Chapter 7 Stability Analysis Techniques 7.2 Stability 7.3 Bilinear Transformation 7.4 Routh-Hurwitz Criterion 7.5 Jury Stability Test 7.6 Root Locus 7.7 Nyquist Criterion 7.8 Bode Diagram 7.9 Interpretation of Frequency Response 7.10 Closed-Loop Frequency Response C.L. Phillips, H. T. Nagle, and A. Chakrabortty, , 4th Edition, Pearson, 2015 7.2 z-Plane Stability Consider the LTI system: Fig. 7-1 (7-1) (7-2) Combining complex pairs of poles: A(r )k cos(qk +f) (6-7) 3 Transient Response (Fig. 6-11) A(r )k cos(qk +f) 4 Stability System stability may be determined from the location of the roots of the characteristic equation. The system characteristic equation is given in various ways: Classic Characteristic Equation: 1+ GH(z) = 0 (7-3) Denominator of pseudo-control function, independent of R(z) |zI - A| = 0 Keep in mind that stability is only a boundary condition in control. Tests for stability: Do not give relative stability Do not give system performance. For relative stability, performance and design we will use: Root locus Nyquist plots Bode plots 5 z-Plane Stability (characteristic equation from the transfer function) R(s) E(s) E* D(z) + - T H(s) ZOH Gp(s) C(s) Fig. 7-1 with added D(z) & ZOH 6 z-Plane Stability (characteristic equation from pseudo-transfer function) R(s) E1* E1(s) + T + G1(z) – – C(s) G2(s) Fig. 7-2(a) From Example 5.3 é G1G2 é é é(z) 2 + G2 é é R é é(1+ G2 )R é é C(z) = é é(z) + é é(z) é G1G2 é é 2 + G2 é é2 + G2 é 1+ é é(z) é2 + G2 é The denominator term, independent of the input R(z), is the characteristic equation: 7 é G1G2 é 1+ é é(z) = 0 é2 + G2 é z-Plane Stability The characteristic equation may also be found by: Opening the loop at a sampler: • Set the input R(s)=0 (stability is independent of the input). • Open the loop at any sampler (label the sampler output as Ei*(s) and its input as Eo(s)). • Solve for the sampler input Eo(s) using Ei*(s) as the source. • Solve for the open-loop transfer fcn (Gop(z) = Eo(z)/Ei(z)). • Characteristic equation is 1 + F(z) = 1 – Gop(z) = 0 (7-5) (Example 7.1: next slide) State space: x(k+1) = A x(k) + B r(k) y(k) = C x(k) + D r(k) Characteristic equation is |zI - A| = 0 8 (7-6) Example 7.1 opening a sampler Open Loop: Eo = -G2 E1* E1 = G1 Ei* - G2 HE1* E1* = G1* Ei* 1+ G2 H Fig. 7-3 E = * o -G1*G2* 1+ G2 H Close the loop ( Eo* = Ei* Þ Gop (z) =1): é G (z)G (z) é 2 é1+ 1 éEo (z) = 0 é 1+ G2 H(z) é 1+ F(z) = 1- Gop (z) = 0 (7-6) * Ei* Eo (z) = Gop (z)Ei (z) Gop (z) = - 1+ G1 (z)G2 (z) + G2 H(z) = 0 * G1 (z)G2 (z) 1+ G2 H(z) Gop(z) is the open-loop transfer function. F(z) is the open-loop function. 9 z-Plane Stability A discrete-time system is stable if its closed-loop roots are inside the unit circle. The text presents two ways to evaluate root locations of the z-plane characteristic equation: The Routh-Hurwitz criterion (Section 7.4): a s-plane method (z-plane mapped to the w-plane which mimics the s-plane): Jury’s Stability Test (Section 7.5): a z-plane method 10 7.3 Bilinear Transformation z-Plane to s-Plane Mapping The obvious question: What mapping method should we use? Consider the mapping we used for s-plane to z-plane mapping in Chapter 6. sT z e z = e sT Thus, Þ sT , z- a e - a Please note that a polynomial method is preferred! The bilinear transformation is another powerful method. 