Radu Grosu Vienna University of Technology
Hybrid Systems Modeling, Analysis and Control
Lecture 3
Solution for Systems with Input
Differential equations: !x = Ax +Bu, y = Cx, x(0)=x0
Output y
Integrator
Next State x
x(t) = x(τ )dτ
0
t
∫ , x0
!x(t) x(t)
y(t)
A x(t)+Bu(t)
C x(t)
u(t)
x(t)
Solution for Systems with Input
!x = Ax +Bu, x(0) = x0
y = Cx
Differential equations:
The solution is: x(t) = eAtx0 + eA(t−τ)
0
t
∫ B u(τ) dτ
Proof:
1) !x = d( eAtx0 + eAt e-Aτ
0
t
∫ B u(τ) dτ ) / dt
2) x(0) = eA0x0 = e0x0 = Ix0 = x0
Fundamental theorem
of calculus
= AeAtx0 + AeAt ( e-Aτ
0
t
∫ B u(τ) dτ ) + eAte-AtB u(t) = Ax +Bu
Derivative Of a product
A =
2 11 2⎡
⎣⎢
⎤
⎦⎥ , B =
11⎡
⎣⎢⎤
⎦⎥ , x0 =
10⎡
⎣⎢
⎤
⎦⎥ , !x = Ax+B, x(0) = x0
Example: Fixpoint Computation
1. Homog. sol.: x(t) = eAtx0 = BeΛtB-1x0 =
et e3t
-et e3t
⎡
⎣⎢
⎤
⎦⎥B-1x0
2. General solution:
x(t) =
et e3t
-et e3t
⎡
⎣⎢
⎤
⎦⎥
10⎡
⎣⎢
⎤
⎦⎥
B
+ et−τ e3( t−τ)
-et−τ e3( t−τ)
⎡
⎣⎢
⎤
⎦⎥
0
t
∫11⎡
⎣⎢⎤
⎦⎥
B
dτ
x(t) = 1
2 et +e3t
-et +e3t
⎡
⎣⎢
⎤
⎦⎥ −
13
e3( t−τ)
e3( t−τ)
⎡
⎣⎢
⎤
⎦⎥
0
t
= 1
6 3et +5e3t - 2-3et +5e3t - 2
⎡
⎣⎢
⎤
⎦⎥
Solution for Systems with Input
Difference equations: x(n+1) = Ax(n) +Bu(n), y(n) = Cx(n), x(0)=x0
Output y
Register
Next State x
x(n +1) = x(n), x0
x(n +1) x(n)
y(n)
A x(n)+Bu(n)
C x(n)
u(n)
x(n)
Solution for Systems with Input
x(n +1) = Ax(n) +Bu(n), x(0) = x0
y = Cx(n)
Difference equations:
The solution is: x(n) = Anx0 + An-1-τ
τ=0
n-1
∑ B u(τ)
Proof:
1) x(n +1) = A ( Anx0 + An-1-τ
B u(τ)
τ=0
n-1
∑ ) +B u(n)
2) x(0) = A0x0 = Ix0 = x0
= Ax(n) +Bu(n)
Matrix: Anx0 = B diag(λ1n,...,λn
n) B−1 x0 = B diag(λ1
n,...,λnn) (x0)
B
A =
2 11 2⎡
⎣⎢
⎤
⎦⎥ , B =
11⎡
⎣⎢⎤
⎦⎥ , x0 =
10⎡
⎣⎢
⎤
⎦⎥ , x(n +1) = Ax(n)+B, x(0) = x0
Example: Fixpoint Computation
1. Homog. sol.:
2. General solution:
x(n) =
1 3n
-1 3n
⎡
⎣⎢
⎤
⎦⎥
10⎡
⎣⎢
⎤
⎦⎥
B
+ 1 3n-1-τ
-1 3n-1-τ
⎡
⎣⎢
⎤
⎦⎥
11⎡
⎣⎢⎤
⎦⎥
τ=0
n-1
∑B
x(n) = 1
2 1+3n
-1 +3n
⎡
⎣⎢
⎤
⎦⎥ + 3n-1 3-τ
3-τ
⎡
⎣⎢
⎤
⎦⎥
τ=0
n-1
∑ =
3n
3n -1
⎡
⎣⎢
⎤
⎦⎥
x(t) = Anx0 = BΛnB-1x0 =
1 3n
-1 3n
⎡
⎣⎢
⎤
⎦⎥B-1x0
Impulse and Delay
Unit-delay function: D(x) (n) ! x(n-1)
Impulse function: δ(n) ! (n = 0) ? 1 : 0
Consequence: x = x(τ)
τ=0
∞
∑ Dτ (δ)
Signal representation: x(n) = x(τ)
τ=0
∞
∑ δ(n-τ)
x(τ) δ(n-τ)
n = τ
x
Time Invariance
Definition: A system f is called time invariant if ∀k,x:
Signal delayed by k
Response delayed by k
f(Dk (x)) = Dk (f(x)) or alternatively (f !Dk )(x) = (Dk ! f)(x)
Theorem: The delay Dk is linear and time invariant:
Dk (ax + by) = aDk (x) + bDk (y) (linearity)Dk (Dn(x)) = Dk+n(x) = Dn(Dk (x)) (time invariance)
The Z-Transform (my definition)
(2) Z{δ}xx = 1-1Z{x} (impulse transform)
where z is a complex variable. Z{x} is the Z-transform of x.
