State Equation

ME 381
Mechanical and
Aerospace Control
Systems
Dr. Robert G. Landers

State Equation Solution

State Equation Solution

Dr. Robert G. Landers

Unforced Response

2

The state equation for an unforced dynamic system is
Assume the solution is

x ( t ) = e At x ( 0 )

The derivative of eAt with respect to time is

d ( e At ) dt Checking the solution

x ( t ) = Ax ( t )

= Ae At

x ( t ) = Ax ( t ) ⇒ Ae At x ( 0 ) = Ae At x ( 0 )

Letting Φ(t) = eAt, the solution is written as

x (t ) = Φ (t ) x ( 0)

The matrix Φ(t) is called the state transition matrix. The state transition matrix transforms the initial conditions.

State Equation Solution

Dr. Robert G. Landers

Properties of State Transition Matrices
State transition matrix evaluated at t = 0

Φ ( 0 ) = eA 0 = I

Inverse of the state transition matrix

Φ (t ) = e = (e
At

)

− At −1

= ⎡Φ ( −t ) ⎤
⎣
⎦

−1

−1

→ ⎡Φ ( t ) ⎤ = Φ ( −t )
⎣
⎦

State transition matrix evaluated at t1 + t2

Φ ( t1 + t2 ) = e A(t1 +t2 ) = e At1 e At2 = Φ ( t1 ) Φ ( t2 ) = Φ ( t2 ) Φ ( t1 )
State transition matrix raised to a power

⎡Φ ( t ) ⎤ = e At e At
⎣
⎦ n e At = e[

A + A + + A ]t

= e nAt = Φ ( nt )

3

State Equation Solution

Dr. Robert G. Landers

Forced Response
The state equation for a forced dynamic system is

Rearranging

Multiply by e–At

4 x ( t ) = Ax ( t ) + Bu ( t )

x ( t ) − Ax ( t ) = Bu ( t )

e

− At

Integrating from τ to t

d − At
⎡ x ( t ) − Ax ( t ) ⎤ = ⎡e x ( t ) ⎤ = e − At Bu ( t )
⎣
⎦ dt ⎣
⎦
λ =t

t

⎡e − Aλ x ( λ ) ⎤ = ∫ e − Aλ Bu ( λ ) d λ
⎣
⎦ λ =τ τ State Equation Solution

Dr. Robert G. Landers

Forced Response

5 t Evaluating the left hand side

e − At x ( t ) − e − Aτ x (τ ) = ∫ e − Aλ Bu ( λ ) d λ τ t

Solving for x(t)

x ( t ) = e A( t −τ ) x (τ ) + ∫ e A( t −λ ) Bu ( λ ) d λ free response

τ

forced response

Solving for y(t)

y ( t ) = Cx (