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State space advantages; state model of (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.

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).

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)=2s2+3s+1s3+5s2+6s+7G(s) = \dfrac{2s^2+3s+1}{s^3+5s^2+6s+7}

Use the controllable canonical (phase-variable) form. Let

X(s)U(s)=1s3+5s2+6s+7,Y(s)X(s)=2s2+3s+1\frac{X(s)}{U(s)} = \frac{1}{s^3+5s^2+6s+7},\qquad \frac{Y(s)}{X(s)} = 2s^2 + 3s + 1

So x...+5x¨+6x˙+7x=u\dddot{x} + 5\ddot{x} + 6\dot{x} + 7x = u and y=2x¨+3x˙+xy = 2\ddot{x} + 3\dot{x} + x.

Choose x1=xx_1 = x, x2=x˙x_2 = \dot{x}, x3=x¨x_3 = \ddot{x}:

x˙1=x2x˙2=x3x˙3=−7x1−6x2−5x3+uy=x1+3x2+2x3\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:

[x˙1x˙2x˙3]=[010001−7−6−5][x1x2x3]+[001]u\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=[132][x1x2x3]y = \begin{bmatrix} 1 & 3 & 2 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}

The last row of AA holds the negated denominator coefficients (7, 6, 5) and CC holds the numerator coefficients in ascending powers (1, 3, 2). Check: C(sI−A)−1B=2s2+3s+1s3+5s2+6s+7C(sI-A)^{-1}B = \dfrac{2s^2+3s+1}{s^3+5s^2+6s+7}.

✓ Answer checked by Claude Opus 5.5 (high effort)

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