Problems

1. There are two good approaches to finding equilibrium. The longer way is to perform a state-space realization in which the i/o ODE is converted to a state-space model. Equilibrium for this state-space model was derived in and the results can be applied directly: \[\begin{aligned} \overline{\bm{x}} &= -A^{-1} B \overline{\bm{u}}. \end{aligned}\] The only caveat here is to realize that \(\dot{u} = 0\) for the purposes of equilibrium.¹ With this state-space realization approach, one must show how the equilibrium state variables are related to \(y\). A canonical choice would be \[\begin{aligned} \bm{x} = \begin{bmatrix} y \\ \dot{y} \end{bmatrix}. \end{aligned}\]

   Conversely, the short way, and this is the way I expect most students will solve this problem,² is to realize that, by the definition of equilibrium in , \(\dot{\bm{x}} = 0\). If we realize that we could choose \(y\) as a state variable, then \(\dot{y} = \ddot{y} = 0\) in equilibrium. As discussed above, \(\dot{u} = 0\), so substituting into the original ODE, \[\begin{aligned} 2\, (0) + 12\, (0) + 50\, \overline{y} &= -10 (0) + 4 \overline{u} \Rightarrow \\ \overline{y} &= \frac{4} {50} \overline{u} \\ &= \frac{4} {50} (6) \\ &= 24/50 \\ &= 0.48. \end{aligned}\]

2. From the ODE, the characteristic equation is \[\begin{aligned} \lambda^2 + 6\, \lambda + 25 &= 0 \Rightarrow\\ \lambda_{1,2} &= -3 \pm j 4. \end{aligned}\] Clearly all (both) roots have negative real part, so the system is stable or assymptotically stable.³

From superposition, the response \(y(t)\) of the system to the initial condition and the summed inputs is \[\begin{aligned} y(t) = y_\text{fr}(t) + y_{\text{fo}_1}(t) + y_{\text{fo}_2}(t). \end{aligned}\]

clear;close all

1. The lti.equistab lecture shows that the equilibrium state is $$\begin{align} \overline{\bm{x}} &= -A^{-1} B \overline{\bm{u}} \\ &= -A^{-1} B \begin{bmatrix} 0 \end{bmatrix} \\ &= \begin{bmatrix} 0 \\ 0 \end{bmatrix}. \tag{$A$ is invertable} \end{align}$$

2. From ss.ss2tf2io, the transfer function \(H(s)\) is \[\begin{aligned} H(s) &= C (sI - A)^{-1} B + D \\ &= \begin{bmatrix} 1 & 0 \end{bmatrix} \left( \begin{bmatrix} s & 0\\ 0 & s \end{bmatrix} - \begin{bmatrix} -4 & 11\\ 3 & -12 \end{bmatrix} \right)^{-1}\begin{bmatrix} 0 \\ 2 \end{bmatrix} + \begin{bmatrix} 0 \end{bmatrix}. \end{aligned}\] In Matlab,

A = [-4, 11; 3, -12];
B = [0; 2];
C = [1, 0];
D = [0];
n = length(A);
syms s % Lapace, complex s
H(s) = C*(s*eye(n) - A)^-1*B + D

\(H(s) =\frac{22}{s^2 +16\,s+15}\)

The input-output ODE can be identified by inspection, or we can use Matlab to define it as follows.

syms t y(t) u(t) s Y(s) U(s)
[num,den] = numden(H(s)); % numerator and denominator
p = solve(den == 0,s) % poles
assume(s ~= p); % for simplification purposes
eq1 = expand(... % Y = H U
    simplify(...
        Y(s) == H(s)*U(s) ...
    ) ...
)
u_c = coeffs(lhs(eq1));
y_c = coeffs(rhs(eq1));
u_d = [u(t)];
for iu = 2:length(u_c)
    u_d(iu) = diff(u(t),t,iu-1);
end
y_d = [y(t)];
for iy = 2:length(y_c)
    y_d(iy) = diff(y(t),t,iy-1);
end
ode = dot(y_d,y_c) == dot(u_d,u_c);
disp(ode)

\(p =\left(\begin{array}{c} -15\\ -1 \end{array}\right)\)

\(eq1 =22\,U\left(s\right)=15\,Y\left(s\right)+16\,s\,Y\left(s\right)+s^2 \,Y\left(s\right)\)

\(16\,\bar{\frac{\partial }{\partial t}\;y\left(t\right)} +15\,\bar{y\left(t\right)} +\bar{\frac{\partial^2 }{\partial t^2 }\;y\left(t\right)} =22\,\bar{u\left(t\right)}\)

1. Stability can be investigated by finding the roots of the characteristic equation. The characteristic equation and its solution are

syms L % L for lambda
odelhs = lhs(ode); % left-hand side of the ODE
for in = 1:n
    odelhs = subs(odelhs,diff(y,t,in),L^in); % substitute for derivatives
end
odelhs = subs(odelhs,y(t),1)
ce = odelhs == 0
r = solve(ce,L)

\(odelhs ={\left(\bar{L} \right)}^2 +16\,\bar{L} +15\)

\(ce ={\left(\bar{L} \right)}^2 +16\,\bar{L} +15=0\)

\(r =\left(\begin{array}{c} -15\\ -1 \end{array}\right)\)

Although we have not considered the transfer function in detail, we see that its poles are equal to the roots of the characteristic equation. Furthermore, these equal the eigenvalues of \(A\):

disp(eig(A))

    -1
   -15

Anyhow, both roots of the characteristic equation have negative real parts, so the system is asymptotically stable.