STABILITY ANALYSIS TECHNIQUES


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1 ECE4540/5540: Digital Control Systems 4 1 STABILITY ANALYSIS TECHNIQUES 41: Bilinear transformation Three main aspects to controlsystem design: 1 Stability, 2 Steadystate response, 3 Transient response Here, we look at determining system stability using various methods DEFINITION: AsystemisBIBOstableiffaboundedinputproducesa bounded output Check by first writing system input output relationship as G(z) Y (z) = 1 + GH(z) R(z) = K m (z z i ) n (z p i ) R(z) Assume for now that all the poles {pi } are distinct and different from the poles in R(z) Then, Y (z) = k 1z + + k nz } z p 1 {{ z p n} Response to initial conditions + Y R (z) }{{} Response to R(z) If the system is stable, the response to initial conditions must decay to zero as time progresses Z 1 [ ki z z p i So, the system is stable if p i < 1 ] = k i (p i ) k 1[k]
2 ECE4540/5540, STABILITY ANALYSIS TECHNIQUES 4 2 {pi } are the roots of 1 + GH(z) = 0 So,therootsof1 + GH(z) = 0 must lie within the unit circle of the zplane Same result even if poles are repeated, but harder to show If the magnitude of a pole pi =1, thenthesystemismarginally stable The unforced response does not decay to zero but also does not increase to However, itispossibletodrivethesystemwitha bounded input and have the output go to Therefore,amarginally stable system is unstable Bilinear transformation The stability criteria for a discretetime system is that all itspoleslie within the unit circle on the zplane Stability criteria for ctstime systems is that the poles be inthelhp Simple tool to test for continuoustime stability Routh test Can we use the Routh test to determine stability of a discretetime system (either directly or indirectly)? To use the Routh test, we need to do a zplane to splane conversion that retains stability information The splane version of the zplane system does NOT need to correspond in any other way That is, The frequency responses may be different The step responses may be different zplane wplane
3 ECE4540/5540, STABILITY ANALYSIS TECHNIQUES 4 3 Since only stability properties are maintained by the transform, it is not accurate to label the destination plane the splane It is often called the wplane, and the transformation between the zplane and the wplane is called the wtransform Atransformthatsatisfiestheserequirementsisthebilineartransform Recall: H(w) = H(z) z= 1+(T/2)w 1 (T/2)w and H(z) = H(w) w= 2 z 1 T z+1 Three things to check: 1 Unit circle in zplane jωaxis in wplane 2 Inside unit circle in zplane LHP in wplane 3 Outside unit circle in zplane RHP in wplane If true, 1 Take H(z) H(w) via the bilinear transform 2 Perform Routh test on H(w) CHECK: Let z = re jωt Then,zis on the unit circle if r = 1, z is inside the unit circle if r < 1 and z is outside the unit circle if r > 1 z = re jωt w = 2 z 1 T z + 1 z=re jωt = 2 T re jωt 1 re jωt + 1 Expand e jωt = cos(ωt ) + j sin(ωt ) and use the shorthand c = cos(ωt ) and s = sin(ωt ) Alsonotethats 2 + c 2 = 1 w = 2 [ ] rc + jrs 1 T rc + jrs + 1
4 ECE4540/5540, STABILITY ANALYSIS TECHNIQUES 4 4 ] +1 = 2 T = 2 T = 2 T [ ][ (rc 1) + jrs (rc + 1) jrs (rc + 1) + jrs (rc + 1) jrs [ (r 2 c 2 1) + j (rs)(rc + 1) j (rs)(rc 1) + r 2 s 2 ] (rc + 1) 2 + (rs) 2 [ r 2 ] 1 + j 2 [ ] 2rs r 2 + 2rc + 1 T r 2 + 2rc + 1 Notice that the real part of w is 0 when r = 1 (w is on the imaginary axis), the real part of w is negative when r < 1 (w in LHP), and that the real part of w is positive when r > 1 (w in RHP) Therefore, the bilinear transformation does exactly what we want When r = 1, w = j 2 T which will be useful to know 2sin(ωT ) 2 + 2cos(ωT ) = j 2 ( ) ωt T tan, 2 The following diagram summarizes the relationship between the splane, zplane, and wplane: ➃ ➄ splane zplane wplane j ω s 2 j ω s 4 j ω s j ω 4 s 2 ➂ ➁ ➀ ➆ ➅ ➂ ➅ ➃ ➄ ➁ ➆ R = 1 ➀ ➂ ➅ ➃ R = 2 ➄ T ➁ j 2 T ➀ ➆
