Linear System Theory and Design Chapter 7

Chapter 7 hat hat f (s) = (s - 1)/(s ^ 2 + 2s ^ 2 - s - 2) hat x = [[- 2, tau, 2], [i, o, 0], [0, i, 0]] * x + [[i], [0], [0]] * u y =[0|-1]x Q = [[0, i, - i], [i, - i, 0], [- 3, i, 2]] der 0-1-3-20 P(0) < 3 Not observable 7.2 tilde x = [[- 2, i, 0], [i, 0, i], [a, 0, 0]] * x + [[0], [i], [- i]] * u y = [100]x \mathcal{C} = [[0, 1, - 3], [1, - 1, 1], [- 1, 0, 2]] p(c) < 3 Not controllable 7.3 We me (6.40) to add an uncontrollabl part to the realization in Problem 7.1 dot x = [[- 2, i, 2, q_{1}], [i, 0, 0, - epsilon_{2}], [0, 0, - epsilon_{2}, 0], [0, 0, - epsilon_{1} * q_{2}, 0], [0, 0, - epsilon_{1} * q_{2}, 0], [0, 0, - epsilon_{2}, 0], [0, 0, - epsilon_{4}, 0]] * x + [[epsilon], [0], [0], [- epsilon_{1}], [0]] * 4 y=[ matrix 0&1-1& c_{y}*Ix where a * and c_{e} are arbitrary. This is an uncontrollable and unobservable realization of (s - 1) / (s ^ 3 + 2s ^ 1 - j - 2) hat g(1) = (s - 1)/((s ^ 2 - 1)(s + 2)) = 1/((s + 1)(s + 1)) = 1/(s ^ 1 + 3s + 2) x' = [[- 3, - 2], [1, 0]] * x + [[1], [0]] * 4 y = [[0, 1]] * x The in a controllatile and observable realization of lus. 1.4 First we form the Sylvester resultant -2-1:00:00 12-100 2012-1 1020 001020 0000110 We use MATLAB to rearch emearly independent columns from left to right we type d =[-2-1 2 +]j pi < [- 1] ]; s> [d00;no0j0d0;onojood; 0001; [r]=gr(s) F = [[d, u, x, z, x, 0, 0], [0, u, x, x, x, 0, 0], [0, u, y, x, x, x, 0], [0, y, y, x, z, 0, 0], [0, y, y, x, z, 0, 0], [0, 0, y, y, y, y, z], [0, 0, 0, y, 0, 0, 0]] where 2, d, n devole nonjere entries d alse denotes D-colimm In denote N-column, There are two emearly independent Nicolumns, thus the degree of grs in Prob 7.1 to 2. 7.5 3S - 1 4S^ * -1 = N * +N 1 S D 8 +D 1 S (-N 0 -N 1 3)^ prime -1+43^ 3 )+ (D_{0} + D_{1}*s)(- 1 + 2s) = 0 [[- i, - i, 0, 0, 0, 0, 0, 0], [0, 2, i - i, - i, 0, 0, 0, 0], [4, 0, i, 0, 2, i, 1, 2], [0, 0, i, 0, 2, 0, 0, 0]] * [[- N_{g}, 0, 0, 0, 0], [D_{g}, 0, 0, 0, 0], [- N_{i}, 0, 0, 0, 0], [0, 0, i, 4, 0]] = 0 although D, can be any nonzero constant. it is convenient to regure D_{1} = 1 We Type d=c-10 435pi =[-1 2 0 37 s =[d emptyset; n 0; 0d;0n1'; }= mell (5) which yields 3: [-0,4082 0.40820081651 We normalije che last entry to by typing 3/3 * (4) which yields C-0.5 0.50 17. Thus we have (2s - 1)/(4t ^ 2 - 1) = 0.5/(s + 0.5) = 1/(1.5 + i) -31- 7.6 we form, from [[1, 0, 1, 0, 0], [2, 1, 1, 0, 0], [2, 1, 1, 0, 0], [0, 2, 1, 1, 0], [0, 0, 0, 2, 1], [0, 0, 0, 2, 0]] (0.5 ^ 2 * ps + 2)/(s ^ 2 + 2s + phi) where the coefficients are arranged in descending power of 5. If we search its limearly independent columns in order from left to right, then the recond D-column is linearly clependent on ito LHS columns and there are two linearly independent N-colurms. Thus in this arrangement, it is not true that all D-columns are emearly inde pendent of their LHS columns and zhar the degree of fiss equals the noorbers of linearly independent N-columas. 