连续状态方程
连续状态方程的表达式为：
x
˙
(
t
)
=
A
x
(
t
)
+
B
u
(
t
)
y
(
t
)
=
C
x
(
t
)
+
D
u
(
t
)
\begin{array}{l}\dot x(t) = Ax(t) + Bu(t)\\y(t) = Cx(t) + Du(t)\end{array}
x
˙
(
t
)
=
A
x
(
t
)
+
B
u
(
t
)
y
(
t
)
=
C
x
(
t
)
+
D
u
(
t
)
将以上状态方程离散化以下形式：
x
(
k
+
1
)
=
E
x
(
k
)
+
F
u
(
k
)
y
(
k
)
=
G
x
(
k
)
+
H
u
(
k
)
\begin{array}{l}x(k + 1) = Ex(k) + Fu(k)\\y(k) = Gx(k) + Hu(k)\end{array}
x
(
k
+
1
)
=
E
x
(
k
)
+
F
u
(
k
)
y
(
k
)
=
G
x
(
k
)
+
H
u
(
k
)
第一种方式：解状态方程
求解
对连续状态方程进行变形：
{
x
˙
(
t
)
−
A
x
(
t
)
=
B
u
(
t
)
α
x
˙
(
t
)
−
α
A
x
(
t
)
=
α
B
u
(
t
)
(
α
x
(
t
)
)
′
=
α
B
u
(
t
)
\left\{ \begin{array}{l}\dot x(t) – Ax(t) = Bu(t)\\\alpha \dot x(t) – \alpha Ax(t) = \alpha Bu(t)\\(\alpha x(t))’ = \alpha Bu(t)\end{array} \right.
⎩
⎨
⎧
x
˙
(
t
)
−
A
x
(
t
)
=
B
u
(
t
)
α
x
˙
(
t
)
−
α
A
x
(
t
)
=
α
B
u
(
t
)
(
α
x
(
t
)
)
′
=
α
B
u
(
t
)
求解
α
\alpha
α
为时间t的函数有：
α
=
e
−
A
t
\alpha = {e^{ – At}}
α
=
e
−
A
t
所以有：
[
e
−
A
t
x
(
t
)
]
′
=
e
−
A
t
B
u
(
t
)
{\left[ {
{e^{ – At}}x(t)} \right]^\prime } = {e^{ – At}}Bu(t)
[
e
−
A
t
x
(
t
)
]
′
=
e
−
A
t
B
u
(
t
)
求解有：
{
[
e
−
A
t
x
(
t
)
]
′
=
e
−
A
t
B
u
(
t
)
d
[
e
−
A
t
x
(
t
)
]
d
t
=
e
−
A
t
B
u
(
t
)
∫
t
0
t
d
[
e
−
A
t
x
(
t
)
]
=
∫
t
0
t
e
−
A
t
B
u
(
t
)
d
t
x
(
t
)
=
x
(
t
0
)
e
A
(
t
−
t
0
)
+
B
∫
t
0
t
e
−
F
τ
u
(
τ
)
d
τ
\left\{ \begin{array}{l}{\left[ {
{e^{ – At}}x(t)} \right]^\prime } = {e^{ – At}}Bu(t)\\\frac{
{d\left[ {
{e^{ – At}}x(t)} \right]}}{
{dt}} = {e^{ – At}}Bu(t)\\\int_{
{t_0}}^t d \left[ {
{e^{ – At}}x(t)} \right] = \int_{
{t_0}}^t {
{e^{ – At}}Bu(t)d} t\\x(t) = x({t_0}){e^{A(t – {t_0})}} + B\int_{
{t_0}}^t {
{e^{ – F\tau }}u(\tau )d\tau } \end{array} \right.
⎩
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎧
[
e
−
A
t
x
(
t
)
]
′
=
e
−
A
t
B
u
(
t
)
d
t
d
[
e
−
A
t
x
(
t
)
]
=
e
−
A
t
B
u
(
t
)
∫
t
0
t
d
[
e
−
A
t
x
(
t
)
]
=
∫
t
0
t
e
−
A
t
B
u
(
t
)
d
t
x
(
t
)
=
x
(
t
0
)
e
A
(
t
−
t
0
)
+
B
∫
t
0
t
e
−
F
τ
u
(
τ
)
d
τ
其中设其实时刻为
t
0
{
{t_0}}
t
0
，对于F为矩阵的情况，可以用指数级数展开的公式计算：
e
A
t
=
∑
n
=
0
∞
(
A
t
)
n
n
!
