System Zeros and Transfer Function Matrices

Topic #8 16.30/31 Feedback Control Systems State-Space Systems • System Zeros • Transfer Function Matrices for MIMO systems 16.30/31 8–1 Zeros in State Space Models • Roots of transfer function numerator called the system zeros. • Need to develop a similar way of defining/computing them using a state space model. • Zero: generalized frequency s0 for which the system can have a non-zero input u(t) = u0es0t, but exactly zero output y(t) ≡ 0 ∀t • Note that there is a specific initial condition associated with this response x0, so the state response is of the form x(t) = x0es0t u(t) = u0es0t ⇒ x(t) = x0es0t ⇒ y(t) ≡ 0 • Given ẋ = Ax + Bu, substitute the above to get: x0s0es0t = Ax0es0t + Bu0es0t ⇒ � � � � x0 =0 s0I − A −B u0 • Also have that y = Cx + Du = 0 which gives: Cx0es0t + Du0es0t = 0 → � � � � x0 =0 C D u0 • So we must find the s0 that solves: � �� � s0I − A −B x0 =0 C D u0 • Is a generalized eigenvalue problem that can be solved in MATLAB using eig.m or tzero.m 1 1 MATLAB is a trademark of the Mathworks Inc. 16.30/31 8–2 • There is a zero at the frequency s0 if there exists a non-trivial solution of � � s0I − A −B det =0 C D • Compare with equation on page 6–?? � • Key Point: Zeros have both direction x0 u0 � and frequency s0 • Just as we would associate a direction (eigenvector) with each pole (frequency λi) s+2 • Example: G(s) = s2+7s+12 � � � � � � −7 −12 1 A= B= C= 1 2 D=0 0 1 0 � s I − A −B det 0 C D � ⎡ ⎤ s0 + 7 12 −1 = det ⎣ −1 s0 0 ⎦ 1 2 0 = (s0 + 7)(0) + 1(2) + 1(s0) = s0 + 2 = 0 so there is clearly a zero at s0 = −2, as we expected. For the directions, solve: ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎡ ⎤ s0 + 7 12 −1 x01 5 12 −1 x01 ⎣ −1 s0 0 ⎦ ⎣ x02 ⎦ = ⎣ −1 −2 0 ⎦ ⎣ x02 ⎦ = 0? 1 2 0 s =−2 u0 1 2 0 u0 0 gives x01 = −2x02 and u0 = 2x02 so that with x02 = 1 � � −2 x0 = and u = 2e−2t 1 16.30/31 8–3 • Further observations: apply the specified control input in the fre­ quency domain, so that Y1(s) = G(s)U (s) where u = 2e−2t, so that U (s) = 2 s+2 s+2 2 2 · = s2 + 7s + 12 s + 2 s2 + 7s + 12 Say that s = −2 is a blocking zero or a transmission zero. Y1(s) = • The response Y1(s) is clearly non-zero, but it does not contain a component at the input frequency s = −2. • That input has been “blocked”. • Note that the output response left in Y1(s) is of a very special form – it corresponds to the (negative of the) response you would see from � �T the system with u(t) = 0 and x0 = −2 1 Y2(s) = C(sI − A)−1x0 � � � � � � s + 7 12 −1 −2 = 1 −2 −1 s 1 � �� � � � s −12 1 −2 = 1 −2 1 s+7 1 s2 + 7s + 12 −2 = 2 s + 7s + 12 • So then the total output is Y (s) = Y1(s)+Y2(s) showing that Y (s) = 0 → y(t) = 0, as expected. 16.30/31 8–4 Simpler Test • Simpler test using transfer function matrix: • If z is a zero with (right) direction [ζ T , ũT ]T , then � �� � zI − A − B ζ =0 C D ũ • If z not an eigenvalue of A, then ζ = (zI − A)−1Bũ, which gives � � C(zI − A)−1B + D ũ = G(z)ũ = 0 • Which implies that G(s) loses rank at s = z • If G(s) is square, can find the zero frequencies by solving: det G(s) = 0 • If any of the resulting roots are also eigenvalues of A, need to re-check the generalized eigenvalue matrix condition. • Need to be very careful when we find MIMO zeros that have the same frequency as the poles of the system, because it is not obvious that a pole/zero cancelation will occur (for MIMO systems). • The zeros have a directionality associated with them, and that must “agree” as well, or else you do not get cancelation • More on this topic later when we talk about controllability and observability 16.30/31 8–5 Transfer Function Matrix • Note that the transfer function matrix (TFM) notion is a MIMO generalization of the SISO transfer function • It is a matrix of transfer functions ⎡ ⎤ g11(s) · · · g1m(s) ... ⎦ G(s) = ⎣ gp1(s) · · · gpm(s) • gij (s) relates input of actuator j to output of sensor i. • It is relatively easy to go from a state-space model to a TFM, but not obvious how to go back the other way. • Simplest approach is to develop a state space model for each element of gij (s) in the form Aij , Bij , Cij , Dij , and then assemble (if TFM is p × m) ⎡ ⎢ ⎢ ⎢ ⎢ A=⎢ ⎢ ⎢ ⎣ A11 ⎤ ... A1m A21 ... Apm ⎡ ⎢ ⎢ C=⎢ ⎣ C11 · · · C1m ⎡ ⎤ B11 ⎥ ⎢ ... ⎥ ⎢ ⎥ ⎢ B1m ⎥ ⎢ ⎥B = ⎢ ⎥ ⎢ B21 ⎥ ⎢ ... ⎦ ⎣ Bpm ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ C21 . . . C2m ... Cp1 · · · Cpm ⎥ ⎥ ⎥ D = [Dij ] ⎦ 16.30/31 8–6 • One issue is how many poles are needed - this realization might be inefficient (larger than necessary). • Related to McMillan degree, which for a proper system is the degree of the characteristic polynomial obtained as the least com­ mon denominator of all minors of G(s).2 • Subtle point: consider a m × m matrix A, then the standard minors formed by deleting 1 row and column and taking the de­ terminant of the resulting matrix are called the m − 1th order minors of A. • To consider all minors of A, must consider all possible orders, i.e. by selecting j ≤ m subsets of the rows and columns and taking the resulting determinant. • Given an n × m matrix A with entries aij , a minor of A is the determinant of a smaller matrix formed from its entries by selecting only some of the rows and columns. • Let K = { k1 k2 . . . kp } and L = { l1 l2 . . . lp } be subsets of {1, 2, . . . , n} and {1, 2, . . . , m}, respectively. • Indices are chosen so k1 < k2 · · · < kp and l1 < l2 · · · < lp. • pth order minor defined by K � �a � k1l1 � �a [A]K,L = � k..2l1 � . � � akpl1 and L is the determinant � ak1l2 . . . ak1lp �� � ak2l2 . . . ak2lp � � ... � � akpl2 . . . akplp � 3 • If p = m = n then the minor is simply the determinant of the matrix. • In a nutshell what this means is that a 2 × 2 matrix has 4 order-1 minors and 1 order-2 minor to consider. 2 Lowest 3 See order polynomial that can be divided cleanly by all denominators of the minors of G(s). here for details 16.30/31 8–7 Gilbert’s Realization • One approach: rewrite the TFM as G(s) = H(s) d(s) where d(s) is the least common multiple of the denominators of the entries of G(s). • Note difference from the discussion about the McMillan degree. • d(s) looks like a characteristic equation for this system, but it is not ⇒ it does not accurately reflect number of poles needed. • For proper systems for which d(s) has distinct roots, can use Gilbert’s realization. • Apply a partial fraction expansion to each of the elements of TFM G(s) and collect residues for each distinct pole4. Nm � Ri G(s) = s − pi i where Ri = lim (s − pi)G(s) s→pi • Then sum of the ranks of matrices Ri gives the McMillan degree 4 Generalizations of this Gilbert’s realization approach exist if the gij have repeated roots. 