Solving autonomous LTI Systems in Matrix Form

(1)

Solving autonomous LTI Systems in Matrix Form

(& all their possible behaviors)

•

Laplace transform of n 1^st order linear ODEs in matrix form

•

The inverse transform and matrix exponential

•

Possible system behaviors

•

Discrete-time LTI system (sampling at constant intervals)

1

(2)

Solving n 1 ^st order linear ODEs in matrix form

•

An autonomous (undriven, no u) LTI system with an n dimensional state:

•

Its Laplace transform is:

•

Rearranging:

•

is known as the resolvent.

•

To solve: simply find the inverse transform of the resolvent:

2

matrix exponential LTI •

LTI •

matrix exponential LTI

˙x = Ax

sX (s) − x(0) = AX(s)

(sI − A)X(s) = x (0)

X (s) = (sI − A)

⁻¹

x (0)

x (t) = L

⁻¹

!(sI − A)

⁻¹

" x(0) x (t) = Φ (t)x(0)

state tran- Φ (t) resolvent (sI − A)

⁻¹

t < 0 sition matrix

harmonic oscilator

˙x = Ax =

# 0 1

−1 0

$

x , x ∈ R

²

matrix exponential LTI •

LTI •

matrix exponential LTI

˙x = Ax

sX (s) − x(0) = AX(s)

(sI − A)X(s) = x (0)

X (s) = (sI − A)

⁻¹

x (0)

x (t) = L

⁻¹

!(sI − A)

⁻¹

" x(0) x (t) = Φ (t)x(0)

state tran- Φ (t) resolvent (sI − A)

⁻¹

t < 0 sition matrix

harmonic oscilator

˙x = Ax =

# 0 1

−1 0

$

x , x ∈ R

²

matrix exponential LTI •

LTI •

matrix exponential LTI

˙x = Ax

sX (s) − x(0) = AX(s)

(sI − A)X(s) = x (0)

X (s) = (sI − A)

⁻¹

x (0)

x (t) = L

⁻¹

!(sI − A)

⁻¹

" x(0) x (t) = Φ (t)x(0)

state tran- Φ (t) resolvent (sI − A)

⁻¹

t < 0 sition matrix

harmonic oscilator

˙x = Ax =

# 0 1

−1 0

$

x , x ∈ R

²

matrix exponential LTI •

LTI •

matrix exponential LTI

˙x = Ax

sX(s) − x(0) = AX(s)

(sI − A)X(s) = x(0)

X(s) = (sI − A)⁻¹x(0)

x(t) = L⁻¹ !(sI − A)⁻¹" x(0) x(t) = Φ(t)x(0)

state tran- Φ(t) resolvent (sI − A)⁻¹

t < 0 sition matrix

harmonic oscilator

˙x = Ax =

# 0 1

−1 0

$

x , x ∈ R²

LTI •

matrix exponential LTI

˙x = Ax

sX(s) − x(0) = AX(s)

(sI − A)X(s) = x(0)

X(s) = (sI − A)⁻¹x(0)

x(t) = L⁻¹ !(sI − A)⁻¹" x(0) x(t) = Φ(t)x(0)

t < 0 sition matrix

harmonic oscilator

˙x = Ax =

# 0 1

−1 0

$

x , x ∈ R²

(3)

•

The inverse transform of the resolvent, is known as the state transition matrix because the system is linearly transformed by it from its state at time 0 (either forwards or backwards in time)

•

Note the dependence on A’s characteristic polynomial: |sI-A|

j,i entry of resolvent (jth row, ith column) can be expressed via Crammer’s rule as:

•

So each of the entries in the resolvent matrix is a polynomial fraction in s whose denominator is A’s characteristic polynomial

•

Since |sI-A| has real coefficients, its roots (the eigenvalues of A) are either real or are complex conjugate pairs...