{e^{At}} = \sum\limits_{n = 0}^\infty {\frac{
{
{
{(At)}^n}}}{
{n!}}}
e
A
t
=
n
=
0
∑
∞
n
!
(
A
t
)
n
一般到二阶即可。
离散化
设相邻采样时刻为
t
k
{t_k}
t
k
和
t
k
+
1
{t_{k + 1}}
t
k
+
1
，有
t
k
+
1
−
t
k
=
T
{t_{k + 1}} – {t_k} = T
t
k
+
1
−
t
k
=
T
，
T
T
T
为采样间隔。对与求解的表达式代入有：
x
(
t
k
+
1
)
=
x
(
t
k
)
e
A
(
t
k
+
1
−
t
k
)
+
B
∫
t
k
t
k
+
1
e
A
(
t
−
τ
)
u
(
τ
)
d
τ
x({t_{k + 1}}) = x({t_k}){e^{A({t_{k + 1}} – {t_k})}} + B\int_{
{t_k}}^{
{t_{k + 1}}} {
{e^{ A(t-\tau) }}u(\tau )d\tau }
x
(
t
k
+
1
)
=
x
(
t
k
)
e
A
(
t
k
+
1
−
t
k
)
+
B
∫
t
k
t
k
+
1
e
A
(
t
−
τ
)
u
(
τ
)
d
τ
x
(
t
k
+
1
)
=
x
(
t
k
)
e
A
T
+
B
∫
t
k
t
k
+
1
e
A
(
t
k
+
1
−
τ
)
u
(
τ
)
d
τ
x({t_{k + 1}}) = x({t_k}){e^{AT}} + B\int_{
{t_k}}^{
{t_{k + 1}}} {
{e^{ A(t_{k+1}-\tau) }}u(\tau )d\tau }
x
(
t
k
+
1
)
=
x
(
t
k
)
e
A
T
+
B
∫
t
k
t
k
+
1
e
A
(
t
k
+
1
−
τ
)
u
(
τ
)
d
τ
相邻采样时刻间隔很小，这里采用一个近似所有：
u
(
τ
)
=
u
(
t
k
)
,
t
k
≤
τ
≤
t
k
+
1
u(\tau ) = u({t_k}),{t_k} \le \tau \le {t_{k + 1}}
u
(
τ
)
=
u
(
t
k
)
,
t
k
≤
τ
≤
t
k
+
1
所以有：
x
(
t
k
+
1
)
=
x
(
t
k
)
e
A
T
+
B
∫
t
k
t
k
+
1
e
A
(
t
k
+
1
−
τ
)
u
(
τ
)
d
τ
=
x
(
t
k
)
e
A
T
+
B
u
(
t
k
)
∫
t
k
t
k
+
1
e
A
(
t
k
+
1
−
τ
)
d
τ
\begin{array}{c}x({t_{k + 1}}) = x({t_k}){e^{AT}} + B\int_{
{t_k}}^{
{t_{k + 1}}} {
{e^{A({t_{k + 1}} – \tau )}}u(\tau )d\tau } \\ = x({t_k}){e^{AT}} + Bu({t_k})\int_{
{t_k}}^{
{t_{k + 1}}} {
{e^{A({t_{k + 1}} – \tau )}}d\tau } \end{array}
x
(
t
k
+
1
)
=
x
(
t
k
)
e
A
T
+
B
∫
t
k
t
k
+
1
e
A
(
t
k
+
1
−
τ
)
u
(
τ
)
d
τ
=
x
(
t
k
)
e
A
T
+
B
u
(
t
k
)
∫
t
k
t
k
+
1
e
A
(
t
k
+
1
−
τ
)
d
τ
对比离散方程可以推出
E
=
e
A
T
,
F
=
B
∫
t
k
t
k
+
1
e
A
(
t
k
+
1
−
τ
)
d
τ
E = {e^{AT}},F = B\int_{
{t_k}}^{
{t_{k + 1}}} {
{e^{A({t_{k + 1}} – \tau )}}d\tau }
E
=
e
A
T
,
F
=
B
∫
t
k
t
k
+
1
e
A
(
t
k
+
1
−
τ
)
d
τ
对于
F
F
F
，设$\lambda = {t_{k + 1}} – \tau $，可以推出B：
F
=
B
∫
0
T
e
A
t
d
t
F = B\int_0^T {
{e^{At }}} dt
F
=
B
∫
0
T
e
A
t
d
t
设
t
k
=
k
T
{t_k} = kT
