State-Space System Identification Without a Hold Device by Lagrange Interpolation

Abstract

This paper presents a solution to the state-space system identification problem from discrete-time data without a hold device on the input. We consider the case where the inter-sampled input is approximated by 4-point Lagrange interpolation. The resultant discrete-time state-space model becomes non-standard and non-causal, but a mathematically equivalent discrete-time model is found so that standard system identification can be adapted to identify it. From the identified equivalent model, a continuous-time state-space model of the system is recovered. Results for the special case of 2-point linear interpolation are also provided.

Appendices

Appendix A: Discrete-Time Model by Lagrange Interpolation

Consider a continuous-time state-space model

$$ \label {eq:continuous state-space} \begin{array}{lll} \dot{x}(t) &= A_{c}x(t) + B_{c}u(t) \\ y(t) &= Cx(t) + Du(t) \end{array} $$

(43)

where the continuous excitation input u(t) at any time t between kT and (k + 1)T follows a 4-point Lagrange interpolation according to Eqs. 3–6,

$$ u(t) = {c_{- 1}}(t)u(k - 1) + {c_{0}}(t)u(k) + {c_{1}}(t)u(k + 1) + {c_{2}}(t)u(k + 2) $$

(44)

In this Appendix, we explain how to convert the continuous-time state-space model to discrete time when the input between sampling instants is obtained from a 4-point Lagrange interpolation. Such a continuous-time model has an exact discrete-time representation of the form

$$ \begin{array}{@{}rcl@{}} x(k + 1) &=& Ax(k) + {B_{- 1}}u(k - 1) + {B_{0}}u(k) + {B_{1}}u(k + 1) + {B_{2}}u(k+2)\\ y(k) &=& Cx(k) + Du(k) \end{array} $$

(45)

where B_− 1, B₀, B₁, and B₂ are given in Eq. 10. The discrete-time model is derived by substituting u(t) into Eq. 7 and carrying out the integration

$$ \int\limits_{kT}^{(k + 1)T} {{e^{{A_{c}}((k + 1)T - \tau )}}} {B_{c}}u(\tau )d\tau $$

With the aid of the following integrals involving the matrix exponentials,

$$ \begin{array}{@{}rcl@{}} \int {{e^{{A_{c}}t}}} dt &=& A_{c}^{- 1}{e^{{A_{c}}t}} + c \end{array} $$

(46)

$$ \begin{array}{@{}rcl@{}} \int {{t}{e^{{A_{c}}{t}}}} d{t} &=& {t}A_{c}^{- 1}{e^{{A_{c}}{t}}} - A_{c}^{- 1}\int {{e^{{A_{c}}{t}}}} d{t}\\ &=& A_{c}^{- 1}{t}{e^{{A_{c}}{t}}} - A_{c}^{- 2}{e^{{A_{c}}{t}}} + c \end{array} $$

(47)

$$ \begin{array}{@{}rcl@{}} \int {{{t}^{2}}{e^{{A_{c}}{t}}}} d{t} &=& {{t}^{2}}A_{c}^{- 1}{e^{{A_{c}}{t}}} - 2A_{c}^{- 1}\int {{t}{e^{{A_{c}}{t}}}} d{t}\\ &=& A_{c}^{- 1}{{t}^{2}}{e^{{A_{c}}{t}}} - 2A_{c}^{- 2}{t}{e^{{A_{c}}{t}}} + 2A_{c}^{- 3}{e^{{A_{c}}{t}}} + c \end{array} $$

(48)

$$ \begin{array}{@{}rcl@{}} \int {{{t}^{3}}{e^{{A_{c}}{t}}}} d{t} &=& {{t}^{3}}A_{c}^{- 1}{e^{{A_{c}}{t}}} - 2A_{c}^{- 1}\int {{{t}^{2}}{e^{{A_{c}}{t}}}} d{t}\\ &=& A_{c}^{- 1}{{t}^{3}}{e^{{A_{c}}{t}}} - 3A_{c}^{- 2}{{t}^{2}}{e^{{A_{c}}{t}}} + 6A_{c}^{- 3}{t}{e^{{A_{c}}{t}}} - 6A_{c}^{- 4}{e^{{A_{c}}{t}}} + c \end{array} $$

(49)

it can be shown that for $A = {e^{{A_{c}}T}}$,

$$ {\int\limits_{0}^{T}} {{e^{{A_{c}}t }}dt } = A_{c}^{- 1}{e^{{A_{c}}T}} - A_{c}^{- 1} = A_{c}^{- 1}A - A_{c}^{- 1} $$