7.7 (N(s))/(D(s)) = (beta_{1}*x + beta_{2})/(s ^ 2 + x_{1}*s + x_{2}) Nis) and (3), coprime if and only if mathcal S =[ matrix alpha 2 & beta 2 ; theta&0\\ alpha_{1}*beta_{1} ; alpha 2 & beta 2 \\ +&0&i& alpha 1 & beta 2 \\ 0&0&i&1&0 matrix ] is roningular det Idet a, A. 0 The walization dot x = [[- alpha_{1}, - alpha_{2}], [1, 0]] * [[4], [0]] * omega H= [beta_{1}*beta_{2}] : is cheewable of and only of mathcal O =[ matrix beta 1 & beta 2 \\ beta_{2} - beta , alpha 1 &- beta 1 alpha 2 matrix ] is ronsmgular r*beta = - x_{1} * beta_{1} ^ 2 - beta_{2} ^ 2 + alpha_{1}*beta_{2} = alpha*r * s' Thus the two conditions ons the same 7.5 Consider (N(y))/(D(s)) = (beta_{1} * s ^ 2 + beta_{2}*s + beta_{3})/(s ^ 1 + alpha_{1} * s ^ 2 + alpha_{2}*s + x_{3}) and its controllable form realijation hat z = [[- r_{i}, - r_{a}, - r_{j}], [i, 0, 0], [0, i, 0]] * x + [[l], [0], [0]] * u y = \mathfrak{r}*A_{1} P. Pa Its observability matrix is O=[ matrix beta 1 & beta n & alpha i \\ A z - alpha i beta i &A j - alpha i beta i &A j - alpha i beta i &A j - alpha i ( beta i - alpha i beta i &- alpha i ( beta_{i} - alpha_{i}*beta_{i} - alpha_{i}*beta_{i} - alpha_{j}*beta_{i} - alpha j ( beta i - alpha j beta) matrix ] The recultant of NIS) and D)1( مند [ matrix a_{1}*beta_{2} ;&0&0\\ a 1 & beta 2 & beta 1 & beta 2 & beta 1 & beta 2 \\ a 1 & beta 1 & beta 2 & beta 2 \\ a 1 & beta 1 & beta 2 & beta 2 \\ a 1 & beta 1 & beta 2 & beta 2 & beta 2 \\ a 1 & beta 1 & beta 2 & beta 2 & beta 1 \\ a 1 & beta 1 & beta 2 & beta 2 & beta 1 \\ matrix ] First we me the 1 in the last now to slimnate all entries above it next we we the 1 in the fifed row to eliminats all entries above it to yuld 0:0 00 0 RA Finally we use the in the first column to eliminate all entries above it to Yuld 10 0 000 σ This can be warranged as ist +3rd clumme -32- Thus the resultant is nonsingular if aud only if O is nonsingular or, exmivalently, the realization is obeer-vable if and only if D(s) cael M(3) are coprime This is Theorem 7.1. hat g(s) = i/((x + i) ^ n) = 0s ^ - 1 + x ^ - 1 - 2z ^ - 1 * z ^ (- varphi) -* s ^ (- pi) * f * s ^ - 4 =*** T(z, z) = [[0, f], [I, - 2]] P(Gamma(x, z)) = 2 7(3, 3) = [[8, 1, - 2], [1, - 2, 1], [- 3, 3, - 6]] Ird columy =-(1st column) -2(28d colemen) TINA P( T(2, 3) = 2 Proceeding forward, we can establish Theorem 7.7 for 1/(5+1) 7.10 21, h(1)=0, 12)=1 Thus, from 17.56), we have x^ prime = [[0, t], [- 2, - i]] ne+[ matrix 0\\ i matrix ]4 y = r This is a companion form realijation. T = [[0, 1], [1, - 2]] tilde T / ([[1, - 2], [- 2, 3]]) We war MATLAB. Typing 20131-21;t=[1-2-231; Ch, S. 21 send (t); 51=sgat (s); w = f + sic =s1+t^ prime j ατί (0) τε (c) b = [c(1, 1); c(21)] C=[ O(l, 1) * O ((,2)] guld chi = [[- 1.7071, 0.7071], [- 0.7871, - 0.2929]] * x + [[0.59446], [- 0.5946]] * 4 y = [- w * 0.5946 - w * 0.574f] * x This so balanced realization. y =C2 form of (640) and can lis y = 2x hat y(4) = 2/(s - 4) This is in thu reduced to dot x = 2pi + 4 Its transfer function in hat x = [[2, 0], [- i, - i]] * x + [[i], [2]] * u This is in the form of (delta, psi*psi) and can be reduced to x' = 2x + u y = 2x Its transfer function is also oint (s) = 2/(s - 2) 2s+1, s ^ 2 - s - 2 = (2(s + 1))/((s - 2)(s + 1)) = 1/(s - 2) The original two state equations are not missimal realijations; they are not algebraically equivalent because they have different sigen-values. The first equation has egenvalue ( 2 ,1 rfloor. the second 12,1 7.13 hat G 1 (s)=[ matrix 1 2 & I T 2 J+1 J+1 \\ 1/(J + 1) & J I+1 matrix ] G_{1}(1) = 1/(s + 1) - 1/(t + 1) = 0 Delta_{1}(s) <= s(5 + 1)(s + 3) - 4g = 3 G_{2}(s) = [[1/((s + 1) ^ k), 1/((s + 1)(s + 2))], [1/(s + 2), 1/((s*pi)(s + z))]] der G_{2}(s) = 1/((s + i) ^ 3 * (s + 2)) - 1/((s + i) * (s + 3) ^ 2) = s+2-(s^ 2 +2 s+1) (s+1)^ 3 ( 5+2)^ 2 = (- x ^ 2 - s + 1)/((1 + 1) ^ 2 * (s + 2) ^ 2) Delta_{3}(x) = (x + 1) ^ 2 * (x + 2) ^ 2 dy = 5