LTI •

matrix exponential LTI

˙x = Ax

sX(s) − x(0) = AX(s)

(sI − A)X(s) = x(0)

X(s) = (sI − A)⁻¹x(0)

x(t) = L⁻¹ !(sI − A)⁻¹" x(0) x(t) = Φ(t)x(0)

t < 0 sition matrix

harmonic oscilator

˙x = Ax =

# 0 1

−1 0

$

x , x ∈ R²

LTI •

matrix exponential LTI

˙x = Ax

sX(s) − x(0) = AX(s)

(sI − A)X(s) = x(0)

X(s) = (sI − A)⁻¹x(0)

x(t) = L⁻¹ !(sI − A)⁻¹" x(0) x(t) = Φ(t)x(0)

t < 0 sition matrix

harmonic oscilator

˙x = Ax =

# 0 1

−1 0

$

x , x ∈ R²

Eigenvalues of A and poles of resolvent

i, j entry of resolvent can be expressed via Cramer’s rule as (−1)^i+j det ∆_ij

det(sI − A)

where ∆_ij is sI − A with jth row and ith column deleted

• det ∆_ij is a polynomial of degree less than n, so i, j entry of resolvent has form f_ij(s)/X (s) where f_ij is polynomial with degree less than n

• poles of entries of resolvent must be eigenvalues of A

• but not all eigenvalues of A show up as poles of each entry (when there are cancellations between det ∆_ij and X (s))

Solution via Laplace transform and matrix exponential 10–11

Matrix exponential

(I − C)⁻¹ = I + C + C² + C³ + · · · (if series converges)

• series expansion of resolvent:

(sI − A)⁻¹ = (1/s)(I − A/s)⁻¹ = I

s + A

s² + A²

s³ + · · · (valid for |s| large enough) so

Φ(t) = L⁻¹ !(sI − A)⁻¹" = I + tA + (tA)²

2! + · · ·

Eigenvalues of A and poles of resolvent

i, j entry of resolvent can be expressed via Cramer’s rule as (−1)^i+j det ∆_ij

det(sI − A)

where ∆_ij is sI − A with jth row and ith column deleted

• det ∆_ij is a polynomial of degree less than n, so i, j entry of resolvent has form f_ij(s)/X (s) where f_ij is polynomial with degree less than n

• poles of entries of resolvent must be eigenvalues of A

• but not all eigenvalues of A show up as poles of each entry (when there are cancellations between det ∆_ij and X (s))

Matrix exponential

(I − C)⁻¹ = I + C + C² + C³ + · · · (if series converges)

• series expansion of resolvent:

(sI − A)⁻¹ = (1/s)(I − A/s)⁻¹ = I

s + A

s² + A²

s³ + · · · (valid for |s| large enough) so

Φ(t) = L⁻¹ !(sI − A)⁻¹" = I + tA + (tA)²

2! + · · ·

(4)

Example: Harmonic Oscillator

The phase plane looks like:

|sI-A| = s²+1 so the eigenvalues of A are i,-i

We could now use the Laplace transform tables to show that:

(but we show a general solution method instead).

4

matrix exponential LTI •

LTI •

matrix exponential LTI

˙x = Ax

sX (s) − x(0) = AX(s)

(sI − A)X(s) = x (0)

X (s) = (sI − A)

⁻¹

x (0)

x (t) = L

⁻¹

!(sI − A)

⁻¹

" x(0) x (t) = Φ (t)x(0)

state tran- Φ (t) resolvent (sI − A)

⁻¹

t < 0 sition matrix

harmonic oscilator

˙x = Ax =

# 0 1

−1 0

$

x , x ∈ R

²

matrix exponential LTI •

LTI •

matrix exponential LTI

˙x = Ax

sX (s) − x(0) = AX(s)

(sI − A)X(s) = x (0)

X (s) = (sI − A)

⁻¹

x (0)

x (t) = L

⁻¹

!(sI − A)

⁻¹

" x(0) x (t) = Φ (t)x(0)

state tran- Φ (t) resolvent (sI − A)

⁻¹

t < 0 sition matrix

harmonic oscilator

˙x = Ax =

# 0 1

−1 0

$

x , x ∈ R

²

!2 !1.5 !1 !0.5 0 0.5 1 1.5 2

!2

!1.5

!1

!0.5 0 0.5 1 1.5 2

sI− A =

! s −1

1 s

"

resolvent (sI − A)⁻¹ =

! s

s2+1 1 s2+1

−1 s2+1

2 s2+1

"

i,−i

Φ(t) = L⁻¹

#! s

s2+1

1 s2+1

−1 s2+1

2 s2+1

"$

=

! cos t sin t

−sin t cos t

"

t

c ≤ 1 (1 − c)⁻¹ = 1 + c + c² + c³ + . . .

|Re(λ)| < 1 C λ C ∈ R^n×n

(I − C)⁻¹ = I + C + C² + C³ + · · ·

2

!2 !1.5 !1 !0.5 0 0.5 1 1.5 2

!2

!1.5

!1

!0.5 0 0.5 1 1.5 2

sI − A =

! s −1

1 s

"

resolvent (sI − A)

⁻¹

=

!

