We consider the input-state system
\(\seteqnumber{0}{11.}{0}\)\begin{equation} \label {eq:is} \dot {x}(t)=Ax(t)+Bu(t). \end{equation}
Here \(A\in \mR ^{n\times n}\) and \(B\in \mR ^{n\times m}\).
Definition 11.1.
• Let \(T>0\) and \(x^0,x^1\in \mR ^n\). If there exists a control \(u:[0,T]\to \mR ^m\) such that \(x(0)=x^0\) and \(x(T)=x^1\), then \(x^1\) is said to be reachable from \(x^0\) in time \(T\).
• Let \(x^0,x^1\in \mR ^n\). If there exists a \(T>0\) such that \(x^1\) is reachable from \(x^0\) in time \(T\), then \(x^1\) is said to be reachable from \(x^0\).
• Let \(T>0\). The system (11.1) (or equivalently the pair of matrices \((A,B)\)) is said to be controllable in time \(T>0\) if for all \(x^0,x^1\in \mR ^n\), \(x^1\) is reachable from \(x^0\) in time \(T\).
• The system (11.1) (or equivalently the pair of matrices \((A,B)\)) is said to be controllable if there exists a time \(T>0\) such that the system is controllable in time \(T\).
Definition 11.2. The Kalman controllability matrix is the \(n\times mn\) matrix
\[ \bbm {B&AB&A^2B&\ldots &A^{n-1}B}. \]
The controllability Gramian, indexed by \(T>0\), is the \(n\times n\) matrix
\[ Q_T=\int _0^T \e ^{At}BB^*\e ^{A^*t}\,dt. \]
The Hautus controllability matrix, indexed by \(s\in \mC \), is the \(n\times (n+m)\) matrix
\[ \bbm {sI-A&B}. \]
Definition 11.3. A polynomial \(p\) is said to be assignable to the pair \((A,B)\) if there exists a \(m\times n\) matrix \(F\) such that the characteristic polynomial of \(A+BF\) is equal to \(p\).
Remark 11.4. Recall that a polynomial is called monic if the coefficient of the highest power is \(1\) and that the characteristic polynomial of an \(n\times n\) matrix is a monic polynomial of degree \(n\).
Theorem 11.5. The following are equivalent:
1. The input-state system is controllable in time \(T\) for all \(T>0\)
2. The input-state system is controllable
3. The Kalman controllability matrix is surjective
4. The Hautus controllability matrix is surjective for all \(s\in \mathbb {C}\)
5. The Hautus controllability matrix is surjective for all eigenvalues \(s\) of \(A\)
6. The controllability Gramian \(Q_T\) is invertible for some \(T>0\)
7. The controllability Gramian \(Q_T\) is invertible for all \(T>0\)
8. Every monic polynomial of degree \(n\) is assignable to the pair \((A,B)\)
Remark 11.6. In the single-input case (i.e. \(m=1\)) an explicit formula for an \(F\) such that the characteristic polynomial of \(A+BF\) equals the given monic polynomial \(p\) of degree \(n\) is:
\[ F=-\bbm {0&\ldots &0&1}\bbm {B&AB&A^2B&\ldots &A^{n-1}B}^{-1}p(A). \]
This is called Ackermann’s formula.
Remark 11.7. A control which steers the system from state \(x^0\) to state \(x^1\) in time \(T\) is given in terms of the controllability Gramian as:
\[ u(t)=B^*\e ^{A^*(T-t)}Q_T^{-1}(x^1-\e ^{AT}x^0). \]
This control is special in the sense that it is the unique control \(u\) which achieves this transition from \(x^0\) to \(x^1\) that minimizes
\[ \int _0^T \|u(t)\|^2\,dt. \]
The minimum value of this control cost is given in terms of the controllability Gramian by
\[ \ipd {Q_T^{-1}(x^1-\e ^{AT}x^0)}{x^1-\e ^{AT}x^0}. \]
Definition 11.8.
• Let \(T>0\). The set of states reachable from zero in time \(T\) is denoted \(\mathcal {R}_T\).
• The set of states reachable from zero is denoted \(\mathcal {R}\) and is called the reachable subspace.
Proposition 11.11. Let \(A\) be asymptotically stable.
• The input-state system is controllable if and only if the infinite-time controllability gramian \(Q\) is invertible.
• The reachable subspace equals the image of the infinite-time controllability gramian \(Q\).
