被控对象用微分方程
y
(
n
)
=
a
n
−
1
y
(
n
−
1
)
+
⋯
+
a
0
y
+
b
n
−
1
u
(
n
−
1
)
+
⋯
+
b
0
u
(
1.1
)
y^{(n)}=a_{n-1}y^{(n-1)}+\cdots+a_0y+b_{n-1}u^{(n-1)}+\cdots+b_0u\qquad(1.1)
y(n)=an−1y(n−1)+⋯+a0y+bn−1u(n−1)+⋯+b0u(1.1) 描述,其相应的差分方程为
y
(
k
)
=
α
1
y
(
k
−
1
)
+
⋯
+
α
n
y
(
k
−
n
)
+
β
1
u
(
k
−
1
)
+
⋯
+
β
n
u
(
k
−
n
)
(
1.2
)
y(k)=\alpha_1y(k-1)+\cdots+\alpha_ny(k-n) +\beta_1u(k-1)+\cdots+\beta_nu(k-n)\qquad(1.2)
y(k)=α1y(k−1)+⋯+αny(k−n)+β1u(k−1)+⋯+βnu(k−n)(1.2) 对应的系统拉氏变换和z变换分别为
G
(
s
)
=
b
n
−
1
s
n
−
1
+
⋯
+
b
1
s
+
b
0
1
−
a
n
−
1
s
n
−
1
−
⋯
−
a
1
s
−
a
0
G
(
z
)
=
β
n
−
1
z
n
−
1
+
⋯
+
β
1
z
+
β
0
1
−
α
n
−
1
z
n
−
1
−
⋯
−
α
1
z
−
α
0
\begin{aligned} G(s) =& \frac{b_{n-1}s^{n-1}+\cdots+b_1s+b_0}{1-a_{n-1}s^{n-1}-\cdots-a_1s-a_0} \\ G(z) =& \frac{\beta_{n-1}z^{n-1}+\cdots+\beta_1z+\beta_0} {1-\alpha_{n-1}z^{n-1}-\cdots-\alpha_1z-\alpha_0} \\ \end{aligned}
G(s)=G(z)=1−an−1sn−1−⋯−a1s−a0bn−1sn−1+⋯+b1s+b01−αn−1zn−1−⋯−α1z−α0βn−1zn−1+⋯+β1z+β0 式(2)中的
α
\alpha
α 和
β
\beta
β 参数在下面两种情况下等于或近似等于1。 (1) 系统静态增益
D
=
−
b
0
/
a
0
D=-b_0/a_0
D=−b0/a0,则差分方程系数之间满足
∑
i
=
1
n
a
i
−
b
0
a
0
∑
j
=
0
n
−
1
β
j
=
1
\sum_{i=1}^na_i-\frac{b_0}{a_0}\sum_{j=0}^{n-1}\beta_j=1
i=1∑nai−a0b0j=0∑n−1βj=1 且在
−
b
0
/
a
0
=
1
-b_0/a_0=1
−b0/a0=1 时,所有系数之和为1。 拉氏变换和z变换的终值定理分别为,当终值存在时,
f
s
(
∞
)
=
lim
s
→
0
s
F
(
s
)
f
z
(
∞
)
=
lim
z
→
1
(
z
−
1
)
F
(
z
)
\begin{aligned} f_s(\infty) =& \lim_{s\rightarrow 0}sF(s) \\ f_z(\infty) =& \lim_{z\rightarrow 1}(z-1)F(z) \\ \end{aligned}
fs(∞)=fz(∞)=s→0limsF(s)z→1lim(z−1)F(z) 阶跃信号的拉氏变换和z变换分别为
U
(
s
)
=
1
s
,
U
(
z
)
=
z
z
−
1
U(s) = \frac{1}{s},U(z) = \frac{z}{z-1}
U(s)=s1,U(z)=z−1z 所以连续系统和离散系统的终值分别为
f
s
(
∞
)
=
−
b
0
a
0
f
z
(
∞
)
=
∑
β
j
1
−
∑
α
i