5 ECE4540/5540, STABILITY ANALYSIS TECHNIQUES : Discretetime stability via Routh Hurwitz test Review of Routh test Let H(w) = b(w) a(w) is the characteristic polynomial a(w) a(w) = a n w n + a n 1 w n 1 + +a 1 w + a 0 Case 0: Case 1: If any of the a n are negative then the system is unstable (unless ALL are negative) Form Routh array: w n a n a n 2 a n 4 w n 1 a n 1 a n 3 a n 5 w n 2 b 1 b 2 w n 3 c 1 c 2 w 1 j 1 b 1 = 1 a n 1 a n a n 1 w 0 k 1 a n 2 a n 3 b 2 = 1 a n 1 a n a n 1 a n 4 a n 5 c 1 = 1 b 1 a n 1 a n 3 b 1 b 2 c 2 = 1 b 1 a n 1 a n 5 b 1 b 3 TEST: Number of RHP roots = number of sign changes in left column Case 2: Case 3: If one of the left column entries is zero, replace it with ϵ as ϵ 0 Suppose an entire row of the Routh array is zero, the w i 1 th row The w i th row, right above it, has coefficients α 1,α 2,
6 ECE4540/5540, STABILITY ANALYSIS TECHNIQUES 4 6 Then, form the auxiliary equation: α 1 w i + α 2 w i 2 + α 3 w i 4 + =0 This equation is a factor of the characteristic equation and must be tested for RHP roots (it WILL have nonlhp roots we might want to know how many are RHP) EXAMPLE: Consider: r(t) T K 1 e st s 1 s(s + 1) y(t) G(s) = ( 1 e Ts)( ) 1 s s(s + 1) From ztransform tables: ( ) [ ] z 1 1 G(z) = Z z s 2 (s + 1) ( )( z 1 (e T + T 1)z 2 + (1 e T Te T ) )z = z (z 1) 2 (z e T ) Let T = 01s = z (z 1)(z 0905) Perform the bilinear transform G(w) = G(z) z= 1+(T/2)w 1 (T/2)w = G(z) z= 1+005w 1 005w The characteristic equation is: = w w w w
7 ECE4540/5540, STABILITY ANALYSIS TECHNIQUES 4 7 (system 0 = 1 + KG(w) = ( K )w 2 + ( K )w + 381K w 2 ( K ) 381K K < 23, 813 w 1 ( K ) K < 203 w 0 381K K > 0 So, for stability, 0 < K < 203 NOTE: The equivalent continuoustimesystemis: r(t) K 1 s(s + 1) y(t) T (s) = KG(s) 1 + KG(s) Characteristic equation: s(s + 1) + K = 0 s 2 1 K s 1 1 s 0 K Stable for all K > 0 sample and hold destabilizes the system EXAMPLE: Let sdothesameexample,butwitht = 1s(not01s) (math happens) 0 = 1 + KG(w) = ( K )w 2 + ( K )w K w 2 ( K ) 0924K K < 262 w 1 ( K ) K < 239 w K K > 0
8 ECE4540/5540, STABILITY ANALYSIS TECHNIQUES 4 8 So, for stability, 0 < K < 239 This is a much more restrictive range than when T = 01s slow sampling really destabilizes a system
9 ECE4540/5540, STABILITY ANALYSIS TECHNIQUES : Jury s stability test H(z) H(w) Routh is complicated and errorprone Jury made a direct test on H(z) for stability Disadvantage (?) another test to learn Let T (z) = b(z) a(z), a(z) = characteristic polynomial a(z) = an z n + a n 1 z n 1 + +a 1 z + a 0 = 0, a n > 0 Form Jury array: z 0 z 1 z 2 z n k z n 1 z n a 0 a 1 a 2 a n k a n 1 a n a n a n 1 a n 2 a k a 1 a 0 b 0 b 1 b 2 b n k b n 1 b n 1 b n 2 b n 3 b k 1 b 0 c 0 c 1 c 2 c n k c n 2 c n 3 c n 4 c k 2 l 0 l 1 l 2 l 3 l 3 l 2 l 1 l 0 m 0 m 1 m 2 Quite different from Routh array Every row is duplicated in reverse order Final row in table has three entries (always) Elements are calculated differently a 0 a n k b 0 b k = c k = a n a k b n 1 b n 1 k b k d k = c 0 c n 2 c n 2 k c k
10 ECE4540/5540, STABILITY ANALYSIS TECHNIQUES 4 10 Stability criteria is different a(z) z=1 > 0 ( 1) n a(z) z= 1 > 0 n = order of a(z) a 0 < a n b 0 > b n 1 c 0 > c n 2 d 0 > d n 3 m 0 > m 2 First, check that a(1) >0, ( 1) n a( 1) >0 and a 0 < a n (relatively few calculations) If not satisfied, stop Next, construct array Stop if any condition not satisfied EXAMPLE: r(t) T = 1s 1 e st s K s(s + 1) y(t) r[k] K 0368z z z y[k] Characteristic equation: 0 = 1 + KG(z) = 1 + K (0368z ) z z