• The infinite-time controllability gramian \(Q\) is the unique solution of the control Lyapunov equation
\[ AQ+QA^*+BB^*=0. \]
Example 11.12. Consider the first order system
\[ \dot {x}+x=u. \]
We have
\[ A=-1,\qquad B=1. \]
Since \(n=1\), the Kalman controllability matrix is simply \(B\). Since a \(1\times 1\) matrix is surjective if and only if it is nonzero, we have that the Kalman controllability matrix is surjective and therefore the system is controllable.
The controllability Gramian is
\[ Q_T=\int _0^T \e ^{-2t}\,dt =\left [-\frac {1}{2}\e ^{-2t}\right ]_{t=0}^T =\frac {1}{2}\left (1-\e ^{-2T}\right ). \]
From Remark 11.7 we then see that a control which steers the system from state \(x^0\) to state \(x^1\) in time \(T\) is given by
\(\seteqnumber{0}{11.}{1}\)\begin{align*} u(t)&=\e ^{t-T}~\frac {2}{1-\e ^{-2T}}(x^1-\e ^{-T}x^0). \end{align*} and that the control cost to achieve this is
\[ \frac {2}{1-\e ^{-2T}}\left (x^1-\e ^{-T}x^0\right )^2. \]
To understand this better, consider \(x^0=0\) and \(x^1\) with \(|x^1|=1\). Then the control cost is
\[ \frac {2}{1-\e ^{-2T}}, \]
which is a decreasing function in \(T\) with limit as \(T\downarrow 0\) equal to \(\infty \) and limit as \(T\to \infty \) equal to \(2\). We therefore see that although the system is controllable in any positive time \(T\), the cost becomes larger the smaller the given time \(T\) is (this is generally the case).
The control Lyapunov equation is
\[ -2Q+1=0, \]
which gives \(Q=\frac {1}{2}\) as the infinite-time controllability Gramian.
Example 11.13. Consider two first order systems with the same control acting on both:
\(\seteqnumber{0}{11.}{1}\)\begin{align*} \dot {x}_1+x_1=u,\\ \dot {x}_2+\alpha x_2=u, \end{align*} where \(\alpha >0\). This gives
\[ A=\bbm {-1&0\\0&-\alpha },\qquad B=\bbm {1\\1}. \]
If \(\alpha =1\), then it is easy to see from the definition that the system is not controllable. Choose \(x^0=\sbm {0\\0}\). Since the equations are identical (and we have chosen identical initial conditions), we then have \(x_1(t)=x_2(t)\) for all \(t\), no matter what \(u\) is. With \(x^1=\sbm {0\\0}\) we then have that \(x^1\) is not reachable from \(x^0\): assume that it is reachable in time \(T\), then \(1=x_1(T)=x_2(T)=0\), which is a contradition.
The Hautus controllability matrix is
\[ \bbm {sI-A&B}=\bbm {s+1&0&1\\0&s+\alpha &1}. \]
We only have to consider this for \(s\) an eigenvalue of \(A\), so we only have to consider \(s=-1\) and \(s=-\alpha \). This gives the matrices
\[ \bbm {0&0&1\\0&-1+\alpha &1},\qquad \bbm {-\alpha +1&0&1\\0&0&1}. \]
We can ignore the zero columns and therefore we really only have to consider the matrices
\[ \bbm {0&1\\-1+\alpha &1},\qquad \bbm {-\alpha +1&1\\0&1}. \]
We see that if \(\alpha =1\) then these matrices are not surjective (they are square and have a zero column) and if \(\alpha \neq 1\) then these matrices are surjective (because their determinant is nonzero). Therefore the system is controllable if and only if \(\alpha \neq 1\).
The Kalman controllability matrix is
\[ \bbm {B&AB}=\bbm {1&-1\\1&-\alpha }, \]
which has determinant
\[ -\alpha +1, \]
which is zero if and only if \(\alpha =1\). So also from here we see that the system is controllable if and only if \(\alpha \neq 1\). If \(\alpha =1\), then the Kalman controllability matrix equals
\[ \bbm {1&-1\\1&-1} \]
and we see that its image equals the multiples of
\[ \bbm {1\\1}, \]
so that the reachable subspace consists of vectors in \(\mR ^2\) which have their first and second component the same.
