Bộ hồi tiếp trạng thái bằng phương pháp đặt cực

\left\{ \begin{array}{l}
\dot x(t) = Ax(t) + Bu(t)\\
y(t) = Cx(t)
\end{array} \right. (1)

u\left( t \right) = r\left( t \right)-K.x\left( t \right)\ (2)

\left\{ \begin{array}{l}
\dot x\left( t \right) = Ax\left( t \right) + B\left[ {r\left( t \right) - Kx\left( t \right)} \right]\\
y\left( t \right) = Cx\left( t \right)
\end{array} \right.

\left\{ \begin{array}{l}
\dot x\left( t \right) = \left[ {A - BK} \right].x\left( t \right) + B.r\left( t \right)\\
y\left( t \right) = C.x\left( t \right)
\end{array} \right.

det\left[ {sI-A + BK} \right] = 0

s = {\mu _1},s = {\mu _2},....s = {\mu _n}

det \left| {sI - A + BK} \right| = \left( {s - {\mu _1}} \right)\left( {s - {\mu _2}} \right)...\left( {s - {\mu _n}} \right)

\det \left| {sI - A + BK} \right| = {s^n} + {\alpha _1}{s^{n - 1}} + {\alpha _2}{s^{n - 2}} + ... + {\alpha _{n - 1}}s + {\alpha _n}

\frac{{Y\left( s \right)}}{{U\left( s \right)}} = \frac{1}{{{s^3} + 6{s^2} + 5s + 1}}

A=\left[\begin{matrix}0&1&0\\0&0&1\\-1&-5&-6\\\end{matrix}\right];B=\left[\begin{matrix}0\\0\\1\\\end{matrix}\right];C=\left[\begin{matrix}1&0&0\\\end{matrix}\right];D=0

det \left( {sI - A + Bk} \right) = \left( {s + 2 + j4} \right)\left( {s + 2 - j4} \right)\left( {s + 10} \right) = {s^3} + 14{s^2} + 60s + 200

det \left(s\left[\begin{matrix}1&0&0\\0&1&0\\0&0&1\\\end{matrix}\right]-\left[\begin{matrix}0&1&0\\0&0&1\\-1&-5&-6\\\end{matrix}\right]+\left[\begin{matrix}0\\0\\1\\\end{matrix}\right]\left[\begin{matrix}K_1&K_2&K_3\\\end{matrix}\right]\right)\\
=s^3+14s^2+60s+200

\Leftrightarrow det\left(\left[\begin{matrix}s&-1&0\\0&s&-1\\1&5&s+6\\\end{matrix}\right]+\left[\begin{matrix}0&0&0\\0&0&0\\K_1&K_2&K_3\\\end{matrix}\right]\right)=s^3+14s^2+60s+200

\Leftrightarrow det\left(\left[\begin{matrix}s&-1&0\\0&s&-1\\K_1+1&K_2+5&K_3+s+6\\\end{matrix}\right]\right)=s^3+14s^2+60s+200

\Leftrightarrow s.s.\left(K_3+s+6\right)+K_1+1+\left(K_2+5\right).s=s^3+14s^2+60s+200

\Leftrightarrow s^3+K_3s^2+6s^2+\left(K_2+5\right)s+K_1+1=s^3+14s^2+60s+200

\left\{ \begin{array}{l}
{K_1} + 1 = 200\\
{K_2} + 5 = 60\\
{K_3} + 6 = 14
\end{array} \right.

\left\{ \begin{array}{l}
{K_1} = 199\\
{K_2} = 55\\
{K_3} = 8
\end{array} \right.

s = {\mu _1},s = {\mu _2},....s = {\mu _n}

\left( {s - {\mu _1}} \right)\left( {s - {\mu _2}} \right)...\left( {s - {\mu _n}} \right) = {s^n} + {\alpha _1}{s^{n - 1}} + {\alpha _2}{s^{n - 2}} + ... + {\alpha _{n - 1}}s + {\alpha _n}

M=\left[\begin{matrix}B&AB&A^2B&\begin{matrix}...&A^{n-1}B\\\end{matrix}\\\end{matrix}\right]

W=\left[\begin{matrix}a_{n-1}&a_{n-2}&...&\begin{matrix}a_1&1\\\end{matrix}\\a_{n-2}&a_{n-3}&...&\begin{matrix}1&0\\\end{matrix}\\...&...&...&\begin{matrix}...&...\\\end{matrix}\\\begin{matrix}a_1\\1\\\end{matrix}&\begin{matrix}1\\0\\\end{matrix}&\begin{matrix}...\\0\\\end{matrix}&\begin{matrix}\begin{matrix}0\\0\\\end{matrix}&\begin{matrix}0\\0\\\end{matrix}\\\end{matrix}\\\end{matrix}\right]

