Solution of LTI State Equation

8.2 Solution of LTI State Equation

8.2.1 State Transition Matrix

For later use, we will define the continuous-time/discrete-time LTI state transition matrices and examine their properties.

Definition 8.2 LTI State Transition Matrix – Fundamental Matrix For an LTI system described by the state equations (8.1.3a)/(8.1.3b), the LTI state transition matrix or fundamental matrix φ (t )/φ[n] is an N × N matrix, which is multiplied with the initial state x (0)/x[0] to make the state x(t)/x[n] at any time t /n ≥ 0 as

x(t) = φ(t)x(0) x[n] = φ[n]x[0] and satisfies the homogeneous state equation with zero input x ′ (t ) = Ax(t)

x[n + 1] = Ax[n] ;φ ′ (t )x(0) = Aφ(t)x(0)

; φ[n + 1]x[0] = Aφ[n]x[0] ;

φ ′ (t ) = Aφ(t)

(8.2.1a)

φ [n + 1] = Aφ[n] (8.2.1b)

where the initial condition is φ (0)/φ[0] = I (an N × N identity matrix). To find φ(t ), we make use of Tables A.2(5)/B.7(2) (with n 1 = 1) to take the

(unilateral) Laplace/z -transform of both sides of Eqs. (8.2.1a)/(8.2.1b) as sΦ (s) − φ(0) = AΦ(s)

zΦ [z] − zφ[0] = AΦ[z] and solve this for Φ(s)/Φ[z] as Φ (s) = [s I − A] −1 φ (0) = [s I − A] −1

Φ [z] = [z I − A] −1 zφ [0] = [z I − A] −1 z = [I − As −1 ] −1 s −1

= [I − z −1 A ] −1

=Is 2 + As +A +··· = I + Az −1 +A z −2 +··· Now we use Table A.1(4)/Eq. (4.1.1) to take the inverse transform of Φ(s)/Φ[z]

{Φ(s)} = L −1 {[s I − A] } φ [n] = Z −1 {Φ[z]} = Z −1 {[z I − A] −1 z}

(8.2.2b) = I + At +

A 2 2 n At =A

t 2! +···=e

(8.2.2a)

which is the continuous-time/discrete-time LTI state transition or fundamental matrix . This result can be verified by substituting Eqs. (8.2.2a)/(8.2.2b) into Eqs. (8.2.1a)/(8.2.1b), respectively.

8.2 Solution of LTI State Equation 365 The LTI state transition matrices possess the following properties:

< Properties of the LTI state transition matrix>

1) φ(t 1 )φ(t 2 ) = φ(t 1 +t 2 )∀t 1 , t 2 1) φ[n 1 ]φ[n 2 ] = φ[n 1 +n 2 ]∀n 1 , n 2

2) φ(−t) = φ −1 (t ); φ(0) = I

2) φ[−n] = φ −1 [n]; φ[0] = I if φ[n]

3) φ(t ) = e At is nonsingular ∀ t < ∞ is nonsingular.

8.2.2 Transformed Solution

To solve the LTI state equations (8.1.3a)/(8.1.3b), we make use of Tables A.2(5)/ B.7(2) (with n 1 = 1) to take the (unilateral) Laplace/z -transform of both sides and write

sX (s) − x(0) = AX(s) + BU (s) zX [z] − zx[0] = AX[z] + BU [z] ; [s I − A]X(s) = x(0) + BU (s)

; [z I − A]X[z] = zx[0] + BU [z] which can be solved for X (s)/ X [z] as

X (s) = [s I − A] −1 x(0)

X [z] = [z I − A] −1 zx [0] +[s I − A] −1 −1 BU (s) (8.2.3a) +[z I − A] BU [z] (8.2.3b)

Now, we will find the inverse transform of this transformed solution: x (t ) = L −1 {[s I − A] −1 }x(0)

x[n] = Z −1 {[z I − A] −1 z}x[0] +L −1 {[s I − A] −1 BU (s)}

+Z −1 {[z I − A] −1 BU [z]}

(8.2.4b) We can use Eq. (8.2.2), the convolution property, and the causality assumption

(8.2.4a)

to write

L (8.2.2b) −1 At n {[s I − A] −1 } =e (8.2.5a) Z −1 −1 {[z I − A] z} =A (8.2.5b) L −1