The control Lyapunov equation is (using that we know that \(Q\) is symmetric)
\[ \bbm {-1&0\\0&-\alpha }\bbm {Q_1&Q_0\\Q_0&Q_2} +\bbm {Q_1&Q_0\\Q_0&Q_2}\bbm {-1&0\\0&-\alpha } +\bbm {1\\1}\bbm {1&1} =\bbm {0&0\\0&0}, \]
which is
\[ \bbm {-2Q_1+1&Q_0\left (-1-\alpha \right )+1\\ Q_0\left (-1-\alpha \right )+1&-2\alpha Q_2+1}=\bbm {0&0\\0&0}. \]
Solving this (we really just have three separate equations in the three unknowns \(Q_1\), \(Q_2\) and \(Q_0\)) gives
\[ Q=\bbm {\frac {1}{2}&\frac {1}{1+\alpha }\\\frac {1}{1+\alpha }&\frac {1}{2\alpha }}. \]
The determinant of \(Q\) equals
\[ \frac {1}{4\alpha }-\frac {1}{(1+\alpha )^2}, \]
which is zero if and only if \((1+\alpha )^2=4\alpha \), i.e. when \((1-\alpha )^2=0\), i.e. when \(\alpha =1\). So we again see that the system is controllable if and only if \(\alpha \neq 1\). When \(\alpha =1\) we have
\[ Q=\bbm {\frac {1}{2}&\frac {1}{2}\\\frac {1}{2}&\frac {1}{2}}, \]
which has image the scalar multiples of
\[ \bbm {1\\1}, \]
so that we again obtain the reachable subspace as obtained using a different method above.
Example 11.14. Consider the second order system
\[ \ddot {q}+2\zeta \dot {q}+q=u, \]
where \(\zeta \geq 0\), with state \(x=\sbm {q\\\dot {q}}\). We then have
\[ A=\bbm {0&1\\-1&-2\zeta },\qquad B=\bbm {0\\1}. \]
The Kalman controllability matrix is
\[ \bbm {B&AB}=\bbm { 0&1\\ 1&-2\zeta }. \]
The determinant equals \(-1\) which is nonzero so that the Kalman controllability matrix is invertible and therefore surjective and therefore the system is controllable.
Example 11.15. We again consider the second order system from Example 11.14 (now with \(\zeta >0\)) and aim to find the infinite-time controllability Gramian. The control Lyapunov equation (using that \(Q\) is symmetric) is
\[ \bbm {0&1\\-1&-2\zeta }\bbm {Q_1&Q_0\\Q_0&Q_2} +\bbm {Q_1&Q_0\\Q_0&Q_2}\bbm {0&-1\\1&-2\zeta } +\bbm {0\\1}\bbm {0&1}=\bbm {0&0\\0&0}, \]
which is
\[ \bbm {2Q_0&Q_2-Q_1-2\zeta Q_0\\ Q_2-Q_1-2\zeta Q_0&-2Q_0-4\zeta Q_2+1} =\bbm {0&0\\0&0}. \]
From the top-left corner we obtain \(Q_0=0\) and subsequently from the bottom-right corner we obtain \(Q_2=\frac {1}{4\zeta }\) and from the off-diagonal entry we obtain \(Q_1=\frac {1}{4\zeta }\). Hence
\[ Q=\frac {1}{4\zeta }\bbm {1&0\\0&1}. \]
Since this is invertible, we see (once again) that the system is controllable.
Example 11.16. Consider the second order system without damping (and gain an natural frequency both normalized to 1)
\[ \ddot {q}+q=u, \]
with state \(x=\sbm {q\\\dot {q}}\), so that
\[ A=\bbm {0&1\\-1&0},\qquad B=\bbm {0\\1}. \]
The aim is to, for a given \(\zeta >0\), find a \(F\) such that the characteristic polynomial of \(A\) equals \(s^2+2\zeta s+1\). From our knowledge of second order systems, we know that \(u=-2\zeta \dot {q}\) will achieve this. We want to check that Ackermann’s formula also leads to this.
Ackermann’s formula tells us that
\[ F=-\bbm {0&1}\bbm {B&AB}^{-1}\left (A^2+2\zeta A+I\right ), \]
will achieve the objective. From Example 11.14 we have
\[ \bbm {B&AB}=\bbm { 0&1\\ 1&0 }, \]
so that
\[ \bbm {B&AB}^{-1}=\bbm { 0&1\\ 1&0 }. \]
We further have
\[ A^2=\bbm { -1&0\\0&-1 }, \]
so that Ackermann’s formula is
\(\seteqnumber{0}{11.}{1}\)\begin{align*} F&= -\bbm {0&1} \bbm {0&1\\1&0} \left (\bbm {-1&0\\0&-1}+2\zeta \bbm {0&1\\-1&0}+\bbm {1&0\\0&1}\right ) \\&= -\bbm {0&1} \bbm {0&1\\1&0} 2\zeta \bbm {0&1\\-1&0} \\&= -\bbm {0&1}2\zeta \bbm {-1&0\\0&1} \\&=-2\zeta \bbm {0&1}, \end{align*} which indeed is equivalent to the formula \(u=-2\zeta \dot {q}\) that we obtained above.