(50)

$$ {\int\limits_{0}^{T}} {t {e^{{A_{c}}t }}dt } = TA_{c}^{- 1}A - A_{c}^{- 2}A + A_{c}^{- 2} $$

(51)

$$ {\int\limits_{0}^{T}} {{t^{2}}{e^{{A_{c}}t }}dt } = {T^{2}}A_{c}^{- 1}A - 2TA_{c}^{- 2}A + 2A_{c}^{- 3}A - 2A_{c}^{- 3} $$

(52)

$$ {\int\limits_{0}^{T}} {{t^{3}}{e^{{A_{c}}t }}dt } = {T^{3}}A_{c}^{- 1}A - 3{T^{2}}A_{c}^{- 2}A + 6TA_{c}^{- 3}A - 6A_{c}^{- 4}A + 6A_{c}^{- 4} $$

(53)

The following provides the details of the derivation.

$$ {B_{- 1}} = \left[ {\int\limits_{kT}^{(k + 1)T} {{e^{{A_{c}}((k+1)T - \tau )}}} {c_{- 1}}(\tau )d\tau } \right]{B_{c}} $$

(54)

where

$$ {c_{- 1}(\tau)} = \frac{{(\tau - kT)(\tau - (k + 1)T)(\tau - (k + 2)T)}}{{(- T)(- 2T)(- 3T)}} $$

(55)

Changing the variable of integration from τ to t where t = (k + 1)T − τ reveals that B_− 1 is time-invariant. The expression for B_− 1 becomes

$$ \begin{array}{lll} {B_{- 1}} &= - \frac{1}{{6{T^{3}}}}\left[ {{\int\limits_{0}^{T}} {{e^{{A_{c}}t }}} \left( {T - t } \right)\left( { - t } \right)\left( { - T - t } \right)dt } \right]{B_{c}}\\ &= - \frac{1}{{6{T^{3}}}}\left[ {{\int\limits_{0}^{T}} {\left( {{T^{2}}t - {t^{3}}} \right){e^{{A_{c}}t }}dt } } \right]{B_{c}}\\ &= \frac{1}{{6{T^{3}}}}\left( { - 2{T^{2}}A_{c}^{- 2}A + 6TA_{c}^{- 3}A - 6A_{c}^{- 4}A - {T^{2}}A_{c}^{- 2} + 6A_{c}^{- 4}} \right) B_{c} \end{array} $$

(56)

Similarly,

$$ {B_{0}} = \left[ {\int\limits_{kT}^{(k + 1)T} {{e^{{A_{c}}((k+1)T - \tau )}}} {c_{0}}(\tau )d\tau } \right]{B_{c}} $$

(57)

$$ {B_{1}} = \left[ {\int\limits_{kT}^{(k + 1)T} {{e^{{A_{c}}((k+1)T - \tau )}}} {c_{1}}(\tau )d\tau } \right]{B_{c}} $$

(58)

$$ {B_{2}} = \left[ {\int\limits_{kT}^{(k + 1)T} {{e^{{A_{c}}((k+1)T - \tau )}}} {c_{2}}(\tau )d\tau } \right]{B_{c}} $$

(59)

where

$$ {c_{0}(\tau)} = \frac{{(\tau - (k - 1)T)(\tau - (k + 1)T)(\tau - (k + 2)T)}}{{(T)(-T)(- 2T)}} $$

(60)

$$ {c_{1}}(\tau) = \frac{{(\tau - (k - 1)T)(\tau - kT)(\tau - (k + 2)T)}}{{(2T)(T)(- T)}} $$

(61)

$$ {c_{2}}(\tau) = \frac{{(\tau - (k - 1)T)(\tau - kT)(\tau - (k + 1)T)}}{{(3T)(2T)(T)}} $$

(62)

Changing the integration variable from τ to t where t = (k + 1)T − τ eliminates k from the integrals. The expressions for B₀, B₁, B₂ become

$$ \begin{array}{@{}rcl@{}} {B_{0}} &=& \frac{1}{{2{T^{3}}}}\left[ {{\int\limits_{0}^{T}} {{e^{{A_{c}}t }}} (2T - t )(- t )(- T - t )dt } \right]{B_{c}}\\ &=& - \frac{1}{{2{T^{3}}}}\left[ {{\int\limits_{0}^{T}} {({t^{3}} - T{t^{2}} - 2{T^{2}}t ){e^{{A_{c}}t }}} dt } \right]{B_{c}} \end{array} $$

