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Infinite Horizon Optimal Output Feedback Control for Linear Systems with State Equality Constraints

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Abstract

This note deals with the infinite horizon linear regulation problem using output feedback with state equality constraints. Similar to the corresponding state feedback problem of a previous work, an existence condition for the output feedback gain and, if it exists, all constrainable output feedback gains are determined. However, different from the fore-mentioned state feedback case, only the necessary conditions for the optimal output feedback gain which minimizes the given standard cost function are determined. The performance of the developed algorithm is demonstrated using numerical simulations for a simple model of divert control system and the lateral dynamics of an F-16 aircraft.

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Abbreviations

\( {\mathbf{A}} \) :

Matrices (upper case boldface)

\( {\mathbf{A}}^{\text{T}} \) :

Transpose matrix of \( {\mathbf{A}} \)

\( {\mathbf{A}}^{\dag } \) :

Moore–Penrose inverse of \( {\mathbf{A}} \)

\( {\text{tr}}({\mathbf{A}}) \) :

Trace of \( {\mathbf{A}} \)

\( {\mathbf{P}} > {\mathbf{0}} \) :

Positive definite matrices

\( {\mathbf{P}} \ge {\mathbf{0}} \) :

Positive semidefinite matrices

\( {\mathcal{A}} \) :

Linear spaces (calligraphic uppercase)

\( {\mathcal{N}}({\mathbf{A}}) \) :

Null-space of \( {\mathbf{A}} \)

\( {\mathbf{x}} \) :

Column vectors (lower case boldface)

\( E\{ {\mathbf{x}}\} \) :

Expectation of a random vector \( {\mathbf{x}} \)

\( y, \, Y \) :

Scalars (lower or upper case)

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Acknowledgements

This work was supported by 2018 Korea Aerospace University Faculty Research Grant.

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Correspondence to Sangho Ko.

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Appendix 1

Appendix 1

To prove that the condition of (28), (29), and (30) becomes, respectively, (32), (33), and (34), the following lemma [8] is useful for later use.

Let us denote the \( ij \)-element of a matrix \( {\mathbf{X}} \) as \( x_{ij} \), the square unit matrix such that all their elements are zero except the one located at the \( i \)th row and \( j \)th column which is unity as \( {\mathbf{E}}_{ij} \). Then, the following hold:

  • If \( \frac{\partial }{{\partial x_{ij} }}\text{tr} [{\mathbf{AX}}] = \text{tr} [{\mathbf{AE}}_{ij} ], \) then \( \frac{\partial }{{\partial {\mathbf{X}}}}\text{tr} [{\mathbf{AX}}] = {\mathbf{A}}^{\text{T}} . \)

  • If \( \frac{\partial }{{\partial x_{ij} }}\text{tr} [{\mathbf{AX}}^{\text{T}} ] = \text{tr} [{\mathbf{AE}}_{ji} ], \) then \( \frac{\partial }{{\partial {\mathbf{X}}}}\text{tr} [{\mathbf{AX}}^{\text{T}} ] = {\mathbf{A}}. \)

The relations are also needed [7, 8, 12]:

$$ \frac{\partial }{{\partial {\mathbf{B}}}}{\text{tr}}({\mathbf{ABC}}) = {\mathbf{A}}^{\text{T}} {\mathbf{C}}^{\text{T}} , $$
(51)
$$ \frac{\partial }{{\partial {\mathbf{B}}}}{\text{tr}}({\mathbf{AB}}^{\text{T}} {\mathbf{C}}) = {\mathbf{CA}}\quad {\text{or}}\quad \frac{\partial y}{{\partial {\mathbf{B}}^{\text{T}} }} = \left[ {\frac{\partial y}{{\partial {\mathbf{B}}}}} \right]^{\text{T}} . $$
(52)

The proof that the condition of (28), (29), and (30) corresponds to (32), (33), and (34) respectively, is as follows:

Proof for (32)

By differentiating (27) with respect to \( {\mathbf{S}} \), we have

$$ \frac{{\partial {\mathcal{H}}}}{{\partial {\mathbf{S}}}} = \frac{1}{2}\frac{{\partial \text{tr} [{\mathbf{PX}}]}}{{\partial {\mathbf{S}}}} + \frac{1}{2}\frac{{\partial \text{tr} [{\mathbf{gS}}]}}{{\partial {\mathbf{S}}}}, $$
(53)

and determine each term separately.

