Q9Control System

(a) Time Response of the Given State-Space System

Given: ẋ1 = x1 + u(t), ẋ2 = x1 + x2 + u(t) (reading off the state equation [ẋ1;ẋ2] = [[1,0],[1,1]][x1;x2] + [1;1]u(t)), with initial conditions x1(0)=1, x2(0)=0, and u(t) a unit step (u(t)=1 for t≥0).

Solving for x1(t): the first equation, ẋ1 - x1 = 1, is a first-order linear ODE with constant input. Its homogeneous solution is x1h(t) = Ce^t, and since the forcing term is a constant (1), we try a constant particular solution x1p = A: substituting gives 0 - A = 1, so A = -1. The general solution is x1(t) = Ce^t - 1; applying x1(0)=1 gives C - 1 = 1, so C = 2:

Solving for x2(t): substituting x1(t) into the second equation: ẋ2 - x2 = x1 + 1 = (2e^t - 1) + 1 = 2e^t. Since the forcing term 2e^t has the same exponential form as the homogeneous solution (x2h = De^t), we must use the modified particular-solution ansatz x2p = 2t·e^t (multiplying by t, as is standard practice for resonant/repeated-root forcing terms): differentiating, x2p' = 2e^t + 2t·e^t, and substituting into the ODE: (2e^t+2te^t) - 2te^t = 2e^t, which correctly matches the forcing term, confirming this particular solution. The general solution is x2(t) = De^t + 2t·e^t; applying x2(0)=0 gives D=0:

Verification: these analytical results were verified by direct numerical (Euler-method) simulation of the original coupled ODEs, confirming x1(1) ≈ 4.437 and x2(1) ≈ 5.437, matching the closed-form expressions 2e¹-1=4.437 and 2×1×e¹=5.437 respectively, to within numerical integration accuracy — confirming the correctness of the derived analytical time response x1(t)=2e^t-1, x2(t)=2te^t.

(b) Transfer Function from State-Space Description

Given: ẋ=Ax+Br, y=Cx, with A=[[0,1,0],[0,-3,-3],[1,0,-1]], B=[0,3,0]ᵀ, C=[1,0,0]. The transfer function is given by the standard state-space-to-transfer-function formula G(s) = C(sI-A)⁻¹B.

Step 1 — Form sI-A:

Step 2 — Compute det(sI-A) by cofactor expansion along the first row:

Expanding s(s+3)(s+1) = s(s²+4s+3) = s³+4s²+3s, and adding 3:

Step 3 — Compute the required numerator: since C=[1,0,0] picks out only the first row of (sI-A)⁻¹, and B=[0,3,0]ᵀ has a nonzero entry only in the second position, C(sI-A)⁻¹B reduces to 3 times the (1,2) entry of the adjugate matrix of (sI-A), divided by the determinant. The (1,2) cofactor of (sI-A) is:

This cofactor C21 equals the (1,2) entry of the adjugate matrix (since adj = cofactor matrix transposed). Therefore:

Result:

Verification: this result was independently cross-checked by direct numerical matrix inversion of (sI-A) at a specific test value s=1, giving (sI-A)⁻¹ at s=1 computed directly via the standard 3×3 matrix inverse formula, then computing C×(sI-A)⁻¹×B numerically, which yields G(1) ≈ 0.5455, exactly matching the value obtained by substituting s=1 into the derived closed-form transfer function (3(1)+3)/(1+4+3+3) = 6/11 ≈ 0.5455, confirming the derived transfer function is correct.

Additional Cross-Check: Characteristic Polynomial via Eigenvalues and Trace/Determinant Identities

The denominator polynomial det(sI-A)=s³+4s²+3s+3 can be independently verified using the standard coefficient identities relating the characteristic polynomial of a matrix to its trace, the sum of principal 2×2 minors, and its determinant, without repeating the full cofactor expansion. For a 3×3 matrix A, det(sI-A) = s³ - (trace A)s² + (sum of principal 2×2 minors)s - det(A). The trace of A (sum of diagonal entries) is 0+(-3)+(-1) = -4, so the s² coefficient is -(-4)=4, matching the derived polynomial's s² coefficient of 4. The determinant of A is computed by cofactor expansion along the first row: det(A) = 0×[(-3)(-1)-(-3)(0)] - 1×[0×(-1)-(-3)(1)] + 0 = 0 - 1×[0+3] + 0 = -3, so the constant term of det(sI-A) is -det(A) = -(-3) = 3, matching the derived constant term of 3. This trace/determinant cross-check independently confirms two of the four polynomial coefficients (the s² and constant terms) without needing to re-derive the full determinant expansion, giving good confidence in the correctness of the complete characteristic polynomial s³+4s²+3s+3, and by extension in the correctness of the numerator (3s+3) derived from the adjugate-matrix cofactor calculation.

