--- title: "State model from 20(4s+2)/(s³+5s²+8s+2)" subject: "Control Systems" chapter: 8 key: "semester_3/control_systems:ss-from-tf-20-4s2" source: https://www.bctengineer.com/q/semester_3/control_systems/ss-from-tf-20-4s2 site: BCT Engineering --- # State model from 20(4s+2)/(s³+5s²+8s+2) Control Systems · Chapter 8: State Space Analysis - Asked in: 2075 Asoj, 8 marks; 2079 Bhadra, 8 marks (asked 2 times) - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/control_systems/ss-from-tf-20-4s2 - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-8 ## Question A system is characterized by the equation Y(s)/U(s) = 20(4s+2)/(s³ + 5s² + 8s + 2). Find its state and output equation and express in matrix form. Then using your matrix, how do you get characteristic equation? ## Short answer Using phase-variable form: A = [0 1 0; 0 0 1; -2 -8 -5], B = [0; 0; 1], C = [40 80 0]. Characteristic equation is |sI-A| = s^3+5s^2+8s+2 = 0. ## Answer $$ \frac{Y(s)}{U(s)} = \frac{20(4s+2)}{s^3+5s^2+8s+2} = \frac{80s + 40}{s^3 + 5s^2 + 8s + 2} $$ Since the numerator has $s$ terms, use the **phase-variable (controllable canonical) form** with an intermediate variable $X(s)$. #### Step 1: Split the transfer function $$ \frac{Y(s)}{U(s)} = \frac{X(s)}{U(s)}\cdot\frac{Y(s)}{X(s)},\qquad \frac{X(s)}{U(s)} = \frac{1}{s^3+5s^2+8s+2},\qquad \frac{Y(s)}{X(s)} = 80s + 40 $$ So $\dddot{x} + 5\ddot{x} + 8\dot{x} + 2x = u$ and $y = 80\dot{x} + 40x$. #### Step 2: Choose state variables $x_1 = x$, $x_2 = \dot{x}$, $x_3 = \ddot{x}$: $$ \begin{aligned} \dot{x}_1 &= x_2 \\ \dot{x}_2 &= x_3 \\ \dot{x}_3 &= -2x_1 - 8x_2 - 5x_3 + u \\ y &= 40x_1 + 80x_2 \end{aligned} $$ #### Step 3: Matrix form $$ \begin{bmatrix} \dot{x}_1 \\ \dot{x}_2 \\ \dot{x}_3 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -2 & -8 & -5 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}u $$ $$ y = \begin{bmatrix} 40 & 80 & 0 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} $$ #### Step 4: Characteristic equation from the matrix The characteristic equation is $|sI - A| = 0$: $$ sI - A = \begin{bmatrix} s & -1 & 0 \\ 0 & s & -1 \\ 2 & 8 & s+5 \end{bmatrix} $$ $$ \begin{aligned} |sI-A| &= s\left[s(s+5) + 8\right] + 1\left[0 + 2\right] \\ &= s^3 + 5s^2 + 8s + 2 = 0 \end{aligned} $$ This equals the denominator of the transfer function, as it must. Its roots (eigenvalues of $A$) are $s = -0.304$ and $-2.348 \pm j1.029$, all in the left half-plane, so the system is stable. **Answer:** $A = \begin{bmatrix} 0&1&0\\0&0&1\\-2&-8&-5 \end{bmatrix}$, $B = \begin{bmatrix}0\\0\\1\end{bmatrix}$, $C = \begin{bmatrix}40&80&0\end{bmatrix}$; characteristic equation $s^3 + 5s^2 + 8s + 2 = 0$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The numerator has s terms. How is the realisation obtained? A. Set all numerator terms to zero B. Split into X/U = 1/denominator and Y/X = numerator, using phase variables C. Use only the denominator and ignore the numerator D. Differentiate the output twice Answer: B. An intermediate variable X(s) makes the denominator define the states and the numerator define the output. (Key point: Why phase-variable form) 2. Which last row does A have for the denominator s^3 + 5s^2 + 8s + 2? A. [-5 -8 -2] B. [2 8 5] C. [-2 -8 -5] D. [-1 -5 -8] Answer: C. The last row is the negative coefficients in ascending order: -a0, -a1, -a2 = -2, -8, -5. (Key point: A matrix) 3. The numerator is 80s + 40. What is C? A. [80 40 0] B. [0 40 80] C. [20 4 0] D. [40 80 0] Answer: D. y = 40x + 80x' = 40x1 + 80x2, so C = [40 80 0]. (Key point: Output matrix) 4. How is the characteristic equation obtained from the model? A. Expand |sI - A| = 0, which gives s^3 + 5s^2 + 8s + 2 = 0 B. Expand |A| = 0 C. Expand |B C| = 0 D. Set the numerator 80s + 40 = 0 Answer: A. |sI - A| = 0 reproduces the denominator of the transfer function. 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