t
k
=
k
T
，代入到
x
(
t
)
x(t)
x
(
t
)
的表达式中有：
x
[
(
k
+
1
)
T
]
=
x
(
k
T
)
e
A
T
+
B
∫
0
T
e
A
t
d
t
⋅
u
(
k
T
)
x\left[ {\left( {k + 1} \right)T} \right] = x\left( {kT} \right){e^{AT}} + B\int_0^T {
{e^{At}}dt} \cdot u(kT)
x
[
(
k
+
1
)
T
]
=
x
(
k
T
)
e
A
T
+
B
∫
0
T
e
A
t
d
t
⋅
u
(
k
T
)
因为
(
k
+
1
)
T
({k + 1})T
(
k
+
1
)
T
只表示序列点出现的时刻，所以拿掉
T
T
T
有：
x
(
k
+
1
)
=
x
(
k
)
e
A
T
+
B
∫
0
T
e
A
t
d
t
⋅
u
(
k
)
x {\left( {k + 1} \right)} = x\left( {k} \right){e^{AT}} + B\int_0^T {
{e^{At}}dt} \cdot u(k)
x
(
k
+
1
)
=
x
(
k
)
e
A
T
+
B
∫
0
T
e
A
t
d
t
⋅
u
(
k
)
对于
e
A
t
{
{e^{At}}}
e
A
t
有另一种计算方式，对状态方程进行Laplace变化：
s
x
(
s
)
−
x
(
0
)
=
A
x
(
s
)
+
B
u
(
s
)
sx(s) – x(0) = Ax(s) + Bu(s)
s
x
(
s
)
−
x
(
0
)
=
A
x
(
s
)
+
B
u
(
s
)
有：
x
(
s
)
=
(
s
I
−
A
)
−
1
x
(
0
)
+
(
s
I
−
A
)
−
1
B
u
(
s
)
x(s) = {(sI – A)^{ – 1}}x(0) + {(sI – A)^{ – 1}}Bu(s)
x
(
s
)
=
(
s
I
−
A
)
−
1
x
(
0
)
+
(
s
I
−
A
)
−
1
B
u
(
s
)
其中
I
I
I
为单位矩阵。对比之前求解的
x
(
t
)
x(t)
x
(
t
)
的表达式有：
e
A
t
=
L
−
1
[
(
s
I
−
A
)
−
1
]
{e^{At}} = {L^{ – 1}}\left[ {
{
{\left( {sI – A} \right)}^{ – 1}}} \right]
e
A
t
=
L
−
1
[
(
s
I
−
A
)
−
1
]
第二种方法：欧拉法
写出
x
˙
(
t
)
\dot x(t)
x
˙
(
t
)
的表达式：
x
˙
(
t
)
=
x
(
k
+
1
)
−
x
(
k
)
T
\dot x(t) = \frac{
{x(k + 1) – x(k)}}{T}
x
˙
(
t
)
=
T
x
(
k
+
1
)
−
x
(
k
)
代入到连续状态方程中有：
{
x
(
k
+
1
)
−
x
(
k
)
T
=
A
x
(
k
)
+
B
u
(
k
)
⇒
x
(
k
+
1
)
−
x
(
k
)
=
A
T
x
(
k
)
+
B
T
u
(
k
)
⇒
x
(
k
+
1
)
=
(
A
T
+
1
)
x
(
k
)
+
B
T
u
(
k
)
\left\{ \begin{array}{l}\frac{
{x(k + 1) – x(k)}}{T} = Ax(k) + Bu(k)\\ \Rightarrow x(k + 1) – x(k) = ATx(k) + BTu(k)\\ \Rightarrow x(k + 1) = \left( {AT + 1} \right)x(k) + BTu(k)\end{array} \right.
⎩
⎨
⎧
T
x
(
k
+
1
)
−
x
(
k
)
=
A
x
(
k
)
+
B
u
(
k
)
⇒
x
(
k
+
1
)
−
x
(
k
)
=
A
T
x
(
k
)
+
B
T
u
(
k
)
⇒
x
(
k
+
1
)
=
(
A
T
+
1
)
x
(
k
)
+
B
T
u
(
k
)
所以有
E
=
(
A
T
+
I
)
,
F
=
B
T
E = \left( {AT + I} \right),F = BT
E
=
(
A
T
+
I
)
,
F
=
B
T