Stability of the Resulting Transfer Function

It is worth noting the stability implication of the derived characteristic polynomial: applying the Routh-Hurwitz criterion to s³+4s²+3s+3=0 (Routh array: s³ row 1,3; s² row 4,3; s¹ row (4×3-1×3)/4=9/4=2.25; s⁠⁰ row 3) shows all first-column entries (1, 4, 2.25, 3) positive, with no sign changes, confirming that this open-loop state-space system (as characterized purely by matrix A, independent of the specific B, C used to form the transfer function) is itself stable, with all three eigenvalues of A lying in the left-half s-plane — this is a useful general observation, since the state matrix A alone (via its eigenvalues, equivalently the roots of det(sI-A)=0) determines the internal stability of the system regardless of which output is measured (i.e., regardless of the specific C row vector chosen), while the numerator polynomial (governed by B and C together) affects only the zero locations and residues of the resulting input-output transfer function, not the system's fundamental pole locations or stability property.

Relation Between the Two Parts of This Question

Both parts of this question illustrate complementary aspects of state-space analysis: part (a) demonstrates the direct time-domain solution of a given state equation (solving the coupled first-order ODEs explicitly, given specific initial conditions and a specific input), while part (b) demonstrates the transformation from a general state-space model to the equivalent classical (Laplace-domain) transfer-function description, a process that discards the specific initial-condition information (since transfer functions describe only the zero-initial-condition input-output relationship) but reveals the system's pole-zero structure directly. In fact, the same Laplace-transform technique used implicitly in part (a) — taking Laplace transforms of the state equations and solving algebraically for the transformed state variables — is exactly the general method by which the transfer-function formula G(s)=C(sI-A)⁻¹B used in part (b) is itself derived: writing ẋ=Ax+Bu in the Laplace domain gives sX(s)-x(0)=AX(s)+BU(s), so (sI-A)X(s)=x(0)+BU(s), giving X(s)=(sI-A)⁻¹[x(0)+BU(s)]; setting x(0)=0 (zero initial conditions, as required for a transfer-function definition) and Y(s)=CX(s) then directly yields Y(s)=C(sI-A)⁻¹BU(s), i.e., G(s)=Y(s)/U(s)=C(sI-A)⁻¹B — showing that the two parts of this question, despite superficially different presentations, both rest on the same underlying Laplace-transform solution technique for linear state-space systems.

Applying the Same Laplace Technique to Re-Derive Part (a) Directly

As a further cross-check linking the two parts of this question, part (a)'s result can be re-derived using this same Laplace-domain state-space method rather than the sequential-ODE-solving approach used earlier. Writing the part (a) system in matrix form, ẋ=Ax+Bu with A=[[1,0],[1,1]], B=[1,1]ᵀ, x(0)=[1,0]ᵀ, and taking Laplace transforms: (sI-A)X(s)=x(0)+BU(s), where sI-A=[[s-1,0],[-1,s-1]]. Its inverse is (1/det)×adj, with det(sI-A)=(s-1)², so (sI-A)⁻¹ = [[1/(s-1), 0],[1/(s-1)², 1/(s-1)]]. For a unit step input, U(s)=1/s, so BU(s)=[1/s,1/s]ᵀ, and x(0)+BU(s)=[1+1/s, 1/s]ᵀ. Multiplying: X1(s) = (1/(s-1))×(1+1/s) = (s+1)/[s(s-1)]. Partial fractions: (s+1)/[s(s-1)] = A/s+B/(s-1); A=(0+1)/(0-1)=-1, B=(1+1)/1=2, so X1(s)=-1/s+2/(s-1), giving x1(t) = -1+2e^t = 2e^t-1, exactly matching the result found earlier by the sequential-ODE method. This agreement between the two independent solution routes (direct sequential ODE integration in part (a), and the general matrix Laplace-transform state-space method used to justify part (b)) provides strong confirmation that both derivations in this answer are mutually consistent and correct.