# 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 control-system design: 1 Stability, 2 Steady-state 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 z-plane 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 discrete-time system is that all itspoleslie within the unit circle on the z-plane Stability criteria for cts-time systems is that the poles be inthelhp Simple tool to test for continuous-time stability Routh test Can we use the Routh test to determine stability of a discrete-time system (either directly or indirectly)? To use the Routh test, we need to do a z-plane to s-plane conversion that retains stability information The s-plane version of the z-plane system does NOT need to correspond in any other way That is, The frequency responses may be different The step responses may be different z-plane w-plane

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 s-plane It is often called the w-plane, and the transformation between the z-plane and the w-plane is called the w-transform 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 z-plane jω-axis in w-plane 2 Inside unit circle in z-plane LHP in w-plane 3 Outside unit circle in z-plane RHP in w-plane 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 s-plane, z-plane, and w-plane: ➃ ➄ s-plane z-plane w-plane 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 : Discrete-time 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 non-lhp 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 z-transform 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 continuous-timesystemis: 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 error-prone 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: Supposethatthecharacteristicequationforaclosed-loop discrete-time 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: Root-locus and Nyquist tests For cts-time control, we examined the locations of the roots ofthe closed-loop 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 Laplace-transform 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([1-1],[1-0368]); d=tf(numd,dend,-1); rlocus(d); The Nyquist test In continuous-time 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 jω-axis

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 open-loop unstable poles (poles inside D shape) Z=No of closed-loop unstable poles Z = N + P, EXAMPLE: r(t) K s(s + 1) Z = 0 for stable closed-loop 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 z-transforms The stable region is the unit circle The z-domain 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 = # closed-loop unstable poles I(z) P = # open-loop 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 jω-axis in w-plane R(w) Use standard continuous-time test in w-plane Summary: Open-loop 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 cts-time 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 continuous-time and discrete-time control design methods Recall: Bode plots are plots of frequency response of a system: Magnitude and Phase In s-plane, H(s) s= jω is frequency response for 0 ω< In z-plane, H(z) z=e jωt is frequency response for 0 ω ω s /2 Straight-line tools of s-plane analysis DON T WORK! They are based on geometry and geometry has changed jω-axis to z-unit circle BUT in w-plane, H(w) w= jωw is the frequency response for 0 ω w < Straight-linetoolswork,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 rational-polynomial 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 s-plane Bode plot Note: Plots are versus log 10 ω w ω w = 2 ( ) ωt T tan Can 2 re-scale 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 frequency-response methods without verifying that discrete-time frequency response means the same thing as continuous-time 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 partial-fraction 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 Closed-loop frequency response We have looked at open-loop concepts and how they apply to closed loop systems our end product Closed-loop 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 ) Closed-loop bandwidth similar to open-loop bandwidth If PM = 90,thenCLBW= OL BW If PM = 45,thenCLBW= 2 OL BW

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