这里存放了一些自动控制原理相关的知识。
| f(t) | F(s) | f(kT) or f(k) | F(z) | name |
---|
1 | δ(t) | 1 | δ(k) | 1 | 单位冲激 (unit impulse) |
2 | δ(t−td) | e−tds | δ(k−n) | z−n | 时延单位冲激(delayed unit impluse) |
3 | u(t)=1 | s1 | u(k)=1 | z−1z | 单位阶跃(unit step) |
4 | t | s21 | kT | (z−1)2zT | 斜坡函数(ramp) |
5 | t2 | s32 | (kT)2 | (z−1)3z(z+1)T2 | 抛物线函数(parabola) |
6 | | | ak | z−az | 几何序列(geometric sequence) |
7 | e−at | s+a1 | e−akT | z−e−aTz | 自然指数(natural exponential) |
8 | 1−e−at | s(s+a)1 | 1−e−akT | (z−1)(z−e−aT)(1−e−aT)z | 单位阶跃指数(unit step exponential) |
9 | te−at | (s+a)21 | kTe−akT | (z−e−aT)2Tze−aT | |
10 | | | kak−1 | (z−a)2z | |
11 | | | ak−bk | (z−a)(z−b)(a−b)z | |
12 | sinωt | s2+ω2ω | sinωkT | z2−2zcosωT+1zsinωT | sine |
13 | cosωt | s2+ω2s | cosωkT | z2−2zcosωT+1z(z−cosωT) | cosine |
14 | e−atsinωt | (s+a)2+ω2s | e−akTsinωkT | z2−2ze−aTcosωT+e−2aTze−aTsinωT | 自然欠阻尼(natural underdamped) |
15 | dtdf | sF(s)−f(0) | (空置) | (空置) | 一阶导(1st derivative) |
16 | dt2d2f | s2F(s)−sf(0)−dtdf(0) | (空置) | (空置) | 二阶导(2nd derivative) |
17 | | | f(k−n) | z−nF(z) | 左移n时位(backward shift of n) |
18 | f(t−td) | etdsF(s) | | | 时延(time delay) |
19 | (空置) | (空置) | f(k+1) | zF(s)−f(0) | 右移1时位(forward shift of one) |
20 | (空置) | (空置) | f(k+2) | sF(s)−f(0) | 右移2时位(forward shift of two) |
21 | f(0) | s→∞limsF(s) | f(0) | z→∞limF(z) | 初值(initial value) |
22 | f(∞) | s→0limsF(s) | f(∞) | z→1lim(z−1)F(z) | 终值(final value) |
其中:
- 对于t<0的函数f(t)=0
- 对于k<0的函数f(kT)=f(k)=0
- k=0,1,2,…
例子:已知 m(k)={1,0,k为偶数k为奇数 ,当 k=0,1,…,4 时,
求函数 y(k)=m(k)−m(k−1)−y(k−1) 的值。
解:
y(k)=m(k)−m(k−1)−y(k−1)⇒Y(z)=M(z)−z−1M(z)−z−1Y(z)
⇒M(z)Y(z)=1+z−11−z−1=z+1z−1
∵m(k)={1,0,k为偶数k为奇数
⇒M(z)=1+0+z−2+0+z−4+⋯=1−z−21=z2−1z2
∴Y(z)=M(z)Y(z)×M(z)=z+1z−1×z2−1z2=(z−1)(z+1)2(z−1)z2=(z+1)2z2
∵Z[y(k)]=Y(z)⇒y(k)=Z−1[Y(z)]
∴y(k)=Z−1[(z+1)2z2]