(63)

$$ \begin{array}{@{}rcl@{}} {B_{1}} &=& - \frac{1}{{2{T^{3}}}}\left[ {{\int\limits_{0}^{T}} {{e^{{A_{c}}t }}} (2T - t )(T - t )(- T - t )dt } \right]{B_{c}}\\ &=& \frac{1}{{2{T^{3}}}}\left[ {{\int\limits_{0}^{T}} {({t^{3}} - 2T{t^{2}} - T^{2}t + 2{T^{3}}){e^{{A_{c}}t }}} dt } \right]{B_{c}} \end{array} $$

(64)

$$ \begin{array}{@{}rcl@{}} {B_{2}} &=& \frac{1}{{6{T^{3}}}}\left[ {{\int\limits_{0}^{T}} {{e^{{A_{c}}t }}} (2T - t )(T - t )(- t )dt } \right]{B_{c}}\\ &=& - \frac{1}{{6{T^{3}}}}\left[ {{\int\limits_{0}^{T}} {({t^{3}} - 3T{t^{2}} + 2{T^{2}}t ){e^{{A_{c}}t }}} dt } \right]{B_{c}} \end{array} $$

(65)

Substituting the definite integrals into these expressions produces the desired results.

Appendix B: Equivalent Discrete-Time State-Space Models

In this Appendix, we show that the input-output relationship of the system

$$ \begin{array}{@{}rcl@{}} x(k + 1) &=& Ax(k) + {B_{- 1}}u(k - 1) + {B_{0}}u(k) + {B_{1}}u(k + 1) + {B_{2}}u(k+2)\\ y(k) &=& Cx(k) + Du(k) \end{array} $$

(66)

with initial condition x(1) is identical to the input-output relationship of the system

$$ \begin{array}{@{}rcl@{}} w(k + 1) &=& Aw(k) + Bu(k)\\ y(k) &=& Cw(k) + {D_{- 1}}u(k - 1) + {D_{0}}u(k) + {D_{1}}u(k + 1) \end{array} $$

(67)

with initial condition w(1) where

$$ w(1)=x(1)+A^{-1}B_{-1}u(0)-B_{1}u(1)-B_{2}u(2)-AB_{2}u(1) $$

(68)

The equivalent model is obtained with the same A matrix and without increasing the state dimension. The coefficients of the two state-space models are related to each other by

$$ \begin{array}{@{}rcl@{}} B &=& A^{-1}B_{-1}+B_{0}+AB_{1}+A^{2}B_{2}\\ D_{-1} &=& -CA^{-1}B_{-1}\\ D_{0} &=& D+C(B_{1}+AB_{2})\\ D_{1} &=& CB_{2} \end{array} $$

(69)

The state of the original system at any time step k + 1 is the sum of the contribution due to the initial state x(1) and the contribution due to each of the four input terms B_− 1u(k − 1), B₀u(k), B₁u(k + 1), B₂u(k + 2) denoted by X_− 1, X₀, X₁, X₂,

$$ x(k + 1) = {A^{k}}x(1) + {X_{- 1}} + {X_{0}} + {X_{1}} + {X_{2}} $$

(70)

where

$$ \begin{array}{@{}rcl@{}} {X_{- 1}} &=& \sum\limits_{n = 1}^{k} {{A^{k - n}}} {B_{-1}}u(n - 1) = \sum\limits_{n = 0}^{k - 1} {{A^{k - n - 1}}} {B_{-1}}u(n)\\ &=& {A^{k - 1}}{B_{-1}}u(0) + \sum\limits_{n = 1}^{k - 1} {{A^{k - n - 1}}} {B_{-1}}u(n) + {A^{- 1}}{B_{-1}}u(k) - {A^{- 1}}{B_{-1}}u(k)\\ &=& {A^{k - 1}}{B_{-1}}u(0) + \sum\limits_{n = 1}^{k} {{A^{k - n - 1}}} {B_{-1}}u(n) - {A^{- 1}}{B_{-1}}u(k)\\ &=& {A^{k}}\left( {{A^{- 1}}{B_{-1}}u(0)} \right) + \sum\limits_{n = 1}^{k} {{A^{k - n}}} \left( {{A^{- 1}}{B_{-1}}} \right)u(n) - {A^{- 1}}{B_{-1}}u(k) \end{array} $$