  • Calculation of \( \frac{1}{2}\frac{{\partial \text{tr} [{\mathbf{PX}}]}}{{\partial {\mathbf{S}}}} \).

Since the variable \( {\mathbf{S}} \) is an independent variable in the term, we obtain

$$ \frac{1}{2}\frac{{\partial \text{tr} [{\mathbf{PX}}]}}{{\partial {\mathbf{S}}}}{\mathbf{ = 0}}. $$
(54)
  • Calculation of \( \frac{1}{2}\frac{{\partial \text{tr} [{\mathbf{gS}}]}}{{\partial {\mathbf{S}}}} \).

From (22), we follow the approach as that used in (i).

$$ \frac{1}{2}\frac{{\partial {\text{tr[}}{\mathbf{gS}} ]}}{{\partial {\text{S}}}} = \frac{1}{2}{\mathbf{g}}^{\text{T}} = \frac{1}{2}\left( {{\mathbf{\check{A} }}_{\text{c}}^{\text{T}} {\mathbf{P}}^{\text{T}} + {\mathbf{P}}^{\text{T}} {\mathbf{\check{A} }}_{\text{c}} + {\mathbf{\check{Q} }}_{\text{c}} - {\mathbf{\check{H} }}_{\text{c}} {\mathbf{\check{G} C}} - {\mathbf{C}}^{\text{T}} {\mathbf{\check{G} }}^{\text{T}} {\mathbf{\check{H} }}_{\text{c}}^{\text{T}} + {\mathbf{C}}^{\text{T}} {\mathbf{\check{G} }}^{\text{T}} {\mathbf{\check{R} }}_{\text{c}} {\mathbf{\check{G} C}}} \right). $$
(55)

Finally, by adding the two terms (54) and (55), we obtain

$$ 0 = \frac{{\partial {\mathcal{H}}}}{{\partial {\mathbf{S}}}} = \frac{1}{2}\left( {{\mathbf{\check{A} }}_{\text{c}}^{\text{T}} {\mathbf{P}}^{\text{T}} + {\mathbf{P}}^{\text{T}} {\mathbf{\check{A} }}_{\text{c}} + {\mathbf{\check{Q} }}_{\text{c}} - {\mathbf{\check{H} }}_{\text{c}} {\mathbf{\check{G} C}} - {\mathbf{C}}^{\text{T}} {\mathbf{\check{G} }}^{\text{T}} {\mathbf{\check{H} }}_{\text{c}}^{\text{T}} + {\mathbf{C}}^{\text{T}} {\mathbf{\check{G} }}^{\text{T}} {\mathbf{\check{R} }}_{\text{c}} {\mathbf{\check{G} C}}} \right). $$
(56)

Since \( {\mathbf{P}} \) and \( {\mathbf{\check{R} }}_{\text{c}} \) are symmetric, positive semidefinite matrix by (5), (18), and (19), we have (32)

$$ 0 = {\mathbf{\check{A} }}_{\text{c}}^{\text{T}} {\mathbf{P}}^{\text{T}} + {\mathbf{P}}^{\text{T}} {\mathbf{\check{A} }}_{\text{c}} + {\mathbf{\check{Q} }}_{\text{c}} - {\mathbf{\check{H} }}_{\text{c}} {\mathbf{\check{G} C}} - {\mathbf{C}}^{\text{T}} {\mathbf{\check{G} }}^{\text{T}} {\mathbf{\check{H} }}_{\text{c}}^{\text{T}} + {\mathbf{C}}^{\text{T}} {\mathbf{\check{G} }}^{\text{T}} {\mathbf{\check{R} }}_{\text{c}} {\mathbf{\check{G} C}}. $$
(57)

