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State-space systems
Solving the state-space differential equation (Laplace approach)
Now consider the case where there is an input.
dx(t)
dt
= A x(t) + B u(t).
Taking Laplace transforms gives,
sx(s) − x(0) = A x(s) + B u(s),
which means that,
x(s) = (sI − A)
− 1 x(0) + (sI − A)
− 1 B u(s).
Now the inverse Laplace gives (recall that Φ(t) is defined as L−^1
(sI − A)−^1
x(t) = Φ(t) x(0) ︸ ︷︷ ︸
∫ (^) t
0
Φ(t − τ )Bu(τ ) dτ.
︸ ︷︷ ︸
zero-input solution convolution of Φ(t) and B u(t)
We can also use this equation for an arbitrary initial time,
x(t) = Φ(t − t 0 ) x(t 0 ) +
∫ (^) t
t 0
Φ(t − τ ) B u(τ ) dτ.
State-space systems
Solving the state-space differential equation (Laplace approach)
Consider a continuous time state-space representation.
dx(t)
dt
= A x(t) + B u(t),
y(t) = C x(t) + D u(t).
To begin, look at the zero-input solution
dx(t)
dt
= A x(t).
Taking unilateral Laplace transforms gives,
sx(s) − x(0) = A x(s) where x(0) = x(t) at t = 0.
This gives,
(sI − A)x(s) = x(0), or x(s) = (sI − A)
− 1 x(0).
Taking an inverse Laplace transform gives,
x(t) = L−^1
(sI − A)−^1
x(0) = Φ(t)x(0).
Φ(t), is also known as the “State Transition Matrix”.
Roy Smith: ECE 147b 9 : 1
State-space systems
Example (continued)
Now look at the State Transition Matrix,
Φ(t) = L
− 1
(sI − A)
− 1
= L
− 1
s + α
= e
−αt (this looks just like the impulse reponse)
So we can calculate ,
x(t) = e
−αt x(0) +
∫ (^) t
0
e
−α(t−τ ) α u(τ ) dτ.
Step response: Zero initial condition (x(0) = 0).
As u(t) = 1 for t ≥ 0,
y(t) = x(t) = e
−αt 0 +
∫ (^) t
0
e
−α(t−τ ) α dτ, = e
−αt
∫ (^) t
0
e
ατ α dτ,
= e
−αt [e
ατ |τ =t −e
ατ |τ =0] = e
−αt ( e
αt − 1
= 1 − e
−αt .
State-space systems
A simple example: A first order system (α > 0).
α
s + α
y(s) u(s)
Take the initial conditions to be zero. The differential equation is,
dy(t)
dt
Define the state as, x(t) = y(t), then,
dx(t)
dt
= −α x(t) + αu(t),
y(t) = x(t)
or
dx(t)
dt
[
−α
]
x(t) +
[
α
]
u(t),
y(t) =
[
]
x(t) +
[
]
u(t).
So the state-space representation is: A = −α, B = α, C = 1 and D = 0.
Roy Smith: ECE 147b 9 : 3
Solving state-space differential equations
Matrix exponential approach to solving state-space differential equations.
We can begin by “guessing” a solution of the form,
x(t) = e
At v(t), where v(t) is a time-varying vector.
Differentiate this to get,
dx(t)
dt
= Ae
At v(t) + e
At dv(t) dt = A x(t) + B u(t). (from the differential equation)
= Ae
At v(t) + B u(t) (by substituting for x(t))
and by equating the first & third lines,
e
At dv(t) dt
= B u(t)
dv(t)
dt
= e
−At Bu(t).
Solve this by integrating to get:
v(t) − v(0) =
∫ (^) t
0
e
−Aτ Bu(τ ) dτ.
Matrix exponentials
We can keep differentiating, substituting t = t 0 , and solving for the vi terms.
This eventually gives,
x(t) = x(t 0 ) + A(t − t 0 ) x(t 0 ) +
A^2 2 (t^ −^ t^0 )
2 x(t 0 ) +
A^3 3! (t^ −^ t^0 )
3 x(t 0 ) + · · ·
[
I + A(t − t 0 ) +
A^2
(t − t 0 )
2
A^3
(t − t 0 )
3
]
x(t 0 ).
define this as e
A(t−t 0 )
If t 0 = 0 (as is usually the case) we have,
x(t) = eAt^ x(0) and so Φ(t) = eAt.
In Matlab the command expm calculates the matrix exponential.
Caveat emptor. This is not the same as the exponential of the individual elements of a
matrix (which is calculated by the Matlab command: exp).
Properties:
e
A× 0 = I, e
A(s+t) = e
As e
At , e
−At e
At = I,
d eAt
dt
= Ae
At .
Roy Smith: ECE 147b 9 : 7
State transition matrix
Example:
P (s) =
(s − 1)
(s + 1)(s + 2)
This system has a state-space representation:
A =
[
]
, B =
[
]
, C =
[
]
, D = 0.
