained from thetransfer function G(s) of the system as G ( jω ) = G ( s ) | s → jω Plotting of the frequency response G ( jω ) = M G ∠φG (ω ) • As a function of frequency, with separate magnitude and phase plots Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
    • 5Problem Find the analytical expression for the magnitude frequency responseand the phase frequency response for a system G(s) = 1 / ( s + 2). Also, plot theseparate magnitude and phase diagrams.Solution 1 Substitute s = jω into G ( s) = s+2 1 2 − jω G ( jω ) = = 2 jω + 2 ω + 4 1 G ( jω ) = M (ω ) = ω2 + 4 ⎛ω ⎞ φ (ω ) = − tan⎜ ⎟ ⎝2⎠ Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
    • 6 120 log M (ω ) = 20 log ω2 + 4 Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
    • 7 Asymptotic Approximations: Bode Plots• Bode plots or Bode diagrams: curves of the log-magnitude and phase frequency response as functions of log ω Consider: Then, the magnitude response is given by: Converting the magnitude response into dB, we obtain: Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
    • 8Phase frequency response- is the sum of the phase frequency response curves of the zero terms minusthe sum of the phase frequency response curves of the pole terms. Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
    • 9 Bode Plots for G(s) = (s + a)At low frequencies when ω approaches zero:The magnitude response in dB is: - low-frequency asymptote Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
    • 10At high frequencies, where ω>>a : - high-frequency asymptote Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
    • 11Phase response From the above equation: • at the break frequency, a, phase is 450 • at low frequencies the phase is 00 • at high frequencies the phase is 900 Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
    • 12• It is often convenient to normalize the magnitude and scale the frequency so that the log-magnitude plot will be 0 dB at a break frequency of unity.• Normalization and scaling makes it easier to add components to obtain the Bode plot of a function such as• To normalize (s + a), factor out the quantity a and form a[(s/a)+ 1].• The frequency is scaled by defining a new frequency variable, s1 = s/a.• The magnitude is divided by the quantity a to yield 0 dB at the break frequency.• Hence, the normalized and scaled function is (sl + 1). Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
    • 13 … original magnitude response… normalized and scaled magnitude response Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
    • 14 … original phase response… normalized and scaled phase response Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
    • 15Similarly, we can find the normalized and scaled Bode plots for: G(s) = s G(s) = 1/s Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
    • 16G(s) = (s + a) G(s) = 1/(s + a) Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
    • 17Problem Draw the Bode plots for the system G(s) = K(s + 3)/[s(s + l)(s + 2)].Solution First we normalize G(s), • the break frequencies are at 1, 2, and 3 To find value of G(s) at ω=0.1 we substitute s=0 for all (s/a+1) terms and 3 K the actual value of s=j0.1 for 2 term. s 3 Hence G ( j 0.1) ≈ K / 0.1 = 15 K 2 For K=1: 20 log(G ( j 0.1)) = 20 log 15 = 23.52dB Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
    • 18 Dr Branislav HredzakControl Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
    • 191 Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
    • 20 Bode Plots for G(s) = s2 + 2ζωns + ω2nAt low frequencies:At high frequencies: Break frequency: ωnAfter scaling and normalization Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
    • 21At low frequencies: Phase = 00At high frequencies: Phase = 1800Phase at ωn : 2 G ( jω n ) = j 2ζω n - hence phase at ωn is 900 Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
    • 22Corrections to Second-Order Bode Plots G(s) = s2 + 2ζωns + ω2n• there is an error between the actual response and the asymptotic approximation of the second-order polynomial Actual response: a magnitude correction of +20log2ζ can be made at the natural, or break, frequency Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
    • 23Bode Plots for G(s) = 1/ (s2 + 2ζωns + ω2n) a magnitude correction of -20log2ζ can be made at the natural, or break, frequency Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
    • 24Problem Draw the Bode log-magnitude and phase plots of ( s + 3) G(s) = ( s + 2)( s 2 + 2 s + 25)Solution Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
    • 25 20 log(3 / 50) = −24.4dB( s 2 + 2 s + 25) ↔ s 2 + 2ζω n + ωn 2ωn = 5 → ζ = 0.2 -the correction can be found by plotting a point -20log(2x0.2)=7.96 dB above the asymptotes at ωn=5 Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