^s

s²+1

1 s²+1

−1 s²+1

2 s²+1

"

i, −i

Φ (t) = L

⁻¹

#!

^s

s²+1

1 s²+1

−1 s²+1

2 s²+1

"$

=

! cos t sin t

−sin t cos t

"

t

c ≤ 1 (1 − c)

⁻¹

= 1 + c + c

²

+ c

³

+ . . .

|Re(λ)| < 1 C λ C ∈ R

^n×n

(I − C)

⁻¹

= I + C + C

²

+ C

³

+ · · ·

2

The resolvent:

A rotation matrix

!2 !1.5 !1 !0.5 0 0.5 1 1.5 2

!2

!1.5

!1

!0.5 0 0.5 1 1.5 2

sI − A =

! s −1 1 s

"

(sI − A)

⁻¹

=

!

_s

s²+1

1 s²+1 s−1²+1

s s²+1

"

i, −i A

Φ(t) = L

⁻¹

#!

_s

s²+1

1 s²+1 s−1²+1

s s²+1

"$

=

! cos t t

− t cos t

"

t

c ≤ 1 (1 − c)

⁻¹

= 1 + c + c

²

+ c

³

+ . . .

|Re(λ)| < 1 C λ C ∈ R

ⁿ^×n

(I − C)

⁻¹

= I + C + C

²

+ C

³

+ · · ·

!2 !1.5 !1 !0.5 0 0.5 1 1.5 2

!2

!1.5

!1

!0.5 0 0.5 1 1.5 2

sI − A =

! s −1

1 s

"

(sI − A)⁻¹ =

! _s

s²+1

1 s²+1 s−1²+1

s s²+1

"

i, −i A

Φ(t) = L⁻¹

#! _s

s²+1

1 s²+1 s−1²+1

s s²+1

"$

=

! cos t t

− t cos t

"

t

c ≤ 1

(1 − c)⁻¹ = 1 + c + c² + c³ + . . .

|Re(λ)| < 1 C λ C ∈ Rⁿ^×n

(I − C)⁻¹ = I + C + C² + C³ + · · ·

(5)

•

For c < 1

•

Similarly, for (with small enough real parts of its eigenvalues)

•

Plug in and (for large enough s)

•

We’ve seen that

•

So (due to the linearity of Laplace tr.) we can apply the inverse transform and get..

5

Matrix Exponential

!2 !1.5 !1 !0.5 0 0.5 1 1.5 2

!2

!1.5

!1

!0.5 0 0.5 1 1.5 2

sI − A =

! s −1

1 s

"

! ^s

s2+1

1 s2+1

−1 s2+1

2 s2+1

"

i, −i

Φ(t) = L⁻¹

#! ^s

s2+1

1 s2+1

−1 s2+1

2 s2+1

"$

=

! cos t sin t

−sin t cos t

"

t

c ≤ 1 (1 − c)⁻¹ = 1 + c + c² + c³ + . . .

|Re(λ)| < 1 C λ C ∈ R^n×n

(I − C)⁻¹ = I + C + C² + C³ + · · ·

2

!2 !1.5 !1 !0.5 0 0.5 1 1.5 2

!2

!1.5

!1

!0.5 0 0.5 1 1.5 2

sI − A =

! s −1

1 s

"

! ^s

s²+1

1 s²+1

−1 s²+1

2 s²+1

"

i, −i

Φ(t) = L⁻¹

#! ^s

s²+1

1 s²+1

−1 s²+1

2 s²+1

"$

=

! cos t sin t

−sin t cos t

"

t

c ≤ 1 (1 − c)⁻¹ = 1 + c + c² + c³ + . . .

|Re(λ)| < 1 C λ C ∈ R^n×n

(I − C)⁻¹ = I + C + C² + C³ + · · ·

2

!2 !1.5 !1 !0.5 0 0.5 1 1.5 2

!2

!1.5

!1

!0.5 0 0.5 1 1.5 2

sI − A =

! s −1 1 s

"

! ^s

s²+1

1 s²+1

−1 s²+1

2 s²+1

"

i, −i

Φ(t) = L⁻¹

#! ^s

s²+1

1 s²+1

−1 s²+1

2 s²+1

"$

=

! cos t sin t

−sin t cos t

"

t

c ≤ 1 (1 − c)⁻¹ = 1 + c + c² + c³ + . . .

|Re(λ)| < 1 C λ C ∈ R^n×n

(I − C)⁻¹ = I + C + C² + C³ + · · ·

2

s C = ^A_s

(sI − A)⁻¹ = 1 s

!