Proof for (33)

By differentiating (27) with respect to \( {\mathbf{P}} \), we have

$$ \frac{{\partial {\mathcal{H}}}}{{\partial {\mathbf{P}}}}{\mathbf{ = }}\frac{1}{2}\frac{{\partial \text{tr} [{\mathbf{PX}}]}}{{\partial {\mathbf{P}}}} + \frac{1}{2}\frac{{\partial \text{tr} [{\mathbf{gS}}]}}{{\partial {\mathbf{P}}}}, $$
(58)

and determine each term separately.

  • Calculation of \( \frac{1}{2}\frac{{\partial \text{tr} [{\mathbf{PX}}]}}{{\partial {\mathbf{P}}}} \).

We follow the approach (50)

$$ \frac{1}{2}\frac{{\partial \text{tr} [{\mathbf{PX}}]}}{{\partial {\mathbf{P}}}} = \frac{1}{2}{\mathbf{X}}^{\text{T}} . $$
(59)
  • Calculation of \( \frac{1}{2}\frac{{\partial \text{tr} [{\mathbf{gS}}]}}{{\partial {\mathbf{P}}}}. \)

From (22),

$$ \begin{aligned} \frac{1}{2}\frac{{\partial {\text{tr[}}{\mathbf{gS}} ]}}{{\partial {\mathbf{P}}}} & = \frac{1}{2}\frac{\partial }{{\partial {\mathbf{P}}}}{\text{tr}}\left[ {{\mathbf{\check{A} }}_{\text{c}}^{\text{T}} {\mathbf{P}}^{\text{T}} {\mathbf{S}} + {\mathbf{P}}^{\text{T}} {\mathbf{\check{A} }}_{\text{c}} {\mathbf{S}} + {\mathbf{\check{Q} }}_{\text{c}} {\mathbf{S}} - {\mathbf{\check{H} }}_{\text{c}} {\mathbf{\check{G} CS}} - {\mathbf{C}}^{\text{T}} {\mathbf{\check{G} }}^{\text{T}} {\mathbf{\check{H} }}_{\text{c}}^{\text{T}} {\mathbf{S}} + {\mathbf{C}}^{\text{T}} {\mathbf{\check{G} }}^{\text{T}} {\mathbf{\check{R} }}_{\text{c}} {\mathbf{\check{G} CS}}} \right] \\ & = \frac{1}{2}\left( {{\mathbf{\check{A} }}_{\text{c}} {\mathbf{S}}^{\text{T}} + {\mathbf{S}}^{\text{T}} {\mathbf{\check{A} }}_{\text{c}}^{\text{T}} } \right). \\ \end{aligned} $$
(60)

Finally, by adding the two terms (59) and (60), we obtain

$$ 0 = \frac{{\partial {\mathcal{H}}}}{{\partial {\mathbf{S}}}} = \frac{1}{2}\left( {{\mathbf{X}}^{\text{T}} + {\mathbf{\check{A} }}_{\text{c}} {\mathbf{S}}^{\text{T}} + {\mathbf{S}}^{\text{T}} {\mathbf{\check{A} }}_{\text{c}}^{\text{T}} } \right). $$
(61)

Since \( {\mathbf{X}} \) and \( {\mathbf{S}} \) are symmetric by (25) and (26), we have (33)

$$ 0 = {\mathbf{\check{A} }}_{\text{c}} {\mathbf{S}} + {\mathbf{S\check{A} }}_{\text{c}}^{\text{T}} + {\mathbf{X}}. $$
(62)

Proof for (34)

By differentiating (27) with respect to \( {\mathbf{\check{G} }} \), we have

$$ \frac{{\partial {\mathcal{H}}}}{{\partial {\mathbf{\check{G} }}}} = \frac{1}{2}\frac{{\partial {\text{tr[}}{\mathbf{PX}} ]}}{{\partial {\mathbf{\check{G} }}}} + \frac{1}{2}\frac{{\partial {\text{tr[}}{\mathbf{gS}} ]}}{{\partial {\mathbf{\check{G} }}}}, $$
(63)

and determine each term separately.