Consequences of this definition:
Z{x} = x(n) Z{Dn(δ)} = x(n) z−n
n=0
∞
∑n=0
∞
∑ standard definition of Z
Definition: A linear, continuous function Z satisfying:
(1) Z{D(x)} = z-1Z{x} (delay transform)
Z{x} = F(z)xxxxx is a function of the complex variable z
Z{x} = Xxxxxx Z{x} is also written simply as X
Convergence
Theorem: Assume that |x(n)| ≤ kσn for n = 0,1,2,..., for some real constants k > 0 and σ ≥ 0. Then F(z) exists for all σ < |z|.
= k 1
1− (σ / |z|)
Z{x} = x(n) z−n
n=0
∞
∑ ≤ k (σ / |z|)n
n=0
∞
∑
Proof:
= k(σ / | z |)*
Z-Transform Properties
PowerMult: Z{ax(n) + by(n)} = zkF(z) - zkx(0) - zk-1x(1) - ... - zx(k -1)
Linearity: Z{ax(n) + by(n)} = aZ{x(n)} + bZ{y(n)}
PowerMult: Z{x(n)an} = F(z / a)
LeftShift: Z{x(n +k)} = zkF(z) - zkx(0) - zk-1x(1) - ... - zx(k -1)
RightShift: Z{x(n - k)} = z-kF(z)
Convolution: Z{x(n)∗y(n)} = Z{x(n)} ⋅ Z{y(n)}
Z-Transform Properties
PowerMult: Z{ax(n) + by(n)} = zkF(z) - zkx(0) - zk-1x(1) - ... - zx(k -1)
Linearity: Z{ax(n) + by(n)} = aZ{x(n)} + bZ{y(n)}
PowerMult: Z{x(n)an} = F(z / a)
LeftShift: Z{x(n +k)} = zkF(z) - zkx(0) - zk-1x(1) - ... - zx(k -1)
RightShift: Z{x(n - k)} = z-kF(z)
Convolution: Z{x(n)∗y(n)} = Z{x(n)} ⋅ Z{y(n)}
Z-Transform Properties
PowerMult: Z{ax(n) + by(n)} = zkF(z) - zkx(0) - zk-1x(1) - ... - zx(k -1)
Linearity: Z{ax(n) + by(n)} = aZ{x(n)} + bZ{y(n)}
PowerMult: Z{x(n)an} = F(z / a)
LeftShift: Z{x(n +k)} = zkF(z) - zkx(0) - zk-1x(1) - ... - zx(k -1)
RightShift: Z{x(n - k)} = z-kF(z)
Convolution: Z{x(n)∗y(n)} = Z{x(n)} ⋅ Z{y(n)}
Z-Transform Properties
PowerMult: Z{ax(n) + by(n)} = zkF(z) - zkx(0) - zk-1x(1) - ... - zx(k -1)
Linearity: Z{ax(n) + by(n)} = aZ{x(n)} + bZ{y(n)}
PowerMult: Z{x(n)an} = F(z / a)
LeftShift: Z{x(n +k)} = zkF(z) - zkx(0) - zk-1x(1) - ... - zx(k -1)
RightShift: Z{x(n - k)} = z-kF(z)
Convolution: Z{x(n)∗y(n)} = Z{x(n)} ⋅ Z{y(n)}
Z-Transform Properties
PowerMult: Z{ax(n) + by(n)} = zkF(z) - zkx(0) - zk-1x(1) - ... - zx(k -1)
Linearity: Z{ax(n) + by(n)} = aZ{x(n)} + bZ{y(n)}
PowerMult: Z{x(n)an} = F(z / a)
LeftShift: Z{x(n +k)} = zkF(z) - zkx(0) - zk-1x(1) - ... - zx(k -1)
RightShift: Z{x(n - k)} = z-kF(z)
Convolution: Z{x(n)∗y(n)} = Z{x(n)} Z{y(n)}
Convolution: x ∗y ! x(n) Dn(y)n=0
∞
∑ = Dn(x) y(n)n=0
∞
∑
List of Z-Transforms
x(n) Z{x(n)}
(1) 1 z / (z -1)
(2) an z / (z - a)
(3) n z / (z -1)2
(4) nan z / (z - a)2
(5) n ≥ 1 ? an-1 : 0 1/ (z - a)
(6) n ≥ 1 ? Cn−1k−1an-1 : 0 1/ (z - a)k
(7) 0 ≤ n ≤ m ? 1 : 0 (1− zm) / zm−1(1- z)
(8) δ 1
Homogeneous Equation
Reduce the solution of LTI difference equations:
x(n +1) = Ax(n), y(n) = Cx(n), x(0) = x0
The choice of z and its power in Z{D(x)} = z-1Z{x}: • An algebraically closed field that includes x(k), A,B,C. • For example, if they belong to field R, choose field C.