I − A s

"⁻1

= I

s + A

s² + A²

s³ + · · · L⁻¹ # ₁

sⁿ $ = _(n−1)!^tⁿ⁻¹ n ∈ Z L⁻¹ #(sI − A)⁻¹$ = I + tA + (tA)²

2! + · · ·

e^at = 1 + ta + (ta)²

2! + · · · matrix exponent e^M ≡ I + M + M²

2! + · · · =

∞

%

i=0

Mⁱ i! LTI

x(t) = Φ(t)x(0) = L⁻¹ #(sI − A)⁻¹$ x(0) = e^t^Ax(0)

LTI

x(0) A ∈ R^n×n ˙x = Ax

P ∈ C^n×n A

D =







λ₁ 0

. ..

0 λn





 ∈ C^n×n

P⁻¹AP = D A = PDP⁻¹

Aⁿ = PDP⁻¹PDP⁻¹ · · · PDP⁻¹ = PDⁿP⁻¹

state transition matrix Φ(t) = e^At = I + tA + (tA)²

2! + (tA)³

3! · · ·

= PIP⁻¹ + PtDP⁻¹ + P(tD)²P⁻¹

2! + P(tD)³P⁻¹

3! · · ·

= Pe^DtP⁻¹

= Pe







λ₁t

. ..

λnt







P⁻¹ 3

s C = ^A_s

(sI − A)⁻¹ = 1 s

!

I − A s

"⁻¹

= I

s + A

s² + A²

s³ + · · · L⁻¹ # ₁

2! + · · ·

e^at = 1 + ta + (ta)²

2! + · · · =

∞

%

i=0

Mⁱ i! LTI

x(t) = Φ(t)x(0) = L⁻¹ #(sI − A)⁻¹$ x(0) = e^t^Ax(0)

LTI

x(0) A ∈ R^n×n ˙x = Ax

P ∈ C^n×n A

D =







λ₁ 0

. ..

0 λⁿ





 ∈ C^n×n

2! + (tA)³

3! · · ·

= PIP⁻¹ + PtDP⁻¹ + P(tD)²P⁻¹

2! + P(tD)³P⁻¹

3! · · ·

= Pe^DtP⁻¹

= Pe







λ₁t

. ..

λⁿt







P⁻¹ 3

s C =

^A_s

(sI − A)

⁻¹

= 1 s

!

I − A s

"

⁻1

= I

s + A

s

²

+ A

²

s

³

+ · · · L

⁻¹

#

₁

sⁿ

$ =

_(n−1)!^tⁿ⁻¹

n ∈ Z L

⁻¹

#(sI − A)

⁻¹

$ = I + tA + (tA)

²

2! + · · ·

e

^at

= 1 + ta + (ta)

²

2! + · · · matrix exponent e

^M

≡ I + M + M

²

2! + · · · =

∞

%

i=0

M

ⁱ

i ! LTI

x (t) = Φ(t)x(0) = L

⁻¹

#(sI − A)

⁻¹

$ x(0) = e

^t^A

x (0)

LTI

x (0) A ∈ R

^n×n

˙x = Ax

P ∈ C

^n×n

A

D =







λ

₁

0 . ..

0 λ

ⁿ





 ∈ C

^n×n

P

⁻¹

AP = D A = PDP

⁻¹

A

ⁿ

= PDP

⁻¹

PDP

⁻¹

· · · PDP

⁻¹

= PD

ⁿ

P

⁻¹

state transition matrix Φ (t) = e

^At

= I + tA + (tA)

²

2! + (tA)

³

3! · · ·

= PIP

⁻¹

+ PtDP

⁻¹

+ P (tD)

²

P

⁻¹

2! + P (tD)

³

P

⁻¹

3! · · ·

= Pe

^Dt

P

⁻¹

= Pe







λ

₁

t

. ..

λ

ⁿ

t







P

⁻¹

3

(6)

•

This (infinite series), multiplied by x(0), is a general solution to LTI systems !