1+ (T / 2)w 2 z- 1 z= or w = 1- (T / 2)w T z+ 1 (7-7) & (7-8) This transformation maps the z-plane to the w-plane. How is the w-plane related to the s-plane? 11 z-Plane to w-Plane Mapping If we let z= e sT , we find from the 1st term of the expansion: 1 1 s = ln z = T T é z-1 1 (z-1)3 ù 2ê + + .......ú ë z+1 3 z+1 û 2 z-1 2 e sT -1 2 e jw T -1 w= = = sT T z+1 z=e sT T e +1 s= jw T e jw T +1 2e = Te - e - jw T /2 2 wT = j tan = jw w jw T /2 - jw T /2 +e T 2 Bilinear Transform jw T /2 (7-9) To interpret this, we observe that the imaginary or frequency axis in the wplane is related to the imaginary or frequency axis in the s-plane by: 2 wT ww = tan T 2 (7-10) when wT <<1, 2 ww » w (7-11) 12 Bilinear Transformation w-plane to z-plane 2 z-1 1) w= T z+1 z-plane to w-plane 2) 2 w+ T z= 2 wT = éT é 1+ é é w é2 é éT é 1- é é w é2 é s-plane frequency, , to w-plane frequency, w 2 wT wT wT wT 3) w w = tan , when << 1, tan » , ww » w T 2 2 2 2 13 z-Plane to w-Plane Mapping jw Fig. 7-4 s-Plane D C ws / 4 B ∞ frequency ws / 2 A 0 ww = For G -w s / 4 E -w s / 2 F 2 wT tan T 2 wT << 1, 2 w-Plane j w ww » w C C ∞ B j2/T C z = e sT D E -2/T F Im z z-Plane 2 z- 1 w= T z+ 1 B 1 C F D E 0 A Re z z = 1 gives w = 0 z = -1 gives w = ∞ z = 0 gives w = –2/T G 14 A 0 G -j2/T ∞ F F w Why map to the w-plane? 1) This is a first-order approximation of the mapping from the z-plane to the s-plane. 2) We can apply the Routh-Hurwitz criterion for stability analysis. 3) We can do compensator design on the vertical j w axis using continuous-time Bode plot design methods (Chapter 8). (lag, lead, lag/lead, PI, PD, PID, etc.) 4) Accurate for small w [tan ( T/2) ≅ T/2] For example, for: T/2 < /10 or < 2/10T or < s/10 (error < 4%) At ω = ωs/10, a ZOH introduces 18º phase lag. 15 (7-12) Example 7.2 Fig. 7-5 (from Example 6.7): The characteristic equation is a function of the sampling interval T. As T increases beyond 0.5, the system goes unstable. ( G(z) = K ) ( e -T + T - 1) z+ 1- (1+ T)e -T (z- 1)(z- e 1+ wT / 2 Let z = , 1- wT / 2 -T ) ) which leads to æ1+ wT / 2 æ -T e + T - 1) æ + 1(1+ T)e æ1- wT / 2 æ æ G(w) = K ææ1+ wT / 2 æ æææ1+ wT / 2 æ -T æ - 1æææ -e æ ææ æ æ ææ1- wT / 2 æ æææ1- wT / 2 æ æ ( -T ) ( 16 ) Bilinear Transformation 1+ G(w) = 0, Next form getting ù éT 2 -T ùù -T -T é é w ê ë2(1+ e ) + K ë2(1- e - T(1+ e )ûûú + û ë 4 wéëT (1- e -T ) - KT éë1- e -T (1+ T )ùûùû + KT (1- e -T ) = 0 2 or aw + bw+ c = 0 2 -b± b2 - 4ac -b ±j = w1,2 = 2a 2a 4ac - b2 2a We can determine stability for a specific gain, say K = 4.25, by increasing T and examining the “-b/2a” term to test for stability. T K - b / 2a 0.1 4.25 - 0.395 0.2 4.25 - 0.294 0.5 4.25 - 0.013 0.5145 4.25 0 This result shows that instability results when T is greater than 0.5145. We can also set the sampling period (say T = 0.1) and examine stability for values of K. 17 7.4 The Routh-Hurwitz Criterion • Find the characteristic equation (discrete). • Use the bilinear transformation to map the function from the z-plane to the w-plane. • Construct the Routh array. • Examine the entries to determine stability. 