We consider the tape drive with only the control input. This gives an input-state system with
\[ A=\bbm { -\frac {d_1}{M_1}&0&\frac {1}{M_1}\\ 0&-\frac {d_2}{M_2}&-\frac {1}{M_2}\\ -k&k&0 },\qquad B=\bbm {\frac {1}{M_1}&0\\0&\frac {1}{M_2}\\0&0}. \]
We show that this system is controllable by considering the Kalman controllability matrix. We have
\[ AB=\bbm { -\frac {d_1}{M_1^2}&0\\ 0&-\frac {d_2}{M_2^2}\\ -\frac {k}{M_1}&\frac {k}{M_2} }. \]
We see that \(\bbm {B&AB}\) already has rank 3 as the first three colums are
\[ \bbm {\frac {1}{M_1}&0&-\frac {d_1}{M_1^2}\\0&\frac {1}{M_2}&0\\0&0&-\frac {k}{M_1}}, \]
which is upper-triangular with non-zero diagonal entries and therefore invertible. Hence the Kalman controllability matrix \(\bbm {B&AB&A^2B}\) is surjective.
Consider the suspension system with only the control inputs. This gives an input-state system with
\[ A=\bbm {0&k_{us}&0&0\\ \frac {-1}{m_{us}}&0&0&0\\ 0&1&0&-1\\ 0&0&0&0 },\qquad B=\bbm {0\\\frac {1}{m_{us}}\\0\\\frac {-1}{m_s}}. \]
We show that this system is controllable by considering the Kalman controllability matrix. We have
\[ \bbm {B&AB&A^2B&A^3B}=\bbm { 0&\frac {1}{m_{us}}&0&\frac {-k_s}{m_{us}^2}\\ \frac {1}{m_{us}}&0&\frac {-k_s}{m_{us}^2}&0\\ 0&\frac {-1}{m_s}+\frac {-1}{m_{us}}&0&\frac {k_s}{m_{us}^2}\\ \frac {-1}{m_s}&0&0&0 }. \]
To see that this matrix is invertible, we calculate its determinant. We develop by the last row to obtain
\[ \frac {1}{m_s}\det \bbm { \frac {k_{us}}{m_{us}}&0&\frac {-k_{us}^2}{m_{us}^2}\\ 0&\frac {-k_{us}}{m_{us}^2}&0\\ \frac {1}{m_{us}}+\frac {1}{m_s}&0&\frac {-k_{us}}{m_{us}^2} }, \]
which we develop by the middle row to obtain
\[ \frac {-k_{us}}{m_{us}^2m_s}\det \bbm { \frac {k_{us}}{m_{us}}&\frac {-k_{us}^2}{m_{us}^2}\\ \frac {1}{m_{us}}+\frac {1}{m_s}&\frac {-k_{us}}{m_{us}^2} }, \]
which gives
\[ \frac {-k_{us}}{m_{us}^2m_s}\left (\frac {-k_{us}^2}{m_{us}^3}+\frac {k_{us}^2}{m_{us}^3}+\frac {k_{us}^2}{m_{us}^2m_s}\right )=\frac {-k_{us}^3}{m_{us}^4m_s^2}, \]
which is nonzero. Hence the Kalman controllability matrix is surjective and therefore our system is controllable.
We consider the Hautus controllability matrix. The characteristic polynomial of \(A\) is \(s^2\left (s^2+\frac {k_{us}}{m_{us}}\right )\) and therefore the eigenvalues of \(A\) are \(0\) (with multiplicity \(2\)) and \(\pm i\sqrt {\frac {k_{us}}{m_{us}}}\). We therefore consider the matrix \(\bbm {sI-A&B}\) with \(s=0\) and with \(s=i\sqrt {\frac {k_{us}}{m_{us}}}\) (for \(s=-i\sqrt {\frac {k_{us}}{m_{us}}}\) we obtain the complex conjugate matrix, which has the same rank). These matrices are
\[ \bbm { 0&-1&0&0&0\\ \frac {k_{us}}{m_{us}}&0&0&0&\frac {1}{m_{us}}\\ 0&1&0&-1&0\\ 0&0&0&0&\frac {-1}{m_s} }, \]
and
\[ \bbm { i\sqrt {\frac {k_{us}}{m_{us}}}&-1&0&0&0\\ \frac {k_{us}}{m_{us}}&i\sqrt {\frac {k_{us}}{m_{us}}}&0&0&\frac {1}{m_{us}}\\ 0&1&i\sqrt {\frac {k_{us}}{m_{us}}}&-1&0\\ 0&0&0&i\sqrt {\frac {k_{us}}{m_{us}}}&\frac {-1}{m_s} }. \]
We see that both of these matrices have rank \(4\) and therefore the pair \((A,B)\) is controllable.