\begin{aligned} f_s(\infty) =& -\frac{b_0}{a_0} \\ f_z(\infty) =& \frac{\sum\beta_j}{1-\sum\alpha_i} \\ \end{aligned}
fs(∞)=fz(∞)=−a0b01−∑αi∑βj 当终值
f
s
(
∞
)
=
f
z
(
∞
)
=
1
f_s(\infty)=f_z(\infty)=1
fs(∞)=fz(∞)=1 时,
∑
α
i
+
∑
β
j
=
1
\sum\alpha_i+\sum\beta_j=1
∑αi+∑βj=1。 (2) 采样周期
Δ
t
→
0
\Delta t\rightarrow 0
Δt→0,且
a
0
,
b
0
a_0,b_0
a0,b0 为有限值,则所有参数的和
lim
Δ
t
→
0
∑
α
i
+
∑
β
j
=
lim
Δ
t
→
0
[
1
+
(
D
−
1
)
(
Δ
t
T
′
)
n
]
=
1
\lim_{\Delta t\rightarrow 0}\sum\alpha_i+\sum\beta_j =\lim_{\Delta t\rightarrow 0}\left[1+(D-1) \left(\frac{\Delta t}{T'}\right)^n\right]=1
Δt→0lim∑αi+∑βj=Δt→0lim[1+(D−1)(T′Δt)n]=1 其中
T
′
T'
T′ 为等效时间常数,且
−
a
0
=
(
1
T
′
)
n
-a_0=(\frac{1}{T'})^n
−a0=(T′1)n。 证明:当
Δ
t
→
0
\Delta t\rightarrow 0
Δt→0 时,
y
˙
=
y
(
k
)
−
y
(
k
−
1
)
Δ
t
y
¨
=
y
(
k
)
−
2
y
(
k
−
1
)
+
y
(
k
−
2
)
(
Δ
t
)
2
y
(
3
)
=
y
(
k
)
−
3
y
(
k
−
1
)
+
3
y
(
k
−
2
)
−
y
(
k
−
3
)
(
Δ
t
)
3
⋮
y
(
n
)
=
∑
i
=
0
n
(
−
1
)
i
C
n
i
y
(
k
−
i
)
(
Δ
t
)
n
\begin{aligned} \dot{y} =& \frac{y(k)-y(k-1)}{\Delta t} \\ \ddot{y} =& \frac{y(k)-2y(k-1)+y(k-2)}{(\Delta t)^2} \\ y^{(3)} =& \frac{y(k)-3y(k-1)+3y(k-2)-y(k-3)}{(\Delta t)^3} \\ \vdots& \\ y^{(n)} =& \frac{\displaystyle\sum_{i=0}^n(-1)^iC_n^iy(k-i)} {(\Delta t)^n} \\ \end{aligned}
y˙=y¨=y(3)=⋮y(n)=Δty(k)−y(k−1)(Δt)2y(k)−2y(k−1)+y(k−2)(Δt)3y(k)−3y(k−1)+3y(k−2)−y(k−3)(Δt)ni=0∑n(−1)iCniy(k−i) 分子的各系数呈杨辉三角。将上述关系代入式(1)得(先以二阶系统为例)
y
¨
=
a
1
y
˙
+
a
0
y
+
b
1
u
˙
+
b
0
u
y
(
k
)
−
2
y
(
k
−
1
)
+
y
(
k
−
2
)
(
Δ
t
)
2
=
a
1
y
(
k
−
1
)
−
y
(
k
−
2
)
Δ
t
+
a
0
y
(
k
−
2
)
+
b
1
u
(
k
−
1
)
−
u
(
k
−
2
)
Δ
t
+
b
0
u
(
k
−
2
)
\begin{aligned} & \ddot{y} = a_1\dot{y}+a_0y+b_1\dot{u}+b_0u \\ & \frac{y(k)-2y(k-1)+y(k-2)}{(\Delta t)^2} = a_1\frac{y(k-1)-y(k-2)}{\Delta t} +a_0y(k-2)+b_1\frac{u(k-1)-u(k-2)}{\Delta t}+b_0u(k-2) \\ \end{aligned}
