In control theory, the state-transition matrix is a matrix whose product with the state vector x {\displaystyle x} at an initial time t 0 {\displaystyle t_{0}} gives x {\displaystyle x} at a later time t {\displaystyle t} . The state-transition matrix can be used to obtain the general solution of linear dynamical systems.
The state-transition matrix is used to find the solution to a general state-space representation of a linear system in the following form
where x ( t ) {\displaystyle \mathbf {x} (t)} are the states of the system, u ( t ) {\displaystyle \mathbf {u} (t)} is the input signal, A ( t ) {\displaystyle \mathbf {A} (t)} and B ( t ) {\displaystyle \mathbf {B} (t)} are matrix functions, and x 0 {\displaystyle \mathbf {x} _{0}} is the initial condition at t 0 {\displaystyle t_{0}} . Using the state-transition matrix Φ ( t , τ ) {\displaystyle \mathbf {\Phi } (t,\tau )} , the solution is given by:[1][2]
The first term is known as the zero-input response and represents how the system's state would evolve in the absence of any input. The second term is known as the zero-state response and defines how the inputs impact the system.
The most general transition matrix is given by a product integral, referred to as the Peano–Baker series
where I {\displaystyle \mathbf {I} } is the identity matrix. This matrix converges uniformly and absolutely to a solution that exists and is unique.[2] The series has a formal sum that can be written as
where T {\displaystyle {\mathcal {T}}} is the time-ordering operator, used to ensure that the repeated product integral is in proper order. The Magnus expansion provides a means for evaluating this product.
The state transition matrix Φ {\displaystyle \mathbf {\Phi } } satisfies the following relationships. These relationships are generic to the product integral.
1. It is continuous and has continuous derivatives.
2, It is never singular; in fact Φ − 1 ( t , τ ) = Φ ( τ , t ) {\displaystyle \mathbf {\Phi } ^{-1}(t,\tau )=\mathbf {\Phi } (\tau ,t)} and Φ − 1 ( t , τ ) Φ ( t , τ ) = I {\displaystyle \mathbf {\Phi } ^{-1}(t,\tau )\mathbf {\Phi } (t,\tau )=\mathbf {I} } , where I {\displaystyle \mathbf {I} } is the identity matrix.
3. Φ ( t , t ) = I {\displaystyle \mathbf {\Phi } (t,t)=\mathbf {I} } for all t {\displaystyle t} .[3]
4. Φ ( t 2 , t 1 ) Φ ( t 1 , t 0 ) = Φ ( t 2 , t 0 ) {\displaystyle \mathbf {\Phi } (t_{2},t_{1})\mathbf {\Phi } (t_{1},t_{0})=\mathbf {\Phi } (t_{2},t_{0})} for all t 0 ≤ t 1 ≤ t 2 {\displaystyle t_{0}\leq t_{1}\leq t_{2}} .
5. It satisfies the differential equation ∂ Φ ( t , t 0 ) ∂ t = A ( t ) Φ ( t , t 0 ) {\displaystyle {\frac {\partial \mathbf {\Phi } (t,t_{0})}{\partial t}}=\mathbf {A} (t)\mathbf {\Phi } (t,t_{0})} with initial conditions Φ ( t 0 , t 0 ) = I {\displaystyle \mathbf {\Phi } (t_{0},t_{0})=\mathbf {I} } .
6. The state-transition matrix Φ ( t , τ ) {\displaystyle \mathbf {\Phi } (t,\tau )} , given by
where the n × n {\displaystyle n\times n} matrix U ( t ) {\displaystyle \mathbf {U} (t)} is the fundamental solution matrix that satisfies
7. Given the state x ( τ ) {\displaystyle \mathbf {x} (\tau )} at any time τ {\displaystyle \tau } , the state at any other time t {\displaystyle t} is given by the mapping
In the time-invariant case, we can define Φ {\displaystyle \mathbf {\Phi } } , using the matrix exponential, as Φ ( t , t 0 ) = e A ( t − t 0 ) {\displaystyle \mathbf {\Phi } (t,t_{0})=e^{\mathbf {A} (t-t_{0})}} . [4]
In the time-variant case, the state-transition matrix Φ ( t , t 0 ) {\displaystyle \mathbf {\Phi } (t,t_{0})} can be estimated from the solutions of the differential equation u ˙ ( t ) = A ( t ) u ( t ) {\displaystyle {\dot {\mathbf {u} }}(t)=\mathbf {A} (t)\mathbf {u} (t)} with initial conditions u ( t 0 ) {\displaystyle \mathbf {u} (t_{0})} given by [ 1 , 0 , … , 0 ] T {\displaystyle [1,\ 0,\ \ldots ,\ 0]^{\mathrm {T} }} , [ 0 , 1 , … , 0 ] T {\displaystyle [0,\ 1,\ \ldots ,\ 0]^{\mathrm {T} }} , ..., [ 0 , 0 , … , 1 ] T {\displaystyle [0,\ 0,\ \ldots ,\ 1]^{\mathrm {T} }} . The corresponding solutions provide the n {\displaystyle n} columns of matrix Φ ( t , t 0 ) {\displaystyle \mathbf {\Phi } (t,t_{0})} . Now, from property 4, Φ ( t , τ ) = Φ ( t , t 0 ) Φ ( τ , t 0 ) − 1 {\displaystyle \mathbf {\Phi } (t,\tau )=\mathbf {\Phi } (t,t_{0})\mathbf {\Phi } (\tau ,t_{0})^{-1}} for all t 0 ≤ τ ≤ t {\displaystyle t_{0}\leq \tau \leq t} . The state-transition matrix must be determined before analysis on the time-varying solution can continue.