11 ECE4540/5540, STABILITY ANALYSIS TECHNIQUES 4 11 The Jury array is: = z 2 + (0368K 1368)z + ( K ) z 0 z 1 z K 0368K The constraint a(1) >0 yields K K = 0632K > 0 K > 0 The constraint ( 1) 2 a( 1) >0 yields K K = 0104K > 0 K < 263 The constraint a0 < a 2 yields K < 1 K < = 239 So, 0 < K < 239 (Same result as on pg 4 8 using bilinear rule) EXAMPLE: Supposethatthecharacteristicequationforaclosedloop discretetime system is given by the expression: a(z) = z 3 18z z 020 = 0 a(1) = = 005 > 0 ( 1) 3 a( 1) = [ ] > 0 a0 =02 < a 3 = 1 Jury array: z 0 z 1 z 2 z
12 ECE4540/5540, STABILITY ANALYSIS TECHNIQUES 4 12 b 0 = = 096 b 1 = = 159 b 2 = = 069 b0 =096 > b 2 =069 The system is stable
13 ECE4540/5540, STABILITY ANALYSIS TECHNIQUES 4 13 ŷ 44: Rootlocus and Nyquist tests For ctstime control, we examined the locations of the roots ofthe closedloop system as a function of the loop gain K Root locus r(t) K D(s) G(s) y(t) H(s) T (s) = KD(s)G(s) 1 + KD(s)G(s)H(s) Let L(s) = D(s)G(s)H(s) (The looptransferfunction ) Developed rules for plotting the roots of the equation Root Locus Drawing Rules Applied them to plotting roots of 1 + K b(s) a(s) = KL(s) = 0 Now, we have the digital system: r(t) K D(z) T G(s) {}}{ 1 e st G p (s) s H(s) y(t) T (z) = So, we let L(z) = D(z)GH(z) Poles are roots of 1 + KL(z) = 0 KD(z)G(z) 1 + KD(z)GH(z),
14 ECE4540/5540, STABILITY ANALYSIS TECHNIQUES 4 14 This is exactly the same form as the Laplacetransform root locus Plot roots in exactly the same way EXAMPLE: GH(z) = 0368(z ) (z 1)(z 0368) D(z) = 1 K = 239 numd=0368*[1 0717]; dend=conv([11],[10368]); d=tf(numd,dend,1); rlocus(d); The Nyquist test In continuoustime control we also used the Nyquist test to assess stability I(s) I(s) Zoom II III ρ 0 I R(s) θ R(s) IV The Nyquist D path encircles the entire (unstable) RHP The Nyquist plot is a polar plot of L(s) evaluated on the D path Adjustments to D shape are made if pole on the jωaxis
15 ECE4540/5540, STABILITY ANALYSIS TECHNIQUES 4 15 The Nyquist test evaluated stability by looking at the Nyquist plot N=No of CW encirclements of 1 in Nyquist plot P=No of openloop unstable poles (poles inside D shape) Z=No of closedloop unstable poles Z = N + P, EXAMPLE: r(t) K s(s + 1) Z = 0 for stable closedloop system y(t) This gives: L(s) = 1 s(s + 1) Pole at origin: Need detour s = ρe jθ, ρ 1 Resulting Nyquist map has infinite radius Cannot draw to scale No poles inside modified D curve: P = 0 Z = N + P = 0 Stable system Note that increasing the gain K onlymagnifiestheentireplotthe 1 point is not encircled for K > 0 (infinite gain margin) Nyquist test for discrete systems Three different ways to do the Nyquist test for discrete systems Based on three different representations of the characteristic eqn L (s) = 0 L = DGH L(z) = L(w) = 0
16 ECE4540/5540, STABILITY ANALYSIS TECHNIQUES L (s) = 0 We know that L (s) is periodic in jω s Therefore, the D curve does not need to encircle the entire RHP to encircle all unstable poles [If there were any, there would be an infinite number] Modify D curve to be: j ω s 2 j ω s 2 I(s) R(s) Evaluate L (s) on new contour and plot polar plot Same Nyquist test as before L(z) = 0 We can do the Nyquist test directly using ztransforms The stable region is the unit circle The zdomain Nyquist plot is done using a Nyquist curve which is the unit circle Nyquist test changes because we are now encircling the STABLE region (albeit CCW) Z = # closedloop unstable poles I(z) P = # openloop unstable poles N = # CCW encirclements of 1 in Nyquist plot R(z) Z = P N Probably difficult to evaluate L(z) z=e jθ for π θ π unless using adigitalcomputer