\left| {sI - A} \right| = {s^n} + {a_1}{s^{n - 1}} + {a_2}{s^{n - 2}} + ... + {a_{n - 1}}s + {a_n}

K=\left[\begin{matrix}\alpha_n-a_n&\alpha_{n-1}-a_{n-1}&...&\alpha_1-a_1\\\end{matrix}\right].T^{-1}

\frac{{Y\left( s \right)}}{{U\left( s \right)}} = \frac{1}{{{s^3} + 6{s^2} + 5s + 1}}

A=\left[\begin{matrix}0&1&0\\0&0&1\\-1&-5&-6\\\end{matrix}\right];B=\left[\begin{matrix}0\\0\\1\\\end{matrix}\right];C=\left[\begin{matrix}1&0&0\\\end{matrix}\right];D=0

\left|sI-A\right|=\left|\left[\begin{matrix}s&0&0\\0&s&0\\0&0&s\\\end{matrix}\right]-\left[\begin{matrix}0&1&0\\0&0&1\\-1&-5&-6\\\end{matrix}\right]\right|=s^3+6s^2+5s+1

M=\left[\begin{matrix}B&AB&A^2B\\\end{matrix}\right]=\left[\begin{matrix}0&0&1\\0&1&-6\\1&-6&31\\\end{matrix}\right]

W=\left[\begin{matrix}5&6&1\\6&1&0\\1&0&0\\\end{matrix}\right]

T=M.W=\left[\begin{matrix}0&0&1\\0&1&-6\\1&-6&31\\\end{matrix}\right]\left[\begin{matrix}5&6&1\\6&1&0\\1&0&0\\\end{matrix}\right]=\left[\begin{matrix}1&0&0\\0&1&0\\0&0&1\\\end{matrix}\right]

T^{-1}=\left[\begin{matrix}1&0&0\\0&1&0\\0&0&1\\\end{matrix}\right]

\begin{matrix}\left(s-\left(-2+j4\right)\right)\left(s-\left(-2-j4\right)\right)\left(s-\left(-10\right)\right)=s^3+14s^2+60s+200\\\Rightarrow\alpha_1=14,\alpha_2=60,\alpha_3=200\\\end{matrix}

\begin{matrix}K=\left[\begin{matrix}\alpha_n-a_n&\alpha_{n-1}-a_{n-1}&...&\alpha_1-a_1\\\end{matrix}\right].T^{-1}\\K=\left[\begin{matrix}200-1&60-5&14-6\\\end{matrix}\right].\left[\begin{matrix}1&0&0\\0&1&0\\0&0&1\\\end{matrix}\right]\\K=\left[\begin{matrix}199&55&8\\\end{matrix}\right]\\\end{matrix}

\det \left( {sI - A + BK} \right) = \left( {s - {p_1}} \right)\left( {s - {p_2}} \right)...\left( {s - {p_n}} \right)

K=\left[\begin{matrix}0&...&1\\\end{matrix}\right].M^{-1}.\varphi\left(A\right)

\frac{{Y\left( s \right)}}{{U\left( s \right)}} = \frac{1}{{{s^3} + 6{s^2} + 5s + 1}}

A=\left[\begin{matrix}0&1&0\\0&0&1\\-1&-5&-6\\\end{matrix}\right];B=\left[\begin{matrix}0\\0\\1\\\end{matrix}\right];C=\left[\begin{matrix}1&0&0\\\end{matrix}\right];D=0

\det \left( {sI - A + BK} \right) = \left( {s + 2 + j4} \right)\left( {s + 2 - j4} \right)\left( {s + 10} \right) = {s^3} + 14{s^2} + 60s + 200

M=\left[\begin{matrix}B&AB&A^2B\\\end{matrix}\right]=\left[\begin{matrix}0&0&1\\0&1&-6\\1&-6&31\\\end{matrix}\right]

\varphi\left(A\right)=A^3+14A^2+60A+200I=\left[\begin{matrix}199&55&8\\-8&159&7\\-7&-43&117\\\end{matrix}\right]

K=\left[\begin{matrix}0&0&1\\\end{matrix}\right]\left[\begin{matrix}0&0&1\\0&1&-6\\1&-6&31\\\end{matrix}\right]^{-1}\left[\begin{matrix}199&55&8\\-8&159&7\\-7&-43&117\\\end{matrix}\right]