(71)

$$ {X_{0}} = \sum\limits_{n = 1}^{k} {{A^{k - n}}} {B_{1}}u(n) $$

(72)

$$ \begin{array}{@{}rcl@{}} {X_{1}} &=& \sum\limits_{n = 1}^{k} {{A^{k - n}}} {B_{1}}u(n + 1) = \sum\limits_{n = 2}^{k + 1} {{A^{k - n + 1}}} {B_{1}}u(n) + {A^{k}}{B_{1}}u(1) - {A^{k}}{B_{1}}u(1)\\ &=& - {A^{k}}{B_{1}}u(1) + {A^{k}}{B_{1}}u(1) + \sum\limits_{n = 2}^{k} {{A^{k - n + 1}}} {B_{1}}u(n) + {B_{1}}u(k + 1)\\ &=& - {A^{k}}{B_{1}}u(1) + \sum\limits_{n = 1}^{k} {{A^{k - n + 1}}} {B_{1}}u(n) + {B_{1}}u(k + 1)\\ &=& {A^{k}}\left( {{B_{1}}u(1)} \right) + \sum\limits_{n = 1}^{k} {{A^{k - n}}} \left( {A{B_{1}}} \right)u(n) + {B_{1}}u(k + 1) \end{array} $$

(73)

$$ \begin{array}{@{}rcl@{}} {X_{2}} &=& \sum\limits_{n = 1}^{k} {{A^{k - n}}} {B_{2}}u(n + 2)\\ &=& \sum\limits_{n = 3}^{k + 2} {{A^{k - n + 2}}} {B_{2}}u(n) + {A^{k}}{B_{2}}u(2) - {A^{k}}{B_{2}}u(2) + {A^{k + 1}}{B_{2}}u(1) - {A^{k + 1}}{B_{2}}u(1)\\ &=& \sum\limits_{n = 1}^{k + 2} {{A^{k - n + 2}}} {B_{2}}u(n) - {A^{k}}{B_{2}}u(2) - {A^{k + 1}}{B_{2}}u(1)\\ &=& \sum\limits_{n = 1}^{k} {{A^{k - n + 2}}} {B_{2}}u(n) + A{B_{2}}u(k + 1) + {B_{2}}u(k + 2)\! \!- {A^{k}}{B_{2}}u(2) - {A^{k + 1}}{B_{2}}u(1)\\ &=& - {A^{k}}\left( {{B_{2}}u(2) + A{B_{2}}u(1)} \right) + \sum\limits_{n = 1}^{k} {{A^{k - n + 2}}} {B_{2}}u(n) + A{B_{2}}u(k + 1) +\! {B_{2}}u(k + 2)\\ \end{array} $$

(74)

The expression for x(k + 1) becomes

$$ \begin{array}{@{}rcl@{}} x(k + 1) &=& {A^{k}}\left[ {x(1) + {A^{- 1}}{B_{- 1}}u(0) - {B_{0}}u(1) - {B_{2}}u(2) - A{B_{2}}u(1)} \right]\\ &&+\sum\limits_{n = 1}^{k} {{A^{k - n}}} \left( {{A^{- 1}}{B_{- 1}} + {B_{0}} + A{B_{2}} + {A^{2}}{B_{2}}} \right)u(n)\\ &&-{A^{- 1}}{B_{- 1}}u(k) + {B_{1}}u(k + 1) + A{B_{2}}u(k + 1) + {B_{2}}u(k + 2) \end{array} $$

(75)

Since y(k + 1) = Cx(k + 1) + Du(k + 1), it follows that

$$ \begin{array}{@{}rcl@{}} y(k + 1) &=& C\left[ {{A^{k}}w(1) + \sum\limits_{n = 1}^{k} {{A^{k - n}}} Bu(n)} \right] + D_{0}u(k + 1)\\ &&- C{A^{- 1}}{B_{- 1}}u(k) + C{B_{2}}u(k + 2) \end{array} $$

(76)

where

$$ w(1)=x(1)+A^{-1}B_{-1}u(0)-B_{1}u(1)-B_{2}u(2)-AB_{2}u(1) $$

(77)

$$ B = {A^{- 1}}{B_{- 1}} + {B_{0}} + A{B_{2}} + {A^{2}}{B_{2}} $$

(78)

$$ {D_{0}} = D + C\left( {{B_{1}} + A{B_{2}}} \right) $$

(79)

The portion defined as

$$ z(k+1)=C\left[ {{A^{k}}w(1) + \sum\limits_{n = 1}^{k} {{A^{k - n}}} Bu(n)} \right] +D_{0}u(k + 1) $$

(80)

can be generated by an (A,B,C,D₀) model with initial state w(1) and a single input term B,

$$ \begin{array}{lll} w(k + 1) &=& Aw(k) + Bu(k)\\ {\kern20pt}z(k) &=& Cw(k) + {D_{0}}u(k) \end{array} $$

(81)

Therefore,

$$ y(k + 1) = z(k + 1) - C{A^{- 1}}{B_{- 1}}u(k) + C{B_{2}}u(k + 2) $$

(82)