to the easier solution of algebraic equations:
z X = AX + zx0, Y = C X ⇒ Y = C (I - z-1A)-1x0 = C(z-1A)*x0
Transfer Function
Conway Expansion
Partition matrix A as: A =
A11 A12
A21 A22
⎡
⎣⎢⎢
⎤
⎦⎥⎥
Then matrix A* is:
A* =
(A11 + A12A22* A21)
* A11* A12(A22 + A21A11
* A12)*
A22* A21 (A11 + A12A22
* A21)* (A22 + A21A11
* A12)*
⎡
⎣⎢⎢
⎤
⎦⎥⎥
Intuition: the entries of A* correspond to paths in
1 2 a11
a12
a21
a22
Conway Expansion in Fields
Partition matrix A as: A =
A11 A12
A21 A22
⎡
⎣⎢⎢
⎤
⎦⎥⎥
Then matrix A* = (I − A)-1 is:
A* =
(I - A11 - A12(I - A22)-1A21)-1 (I - A11)
-1A12(I - A22 - A21(I - A11)-1A12)-1
(I - A22)-1A21(I - A11 - A12(I - A22)-1A21)-1 (I - A22 - A21(I - A11)
-1A12)-1
⎡
⎣⎢⎢
⎤
⎦⎥⎥
Gaussian Elimination and A*
Consider the algebraic equation: X = AX + x0
x1 = a11x1 +a12x2 +1
x2 = a21x1 +a22x2
and vector: x0 =
10⎡
⎣⎢
⎤
⎦⎥ =
01⎡
⎣⎢
⎤
⎦⎥
Let matrix: A =
a11 a12
a21 a22
⎡
⎣⎢⎢
⎤
⎦⎥⎥
x1 = (1- a11 +a12(1- a22)-1a21)-1 = (a11 +a12a22
* a21)*
x2 = (1- a22)-1a21x1 = a22* a21x1
x1 = a11x1 +a12x2
x2 = a21x1 +a22x2 +1
x1 = (1- a11)-1a12x2 = a11
* a12x2
x2 = (1- a22 +a21(1- a11)-1a12)-1 = (a22 +a21a11
* a12)*
A* =
1
1- a11 - a12
11- a22
a21
11- a11
a12
1
1- a22 - a21
11- a11
a12
11- a22
a21
1
1- a11 - a12
11- a22
a21
1
1- a22 - a21
11- a11
a12
⎡
⎣
⎢⎢⎢⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥⎥⎥⎥
A* Determinants And (I - A)-1
(A)* = 1
(1- a11)(1- a22) - a12a21
1- a22 a12
a21 1- a11
⎡
⎣⎢⎢
⎤
⎦⎥⎥
(A)* = adj(I - A)
det(I - A) = (I - A)−1
Consider matrix z-1A =
2z-1 z-1
z-1 2z-1
⎡
⎣⎢
⎤
⎦⎥ . Then matrix (z-1A)* is:
(z−1A)* =
1
1- 2z−1 - z-1 11- 2z−1 z-1
1
1- 2z−1 - z-1 11- 2z−1 z-1
z-1 11- 2z−1
1
1- 2z−1 - z-1 11- 2z−1 z-1
z-1 11- 2z−1 1
1- 2z−1 - z-1 11- 2z−1 z-1
⎡
⎣
⎢⎢⎢⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥⎥⎥⎥
Example
(z−1A)* =
1- 2z−1
(1- 2z-1)2 - z-2 z-1
(1- 2z-1)2 - z-2
z-1
(1- 2z-1)2 - z-2 1- 2z−1
(1- 2z-1)2 - z-2
⎡
⎣
⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥
Rational Function
=
z(z - 2)(z -1)(z - 3)
z(z -1)(z - 3)
z(z -1)(z - 3)
z(z - 2)(z -1)(z - 3)
⎡
⎣
⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥
Strict RF
Partial Fraction Expansion
Let
z(z - 2)(z -1)(z - 3)
Not strict. Divide:
Hence: z(z - 2)
(z -1)(z - 3)=1+ 2z - 3
(z -1)(z - 3)=1+ a
(z -1)+ b
(z - 3)
z2 − 4z + 3 z2 − 2z 1
z2 − 4z + 32z − 3
a = (2z − 3) / (z - 3)
z=1=1/ 2
b = (2z - 3) / (z -1)
z=3= 3 / 2
Therefore:
z(z - 2)(z -1)(z - 3)
=1+ 12(z -1)
+ 32(z - 3)
Partial Fraction Expansion
Let
z(z -1)(z - 3)
Strict so no need to divide.
a = z / (z - 3)
z=1= -1/ 2
b = z / (z -1)
z=3= 3 / 2
Hence z
(z -1)(z - 3)= a
(z -1)+ b
(z - 3)
Therefore
z(z -1)(z - 3)
= −12(z -1)
+ 32(z - 3)
Partial Fraction Expansion
(z−1A)* = 12
2+ 1z -1
+ 3z - 3
-1z -1
+ 3z - 3
-1z -1
+ 3z - 3
2+ 1z -1
+ 3z - 3
⎡
⎣
⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥
Z-1{(z−1A)* } = 1
2(n = 0)?2:0 + (n ≥1):1+ 3n : 0 (n ≥1):3n −1:0 (n ≥1)?3n −1:0 (n = 0)?2:0 + (n ≥1):1+ 3n : 0
⎡
⎣⎢
⎤
⎦⎥
Hence : Z-1{(z−1A)* } = 1
2 3n +1 3n −13n −1 3n +1
⎡
⎣⎢
⎤
⎦⎥
A =
2 11 2⎡
⎣⎢
⎤
⎦⎥ , x0 =
10⎡
⎣⎢
⎤
⎦⎥ , x(n +1) = Ax(n), x(0) = x0
Example: Fixpoint Computation
The algebraic solution in transform domain is:
X = (I - z-1A)-1x0 = (z-1A)*x0
Hence the solution in time domain is:
x(n) = Z-1{(z-1A)*}x0
= 1
2 1+3n -1+3n -1+3n -1+3n
⎡
⎣⎢
⎤
⎦⎥
1 0⎡
⎣⎢
⎤
⎦⎥ = 1
2 1+3n
-1+3n
⎡
⎣⎢
⎤
⎦⎥
Zero Initial State
Reduce the solution of LTI difference equations:
x(n) = Ax(n -1) +Bu(n -1), y(n) = Cx(n), x(0) = 0
to the easier solution of algebraic equations:
X = A z-1 X +B z-1
U, Y = C X ⇒ Y/U = C (I - z−1A)-1 B z-1
Transfer Function
A =
2 11 2⎡
⎣⎢
⎤
⎦⎥ , x0 =
10⎡
⎣⎢
⎤
⎦⎥ , C =
01⎡
⎣⎢
⎤
⎦⎥
T
, x(n +1) = Ax(n), y(n) = Cx(n), x(0) = x0
Block Diagrams
Transform: X = z-1AX + x0
x1
x2
⎡
⎣⎢⎢
⎤
⎦⎥⎥
=z-12 z-1
z-1 z-12
⎡
⎣⎢
⎤
⎦⎥
x1
x2
⎡
⎣⎢⎢
⎤
⎦⎥⎥
+10⎡
⎣⎢
⎤
⎦⎥
x1 = z−12x1 + z−1x2 +1
x2 = z−1x1 + z−12x2
z-12 z-1 z-1 z-12
x1 x2
1
z-12 z-1
z-1 z-12 x1 x2
1
Input and Null Initial State
A =
2 11 2⎡
⎣⎢
⎤
⎦⎥ , B =
10⎡
⎣⎢
⎤
⎦⎥ , C =
01⎡
⎣⎢
⎤
⎦⎥ , x(n +1) = Ax(n) +Bu(n), y(n) = Cx(n)
z-12 z-1 z-1 z-12
x1 x2
z-1
u
z-12 z-1
z-1 z-12 x1 x2
u z-1
Z-Transform and Laplace Transform
• For continuous LTI systems, LT plays the same role as the ZT does in discrete-time systems:
Z{x} = F(z) = x(n) z−n
n=0
∞
∑
Z-transform L-transform
L{x} = F(s) = x(t)e−st dt
0
∞
∫
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