(8.2.2a)

{[s I − A] −1 BU (s)}

Z −1 {[z I − A] −1 BU [z]} =L −1 {[s I − A] −1 }∗L −1 {BU (s)}

B.7(4)

−1 −1 (8.2.5a) At

=Z {[z I − A] }∗Z {BU [z]} =e ∗ Bu(t)

=A n−1 ∗ Bu[n]

(A.17) t

e = A (t−τ ) Bu (τ )dτ

(8.2.6a)

n−1

A n−1−m Bu [m] (8.2.6b)

m=0

Substituting Eqs. (8.2.5) and (8.2.6) into Eq. (8.2.4) yields the solution of the LTI state equation as

366 8 State Space Analysis of LTI Systems

A e n−1−m Bu (τ )dτ Bu [m] x(t) = e x(0) +

t n−1 At

A (t−τ ) n x[n] =A x[0] +

0 m=0 t

= φ[n]x[0]+ = φ(t)x(0) +

φ (t − τ )Bu(τ )dτ

n−1

[n − 1 − m]Bu[m] (8.2.7b)

(8.2.7a)

m=0

which is referred to as the state transition equation. Note that, if the initial time is t 0 / n 0 , then we will have

x(t) = e n−1−m x(t 0 )+

x[n] =A

= φ[n − n 0 ]x[n 0 ]+ = φ(t − t 0 )x(t 0 )+

(t − τ )Bu(τ )dτ

n−1

0 m=n φ 0 [n − 1 − m]Bu[m]

(8.2.8a)

(8.2.8b)

Example 8.1 Solving a State Equation (a) Consider a continuous-time LTI system described by the following differential

equation:

(E8.1.1) Applying the procedure illustrated in Sects. 1.3.4 and 8.1, we can write the state

y ′′ (t ) + y ′ (t ) = u(t)

equation as

0 −1 , B= 1 , C= '10(,D=0 Thus we have

A=

−1 1 −1 −1 1 [s I − A] =

0s −

0 s+1

s (s + 1)

s −1 − (s + 1) −1

(s + 1) −1

(E8.1.4)

8.2 Solution of LTI State Equation 367 so that the state transition matrix is

φ (t ) =L {[s I − A] } B.8(3),(6) =

for t ≥ 0 (E8.1.5)

Therefore, from Eq. (8.2.7a), we can write the solution of the state equation as

1 (t ) (8.2.7a)

−t

1 (0) −(t−τ ) u

x 2 (t ) (E8.1.5) = 0 e −t

x 2 (0) + 0 e −(t−τ ) (τ )dτ for t ≥ 0 (E8.1.6)

(b) Consider a discrete-time LTI system described by the following state equation:

u [n] (E8.1.7)

1 [n]

y [n] = '1 0( (E8.1.8)

C= '10(,D=0

1−e

0 e −T

For this system, we have )

= (z − 1)(z − e −T )

0 z−1

= −T z−1 (z−1)(z−e −T ) z−1 z−1 − z−e

(1−e −T )z

so that the state transition matrix is φ (8.2.2b) [n] =Z −1

(E8.1.9)

{[z I − A] −1 z} B.8(3),(6) = 0 e −nT

−nT

for n ≥ 0 (E8.1.10)

as would be obtained from φ[n] (8.2.2b) n =A . Therefore, from Eq. (8.2.7b), we can write the solution of the state equation as

1 [n] (8.2.7b)

)e −(n−1−m)T

x 2 [n] (E8.1.10) = 0 e −nT

x 2 [0] +

(1 − e −T )e −(n−1−m)T

u [m]

m=0

for n ≥ 0

(E8.1.11)

368 8 State Space Analysis of LTI Systems

8.2.3 Recursive Solution

The discrete-time state equation has a recursive form and therefore is well-suited for a digital computer, which can be programmed to perform the following compu- tation:

x[1] = Ax[0] + Bu[0] x[2] = Ax[1] + Bu[1] = A 2 x[0] + ABu[0] + Bu[1]

3 x[3] = Ax[2] + Bu[2] = A 2 x[0] + A Bu [0] + ABu[1] + Bu[2] (8.2.9) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..

n−1

x[n] = A x[0] +

A n−1−m Bu [m]

m=0

If the continuous-time state equation is somehow discretized, it can be solved recursively, too.