  • Calculation of \( \frac{1}{2}\frac{{\partial {\text{tr[}}{\mathbf{PX}} ]}}{{\partial {\mathbf{\check{G} }}}} \).

Since the variable \( {\mathbf{\check{G} }} \) is an independent variable in the term, we obtain

$$ \frac{1}{2}\frac{{\partial \text{tr} \left[ {{\mathbf{PX}}} \right]}}{{\partial {\mathbf{\check{G} }}}}{\mathbf{ = 0}}, $$
(64)
  • Calculation of \( \frac{1}{2}\frac{{\partial {\text{tr}}\left[ {{\mathbf{gS}}} \right]}}{{\partial {\mathbf{\check{G} }}}}. \)

From (21) and (22)

$$ \begin{aligned} \frac{1}{2}\frac{{\partial {\text{tr[}}{\mathbf{gS}} ]}}{{\partial {\mathbf{\check{G} }}}} & = \frac{1}{2}\frac{\partial }{{\partial {\mathbf{\check{G} }}}}{\text{tr}}\left[ {{\mathbf{\check{A} }}_{\text{c}}^{\text{T}} {\mathbf{P}}^{\text{T}} {\mathbf{S}} + {\mathbf{P}}^{\text{T}} {\mathbf{\check{A} }}_{\text{c}} {\mathbf{S}} + {\mathbf{\check{Q} }}_{\text{c}} {\mathbf{S}} - {\mathbf{\check{H} }}_{\text{c}} {\mathbf{\check{G} CS}} - {\mathbf{C}}^{\text{T}} {\mathbf{\check{G} }}^{\text{T}} {\mathbf{\check{H} }}_{\text{c}}^{\text{T}} {\mathbf{S}} + {\mathbf{C}}^{\text{T}} {\mathbf{\check{G} }}^{\text{T}} {\mathbf{\check{R} }}_{\text{c}} {\mathbf{\check{G} CS}}} \right] \\ & = \frac{1}{2}\frac{\partial }{{\partial {\mathbf{\check{G} }}}}{\text{tr}}\left[ {\left( {{\mathbf{\check{A} }} - {\mathbf{BV}}_{2} {\mathbf{\check{G} C}}} \right)^{\text{T}} {\mathbf{P}}^{\text{T}} {\mathbf{S}} + {\mathbf{P}}^{\text{T}} \left( {{\mathbf{\check{A} }} - {\mathbf{BV}}_{2} {\mathbf{\check{G} C}}} \right){\mathbf{S}} + {\mathbf{\check{Q} }}_{\text{c}} {\mathbf{S}} } \right. \\ & \quad \left. { - {\mathbf{\check{H} }}_{\text{c}} {\mathbf{\check{G} CS}} - {\mathbf{C}}^{\text{T}} {\mathbf{\check{G} }}^{\text{T}} {\mathbf{\check{H} }}_{\text{c}}^{\text{T}} {\mathbf{S}} + {\mathbf{C}}^{\text{T}} {\mathbf{\check{G} }}^{\text{T}} {\mathbf{\check{R} }}_{\text{c}} {\mathbf{\check{G} CS}}} \right] \\ & = \frac{1}{2}\frac{\partial }{{\partial {\mathbf{\check{G} }}}}{\text{tr}}\left[ {{\mathbf{\check{A} }}^{\text{T}} {\mathbf{P}}^{\text{T}} {\mathbf{S}} - {\mathbf{C}}^{\text{T}} {\mathbf{\check{G} }}^{\text{T}} {\mathbf{V}}_{2}^{\text{T}} {\mathbf{B}}^{\text{T}} {\mathbf{P}}^{\text{T}} {\mathbf{S}} + {\mathbf{P}}^{\text{T}} {\mathbf{AS}} - {\mathbf{P}}^{\text{T}} {\mathbf{BV}}_{2} {\mathbf{\check{G} CS}} + {\mathbf{\check{Q} }}_{\text{c}} {\mathbf{S}}} \right. \\ & \quad \left. { - {\mathbf{\check{H} }}_{\text{c}} {\mathbf{\check{G} CS}} - {\mathbf{C}}^{\text{T}} {\mathbf{\check{G} }}^{\text{T}} {\mathbf{\check{H} }}_{\text{c}}^{\text{T}} {\mathbf{S}} + {\mathbf{C}}^{\text{T}} {\mathbf{\check{G} }}^{\text{T}} {\mathbf{\check{R} }}_{\text{c}} {\mathbf{\check{G} CS}}} \right] \\ & = \frac{1}{2}\left( { - {\mathbf{V}}_{2}^{\text{T}} {\mathbf{B}}^{\text{T}} {\mathbf{PSC}}^{\text{T}} - {\mathbf{V}}_{2}^{\text{T}} {\mathbf{B}}^{\text{T}} {\mathbf{PSC}}^{\text{T}} - {\mathbf{\check{H} }}_{\text{c}}^{\text{T}} {\mathbf{SC}}^{\text{T}} } \right. \\ & \quad \left. { - {\mathbf{\check{H} }}_{\text{c}}^{\text{T}} {\mathbf{SC}}^{\text{T}} + {\mathbf{\check{R} }}_{\text{c}} {\mathbf{\check{G} CSC}}^{\text{T}} + {\mathbf{\check{R} }}_{\text{c}} {\mathbf{\check{G} CSC}}^{\text{T}} } \right) \\ & = - {\mathbf{V}}_{2}^{\text{T}} {\mathbf{B}}^{\text{T}} {\mathbf{PSC}}^{\text{T}} - {\mathbf{\check{H} }}_{\text{c}}^{\text{T}} {\mathbf{SC}}^{\text{T}} + {\mathbf{\check{R} }}_{\text{c}} {\mathbf{\check{G} CSC}}^{\text{T}} . \\ \end{aligned} $$
(65)