16.30/31 8–8 • Can develop a state space realization by analyzing each element of the partial fraction expansion • Set Ri = CiBi, and find appropriate Bi and Ci • Form Ai by placing the poles on the diagonal as many times as needed (determined by rank of Ri) • Form state space model: ⎡ A1 ... ẋ = ⎣ ⎤ ⎡ ⎤ B1 ⎦ x + ⎣ ... ⎦ u ANm BNm � � y = C1 · · · CNm x 16.30/31 8–9 Zero Example 1 ⎡ 1 1 ⎢ s+2 s+2 ⎢ • TFM G(s) = ⎢ ⎢ ⎣ 1 s−2 s − 2 (s + 1)(s + 2) ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ • To compute the McMillan degree for this system, form all minors (4 order-1 and 1 order-2): � � 1 1 1 s−2 2 − 7s , , , , s + 2 s + 2 s − 2 (s + 1)(s + 2) (s − 2)(s + 1)(s + 2)2 • To find LCD (least common multiple of denominators), pull out small­ est polynomial that leaves all terms with no denominator: 1 {(s − 2)(s + 1)(s + 2), (s − 2)(s + 1)(s + 2), (s − 2)(s + 1)(s + 2)2 � 2 2 (s + 1)(s + 2 ), (s − 2) (s + 2), 2 − 7s • So we expect a fourth order system with poles at s = 2, s = −2 (two), and s = −1 • Compare with the Gilbert realization, find d(s): ⎡ ⎤ (s + 1)(s − 2) (s + 1)(s − 2) 1 ⎢ ⎥ G(s) = ⎣ ⎦ (s + 1)(s + 2)(s − 2) (s + 1)(s + 2) (s − 2)2 � � � � � � 1 1 1 0 0 0 0 1 1 = + + s + 1 0 −3 (s + 2) 0 4 s−2 1 0 • Note that the rank of the last 2×2 matrix is 2 • So the system order is 4 - we need to have two poles s = −2. 16.30/31 8–10 • So the system model for the example is � A1 = [−1] B1 = 0 � � � −2 0 1 A2 = B2 = 0 −2 0 � A3 = [2] B3 = 1 −3 � 1 4 0 � � � C1 = 0 1 � C2 = I2 � � 0 C3 = 1 • Note, realization model on 8–5 would be 5th order, not 4th. Code: MIMO Models 1 2 3 4 % % basic MIMO TFM to SS % G=tf({1 1;1 [1 −2]},{[1 2] [1 2];[1 −2] [1 3 2]}); 5 6 7 8 9 % find residue matrices of the 3 poles R1=tf([1 1],1)*G;R1=minreal(R1);R1=evalfr(R1,−1) R2=tf([1 2],1)*G;R2=minreal(R2);R2=evalfr(R2,−2) R3=tf([1 −2],1)*G;R3=minreal(R3);R3=evalfr(R3,2) 10 11 12 13 14 % form SS model for 3 poles using the residue matrices A1=[−1];B1=R1(2,:);C1=[0 1]'; A2=[−2 0;0 −2];B2=R2;C2=eye(2); A3=[2];B3=R3(2,:);C3=[0 1]'; 15 16 17 18 19 % combine submodels A=zeros(4);A(1:1,1:1)=A1;A(2:3,2:3)=A2;A(4,4)=A3; B=[B1;B2;B3]; C=[C1 C2 C3]; 20 21 22 syms s Gn=simple(C*inv(s*eye(4)−A)*B); 23 24 25 26 27 28 % alternative is to make a SS model of each g {ij} A11=−2;B11=1;C11=1; A12=−2;B12=1;C12=1; A21=2;B21=1;C21=1; A22=[−3 −2;1 0];B22=[2 0]';C22=[0.5 −1]; 29 30 31 32 33 34 % and then combine AA=zeros(5);AA(1,1)=A11;AA(2,2)=A12;AA(3,3)=A21;AA(4:5,4:5)=A22; BB=[B11 B11*0;B12*0 B12;B21 B21*0;B22*0 B22]; CC=[C11 C12 zeros(1,3);zeros(1,2) C21 C22]; GGn=simple(CC*inv(s*eye(5)−AA)*BB); 35 36 Gn,GGn 16.30/31 8–11 Zero Example 2 ⎡ 1 1 ⎢ s + 1 (s + 1)2 ⎢ • TFM G(s) = ⎢ ⎢ ⎣ 1 1 (s + 1)3 (s + 1)4 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ • McMillan Degree: find all minors of G(s) 1 , s+1 1 , (s + 1)2 1 , (s + 1)3 1 , (s + 1)4 0 • To find LCD (least common multiple of denominators), pull out small­ est polynomial that leaves all terms with no denominator: � � 1 3 2 (s + 1) , (s + 1) , (s + 1), 1 (s + 1)4 • So the LCD is (s + 1)4 and the McMillan degree is 4 – we expect the minimal state space model to have 4 poles at s = −1. • Gilbert approach as given cannot be applied directly since d(s) = 1 has repeated roots (s+1)4 • See Matlab code for model development ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ A=⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ −1.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 −3.00 2.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 � C= 1.00 0.00 0.00 −1.50 0.00 1.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 −0.50 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 1.00 0.00 0.00 0.00 0.00 −2.00 1.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 −1.00 0.00 0.00 0.00 0.00 0.00 1.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 −4.00 4.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 −1.50 0.00 1.