--- title: "State space advantages; state model of (2s²+3s+1)/(s³+5s²+6s+7)" subject: "Control Systems" chapter: 8 key: "semester_3/control_systems:ss-adv-tf-2s2-3s-1" source: https://www.bctengineer.com/q/semester_3/control_systems/ss-adv-tf-2s2-3s-1 site: BCT Engineering --- # State space advantages; state model of (2s²+3s+1)/(s³+5s²+6s+7) Control Systems · Chapter 8: State Space Analysis - Asked in: 2065 Shrawan (old course), 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/control_systems/ss-adv-tf-2s2-3s-1 - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-8 ## Question Discuss the advantages of state space representation. Find the state equation and output equation of state space form for the system represented by transfer function G(s) = (2s² + 3s + 1)/(s³ + 5s² + 6s + 7). ## Short answer Advantages: applies to MIMO, nonlinear/time-varying; includes initial conditions; enables controllability/observability tests. Phase-variable form yields state equation ̇x=[0 1 0;0 0 1;-7 -6 -5]x + [0;0;1]u and output y=[1 3 2]x. ## Answer #### Advantages of state-space analysis 1. Applies to **MIMO** (multi-input multi-output) systems as easily as to single-input single-output systems. 2. Can handle **non-linear and time-varying** systems; the transfer function needs linear time-invariant systems. 3. Includes **initial conditions**; the transfer function assumes zero initial conditions. 4. Gives the **internal behaviour** (all state variables), not only the input–output relation. 5. Allows tests of **controllability and observability**, and hidden (cancelled) modes are not lost. 6. Suited to **computer solution** and simulation, since it uses first-order matrix equations. 7. Forms the basis of modern and optimal control design: state feedback (pole placement), observers, Kalman filter. #### State model of $G(s) = \dfrac{2s^2+3s+1}{s^3+5s^2+6s+7}$ Use the **controllable canonical (phase-variable) form**. Let $$ \frac{X(s)}{U(s)} = \frac{1}{s^3+5s^2+6s+7},\qquad \frac{Y(s)}{X(s)} = 2s^2 + 3s + 1 $$ So $\dddot{x} + 5\ddot{x} + 6\dot{x} + 7x = u$ and $y = 2\ddot{x} + 3\dot{x} + x$. Choose $x_1 = x$, $x_2 = \dot{x}$, $x_3 = \ddot{x}$: $$ \begin{aligned} \dot{x}_1 &= x_2 \\ \dot{x}_2 &= x_3 \\ \dot{x}_3 &= -7x_1 - 6x_2 - 5x_3 + u \\ y &= x_1 + 3x_2 + 2x_3 \end{aligned} $$ **State equation:** $$ \begin{bmatrix} \dot{x}_1 \\ \dot{x}_2 \\ \dot{x}_3 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -7 & -6 & -5 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}u $$ **Output equation:** $$ y = \begin{bmatrix} 1 & 3 & 2 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} $$ The last row of $A$ holds the negated denominator coefficients (7, 6, 5) and $C$ holds the numerator coefficients in ascending powers (1, 3, 2). Check: $C(sI-A)^{-1}B = \dfrac{2s^2+3s+1}{s^3+5s^2+6s+7}$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which statement correctly contrasts state-space analysis with the transfer function approach? A. State space handles MIMO, non-linear, time-varying systems and non-zero initial conditions; the transfer function does not B. State space works only for single-input single-output linear systems C. State space assumes zero initial conditions while the transfer function includes them D. State space gives only the input-output relation, hiding internal variables Answer: A. The transfer function needs LTI systems and zero initial conditions, whereas state-space models cover MIMO, non-linear and time-varying systems and include initial conditions. (Key point: Advantages over transfer function) 2. Why is state-space representation useful for checking hidden (cancelled) modes? A. It gives only the output, so cancelled modes disappear B. It keeps all state variables, so controllability and observability can be tested C. It converts the system into the frequency domain where cancellation is shown D. It removes all poles that are cancelled by zeros Answer: B. Because the internal states are retained, controllability and observability tests reveal modes that pole-zero cancellation hides in a transfer function. (Key point: Controllability and observability) 3. For G(s) = (2s^2+3s+1)/(s^3+5s^2+6s+7), what is the first step in the controllable canonical form? A. Set X/U = 2s^2+3s+1 and Y/X = 1/(s^3+5s^2+6s+7) B. Set X/U = 1/(s^3+5s^2+6s+7) and Y/X = 2s^2+3s+1 C. Set X/U = s^3+5s^2+6s+7 and Y/X = 1 D. Expand G(s) into partial fractions only Answer: B. The denominator is placed in X/U to give the differential equation for x, and the numerator in Y/X to give the output equation. (Key point: Phase-variable setup) 4. What is the last row of matrix A for this system? A. [-5 -6 -7] B. [7 6 5] C. [-7 -6 -5] D. [-1 -3 -2] Answer: C. The last row holds the negated denominator coefficients in ascending order of power: -7, -6, -5. (Key point: Last row of A) 5. What is the output matrix C, with states x1 = x, x2 = x', x3 = x''? A. [2 3 1] B. [1 3 2] C. [0 0 1] D. [1 5 6] Answer: B. y = 2x'' + 3x' + x = x1 + 3x2 + 2x3, so C = [1 3 2], numerator coefficients in ascending powers. 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