y¨=a1y˙+a0y+b1u˙+b0u(Δt)2y(k)−2y(k−1)+y(k−2)=a1Δty(k−1)−y(k−2)+a0y(k−2)+b1Δtu(k−1)−u(k−2)+b0u(k−2)
y
(
k
)
=
2
y
(
k
−
1
)
−
y
(
k
−
2
)
+
a
1
[
y
(
k
−
1
)
−
y
(
k
−
2
)
]
Δ
t
+
b
1
[
u
(
k
−
1
)
−
u
(
k
−
2
)
]
Δ
t
+
[
a
0
y
(
k
−
2
)
+
b
0
u
(
k
−
2
)
]
(
Δ
t
)
2
=
(
a
1
Δ
t
+
2
)
y
(
k
−
1
)
+
(
a
0
Δ
2
t
−
a
1
Δ
t
−
1
)
y
(
k
−
2
)
+
b
1
Δ
t
u
(
k
−
1
)
+
(
b
0
Δ
2
t
−
b
1
Δ
t
)
u
(
k
−
2
)
=
α
1
y
(
k
−
1
)
+
α
2
y
(
k
−
2
)
+
β
1
u
(
k
−
1
)
+
β
2
u
(
k
−
2
)
\begin{aligned} y(k) =& 2y(k-1)-y(k-2)+a_1[y(k-1)-y(k-2)]\Delta t +b_1[u(k-1)-u(k-2)]\Delta t+[a_0y(k-2)+b_0u(k-2)](\Delta t)^2 \\ =& (a_1\Delta t+2)y(k-1) +(a_0\Delta^2t-a_1\Delta t-1)y(k-2) +b_1\Delta tu(k-1) +(b_0\Delta^2t-b_1\Delta t)u(k-2) \\ =& \alpha_1y(k-1)+\alpha_2y(k-2)+\beta_1u(k-1)+\beta_2u(k-2) \end{aligned}
y(k)===2y(k−1)−y(k−2)+a1[y(k−1)−y(k−2)]Δt+b1[u(k−1)−u(k−2)]Δt+[a0y(k−2)+b0u(k−2)](Δt)2(a1Δt+2)y(k−1)+(a0Δ2t−a1Δt−1)y(k−2)+b1Δtu(k−1)+(b0Δ2t−b1Δt)u(k−2)α1y(k−1)+α2y(k−2)+β1u(k−1)+β2u(k−2)
∑
α
i
+
∑
β
j
=
(
a
1
Δ
t
+
2
)
+
(
a
0
Δ
2
t
−
a
1
Δ
t
−
1
)
+
b
1
Δ
t
+
(
b
0
Δ
2
t
−
b
1
Δ
t
)
=
(
a
0
+
b
0
)
Δ
2
t
+
2
−
1
lim
Δ
t
→
0
∑
α
i
+
∑
β
j
=
1
\begin{aligned} \sum\alpha_i+\sum\beta_j =& (a_1\Delta t+2)+(a_0\Delta^2t-a_1\Delta t-1)+b_1\Delta t +(b_0\Delta^2t-b_1\Delta t) \\ =& (a_0+b_0)\Delta^2t+2-1 \\ \lim_{\Delta t\rightarrow 0}\sum\alpha_i+\sum\beta_j =& 1 \end{aligned}
∑αi+∑βj==Δt→0lim∑αi+∑βj=(a1Δt+2)+(a0Δ2t−a1Δt−1)+b1Δt+(b0Δ2t−b1Δt)(a0+b0)Δ2t+2−11 可见不包含
Δ
t
\Delta t
Δt 的一阶项。再推广到n阶,省略包含
Δ
t
\Delta t
Δt 的高阶项
y
(
k
)
=
∑
i
=
1
n
(
−
1
)
i
C
n
i
y
(
k
−
i
)
+
a
n
−
1
Δ
t
(
∑
i
=
0
n
−
1
(
−
1
)
i
C
n
i
y
(
k
−
i
)
)
+
a
n
−
2
Δ
2
t
(
∑
i
=
0
n
−
2
(
−
1
)
i
C
n
i
y
(
k
−
i
)
)
+
⋯
+
b
n
−
1
Δ
t
(
∑
i
=
0
n
−
1
(
−
1
)
i
C
n
i