18 Routh-Hurwitz The Routh-Hurwitz criterion (Table 7-1) identifies the number of rhp poles from the number of sign changes in the leftmost column of the Routh array. F(w) = bnwn + bn-1wn-1 + bn-2 wn-2 + Form the array: n w : n-1 w n-2 w a1 a2 wn : a3 : b1 b2 b3 : c1 c2 c3 + b1w+ b0 1 a1 ai 1 ci b1 b1 bi 1 19 bn bn-2 bn-4 wn-1 : bn-1 bn-3 bn-5 wn-2 : c1 c2 c3 : d1 w1 : j1 w0 : k1 d2 d3 % MATLAB® Fw=[1 9 8 12 21 30] roots(Fw) Routh Array ans = -8.1687 0.6302 0.6302 -1.0458 -1.0458 For the an example open-loop function: F(w) = w5 + 9w4 + 8w3 + 12w2 + 21w+ 30 Forming the array: wn : a1 a2 + + - 1.2968i 1.2968i 0.8204i 0.8204i a3 wn : 1 8 21 wn-1 : b1 b2 b3 1 a1 ai+1 ci = wn-1 : 9 12 30 wn-2 : c1 c2 c3 b1 b1 bi+1 wn-2 : 6.667 17.667 0 wn-3 : -11.85 30 0 1 60 (1*12 9 *8) = = 6.667 wn-4 : 9 9 1 159 53 wn-5 : - (1*30 - 9 * 21) = = =17.667 9 - 9 3 1 79 (9 *17.667 - 6.667*12) = = -11.85 6.667 6.667 The appearance of the minus sign indicates rhp root(s), and thus instability. 20 % MATLAB® Fw=[1 10 38 67 52 12] roots(Fw) Routh Array A second example: F(w) = w5 + 10w4 + 38w3 + 67w2 + 52w+ 12 wn : 1 38 52 wn-1 : 10 67 12 wn-2 : 31.3 50.8 0 wn-3 : 50.8 12 0 wn-4 : 43.4 0 0 wn-5 : 12 0 0 1 a1 ai 1 ci b1 b1 bi 1 - ans = -3.0000 -2.6180 -2.0000 -2.0000 -0.3820 1 313 (1* 67 -10 * 38) = = 31.3 10 10 The process continues to a single number in the bottom two rows. No minus signs appear so the system is stable. The number of rhp poles is equal to the number of sign changes in the left most column. The system is stable. 21 Example 7.2 (Again) Fig. 7-5 Find the bilinear transform: >> T=0.1; numz=[0.00484 0.00468]; denz=[1 -1.905 0.905]; Gz=tf(numz,denz,T); Gw=d2c(Gz,'tustin') Gw = -4.199e-05 s^2 - 0.04913 s + 0.9995 ----------------------------------s^2 + 0.9974 s + 4.429e-15 22 Example 7.2 (continued) The characteristic equation for any value of K The Routh array No sign changes in the first column. K must be in the range 0 < K < 20.3 for stability. 23 7.5 Jury’s Stability Test • Can be applied directly to z-domain functions. • It is not necessary to perform the z-plane to w-plane transformation. • The array is larger and the computations are more complex than in the Routh-Hurwitz test. • First, determine the characteristic equation: Q(z) = 1+ D(z)GH(z) = an z + an-1z n n-1 + + a1 z+ a0 = 0 for an > 0 (7-13) • For the system to be stable, all of the roots of this polynomial must lie within the unit circle. • From the characteristic equation, match terms to the polynomial (and form the array on the next slide): 24 Jury’s Array Q(z) = anz + an-1z + n + a1z+ a0 = 0 for an > 0 n-1 Row z0 z1 z2 z n -k z n -1 1 a0 a1 a2 an - k an -1 an 2 an an -1 an -2 ak a1 a0 3 4 b0 bn -1 b1 bn -2 b2 bn -3 bn -k bk -1 bn -1 b0 0 0 5 c0 c1 c2 c3 cn -2 0 0 6 cn -2 cn -3 cn -4 cn -5 c0 0 0 0 0 0 2n - 5 2n - 4 l0 l3 l1 l2 l2 l1 l3 l0 0 0 0 0 0 0 0 0 2n - 3 m0 m1 m2 0 0 0 0 0 Table 7-2 25 zn Q(z) = anzn + an-1zn-1 + Jury’s Array + a1z+ a0 = 0 for an > 0 The elements of the array are calculated: bk = m0 = a0 an-k an ak l0 l3 l3 l0 ck = m1 = The Jury criterion states that the necessary and sufficient conditions for stability, i.e. that all roots lie inside the unit circle, are: b0 bn-k-1 bn-1 bk l0 l2 l3 l1 m2 = Q(1) > 0 ( -1) Q(-1) > 0 n dk = l0 l1 l3 l2 c0 cn-k-2 cn-2 ck (7-14) order of the polynomial and the following (n- 1) constraints a0 < an b0 > bn-1 first row 26 c0 > cn-2 last row (7-15) m0 > m2 % MATLAB® Qz=[17 29 27 7] abs(roots(Qz)) Jury Example Q(z) = 17z3 + 29z2 + 27z+ 7 = 0 a0 < a3 Þ 7 < 17 Q(1) = 80 > 0 Þ OK ans = 1.0415 1.0415 0.3796 Þ OK (-1)3 Q(-1) = (-1)(-8) = 8 > 0 Þ OK The array becomes 1 7 2 17 3 -240 b0 = 27 29 -304 29 17 27 7 -256 0 b1 = m0 > m2 ? Þ Fails -240 > -256 We thus conclude that the system is unstable. 