Finally, by adding the two terms (64) and (65), we obtain

$$ 0 = \frac{{\partial {\mathcal{H}}}}{{\partial {\mathbf{\check{G} }}}} = - {\mathbf{V}}_{2}^{\text{T}} {\mathbf{B}}^{\text{T}} {\mathbf{PSC}}^{\text{T}} - {\mathbf{\check{H} }}_{\text{c}}^{\text{T}} {\mathbf{SC}}^{\text{T}} + {\mathbf{\check{R} }}_{\text{c}} {\mathbf{\check{G} CSC}}^{\text{T}} . $$
(66)

Since \( {\mathbf{\check{R} }}_{\text{c}} \) is positive definite (i.e., all eigenvalues greater than zero, which implies nonsingularity, \( {\mathbf{\check{R} }}_{\text{c}} > 0 \)) by (5) and (18), we have (34)

$$ {\mathbf{\check{G} CSC}}^{\text{T}} = {\mathbf{\check{R} }}_{\text{c}}^{ - 1} {\mathbf{V}}_{2}^{\text{T}} {\mathbf{B}}^{\text{T}} {\mathbf{PSC}}^{\text{T}} + {\mathbf{\check{R} }}_{\text{c}}^{ - 1} {\mathbf{\check{H} }}_{\text{c}}^{\text{T}} {\mathbf{SC}}^{\text{T}} . $$
(67)

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Cha, J., Kang, S. & Ko, S. Infinite Horizon Optimal Output Feedback Control for Linear Systems with State Equality Constraints. Int. J. Aeronaut. Space Sci. 20, 483–492 (2019). https://doi.org/10.1007/s42405-019-00145-w

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