--- title: "Jordan canonical form of (2z³+6z+5)/((z+2)²(z+3))" subject: "Digital Control System" chapter: 5 key: "semester_6/digital_control_system:jordan-form-2z3-6z-5" source: https://www.bctengineer.com/q/semester_6/digital_control_system/jordan-form-2z3-6z-5 site: BCT Engineering --- # Jordan canonical form of (2z³+6z+5)/((z+2)²(z+3)) Digital Control System · Chapter 5: Discrete Time State Equations - Asked in: 2082 Chaitra, 5 marks; 2082 Kartik, 8 marks (asked 2 times) - Video: [Jodran Canoncial Form and State Space Representation](https://www.youtube.com/watch?v=Ie37sB5i91Y) — Electric Fund - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/digital_control_system/jordan-form-2z3-6z-5 - Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-5 ## Question Obtain the state space representation of the system shown below in Jordan canonical form: Y(z)/U(z)=(2z³+6z+5)/((z+2)²(z+3)). ## Short answer For Y(z)/U(z)=(2z³+6z+5)/((z+2)²(z+3)), separate direct term D=2. Partial fractions of remainder: -23/(z+2)² + 53/(z+2) - 67/(z+3). Jordan form: G=[[-2,1,0],[0,-2,0],[0,0,-3]], H=[0;1;1], C=[-23,53,-67], D=2. Repeated pole at -2 gives Jordan block. ## Answer The pole $z=-2$ is repeated, so the Jordan canonical form is used. The numerator and denominator have the same degree, so there is a direct term $D = b_0$. #### Separate the direct term Denominator: $(z+2)^2(z+3) = z^3 + 7z^2 + 16z + 12$. $$ \frac{2z^3+6z+5}{z^3+7z^2+16z+12} = 2 + \frac{-14z^2 - 26z - 19}{(z+2)^2(z+3)} $$ So $D = b_0 = 2$. #### Partial fractions $$ \frac{Y(z)}{U(z)} = 2 + \frac{c_1}{(z+2)^2} + \frac{c_2}{z+2} + \frac{c_3}{z+3} $$ Let $N(z) = 2z^3 + 6z + 5$, so $N(-2) = -16 - 12 + 5 = -23$ and $N(-3) = -54 - 18 + 5 = -67$. $$ \begin{aligned} c_1 &= \left[\frac{N(z)}{z+3}\right]_{z=-2} = \frac{-23}{1} = -23\\ c_2 &= \frac{d}{dz}\left[\frac{N(z)}{z+3}\right]_{z=-2} = \left[\frac{(6z^2+6)(z+3) - N(z)}{(z+3)^2}\right]_{z=-2} = \frac{30 + 23}{1} = 53\\ c_3 &= \left[\frac{N(z)}{(z+2)^2}\right]_{z=-3} = \frac{-67}{1} = -67 \end{aligned} $$ $$ \frac{Y(z)}{U(z)} = 2 - \frac{23}{(z+2)^2} + \frac{53}{z+2} - \frac{67}{z+3} $$ #### States $X_2 = \dfrac{U}{z+2}$, $X_1 = \dfrac{X_2}{z+2}$, $X_3 = \dfrac{U}{z+3}$: $$ \begin{aligned} x_1(k+1) &= -2x_1(k) + x_2(k)\\ x_2(k+1) &= -2x_2(k) + u(k)\\ x_3(k+1) &= -3x_3(k) + u(k)\\ y(k) &= -23x_1(k) + 53x_2(k) - 67x_3(k) + 2u(k) \end{aligned} $$ #### Jordan canonical form $$ \mathbf{x}(k+1) = \begin{bmatrix} -2 & 1 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & -3 \end{bmatrix}\mathbf{x}(k) + \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}u(k) $$ $$ y(k) = \begin{bmatrix} -23 & 53 & -67 \end{bmatrix}\mathbf{x}(k) + 2u(k) $$ Check: $\mathbf{C}(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H} + 2$ gives back $\dfrac{2z^3+6z+5}{(z+2)^2(z+3)}$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the Jordan canonical form used for Y/U = (2z^3+6z+5)/((z+2)^2(z+3))? A. The pole z = -2 is repeated B. The numerator degree is 3 C. There is a pole at z = -3 D. The system is unstable Answer: A. A repeated pole cannot be fully diagonalised in this way, so a Jordan block is used. (Key point: When Jordan form is used) 2. What is the direct term D, and why does it exist? A. D = 0, because there are no zeros B. D = 2, because numerator and denominator have the same degree C. D = 5, the constant term of the numerator D. D = 12, the constant term of the denominator Answer: B. Equal degrees mean the leading coefficient ratio b0 = 2 passes straight from input to output. (Key point: Direct term) 3. How is c1 (the coefficient of 1/(z+2)^2) found? A. Evaluate N(z)/(z+2)^2 at z = -3 B. Differentiate N(z)/(z+3) at z = -2 C. Evaluate N(z)/(z+3) at z = -2, giving -23 D. Evaluate N(z) at z = 0 Answer: C. The highest-order coefficient is the residue function with the repeated factor removed, evaluated at the pole. The derivative gives c2. (Key point: Partial fraction for repeated pole) 4. In the resulting G matrix, where does the 1 for the repeated pole appear? A. On the subdiagonal, G(2,1) = 1 B. In the entry G(1,3) C. Nowhere; G is purely diagonal D. On the superdiagonal, G(1,2) = 1, between the two -2 entries Answer: D. The Jordan block for -2 has -2 on the diagonal and 1 on the superdiagonal because X1 = X2/(z+2). 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