27 a0 an-0 an a0 7 29 17 27 = 7 17 17 7 b2 = 7 27 17 29 Note, the roots are: - 0.380, - 0.663 j 0.803. The magnitude of the complex roots is 1.045. Example 7.4 Step input R(s) = 1/s R(s) E(s) + - E* T=1sec 1--Ts s The open-loop transfer function is given by: . K . s(s+1) C(s) Fig. 7-5 The unity feedback closed-loop characteristic equation (with open-loop gain K) is given by: (0.368z+ 0.264)K 1+ KG(z) = 1+ 2 =0 z -1.368z+ 0.368 or z2 + (0.368K -1.368)z+ (0.368 + 0.264K ) = 0 30 Example 7.4 (cont.) The Jury array is: z0 z1 0.368 + 0.264 K 0.368 K – 1.368 z2 1 Applying the constraints: a) Q(1) > 0, [z2 +(0.368K − 1.368)z + (0.368 + 0.264)K)] |z = 1 1 +(0.368K − 1.368) + (0.368 + 0.264 K) = 0.632K > 0 => K > 0 b)(-1)2 Q(-1) > 0 1− 0.368K + 1.368 + 0.368 + 0.264K > 0 => K < .632/.104 = 26.3 c)|a0| < an 0.368 + 0.264K < 1 => K < 0.632/0.264 = 2.39 31 % MATLAB® Qz=[1 -0.488 1]; r=roots(Qz) abs(r) angle(r) Example 7.4 (cont.) Then for stability: 0 < K < 2.39 and marginal stability is at K = 2.39, for which: [z2 +(0.368K−1.368)z + (0.368 + 0.264) K)]|K=2.39 r = 0.2440 + 0.9698i 0.2440 - 0.9698i ans = 1.0000 1.0000 ans = 1.3243 -1.3243 = z2 – 0.488 z + 1 with roots: z = 0.244 ± j0.970 = 1Ð (± 75.9o) = Ð 1 = 1 Ð (± T), (± 1.32 rad) for T = 1 sec => = 1.32 rad/sec 32 Jury’s Method “Quick” tests for low-order systems: First-order: a0 <1 a1 Second-order: a2 + a1 + a0 > 0 a2 - a1 + a0 > 0 a0 - a2 < 0 Third-order: a3 + a2 + a1 + a0 > 0 a3 - a2 + a1 - a0 > 0 a0 < a3 2 2 a a ( 0 3 ) < ( a0 a2 - a1a3 ) 33 7.6 Root Locus Determines the location of roots of the characteristic equation of a closed loop control system as the overall system gain (often denoted by K) varies. The characteristic equation may be written in any of the following forms: 1+ KGH = 0 (7-16) Roots of this equation must meet two conditions: KGH(z) = 1, ( ) Ð KGH(z) = (1+ 2k)(180°) A root locus is a plot of the roots of this equation in the z-Plane as K varies. While construction rules for the z-Plane are identical with those for the s-Plane, the interpretation is different. 34 Root Locus DESIGN METHOD: 1. Derive the open loop function KGH(z) . 2. Factor numerator and denominator, linear & quadratic factors. 3. Plot roots in z-Plane. 4. Determine the root loci geometrically, analytically, by computer-any convenient method. 5. Calibrate in terms of K. 6. Find time response, if needed. 7. Refine, redesign, relocate, etc., to meet specifications. 8. Synthesize or realize any necessary compensation networks or digital filters. 35 Root Locus Construction 1+ KGH (z) = 0 GH (z) = N(z) D(z) 1. Loci originate on the poles of KGH(z) and terminate on its zeros. 2. For K > 0, the real axis loci lie to the left of an odd number of poles and zeros. 3. The loci are symmetrical with respect to the real axis. 4. The number of asymptotes is equal to the number of poles of KGH(z), np, minus the number of its zeros, nz. The angles of the asymptotes are found by (1+ 2k)(180 o ) qA = , np - nz k = 0, 1, 2, ... (np - nz -1) 5. The origin of the asymptotes on the real axis is given by: poles of GH(z) - å zeros of GH(z) å s= np - nz 36 Root Locus Construction 6. The breakaway point for the locus between two poles (or the break-in point for the locus between two zeros) is found by break-in dGH(z) = 0. Solve for the roots. dz num(z) d(num(z)) d(den(z)) GH (z) = , den(z) - num(z) =0 den(z) dz dz breakaway 7. The unit circle crossings of the root locus may be determined by the Jury test (the value of K and location of roots). 