u
(
k
−
i
)
)
+
b
n
−
2
Δ
2
t
(
∑
i
=
0
n
−
2
(
−
1
)
i
C
n
i
u
(
k
−
i
)
)
\begin{aligned} y(k) =& \sum_{i=1}^n(-1)^iC_n^iy(k-i) +a_{n-1}\Delta t\left(\sum_{i=0}^{n-1}(-1)^iC_n^iy(k-i)\right) +a_{n-2}\Delta^2t\left(\sum_{i=0}^{n-2}(-1)^iC_n^iy(k-i)\right) \\ &+\cdots+b_{n-1}\Delta t\left(\sum_{i=0}^{n-1}(-1)^iC_n^iu(k-i)\right) +b_{n-2}\Delta^2t\left(\sum_{i=0}^{n-2}(-1)^iC_n^iu(k-i)\right) \\ \end{aligned}
y(k)=i=1∑n(−1)iCniy(k−i)+an−1Δt(i=0∑n−1(−1)iCniy(k−i))+an−2Δ2t(i=0∑n−2(−1)iCniy(k−i))+⋯+bn−1Δt(i=0∑n−1(−1)iCniu(k−i))+bn−2Δ2t(i=0∑n−2(−1)iCniu(k−i)) 进一步观察得到杨辉三角各行元素的和为0,而第一个元素为1,所以等号右边第一项的系数和
∑
i
=
1
n
(
−
1
)
i
C
n
=
1
\displaystyle\sum_{i=1}^n(-1)^iC_n=1
i=1∑n(−1)iCn=1,剩下的每一项中的系数都是完整的杨辉三角中的一行,系数和都是0,只剩下杨辉三角第一行的
a
0
y
(
k
−
n
)
Δ
n
t
a_0y(k-n)\Delta^nt
a0y(k−n)Δnt 和
b
0
u
(
k
−
n
)
Δ
n
t
b_0u(k-n)\Delta^nt
b0u(k−n)Δnt,都是
Δ
t
\Delta t
Δt 的高阶小量,可以忽略。 这种推导方法用的是欧拉离散化方法,文末的第一个参考文献证明了用其他方法离散化的系数和也是1。 根据微分方程系数
a
i
a_i
ai 与极点
p
i
p_i
pi、时间常数
T
i
T_i
Ti 的关系可以知道,当
a
0
≠
0
a_0 \neq 0
a0=0 时
−
a
0
=
∏
i
=
1
n
p
i
=
∏
i
=
1
n
1
T
i
>
0
-a_0 = \prod_{i=1}^n p_i=\prod_{i=1}^n \frac{1}{T_i}>0
−a0=i=1∏npi=i=1∏nTi1>0 可以定义
T
′
=
∏
i
=
1
n
T
i
T'=\sqrt{\prod_{i=1}^n T_i}
T′=i=1∏nTi 为等效时间常数。于是有
a
0
=
−
(
1
T
′
)
n
a_0=-\displaystyle\left(\frac{1}{T'}\right)^n
a0=−(T′1)n,又因
(
−
b
0
/
a
0
)
=
D
(-b_0/a_0)=D
(−b0/a0)=D 为等效静态增益,所以如果不忽略
Δ
t
\Delta t
Δt 的两个高阶小量,则
∑
i
=
1
n
α
i
+
∑
j
=
0
n
−
1
β
j
=
1
+
a
0
Δ
n
t
+
b
0
Δ
n
t
=
1
+
a
0
(
1
−
D
)
Δ
n
t
=
1
+
(
D
−
1
)
(
Δ
t
T
′
)
n
\begin{aligned} \sum_{i=1}^n \alpha_i+\sum_{j=0}^{n-1} \beta_j =& 1+a_0\Delta^nt+b_0\Delta^nt \\ =& 1+a_0(1-D)\Delta^nt \\ =& 1+(D-1)\left(\frac{\Delta t}{T^{\prime}}\right)^n \\ \end{aligned}
i=1∑nαi+j=0∑n−1βj===1+a0Δnt+b0Δnt1+a0(1−D)Δnt1+(D−1)(T′Δt)n