$$
\\ddot{y} + a_1 y + a_2 y = b_2 u,
$$
in which no derivatives of the input appears or, equivalently, a transfer-function
$$
G(s) = \\frac{b_2}{s^2+a_1 s+a_2},
$$
with no zeros, one can calculate relate $y$, $\\dot{y}$ and $\\ddot{y}$ through a chain of integrators, which leads to a state-space realization in which the state vector is comprised of the integrator outputs
$$
\\begin{pmatrix}
x_1 \\\\ x_2
\\end{pmatrix}
=
\\begin{pmatrix}
y \\\\ \\dot{y}
\\end{pmatrix}.
$$
The corresponding simulation block-diagram is in the next figure.<\/p>\n\n\n\n
![\"\"](\"https:\/\/i0.wp.com\/linearcontrol.info\/fundamentals\/wp-content\/uploads\/2022\/02\/chapter5ss.png?resize=840%2C315&ssl=1\")
<\/figure>\n\n\n\nA state-space representation is obtaining by solving for $\\dot{x}_2 = \\ddot{y}$, as in
$$
\\dot{x}_2 =\\ddot{y} = -a_1 y – a_2 y + b_2 u = -a_1 x_1 – a_2 x_2 + b_2 u
$$
and observing that
$$
\\begin{aligned}
\\dot{x}_1 &= \\dot{y} = x_2, &
y &= x_1.
\\end{aligned}
$$
The complete state-space realization is
$$
\\begin{aligned}
\\begin{pmatrix}
\\dot{x}_1 \\\\ \\dot{x}_2
\\end{pmatrix}
&=
\\begin{pmatrix}
x_2 \\\\ -a_1 x_1 – a_2 x_2 + b_2 u
\\end{pmatrix}, \\\\
y &= x_1.
\\end{aligned}
$$
which can be represented by the standard $A,B,C,D$ matrices as
$$
\\begin{aligned}
\\begin{pmatrix}
\\dot{x}_1 \\\\ \\dot{x}_2
\\end{pmatrix}
&=
\\begin{bmatrix}
0 & 1 \\\\ -a_1 & -a_2
\\end{bmatrix}
\\begin{pmatrix}
x_1 \\\\ x_2
\\end{pmatrix}
+
\\begin{bmatrix}
0 \\\\ b_2
\\end{bmatrix}
u \\\\
y &= \\begin{bmatrix}
1 & 0
\\end{bmatrix}
\\begin{pmatrix}
x_1 \\\\ x_2
\\end{pmatrix}
+
0 u.
\\end{aligned}
$$<\/p>\n\n\n\n
The same strategy works for more complex systems, as long as there are no derivatives of the input involved in the equations. For example, the mass-spring-damper system from P5.26<\/strong> is given by the differential equations Having obtained the state-space matrices $(A,B,C,D)$ one can then use MATLAB or another computer tool to calculate properties of the system. For example<\/p>\n\n\n\n creates a state-space system in MATLAB from which one can obtain the poles and zeros of the system with the commands <\/p>\n\n\n\n or the complete transfer-function by the command<\/p>\n\n\n\n Note that state-space systems with different outputs can be easily produced by changing the $C$ matrix. For example, the system with output $y = z_1 = x_1$ has the same $A$ and $B$ matrices above and the matrix We have seen in Chapter 5 how to construct block-diagrams for simulation using integrators and, from there, derive state-space equations. The task was made easier if it was possible to use a chain of integrators, which led to a trivial definition of the state. In many cases that strategy is enough.<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"inline_featured_image":false},"categories":[4],"tags":[52,10,13],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/linearcontrol.info\/fundamentals\/index.php\/wp-json\/wp\/v2\/posts\/1744"}],"collection":[{"href":"https:\/\/linearcontrol.info\/fundamentals\/index.php\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/linearcontrol.info\/fundamentals\/index.php\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/linearcontrol.info\/fundamentals\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/linearcontrol.info\/fundamentals\/index.php\/wp-json\/wp\/v2\/comments?post=1744"}],"version-history