37 G(z) = Root Locus Example (z+ 0.9700) (z- 1)(z- 0.9100) (z+ 0.9700) N(z) 1+ K = 1+ KG(z) = 1+ K =0 (z-1)(z- 0.9100) D(z) To find the breakaway points: den(z) = (z-1)(z- 0.9100) num(z) = z+ 0.9700 d(num(z)) d(den(z)) - num(z) = z2 +1.9400z- 2.7627 = 0. dz dz Solving yields z = 0.9471, - 2.9171 Use the Jury test to find that the locus crosses den(z) the unit circle for K = ?. z2 + (K -1.9100)z+ (0.9700 K + 0.9100) = 0 Q(1) > 0 Þ K > 0; (-1) n Q(-1) > 0 Þ K < 127.33 a2 > a0 Þ 1 > 0.97 K + 0.91, or K < 0.0928. We conclude that the system is stable iff: 38 0 < K < 0.0928 G(z) = Root Locus Example (cont.) (z+ 0.9700) (z- 1)(z- 0.9100) K=0.0928 0 -2.9171 -0.97 x x1 0.91 0.9471 39 num=[1 0.970];den=[1 -1.91 0.91]; Gz=tf(num,den,0.1);rlocus(Gz) axis([-4 1.5 -2 2]) G(z) = 40 (z+ 0.9700) (z- 1)(z- 0.9100) Root Locus Example (cont.) Find the breakaway point. rlocus(Gz) axis([.75 1.1 -1.1 1.1]) Breakaway point (Damping = 1.0) is 0.954 41 G(z) = Root Locus Example (cont.) (z+ 0.9700) (z- 1)(z- 0.9100) Suppose we add some compensation, with a digital filter D(z): D(z) = z- 0.9100 , z- 0.4000 D(z)G(z) = (z+ 0.9700) (z-1)(z- 0.4000) K=0.6200 x -0.97 0 0.40 0 x 1 0.6700 While the root locus has a very similar shape, analysis of the data shows that the system is stable for 0 < K < 0.62. This greatly reduces the steady-state errors in the system. 42 Compensated Example T=0.1; num=[1 0.970]; den=[1 -1.91 0.91]; Gz=tf(num,den,T); Dz=tf([1 -0.91],[1 0.4],T); DzGz=zpk(Dz*Gz) rlocus(Dz*Gz) axis([-1.5 1.5 -1.5 1.5]) DzGz = (z+0.97) (z-0.91) ---------------------(z-1) (z-0.91) (z+0.4) 43 Compensated Example 44 7.7 The Nyquist Criterion Let GH (z) be the open-loop function, Then the characteristic equation for the closed-loop system is 1+ GH (z) = 0 (7-21) Consider GH (z) = -1 We can examine the behavior of GH (z) in relation to the -1 point as we traverse the frequency response contour in the z-Plane. Let Zi = number of zeroes of characteristic equation inside the unit circle Zo = number of zeroes of characteristic equation outside the unit circle Pi = number of poles of the open-loop function that are inside the unit circle. Fig. 7-15 Po = number of poles of the open-loop function that are outside the unit circle. 57 The Nyquist Criterion We can examine the behavior of GH (z) in relation to the -1 point as we traverse the frequency response contour in the z-Plane. The number of encirclements of the –1 point is (7-24) Fig. 7-15 In general, the numerator and denominator of (7-21) are the same, say n (7-25) & (7-26) Zi = number of zeroes of characteristic equation inside the unit circle Zo = number of zeroes of characteristic equation outside the unit circle Pi = number of poles of the openloop function that are inside the unit circle. Po = number of poles of the openloop function that are outside the unit circle. So Dropping the subscripts (7-27) N = number of encirclement of the –1 point. Z = number of zeroes of characteristic equation outside the unit circle P = number of poles of the open-loop function (or characteristic equation) that are outside the unit circle. 