The state transition matrix is Φ(t) = eAt^ = L−^1
(sI − A)−^1
In this example:
e
At = L
− 1
{([
s 0
0 s
]
[
])− 1 }
= L
− 1
{[
s + 3 2
− 1 s
]− 1 }
= L
− 1
s(s + 3) + 2
[
s − 2
1 s + 3
]}
= L
− 1
s + 2
s + 1
s + 2
s + 1
s + 2
s + 1
s + 2
s + 1
2e−^2 t^ − e−t^ 2e−^2 t^ − 2e−t
−e
− 2 t
−t −e
− 2 t
−t
Solving state-space differential equations
From before,
v(t) − v(0) =
∫ (^) t
0
e
−Aτ Bu(τ ) dτ.
but,
x(t) = e
At v(t) so v(t) = e
−At x(t) and v(0) = x(0).
Substituting these gives,
e
−At x(t) − x(0) =
∫ (^) t
0
e
−Aτ Bu(τ ) dτ.
or,
x(t) = e
At x(0) + e
At
∫ (^) t
0
e
−Aτ Bu(τ ) dτ,
= e
At x(0) +
∫ (^) t
0
e
A(t−τ ) Bu(τ ) dτ,
This is exactly the same solution as before where Φ(t) = eAt.
Roy Smith: ECE 147b 9 : 9
Poles and zeros
Poles and zeros
If a system, P (s), has a pole at s = pi, then its partial fraction expansion is,
P (s) =
a(s)
b(s)
(s − z 1 )... (s − zm)
(s − p 1 )... (s − pi)... (s − pn)
E 1
(s − p 1 )
Ei
(s − pi)
En
(s − pn)
For the moment assume that pi is not repeated.
The impulse response will be of the form,
p(t) = E 1 e
p 1 t
pit
pnt .
The zero-input solutions of the corresponding differential equation will have terms of the
form,
y(t) = kie
pit +...
Look at this idea from a state-space point of view.
Transfer functions
Example: (revisted)
P (s) =
(s − 1)
(s + 1)(s + 2)
This system has a state-space representation:
A =
[
]
, B =
[
]
, C =
[
]
, D = 0.
From before we have,
(sI − A)
− 1
s(s + 3) + 2
[
s − 2
1 s + 3
]
so,
C(sI − A)
− 1 B + D =
s(s + 3) + 2
[
]
[
s − 2
1 s + 3
] [
]
s(s + 3) + 2
[
]
[
s
1
]
s − 1
s^2 + 3s + 2
Roy Smith: ECE 147b 9 : 13
Poles and zeros
Example: (yet again)
P (s) =
(s − 1)
(s + 1)(s + 2)
This system has a state-space representation:
A =
[
]
, B =
[
]
, C =
[
]
, D = 0.
We will sometimes abbreviate this to,
P (s) =
(^). Note that the dimensions always make this possible.
Eigenvalue equation:
The eigenvalues of A satisfy,
det(λI − A) = 0.
But here we see that this is simply the roots of the denominator,
det
[
λ + 3 2
− 1 λ
]
= λ(λ + 3) + 2 = λ
2
- 3λ + 2 = (λ + 1)(λ + 2) = 0.
Poles and zeros
State-space point of view:
Consider the zero-input case,
dx(t)
dt
= A x(t), and look at solutions of the form,
x(t) = e
pit x(0).
Differentiating this gives,
dx(t)
dt
= pie
pit x 0 = pi x(t).
As
dx(t)
dt
= A x(t), we have,
A x(t) = pix(t),
and taking t = 0 gives,
A x(0) = pi x(0) ←− This is an eigenvalue equation.
The eigenvalues of A are the poles of P (s).
The poles (eigenvalues) are also called “natural frequencies” or “modes” of P (s).
Roy Smith: ECE 147b 9 : 15
Poles and zeros
Zeros
If P (s) =
y(s)
u(s)
has a zero at s = s 0 , then for all inputs u(s 0 ) the output y(s 0 ) = 0.
This is equivalent to,
0 = C(s 0 I − A)
− 1 B + D
Using a state-space Laplace domain form (at s = s 0 ) we have,
s 0 x(s 0 ) = A x(s 0 ) + B u(s 0 )
0 = C x(s 0 ) + D u(s 0 )
This can be rearranged into matrix form,
[ (s 0 I − A) −B
C D
] [
x(s 0 )
u(s 0 )
]
This means that the matrix,
[ (s 0 I − A) −B
C D
]
is singular, or equivalently det
[
(s 0 I − A) −B
C D
]
Similarity transforms
Similarity transforms
Any invertible n × n matrix, V , can be used to transform a matrix,
Aˆ = V A V −^1.
The matrix, Aˆ is “similar” to (has the same eigenvalues as) A.
This is called a “similarity transform”.
Similarity transformed state-space representations
By defining a new state, ξ(t) = V −^1 x(t),
[ A B
C D
]
[
V
− 1 AV V
− 1 B
CV D
]
So there are obviously an infinite number of equivalent state-space representations.
Exercises:
- Show that the eigenvalues of A are unchanged by a similarity transform.
- Show that the transfer function is unchanged by a similarity transform of A.
Roy Smith: ECE 147b 9 : 19
Poles and zeros
Summary: Continuous time
Poles are given by: det(sI − A) = 0
Zeros are given by: det
[
(sI − A) −B
C D
]
Summary: Discrete time
Poles are given by: det(zI − A) = 0
Zeros are given by: det
[
(zI − A) −B
C D
]