17 ECE4540/5540, STABILITY ANALYSIS TECHNIQUES L(w) = 0 Now, we convert L(z) L(w) I(w) L(w) = L(z) z= 1+(T/2)w 1 (T/2)w Bilinear transform maps unit circle to jωaxis in wplane R(w) Use standard continuoustime test in wplane Summary: Openloop fn Range of variable Rule GH (s) s = jω, ω s /2 ω ω s /2 Z = P + N cw GH(z) z = e jωt, π ωt π Z = P N ccw = P + N cw GH(w) w = jω w, ω w Z = P + N cw All three methods produce identical Nyquist plots Nyquist Diagrams Note that the sampled system does not have gain margin (a = 0418, GM = 239) andhassmaller PM than ctstime system Imaginary Axis PM a Real Axis
18 ECE4540/5540, STABILITY ANALYSIS TECHNIQUES : Bode methods Bode plots are an extremely important tool for analyzing and designing control systems They provide a critical link between continuoustime and discretetime control design methods Recall: Bode plots are plots of frequency response of a system: Magnitude and Phase In splane, H(s) s= jω is frequency response for 0 ω< In zplane, H(z) z=e jωt is frequency response for 0 ω ω s /2 Straightline tools of splane analysis DON T WORK! They are based on geometry and geometry has changed jωaxis to zunit circle BUT in wplane, H(w) w= jωw is the frequency response for 0 ω w < Straightlinetoolswork,butfrequencyaxisiswarped PROCEDURE: 1 Convert H(z) to H(w) by H(w) = H(z) z= 1+(T/2)w 1 (T/2)w 2 Simplify expression to rationalpolynomial in w 3 Factor into zeros and poles in standard Bode Form (Refer to review notes) 4 Plot the response exactly the same way as an splane Bode plot Note: Plots are versus log 10 ω w ω w = 2 ( ) ωt T tan Can 2 rescale axis in terms if ω if we want EXAMPLE: Example seen before with T = 1 second
19 ECE4540/5540, STABILITY ANALYSIS TECHNIQUES 4 19 (1,2) (3) (4) G(w) = = = [ 1+05w 1 05w Let G(z) = 0368z z z [ 1+05w 1 05w] ] 2 [ w 1 05w] (1 + 05w)(1 05w) (1 05w) 2 (1 + 05w) (1 + 05w)(1 05w) (1 05w) (w 2)(w ) w(w ) Magnitude (db) G( jω w ) = ( j ω w 2 1 )( j ω w 1214 ( + 1) jω w j ω w ) Bode Plots Phase (deg) Frequency (warped rads/sec) Gain margin and phase margin work the SAME way we expect
20 ECE4540/5540, STABILITY ANALYSIS TECHNIQUES 4 20 WAIT! We have discussed frequencyresponse methods without verifying that discretetime frequency response means the same thing as continuoustime frequency response Verify X (z) G(z) Y (z) Let x[k] =sin(ωkt) X (z) = z sin ωt (z e jωt )(z e jωt ) Y (z) = G(z)X (z) = G(z)z sin ωt (z e jωt )(z e jωt ) Do partialfraction expansion Y (z) k 1 = z z e + k 2 jωt z e + Y g(z) jωt Yg (z) is the response due to the poles of G(z) IF the system is stable, the response due to Y g (z) 0 as t So, as t we say Y ss(z) z = k 1 = k 1 z e + k 2 jωt z e jωt G(z) sin ωt z e jωt z=e jωt = G(e jωt ) sin ωt e jωt e jωt = G(e jωt ) 2 j = G(e jωt ) e j G(e jωt ) 2 j
21 ECE4540/5540, STABILITY ANALYSIS TECHNIQUES 4 21 Similarly, k 2 = G(e jωt ) e j G(e jωt ) 2( j) = G(e jωt ) e j G(e jωt ) 2 j Combining and solving for yss [k] y ss [k] =k 1 (e jωt ) k + k 2 (e jωt ) k = G(e jωt ) e jωkt+ j G(e jωt ) e jωkt j G(e jωt ) 2 j = G(e jωt ) sin(ωkt + G(e jωt )) Sure enough, G(e jωt ) is magnitude response to sinusoid, and G(e jωt ) is phase response to sinusoid Closedloop frequency response We have looked at openloop concepts and how they apply to closed loop systems our end product Closedloop frequency response usually calculated by computer: G(z) 1 + G(z),forexample In general, if G(e jωt ) large, T (e jωt ) 1 If G(e jωt ) small, T (e jωt ) G(e jωt ) Closedloop bandwidth similar to openloop bandwidth If PM = 90,thenCLBW= OL BW If PM = 45,thenCLBW= 2 OL BW
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