58 Nyquist Diagrams (Stability) gain margin = 1 a phase margin = fm Fig. 7-18 59 Nyquist Diagrams (Example 7.10) 0.368z+ 0.264 Open-loop function: G(z) = 2 z -1.368z+ 0.368 T =1 Frequency contour is plotted in G(z) plane. Closed-loop system is unstable if -1 point in G(z) plane is encircled. Fig. 7-17 60 Nyquist Diagrams (dnyquist) dnyquist: calculates the frequency response of discrete-time LTI systems. %Example 7.10 num=[0 .368 .264]; den=[1 -1.368 .368]; T=1; dnyquist(num,den,T) axis([-1.2 0.2 -0.7 0.7]) 61 7.8 The Bode Diagram - Discrete System Example A review of Bode diagrams for continuous systems is available at the end of this slide set. These continuous-time procedures can be used on sampled-time open-loop transfer functions that have been mapped to the w-plane (which mimics the frequency behavior of the s-plane if the sampling period is chosen appropriately). Example 7.12 Consider again the system of Example 7.3 Numerator break frequencies: 2.0, 12.14 Denominator break frequencies: 0, 0.924 62 Gm = 7.58 dB Pm = 30.4 deg Example 7.12 Straightline Bode diagram Fig. 7-21 63 Example 7.12 (via MATLAB & G(w)) >>Gw = -0.0381*tf([1 -2],[1 0])*tf([1 12.14],[1 0.924]); margin(Gw), grid % margin generates a labeled Bode diag. Gm = 7.58 dB Pm = 30.4 deg 64 Example 7.12 (via MATLAB & G(s) to G(z)) >>Gs=tf([1],[1 1 0]); T=1; Gz=c2d(Gs,T); margin(Gz), grid Gm = 7.58 dB Pm = 30.4 deg 65 Gain-Phase Plots Another way of displaying the relationship between the gain and phase of a frequency response is to plot them on different axes of the same diagram. For Example 7.12, such a gain-phase plot is shown below. This plot is called a Nichols chart and will be examined in more detail in Section 7.10. Origin is 0 dB and –180º Fig. 7-22 66 Example: Bode diagram, root locus, Nyquist plot Compare the gain margins. Gp (s) = 10 10 = s(s2 + 4s+ 8) s3 + 4s2 + 8s >> T=0.1; num=[10]; den=[1 4 8 0]; Gp=tf(num,den); Gz=c2d(Gp,T); Gz=zpk(Gz) [numz,denz]=tfdata(Gz,'v'); figure(1),margin(Gz) % Bode with margins figure(2),rlocus(Gz),grid figure(3),nyquist(numz,denz,T),grid Gz = 0.0015067 (z+3.373) (z+0.2427) -----------------------------(z-1) (z^2 - 1.605z + 0.6703) 67 ) 68 Gz = 0.0015067 (z+3.373) (z+0.2427) ----------------------------(z-1) (z^2 - 1.605z + 0.6703) 69 Gz = 0.0015067 (z+3.373) (z+0.2427) ----------------------------(z-1) (z^2 - 1.605z + 0.6703) 70 Gz = 0.0015067 (z+3.373) (z+0.2427) ----------------------------(z-1) (z^2 - 1.605z + 0.6703) K = 1.0 71 Gz = 0.0015067 (z+3.373) (z+0.2427) ----------------------------(z-1) (z^2 - 1.605z + 0.6703) K = 2.7 >> Gm=2.7; Gmdb=20*log10(Gm) Gmdb = 8.6273 72 Gz = 0.0015067 (z+3.373) (z+0.2427) ----------------------------(z-1) (z^2 - 1.605z + 0.6703) >> A0=-0.365; Gmdb=20*log10(-1/A0) Gmdb = 8.7541 73 7.9 Interpretation of the Frequency Response Consider a discrete-time system described by C(z) whose input is a sampled sine wave E(z) (7-35) Cg(z) is the response originating from the poles of G(z), which tends to zero over time for stable systems. Let Css(z) be the steady-state response. (7-37) (7-36) (7-38) 74 Interpretation of the Frequency Response The steady-state time response css(t) becomes (7-39) If the input is a sinusoid of frequency ω, the steady-state response is also sinusoidal at the same frequency. The amplitude of the response is equal to the amplitude of the input multiplied by the magnitude of and the phase of the response is equal to the phase of the input plus the angle of . 