Equivalently,

$$ \begin{array}{lll} y(k) &= z(k) + {D_{- 1}}u(k - 1) + {D_{1}}u(k + 1)\\ &= Cw(k) + {D_{- 1}}u(k - 1) + {D_{0}}u(k) + {D_{1}}u(k + 1) \end{array} $$

(83)

where

$$ \begin{array}{lll} {D_{- 1}} &= - C{A^{- 1}}{B_{- 1}}\\ {D_{1}} &= C{B_{2}} \end{array} $$

(84)

The state equation for w(k) together with the final output equation define a state-space model with a single input matrix B that has identical input-output relationship to the original state-space model with four input matrices B_− 1, B₀, B₁, and B₂, where the initial state w(1) is related to x(1) by Eq. ??. In system identification we do not need to know that initial state w(1).

Appendix C: Discrete-Time Models by Linear Interpolation

Consider a continuous-time state-space system

$$ \label {eq:continuous state-space} \begin{array}{lll} \dot{x}(t) &= A_{c}x(t) + B_{c}u(t) \\ y(t) &= Cx(t) + Du(t) \end{array} $$

(85)

with a linear interpolation for u(t) between kT and (k + 1)T,

$$ u(t) = u(k) +\left[ \frac{{u(k + 1) - u(k)}}{T}\right] \left( {t - kT} \right)\\ $$

(86)

This relationship can be re-expressed as

$$ u(t) = {c_{0}}(t)u(k) + {c_{1}}(t)u(k + 1) $$

(87)

where the two coefficients c₀(t) and c₁(t) are

$$ {c_{0}}(t) = \frac{{t - \left( {k + 1} \right)T}}{{(- T)}} {c_{1}}(t) = \frac{{\left( {t - kT} \right)}}{T} $$

(88)

The corresponding discrete-time model can be shown to be [28]

$$ \begin{array}{@{}rcl@{}} x(k + 1) &=& Ax(k) + {B_{0}}u(k) + {B_{1}}u(k + 1)\\ y(k) &=& Cx(k) + Du(k) \end{array} $$

(89)

where A,B₀,B₁ are related to the original model A_c,B_c and the sampling interval T by

$$ \begin{array}{lll} A &= {e^{{A_{c}}T}}\\ {B_{0}} &= \frac{1}{T}\left( {TAA_{c}^{- 1} - AA_{c}^{- 2} + A_{c}^{- 2}} \right){B_{c}}\\ {B_{1}} &= \frac{1}{T}\left( {AA_{c}^{- 2} - A_{c}^{- 2} - TA_{c}^{- 1}} \right){B_{c}} \end{array} $$

(90)

Discretization with the linear interpolation assumption results in a state-space model in non-standard form. However, this non-standard two-input model with B₀u(k) and B₁u(k + 1) is identical to the following standard and mathematically equivalent model with a single B,

$$ \begin{array}{lll} w(k + 1) &=& Aw(k) + Bu(k)\\ {\kern20pt} y(k) &=& Cw(k) + {D_{0}}u(k) \end{array} $$

(91)

where B and D₀ are related to the (A,B₀,B₁,C,D) model by

$$ \begin{array}{lll} {\kern14pt}B &=& {B_{0}} + A{B_{1}}\\ {\kern10pt}{D_{0}} &=& D + C{B_{1}} \end{array} $$

(92)

where w(0) = x(0) − B₁u(0). Note that the equivalent model is obtained with the same A matrix and without increasing the state dimension. This single B state-space model can be identified from input-output data, then the original continuous-time state-space model can be recovered. First, A_c can be easily computed from A since $A = {e^{{A_{c}}T}}$. To find B_c, define

$$ \begin{array}{lll} {G_{0}} &= \frac{1}{T}\left( {TAA_{c}^{- 1} - AA_{c}^{- 2} + A_{c}^{- 2}} \right)\\ {G_{1}} &= \frac{1}{T}\left( {AA_{c}^{- 2} - A_{c}^{- 2} - TA_{c}^{- 1}} \right) \end{array} $$

(93)

so that B₀ = G₀B_c, and B₁ = G₁B_c, from which B_c and D can be solved from

$$ \begin{array}{lll} {B_{c}} &= {\left( {{G_{0}} + A{G_{1}}} \right)^{- 1}}B\\ D&=D_{0}-CG_{1}B_{c} \end{array} $$

(94)

Note that the equivalent model is in standard form for which an existing state-space system identification algorithm such as OKID can be directly applied without modification to find A,B,C,D₀ from which the original continuous-time state-space model A_c,B_c,C,D can be recovered.