":[{"count":23,"href":"https:\/\/linearcontrol.info\/fundamentals\/index.php\/wp-json\/wp\/v2\/posts\/1744\/revisions"}],"predecessor-version":[{"id":1772,"href":"https:\/\/linearcontrol.info\/fundamentals\/index.php\/wp-json\/wp\/v2\/posts\/1744\/revisions\/1772"}],"wp:attachment":[{"href":"https:\/\/linearcontrol.info\/fundamentals\/index.php\/wp-json\/wp\/v2\/media?parent=1744"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/linearcontrol.info\/fundamentals\/index.php\/wp-json\/wp\/v2\/categories?post=1744"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/linearcontrol.info\/fundamentals\/index.php\/wp-json\/wp\/v2\/tags?post=1744"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
$$
\\begin{aligned}
m_1 \\ddot{z}_1 + (b_1 + b_2) \\dot{z}_1 +
(k_1 + k_2) z_1 – b_2 \\dot{z}_2 – k_2 z_2 &= 0, \\\\
m_2 \\ddot{z}_2 + b_2 (\\dot{z}_2 – \\dot{z}_1) + k_2 (z_2 – z_1) &= f_2
\\end{aligned}
$$
Note that if the force $u = f_2$ is the input, then no derivatives of the input appear in these equations. Therefore, one should be able to use a chain of integrators to realize the system. The resulting state is the vector
$$
\\begin{pmatrix}
x_1 \\\\ x_2 \\\\ x_3 \\\\ x_4
\\end{pmatrix}
=
\\begin{pmatrix}
z_1 \\\\ \\dot{z}_1 \\\\ z_2 \\\\ \\dot{z}_2
\\end{pmatrix},
$$
and the state-space equations are obtained by solving for the second-order derivatives
$$
\\begin{aligned}
\\dot{x}_2 = \\ddot{z}_1 &= – \\frac{k_1 + k_2}{m_1} z_1 -\\frac{b_1 + b_2}{m_1} \\dot{z}_1 + \\frac{k_2}{m_1} z_2 + \\frac{b_2}{m_1} \\dot{z}_2, \\\\
&= – \\frac{k_1 + k_2}{m_1} x_1 -\\frac{b_1 + b_2}{m_1} x_2 + \\frac{k_2}{m_1} x_3 + \\frac{b_2}{m_1} x_4, \\\\
\\dot{x}_4 = \\ddot{z}_2 &= – \\frac{k_2}{m_2} (z_2 – z_1) -\\frac{b_2}{m_2} (\\dot{z}_2 – \\dot{z}_1) + \\frac{1}{m_2} u, \\\\
&= – \\frac{k_2}{m_2} (x_3 – x_1) -\\frac{b_2}{m_2} (x_4 – x_2) + \\frac{1}{m_2} u.
\\end{aligned}
$$
If the output is $y = z_2 = x_3$ then
$$
\\begin{aligned}
\\dot{x}_1 &= \\dot{z}_1 = x_2, &
\\dot{x}_3 &= \\dot{z}_2 = x_4, &
y &= x_1.
\\end{aligned}
$$
Putting it all together
$$
\\begin{aligned}
\\begin{pmatrix}
\\dot{x}_1 \\\\ \\dot{x}_2 \\\\ \\dot{x}_3 \\\\ \\dot{x}_4
\\end{pmatrix}
=
\\begin{pmatrix}
x_2 \\\\
-\\frac{k_1 + k_2}{m_1} x_1-\\frac{b_1 + b_2}{m_1} x_2 + \\frac{k_2}{m_1} x_3 + \\frac{b_2}{m_1} x_4 + \\\\
x_4 \\\\
\\frac{k_2}{m_2} x_1 + \\frac{b_2}{m_2} x_2 – \\frac{k_2}{m_2} x_3 – \\frac{b_2}{m_2} x_4 + \\frac{1}{m_2} u
\\end{pmatrix}
\\end{aligned}
$$
which can be represented by the standard $A,B,C,D$ matrices as
$$
\\begin{aligned}
\\begin{pmatrix}
\\dot{x}_1 \\\\ \\dot{x}_2 \\\\ \\dot{x}_3 \\\\ \\dot{x}_4
\\end{pmatrix}
&=
\\begin{bmatrix}
0 & 1 & 0 & 0 \\\\
-\\frac{k_1 + k_2}{m_1} & -\\frac{b_1 + b_2}{m_1} & \\frac{k_2}{m_1} & \\frac{b_2}{m_1} \\\\
0 & 0 & 0 & 1 \\\\
\\frac{k_2}{m_2} & \\frac{b_2}{m_2} & -\\frac{k_2}{m_2} & -\\frac{b_2}{m_2}
\\end{bmatrix}
\\begin{pmatrix}
x_1 \\\\ x_2 \\\\ x_3 \\\\ x_4
\\end{pmatrix}
+
\\begin{bmatrix}
0 \\\\ 0 \\\\ 0 \\\\ \\frac{1}{m_2}
\\end{bmatrix}
u \\\\
y &= \\begin{bmatrix}
0 & 0 & 1 & 0
\\end{bmatrix}
\\begin{pmatrix}
x_1 \\\\ x_2 \\\\ x_3 \\\\ x_4
\\end{pmatrix}
+
0 u.
\\end{aligned}
$$<\/p>\n\n\n\nsys = ss(A,B,C,D)<\/code><\/pre>\n\n\n\npole(sys)\nzero(sys)<\/code><\/pre>\n\n\n\ntf(sys)<\/code><\/pre>\n\n\n\n
$$
C = \\begin{bmatrix} 1 & 0 & 0 & 0 \\end{bmatrix}.
$$ <\/p>\n","protected":false},"excerpt":{"rendered":"