75 7.10 Closed-Loop Frequency Response Graphically relating a system’s closed-loop frequency response to its open-loop frequency response. Fig. 7-23 (7-40) At frequency ω1 (7-41) 76 Constant M Circles The locus of points in the plane for which the magnitude of the closed-loop frequency response, M, is a constant is called a constant magnitude locus, or a constant M circle. For (7-42) (7-44) This relationship is the equation of a circle of radius with center at (7-43) 77 For a straight line Constant M Circles Fig. 7-24 78 Constant N Circles The locus of points of constant phase are also circles. This relationship is the equation of a circle of radius with center at (7-45) 79 NOTE: Accuracy is limited near the –1 point in plots of constant M and constant N circles. The Nichols chart is more accurate Constant N Circles Fig. 7-25 80 Nichols Chart These are gain-phase plots the magnitude of on the vertical axis and the angle of on the horizontal. Contours of constant magnitude and constant phase are also included. Here is the plot format. 0 –180 81 Nichols Chart Constant M and N circles are shown as contours. 0 –180 82 Example 7.13 Compute the Nichols chart for >> Gp = tf([1],[1 1 0]); T = 1; Gz=c2d(Gp,T); nichols(Gz), grid 83 Fig. 7-27 84 Example 7.13 (continued) Using Bode plots and step responses, compare the performance of this system with T = 1 and T = 0.1 Verify the gain margin via Nyquist plots. >> num = [1]; den = [1 1 0]; Gp = tf(num,den); T1 = 1; Gz1=c2d(Gp,T1); [Gm1,Pm1] = margin(Gz1); Cz1 = feedback(Gz1,1); T2 = 0.1; Gz2 = c2d(Gp,T2); [Gm2,Pm2] = margin(Gz2); Cz2 = feedback(Gz2,1); % Plot the closed-loop frequency response figure(1), bode(Cz1,'--k',Cz2,'-k'), axis([0.25 2.5 -225 10]), grid % Plot the step response figure(2), step(Cz1,'--k',Cz2,'-k'), axis([0 20 0 1.5]), grid % List the step response and margin parameters StepT1=stepinfo(Cz1),Pm1,Gm1=20*log10(Gm1) % Convert to dB StepT2=stepinfo(Cz2),Pm2,Gm2=20*log10(Gm2) % Convert to dB % Compute Nyquist plots figure(3),nyquist(Gz1,Gz2), axis([-0.5 0.05 -0.5 0.5]), grid figure(4),nyquist(Gz1,Gz2), axis([-0.08 0.01 -0.08 0.08]), grid 85 Example 7.13 (Bode plots) T = 1: peak mag. is 6.06 dB at 0.87 rad/s T = 0.1: peak mag. is 1.56 dB at 0.74 rad/s Fig. 7-28 86 Example 7.13 (Step responses) T = 1: peak mag. is 1.4 at 3 s T = 0.1: peak mag. is 1.18 at 3.5 s Fig. 7-29 87 Example 7.13 (step response parameters and gain margins) T = 0.1 T=1 StepT2 = StepT1 = RiseTime: SettlingTime: SettlingMin: SettlingMax: Overshoot: Undershoot: Peak: PeakTime: RiseTime: SettlingTime: SettlingMin: SettlingMax: Overshoot: Undershoot: Peak: PeakTime: 1 16 0.8015 1.3996 39.9576 0 1.3996 3 Pm1 = 30.3747 Pm2 = 49.5807 Gm1 = 7.5779 Gm2 = 26.1666 88 1.6000 8.3000 0.9090 1.1837 18.3712 0 1.1837 3.6000 Example 7.13 (Nyquist plots) T = 1, a = –0.419 @ 1.32 rad/s Gm = 20*log10(1/0.419) = 7.56 dB ✔︎ * Fig. 7-30(a) 89 Example 7.13 (Nyquist plots) T = 0.1, a = –0.0488 Gm = 20*log10(1/0.0488) = 26.2 dB ✔︎ * Fig. 7-30(b) 90 Homework Problems: 7.2-6 7.5-1 7.6-4 a,c (using MATLAB) 7.8-3 7.9-1 91