•

The above series looks a lot like the Tailor expansion of an exponent:

•

So we will borrow the exponent symbol and define a matrix exponent as

•

Using this new symbol we can write the solution

•

For a scalar system, the above solution is what we expect:

6

s C = ^A_s

(sI − A)⁻¹ = 1 s

!

I − A s

"⁻1

= I

s + A

s² + A²

s³ + · · · L⁻¹ # ₁

2! + · · ·

e^at = 1 + ta + (ta)²

2! + · · · =

∞

%

i=0

Mⁱ i! LTI

x(t) = Φ(t)x(0) = L⁻¹ #(sI − A)⁻¹$ x(0) = e^t^Ax(0)

LTI

x(0) A ∈ R^n×n ˙x = Ax

P ∈ C^n×n A

D =







λ₁ 0

. ..

0 λⁿ





 ∈ C^n×n

2! + (tA)³

3! · · ·

= PIP⁻¹ + PtDP⁻¹ + P(tD)²P⁻¹

2! + P(tD)³P⁻¹

3! · · ·

= Pe^DtP⁻¹

= Pe







λ₁t

. ..

λⁿt







P⁻¹ 3

s C =

^A_s

(sI − A)

⁻¹

= 1 s

!

I − A s

"

⁻1

= I

s + A

s

²

+ A

²

s

³

+ · · · L

⁻¹

#

₁

sⁿ

$ =

_(n−1)!^tⁿ⁻¹

n ∈ Z L

⁻¹

#(sI − A)

⁻¹

$ = I + tA + (tA)

²

2! + · · ·

e

^at

= 1 + ta + (ta)

²

2! + · · · matrix exponent e

^M

≡ I + M + M

²

2! + · · · =

∞

%

i=0

M

ⁱ

i ! LTI

x (t) = Φ(t)x(0) = L

⁻¹

#(sI − A)

⁻¹

$ x(0) = e

^t^A

x (0)

LTI

x (0) A ∈ R

^n×n

˙x = Ax

P ∈ C

^n×n

A

D =







λ

₁

0 . ..

0 λ

ⁿ





 ∈ C

^n×n

P

⁻¹

AP = D A = PDP

⁻¹

A

ⁿ

= PDP

⁻¹

PDP

⁻¹

· · · PDP

⁻¹

= PD

ⁿ

P

⁻¹

state transition matrix Φ (t) = e

^At

= I + tA + (tA)

²

2! + (tA)

³

3! · · ·

= PIP

⁻¹

+ PtDP

⁻¹

+ P (tD)

²

P

⁻¹

2! + P (tD)

³

P

⁻¹

3! · · ·

= Pe

^Dt

P

⁻¹

= Pe







λ

₁

t

. ..

λ

ⁿ

t







P

⁻¹

3

s C = ^A_s

(sI − A)⁻¹ = 1 s

!

I − A s

"⁻1

= I

s + A

s² + A²

s³ + · · · L⁻¹ # ₁

2! + · · ·

e^at = 1 + ta + (ta)²

2! + · · · =

∞

%

i=0

Mⁱ i! LTI

x(t) = Φ(t)x(0) = L⁻¹ #(sI − A)⁻¹$ x(0) = e^t^Ax(0)

LTI

x(0) A ∈ R^n×n ˙x = Ax

P ∈ C^n×n A

D =







λ₁ 0

. ..

0 λn





 ∈ C^n×n

2! + (tA)³

3! · · ·

= PIP⁻¹ + PtDP⁻¹ + P(tD)²P⁻¹

2! + P(tD)³P⁻¹

3! · · ·

= Pe^DtP⁻¹

= Pe







λ₁t

. ..

λⁿt







P⁻¹ 3

s C = ^A_s

(sI − A)⁻¹ = 1 s

!

I − A s

"⁻1

= I

s + A

s² + A²

s³ + · · · L⁻¹ # ₁

2! + · · ·

e^at = 1 + ta + (ta)²

2! + · · · =

∞

%

i=0

Mⁱ i! LTI

x(t) = Φ(t)x(0) = L⁻¹ #(sI − A)⁻¹$ x(0) = e^t^Ax(0)

LTI

x(0) A ∈ R^n×n ˙x = Ax

P ∈ C^n×n A

D =







λ₁ 0

. ..

0 λⁿ





 ∈ C^n×n

2! + (tA)³

3! · · ·

= PIP⁻¹ + PtDP⁻¹ + P(tD)²P⁻¹

2! + P(tD)³P⁻¹

3! · · ·

= Pe^DtP⁻¹

= Pe







λ₁t

. ..