    • 26 Dr Branislav HredzakControl Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
    • 27 lntroduction to the Nyquist Criterion• relates the stability of a closed-loop system to open-loop poles location • the Nyquist criterion can tell us how many closed-loop poles are in the right-half of the complex plane • it is based on mapping of a contour through the function G(s)H(s) Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
    • 28Mapping Contour A Q -1• Contour A can be mapped through F(s)=G(s)H(s) into contour B (Nyquist diagram) by substituting each point of contour A into the function F(s) and plotting the resulting complex numbers. Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
    • Nyquist Criterion Contour A -1 • Stability can be determined using equation Z = P – N where P… the number of open-loop poles of G(s)H(s) enclosed by the contour A N…the number of encirclements the Nyquist diagram (contour B) makes about (– 1) Z…the number of right-half-plane poles of the closed-loop systemNote: Counterclockwise encirclements are positive, and clockwise encirclements arenegative Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
    • 30 Applying the Nyquist Criterion to Determine StabilityNote: Counterclockwise encirclements are positive, and clockwise encirclements arenegative P=0 N=0 Z=P-N=0 … system is stable P=0 N= - 2 Z=P-N=2 … system is unstable Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
    • 31 Sketching the Nyquist Diagram• The contour that encloses the right half-plane can be mapped through the function G(s)H(s) by substituting points along the contour into G(s)H(s)• Approximations can be made to G(s)H(s) for points around the infinite semicircle by assuming that the vectors originate at the origin. Thus, their length is infinite, and their angles are easily evaluated. Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
    • 32Problem Sketch the Nyquist diagram for the system G(s) = 500/[(s + 1)(s + 3)(s + 10)].Solution ω = 43 ω = 30 / 14 SystemsBranislav Hredzak S. Nise Control Dr Engineering, Fourth Edition by Norman Copyright © 2004 by John Wiley & Sons. All rights reserved.
    • 33• If there are open-loop poles situated along the contour enclosing the right half-plane, then a detour around the poles on the contour is required; otherwise, a complete sketch of the Nyquist diagram cannot be made. • each poles vector rotates through + 1800 as we move around the contour near that pole • detour must be only an infinitesimal distance into the right half- plane, or else some closed-loop, right-half plane poles will be excluded in the count Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
    • 34 • each poles vector rotates through - 1800 as we move around the contour near that pole• detour must be only an infinitesimal distance into the left half- plane, or else some closed-loop, left-half plane poles will be excluded in the count Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
    • 35Problem Sketch the Nyquist diagram of the unity feedback systemG(s) = (s + 2)/s2.Solution Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
    • 36 Stability via the Nyquist Diagram• Stability can be determined using equation Z = P – N where P… the number of open-loop poles of G(s)H(s) enclosed by the contour N…the number of encirclements the Nyquist diagram makes about (– 1) Z…the number of right-half-plane poles of the closed-loop system• If the closed-loop system has a variable gain K in the loop we can visualize the Nyquist diagram as expanding (increased gain) or shrinking (decreased gain)• OR• we can visualize the Nyquist diagram as remains stationary and the (- 1) point is moved along the real axis. First, we set the gain to unity and position the critical point at (- 1/K) rather than (- 1). Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
    • 37 Stability via the Nyquist DiagramP = 2, hence in order to obtain a stable system Z = P – N = 0, therefore N = 2,which means there must be 2 counterclockwise crossings Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
    • 38 Stability via Mapping Only the Positive jω-Axis - this is a simplified approach to determine stability - there must be no encirclements for this system to be stable• the encirclements of the critical point can be determined from the mapping of the positive jω-axis alone• this system is stable for the range of loop gain, K, that ensures that the open-loop magnitude is less than unity at that frequency where the phase angle is 1800 (or equivalently, - 1800) Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