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 −1.00 0.00 0.00 0.50 0.00 1.00 0.00 0.00 0.00 0.00 0.00 0.00 −0.50 0.00 0.00 0.00 � ⎤ ⎡ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ B=⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ � D= 0.00 0.00 1.00 0.50 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 1.00 0.00 0.50 0.00 0.00 0.00 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ � • Note that λi(A) = −1 – there are 10 poles there. So this is clearly not minimal since the order is 10, not the 4 we expected. 16.30/31 8–12 • Matlab command minreal can be used to convert to a minimal realization. � −0.40 −0.16 −1.00 0.01 � � 0.23 −0.02 � 0.32 −1.49 −0.06 1.07 0.36 A= B = −0.97 0.50 −1.06 −1.17 −0.39 −0.05 −0.31 −0.07 � C= 0.18 −1.11 0.16 −1.01 −0.29 0.02 0.35 0.43 −0.94 −0.63 −0.28 � � D= 0.00 0.00 0.01 −0.75 0.00 0.00 � • New model has 6 states removed - so the minimal degree is 4 as expected. Code: Zeros (zero example1.m) 1 2 3 4 5 6 7 8 9 10 11 12 G1=ss(tf({1 1;1 1},{[1 1] conv([1 1],[1 1]);conv([1 1],conv([1 1],[1 1])) ... conv([1 1],conv([1 1],conv([1 1],[1 1])))})); % [a,b,c,d]=ssdata(G1); latex(a,'%.2f','nomath') % latex(b,'%.2f','nomath') % latex(c,'%.2f','nomath') % latex(d,'%.2f','nomath') % G2=minreal(G1);[a2,b2,c2,d2]=ssdata(G2); latex(a2,'%.2f','nomath') % latex(b2,'%.2f','nomath') % latex(c2,'%.2f','nomath') % latex(d2,'%.2f','nomath') % 16.30/31 8–13 Zero Example 3 ⎡ 2s + 3 3s + 5 ⎢ s2 + 3s + 2 s2 + 3s + 2 ⎢ • TFM G(s) = ⎢ ⎢ ⎣ −1 0 (s + 1) ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ • McMillan Degree: find all minors of G(s) 2s + 3 , s2 + 3s + 2 3s + 5 , s2 + 3s + 2 −1 , (s + 1) −(3s + 5) (s + 1)(s2 + 3s + 2) • To find LCD, pull out smallest polynomial that leaves all terms with no denominator: (s2 � 1 (2s + 3)(s + 1), + 3s + 2)(s + 1) (3s + 5)(s + 1), −(s2 + 3s + 2), � −(3s + 5) • So the LCD is (s2 + 3s + 2)(s + 1) = (s + 1)2(s + 2) • The McMillan degree is 3 – we expect the minimal state space model to have 3 poles. • For Gilbert approach, we rewrite � � 2s + 3 3s + 5 −(s + 2) 0 R2 R1 + G(s) = = s+1 s+2 (s + 1)(s + 2) where � R1 = lim (s + 1)G(s) = lim s→−1 s→−1 � R2 = lim (s + 2)G(s) = lim s→−2 s→−2 2s+3 s+2 3s+5 s+2 −1 0 2s+3 s+1 − s+2 s+1 3s+5 s+1 0 � � = � = 1 2 � −1 0 � � 1 1 0 0 which also indicates that we will have a third order system with 2 poles at s = −1 and 1 at s = −2. 16.30/31 8–14 • For the state space model, note that � �� � 1 2 1 0 R1 = = C1B1 −1 0 0 1 � � � 1 � R2 = 1 1 = C2B2 0 giving ⎡ −1 A = ⎣ 0 0 � 1 C = −1 ⎤ 0 0 −1 0 ⎦ 0 −2 � 2 1 0 0 ⎡ ⎤ 1 0 B=⎣0 1⎦ 1 1 • From Matlab you get: ⎡ ⎤ ⎡ ⎤ −1.00 0.00 0.00 0.56 1.12 A = ⎣ 0.00 −2.00 0.00 ⎦ B = ⎣ 0.35 0.35 ⎦ 0.00 0.00 −1.00 0.50 0.00 � � � � 1.79 2.83 0.00 0.00 0.00 C = D= 0.00 0.00 −2.00 0.00 0.00 Code: Zeros (zero example2.m) 1 2 3 4 5 6 7 G1=ss(tf({[2 3] [3 5];−1 G1=canon(G1,'modal') [a,b,c,d]=ssdata(G1); latex(a,'%.2f','nomath') latex(b,'%.2f','nomath') latex(c,'%.2f','nomath') latex(d,'%.2f','nomath') 0},{[1 3 2] [1 3 2];[1 1] 1})); % % % % % 16.30/31 8–15 Summary of Zeros and TFMs • Great feature of solving for zeros using the generalized eigenvalue matrix condition is that it can be used to find MIMO zeros of a system with multiple inputs/outputs. � � s I −A −B det 0 =0 C D • Note: we have to be careful how to analyze these TFM’s. • Just looking at individual transfer functions is not useful. • Need to look at system as a whole – use the singular values of G(s) • Will see later the conditions to determine if the order of a state space model is minimal.