λnt







P⁻¹ 3

• looks like ordinary power series

e

^at

= 1 + ta + (ta)

²

2! + · · · with square matrices instead of scalars . . .

• define matrix exponential as

e

^M

= I + M + M

²

2! + · · · for M ∈ R

^n×n

(which in fact converges for all M )

• with this definition, state-transition matrix is

Φ(t) = L

⁻¹

!(sI − A)

⁻¹

" = e

^tA

Matrix exponential solution of autonomous LDS

solution of ˙x = Ax, with A ∈ R

^n×n

and constant, is x(t) = e

^tA

x(0)

generalizes scalar case: solution of ˙x = ax, with a ∈ R and constant, is x(t) = e

^ta

x(0)

• looks like ordinary power series

e

^at

= 1 + ta + (ta)

²

2! + · · · with square matrices instead of scalars . . .

• define matrix exponential as

e

^M

= I + M + M

²

2! + · · · for M ∈ R

^n×n

(which in fact converges for all M )

• with this definition, state-transition matrix is

Φ(t) = L

⁻¹

!(sI − A)

⁻¹

" = e

^tA

Matrix exponential solution of autonomous LDS

solution of ˙x = Ax, with A ∈ R

^n×n

and constant, is x(t) = e

^tA

x(0)

generalizes scalar case: solution of ˙x = ax, with a ∈ R and constant, is x(t) = e

^ta

x(0)

(7)

•

So for scalars t,s e^{(t+s)A =}e^(tA+sA)= e^tAe^sA since (tA)(sA) = (sA)(tA)

This is useful in understanding the LTI system solution...

• matrix exponential is meant to look like scalar exponential

• some things you’d guess hold for the matrix exponential (by analogy with the scalar exponential) do in fact hold

• but many things you’d guess are wrong

example: you might guess that e^A+B = e^Ae^B, but it’s false (in general)

A =

! 0 1

−1 0

"

, B =

! 0 1 0 0

"

e^A =

! 0.54 0.84

−0.84 0.54

"

, e^B =

! 1 1 0 1

"

e^A+B =

! 0.16 1.40

−0.70 0.16

"

"= e^Ae^B =

! 0.54 1.38

−0.84 −0.30

"

however, we do have e^A+B = e^Ae^B if AB = BA, i.e., A and B commute

thus for t, s ∈ R, e^(tA+sA) = e^tAe^sA

with s = −t we get

e^tAe⁻^tA = e^tA−tA = e⁰ = I so e^tA is nonsingular, with inverse

#e^tA$⁻¹

= e⁻^tA

• matrix exponential is meant to look like scalar exponential

• some things you’d guess hold for the matrix exponential (by analogy with the scalar exponential) do in fact hold

• but many things you’d guess are wrong

example: you might guess that e^A+B = e^Ae^B, but it’s false (in general)

A =

! 0 1

−1 0

"

, B =

! 0 1 0 0

"

e^A =

! 0.54 0.84

−0.84 0.54

"

, e^B =

! 1 1 0 1

"

e^A+B =

! 0.16 1.40

−0.70 0.16

"

"= e^Ae^B =

! 0.54 1.38

−0.84 −0.30

"

however, we do have e^A+B = e^Ae^B if AB = BA, i.e., A and B commute

thus for t, s ∈ R, e^(tA+sA) = e^tAe^sA

with s = −t we get

e^tAe⁻^tA = e^tA−tA = e⁰ = I so e^tA is nonsingular, with inverse

#e^tA$⁻¹

= e⁻^tA

(8)

8

example: let’s find e^A, where A =

! 0 1 0 0

"

we already found

e^tA = L⁻¹(sI − A)⁻¹ =

! 1 t 0 1

"

so, plugging in t = 1, we get e^A =

! 1 1 0 1

"

let’s check power series:

e^A = I + A + A²

2! + · · · = I + A since A² = A³ = · · · = 0

Time transfer property

for ˙x = Ax we know

x(t) = Φ(t)x(0) = e^tAx(0)

interpretation: the matrix e^tA propagates initial condition into state at time t

more generally we have, for any t and τ ,

x(τ + t) = e^tAx(τ )

(to see this, apply result above to z(t) = x(t + τ ))

interpretation: the matrix e^tA propagates state t seconds forward in time (backward if t < 0)

Solving autonomous LTI Systems in Matrix Form