(a) Consider the first order system
\[ \dot {x}+2x=4u. \]
(i) Write this in the standard form \(\dot {x}=Ax+Bu\).
(ii) Determine the Kalman controllability matrix and use this to show that the system is controllable.
(iii) Determine the Hautus controllability matrix and use this to show that the system is controllable.
(iv) Determine the controllability Gramian \(Q_T\) and use this to show that the system is controllable. Moreover, determine a control which steers the system from a given \(x^0\in \mR \) to a given \(x^1\in \mR \) in a given time \(T>0\).
(v) Determine the infinite-time controllability Gramian \(Q\) by solving the control Lyapunov equation and use this to show that the system is controllable.
Solution. This is very similar to Example 11.12 (just different numbers).
We have
\[ A=-2,\qquad B=4. \]
Since \(n=1\), the Kalman controllability matrix is simply \(B\). Since a \(1\times 1\) matrix is surjective if and only if it is nonzero, we have that the Kalman controllability matrix is surjective and therefore the system is controllable.
The Hautus controllability matrix is
\[ \bbm {s+2&4}. \]
This matrix is surjective no matter what \(s\) is (for a given \(b\in \mR \), we have \(b=\bbm {s+2&4}\bbm {0\\b/4}\), so that \(b\) is in the image of the Hautus controllability matrix).
The controllability Gramian is
\[ Q_T=\int _0^T 16\e ^{-4t}\,dt =\left [-4\e ^{-4t}\right ]_{t=0}^T =4\left (1-\e ^{-4T}\right ). \]
From Remark 11.7 we then see that a control which steers the system from state \(x^0\) to state \(x^1\) in time \(T\) is given by
\(\seteqnumber{0}{11.}{1}\)\begin{align*} u(t)&=\e ^{-2(t-T)}~\frac {1}{1-\e ^{-4T}}(x^1-\e ^{-2T}x^0). \end{align*}
The control Lyapunov equation is
\[ -4Q+16=0, \]
which gives \(Q=4\) as the infinite-time controllability Gramian. Since this is invertible, the system is controllable. □
(b) Consider the second order system
\[ \ddot {q}+q=u. \]
(i) Write this in the standard form \(\dot {x}=Ax+Bu\) with \(x:=\sbm {q\\\dot {q}}\).
(ii) Determine the Hautus controllability matrix and use this to show that the system is controllable.
Solution. We have
\[ A=\bbm {0&1\\-1&0},\qquad B=\bbm {0\\1}. \]
The Hautus controllability matrix is
\[ \bbm {sI-A&B}=\bbm {s&-1&0\\1&s&1}. \]
Selecting the last two columns, we obtain a matrix with determinant \(-1\). So we that the Hautus controllability matrix is surjective for all \(s\in \mC \). Therefore, the system is controllable. □
(c) For the input-state system \(\dot {x}=Ax+Bu\) with
\[ A=\bbm {5&4\\-3&-2},\qquad B=\bbm {1\\-1}, \]
determine the reachable subspace.
Solution. The Kalman controllability matrix is
\[ \bbm {B&AB}=\bbm {1&1\\-1&-1}. \]
The image of this is the space spanned by the vector \(\bbm {1\\-1}\). That space is therefore the reachable subspace. □
• We consider the situation as in Section 1.4 but without the external force \(F_e\) and without the control input so that the only input is \(v_e\). This gives
\[ A=\bbm {0&1\\\frac {-k}{m}&\frac {-d}{m}},\quad B=\bbm {-1\\0}, \]
where \(m,k>0\) and \(d\geq 0\). Determine whether or not the pair \((A,B)\) is controllable.
Solution. The Kalman controllability matrix is
\[ \bbm {B&AB}=\bbm {-1&0\\0&\frac {k}{m}}, \]
which is diagonal with nonzero entries on the diagonal and is therefore invertible and therefore surjective. It follows that the system is controllable. □