    • 39 Stability via Mapping Only the Positive jω-Axis - there must be two counterclockwise encirclements for this system to be stable• this system is stable if the open-loop magnitude is greater than unity at that frequency where the phase angle is 1800 (or equivalently, -1800) Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
    • 40Problem Find the range of gain for stability and instability, and the gain formarginal stability, for the unity feedback system G(s) = K / [ ( s2 + 2s + 2)(s+ 2)].For marginal stability find the radian frequency of oscillation. Use the Nyquistcriterion and the mapping of only the positive imaginary axis.Solution First we draw the Nyquist diagram for K = 1 Im[G ( jω ) H ( jω )] = 0 when ω = 6 1 1 G ( jω ) H ( jω ) ω = 6 =− = ∠1800 20 20 Hence, stable for K < 20 unstable K > 20 marginally stable for K = 20, and oscillation is √6 Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
    • 41 Gain Margin and Phase Margin via the Nyquist Diagram• Gain margin and phase margin can be used to determine how stable a system is.• Systems with greater gain and phase margins can withstand greater changes in system parameters before becoming unstable. Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
    • 42• Gain margin, GM , is the change in open-loop gain, expressed in decibels (dB), required at 1800 of phase shift to make the closed-loop system unstable.• Phase margin, ΦM ,is the change in open-loop phase shift required at unity gain to make the closed-loop system unstable. Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
    • 43Problem Find the gain and phase margin for the systemG(s) = K / [ ( s2 + 2s + 2)(s+ 2)] if K = 6 .Solution: Im[G ( jω ) H ( jω )] = 0 when ω = 6 G ( jω ) H ( jω ) ω = 6 = −0.3 Hence GM = 20 log(1 / 0.3) = 10.45dB Using computer it can be found that G ( jω ) H ( jω ) has a unity gain at ω =1.253 rad/s The phase angle at ω = 1.253 rad/s is -112.30 Hence Φ M = −112.30 − (−1800 ) = 67.7 0 Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
    • 44 Stability, Gain Margin, and Phase Margin via Bode Plots• Bode plots are subsets of the complete Nyquist diagram but in another form• They can be easily drawnEvaluating Gain and Phase Margins using Bode Plots Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
    • 45 Relation between Closed - Loop Transient and Closed - Loop Frequency Responses Damping Ratio and Closed-Loop Frequency Response• There will be no peak at frequencies above zero if ζ>0.707 Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
    • 46Since ζ is related to percent overshoot, we can plot Mp vs. percent overshoot − ln(%OS / 100) ζ = π 2 + ln 2 (%OS / 100) Closed-loop frequency response peak vs percent overshoot for a two-pole system Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
    • 47Response Speed and Closed-Loop Frequency Response Bandwidth of the closed-loop frequency response is defined as the frequency, ωBW , at which the magnitude response curve is 3 dB down from its value at zero frequency M = 1 / 2 corresponds to - 3 dB Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
    • 48 Dr Branislav HredzakControl Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
    • 49 Dr Branislav HredzakControl Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
    • 50 Relation between Closed- and Open-Loop Frequency ResponsesDamping Ratio from Phase Margin 2 2 ωn ωnG ( jω ) = = jω ( jω + 2ζω n ) − ω 2 + j 2ζω nω • a phase margin of 65.520 (ζ = 0.707) or larger is required from the open-loop frequency response to ensure there is no peaking in the closed-loop frequency response Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
    • 51 Response Speed from Open-Loop Frequency Response• the closed-loop bandwidth can be estimated from the open-loop frequency response • It can be shown that, approximately, the closed-loop bandwidth, ωBW,equals the frequency at which the open-loop magnitude response is between -6 and -7.5 dB if the open-loop phase response is between -1350 and -2250 ωBW=3.5rad/s Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
    • 52 Steady-State Error Characteristics from Frequency ResponseType 0 system Initial value: For Type 0 system: - which is the same as the initial value Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
    • 53Type 1 system The initial -20 dB/decade slope can be thought of as originating from a function, which crosses the frequency axis (G’=0dB) at For Type 1 system: - which is the same as the frequency intercept Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.
    • 54Type 2 system The initial -40 dB/decade slope can be thought of as originating from a function, which crosses the frequency axis (G’=0dB) at For type 2 system: Dr Branislav Hredzak Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved.