![ECE 595, Discrete-Time Control Systems October 20, 2005 1
Discretization of Continuous Time State Space Systems
Suppose we are given the continuous time state space system
ẋ(t) = Ax(t) + Bu(t) (1)
y(t) = Cx(t) + Du(t) (2)
and apply an input that changes only at discrete (equal) sampling intervals. It would be nice if
we could find matrices G and H, independent of t or k so that we could obtain a discrete time
model of the system,
x ((k + 1)T) = G(T)x(kT) + H(T)u(kT) (3)
y(kT) = Cx(kT) + Du(kT). (4)
We will now determine the values of the matrices G(T) and H(T). It will turn out that while
they are constant for a particular sampling interval, they depend on the value of the sampling
interval, so for that reason I have written them as G(T) and H(T) in (3) above.
We start by using the solution of (1) to calculate the values of the state x at times kT and
(k + 1)T. These are
x ((k + 1)T) = eA(k+1)T
x(0) + eA(k+1)T
Z (k+1)T
0
e−Aτ
Bu(τ)dτ (5)
x(kT) = eAkT
x(0) + eAkT
Z kT
0
e−Aτ
Bu(τ)dτ. (6)
We want to write x((k +1)T) in terms of x(kT) so we multiply all terms of (6) by eAT and solve
for eA(k+1)T x(0), obtaining
eA(k+1)T
x(0) = eAT
x(kT) − eA(k+1)T
Z kT
0
e−Aτ
Bu(τ)dτ. (7)
Substituting for eA(k+1)T x(0) in (5) we obtain
x ((k + 1)T) = eAT
x(kT) + eA(k+1)T
"Z (k+1)T
0
e−Aτ
Bu(τ)dτ −
Z kT
0
e−Aτ
Bu(τ)dτ
#
(8)
which, by linearity of integration, is equivalent to
x ((k + 1)T) = eAT
x(kT) + eA(k+1)T
Z (k+1)T
kT
e−Aτ
Bu(τ)dτ. (9)
Next, we notice that within the interval from kT to (k + 1)T, u(t) = u(kT) is constant, as is the
matrix B, so we can take them out of the integral to obtain
x ((k + 1)T) = eAT
x(kT) + eA(k+1)T
Z (k+1)T
kT
e−Aτ
dτBu(kT) τ ∈ [kT, (k + 1)T). (10)
We can take the eA(k+1)T inside the integral to obtain
x ((k + 1)T) = eAT
x(kT) +
Z (k+1)T
kT
eA[(k+1)T−τ]
dτBu(kT) τ ∈ [kT, (k + 1)T). (11)](https://image.slidesharecdn.com/discretization-211001082152/85/Discretization-1-320.jpg)
Discretization
- 1. ECE 595, Discrete-Time Control Systems October 20, 2005 1 Discretization of Continuous Time State Space Systems Suppose we are given the continuous time state space system ẋ(t) = Ax(t) + Bu(t) (1) y(t) = Cx(t) + Du(t) (2) and apply an input that changes only at discrete (equal) sampling intervals. It would be nice if we could find matrices G and H, independent of t or k so that we could obtain a discrete time model of the system, x ((k + 1)T) = G(T)x(kT) + H(T)u(kT) (3) y(kT) = Cx(kT) + Du(kT). (4) We will now determine the values of the matrices G(T) and H(T). It will turn out that while they are constant for a particular sampling interval, they depend on the value of the sampling interval, so for that reason I have written them as G(T) and H(T) in (3) above. We start by using the solution of (1) to calculate the values of the state x at times kT and (k + 1)T. These are x ((k + 1)T) = eA(k+1)T x(0) + eA(k+1)T Z (k+1)T 0 e−Aτ Bu(τ)dτ (5) x(kT) = eAkT x(0) + eAkT Z kT 0 e−Aτ Bu(τ)dτ. (6) We want to write x((k +1)T) in terms of x(kT) so we multiply all terms of (6) by eAT and solve for eA(k+1)T x(0), obtaining eA(k+1)T x(0) = eAT x(kT) − eA(k+1)T Z kT 0 e−Aτ Bu(τ)dτ. (7) Substituting for eA(k+1)T x(0) in (5) we obtain x ((k + 1)T) = eAT x(kT) + eA(k+1)T "Z (k+1)T 0 e−Aτ Bu(τ)dτ − Z kT 0 e−Aτ Bu(τ)dτ # (8) which, by linearity of integration, is equivalent to x ((k + 1)T) = eAT x(kT) + eA(k+1)T Z (k+1)T kT e−Aτ Bu(τ)dτ. (9) Next, we notice that within the interval from kT to (k + 1)T, u(t) = u(kT) is constant, as is the matrix B, so we can take them out of the integral to obtain x ((k + 1)T) = eAT x(kT) + eA(k+1)T Z (k+1)T kT e−Aτ dτBu(kT) τ ∈ [kT, (k + 1)T). (10) We can take the eA(k+1)T inside the integral to obtain x ((k + 1)T) = eAT x(kT) + Z (k+1)T kT eA[(k+1)T−τ] dτBu(kT) τ ∈ [kT, (k + 1)T). (11)
- 2. ECE 595, Discrete-Time Control Systems October 20, 2005 2 Now we see that as τ ranges from kT to (k + 1)T (the lower to the upper limit of integration) the exponent of e ranges from T to 0. Accordingly, let’s define a new variable λ = (k + 1)T − τ. Then dλ = −dτ and λ ranges from T to 0 as τ ranges from kT to (k + 1)T. Thus we have x ((k + 1)T) = eAT x(kT) − Z 0 T eAλ dλBu(kT) λ ∈ [0, kT), (12) or x ((k + 1)T) = eAT x(kT) + Z T 0 eAλ dλBu(kT) λ ∈ [0, kT). (13) We see that in (13) we have written the state update equation exactly in the form of (3) where G(T) = eAT (14) H(T) = Z T 0 eAλ dλB, (15) so we’re done . . . except that we’d rather not leave the expression for H(T) in the form of an integral. So long as A is invertible, we can easily integrate, using the fact that d dt eAT = AeAT = eAT A (16) to obtain H(T) = A−1 Z T 0 AeAλ dλB = A−1 eAλ |T λ=0B (17) = A−1 (eAT − I)B = (eAT − I)BA−1 . (18) Finally, note that while I restricted the value of τ and λ to lie within a single sampling interval, k appears nowhere in the expressions for G(T) and H(T). Our solution to (3) is thus x(kT) = (G(T))k x(0) + k−1 X j=0 (G(T))k−j−1 H(T)u(jT), k = 1, 2, 3, . . . (19) and we can see that at the sampling instants kT, this has exactly the same value as is obtained using (1). Specifically, (G(T))k = (eAT )k = eAkT , (20) and since the input u(t) is constant on sampling intervals, eAkT Z kT 0 e−Aτ Bu(τ)dτ = k−1 X j=0 eA(k−j−1)T A−1 (eAT − I)Bu(jT) = k−1 X j=0 (G(T))k−j−1 H(T)u(jT). (21)