*本文只介绍计算方法,并不涉及计算原理。
1. 连续系统形式
对于一个线性定常连续系统,其一般形式为
W ( s ) = b m s m + b m − 1 s m − 1 + ⋯ + b 1 s + b 0 a n s n + a n − 1 s n − 1 + ⋯ + a 1 s + a 0 ( n , m ∈ Z , n ≥ m ) W(s) = \frac{b_m s^m + b_{m-1} s^{m-1} + \cdots + b_1 s + b_0}{a_n s^n + a_{n-1} s^{n-1} + \cdots + a_1 s + a_0} \quad \left( n, m \in Z, n \geq m \right) W(s)=ansn+an−1sn−1+⋯+a1s+a0bmsm+bm−1sm−1+⋯+b1s+b0(n,m∈Z,n≥m)设其极点为 s = − d i ( i = 1 , ⋯ , n ) s = -d_i \left( i = 1, \cdots, n \right) s=−di(i=1,⋯,n),则上式可以化为
W ( s ) = ∑ i n c i s + d i (1) W(s) = \sum _i ^ n \frac{c_i}{s+d_i} \tag{1} W(s)=i∑ns+dici(1)其中系数 c i c_i ci可用留数定理计算:
c i = W ( s ) ( s + d i ) ∣ s = − d i c_i = W(s) \left( s+ d_i \right) \Big\rvert _{s = -d_i} ci=W(s)(s+di)
s=−di
2. 极点代换
连续系统为 s s s域,离散系统为 z z z域,两者之间存在变量代换。变量代换具有如下形式:
1 s + d i → 1 1 − z − 1 e − d i T (2) \frac{1}{s+d_i} \rightarrow \frac{1}{1-z^{-1} e^{-d_iT}} \tag{2} s+di1→1−z−1e−diT1(2)其中 T T T为离散化时间单位(如0.01s)。
把式(1)中的每一项代换成式(2)的形式,即可将连续系统传函 W ( s ) W(s) W(s)转换成其极点代换形式 W d ( z ) W_d (z) Wd(z)。
3. 等效离散系统
在得到了极点代换形式 W d ( s ) W_d (s) Wd(s)的基础上,通过如下公式得到原连续系统 W ( s ) W(s) W(s)的等效离散系统 W ( z ) W(z) W(z):
W ( z ) = ∣ W ( j ω ) ∣ ∣ W d ( e j ω T ) ∣ W d ( z ) (3) W(z) = \frac{ \big\lvert W \left( j \omega \right) \big\rvert}{ \big\lvert W_d \left( e^{j \omega T} \right)\big\rvert} W_d (z) \tag{3} W(z)=
Wd(ejωT)
W(jω)
Wd(z)(3)其中 ∣ W ( ⋅ ) ∣ \big\lvert W(\cdot) \big\rvert
W(⋅)
指 W ( ⋅ ) W(\cdot) W(⋅)的模值; T T T为离散化时间单位,根据实际问题取值; ω \omega ω取1。
4. 计算举例
以如下传递函数为例,计算其等效离散系统的传函。离散化时间间隔 T = 0.01 s T = 0.01s T=0.01s。
W ( s ) = s + 1 s 2 + 12 s + 32 = s + 1 ( s + 4 ) ( s + 8 ) W(s) = \frac{s+1}{s^2 + 12s + 32} = \frac{s+1}{(s+4)(s+8)} W(s)=s2+12s+32s+1=(s+4)(s+8)s+1(1) 首先利用留数定理将其写成式(1)形式。
c 1 = W ( s ) ( s + 4 ) ∣ s = − 4 = s + 1 s + 8 ∣ s = − 4 = − 0.75 c 2 = W ( s ) ( s + 8 ) ∣ s = − 8 = s + 1 s + 4 ∣ s = − 8 = 1.75 c_1 = W(s) (s+4) \Big\rvert_{s=-4} = \frac{s+1}{s+8} \Bigg\rvert_{s=-4} = -0.75 \\ c_2 = W(s) (s+8) \Big\rvert_{s=-8} = \frac{s+1}{s+4} \Bigg\rvert_{s=-8} = 1.75 c1=W(s)(s+4)
s=−4=s+8s+1
s=−4=−0.75c2=W(s)(s+8)
s=−8=s+4s+1
s=−8=1.75于是 W ( s ) = c 1 s + d 1 + c 2 s + d 2 = − 0.75 s + 4 + 1.75 s + 8 W(s) = \frac{c_1}{s+d_1} + \frac{c_2}{s+d_2} = \frac{-0.75}{s+4} + \frac{1.75}{s+8} W(s)=s+d1c1+s+d2c2=s+4−0.75+s+81.75
(2) 极点代换。根据式(2),可以得到极点代换形式的传函:
W d ( z ) = c 1 1 − z − 1 e − d 1 T + c 2 1 − z − 1 e − d 2 T = − 0.75 1 − z − 1 e − 4 T + 1.75 1 − z − 1 e − 8 T ∣ T = 0.01 = − 0.75 1 − z − 1 e − 0.04 + 1.75 1 − z − 1 e − 0.08 = − 0.75 1 − 0.96079 z − 1 + 1.75 1 − 0.92312 z − 1 = 1 − 0.98904 z − 1 1 − 1.88391 z − 1 + 0.88692 z − 2 = z 2 − 0.98904 z z 2 − 1.88391 z + 0.88692 \begin{aligned} W_d (z) &= \frac{c_1}{1-z^{-1} e^{-d_1T}} + \frac{c_2}{1-z^{-1} e^{-d_2T}} \\ &= \frac{-0.75}{1-z^{-1} e^{-4T}} + \frac{1.75}{1-z^{-1} e^{-8T}} \Bigg\rvert_{T=0.01} \\ &= \frac{-0.75}{1-z^{-1} e^{-0.04}} + \frac{1.75}{1-z^{-1} e^{-0.08}} \\ &= \frac{-0.75}{1-0.96079z^{-1}} + \frac{1.75}{1-0.92312z^{-1}} \\ &= \frac{1-0.98904z^{-1}}{1-1.88391z^{-1}+0.88692z^{-2}} \\ &= \frac{z^2-0.98904z}{z^2-1.88391z+0.88692} \end{aligned} Wd(z)=1−z−1e−d1Tc1+1−z−1e−d2Tc2=1−z−1e−4T−0.75+1−z−1e−8T1.75
T=0.01=1−z−1e−0.04−0.75+1−z−1e−0.081.75=1−0.96079z−1−0.75+1−0.92312z−11.75=1−1.88391z−1+0.88692z−21−0.98904z−1=z2−1.88391z+0.88692z2−0.98904z
(3) 等效离散系统。根据式(3),需要先计算两个模值 ∣ W ( j ω ) ∣ \big\lvert W \left( j \omega \right) \big\rvert
W(jω)
和 ∣ W d ( e j ω T ) ∣ \big\lvert W_d \left( e^{j \omega T} \right)\big\rvert
Wd(ejωT)
。对于前者:
∣ W ( j ω ) ∣ = ∣ s + 1 ( s + 4 ) ( s + 8 ) ∣ s = j ω = ω 2 + 1 ω 2 + 4 2 ω 2 + 8 2 ∣ ω = 1 = 2 17 65 = 0.04254 \begin{aligned} \Big\lvert W \left( j \omega \right) \Big\rvert &= \Bigg\lvert \frac{s+1}{(s+4)(s+8)} \Bigg\rvert_{s=j \omega} \\ &= \frac{ \sqrt{\omega^2 + 1} }{ \sqrt{\omega^2 + 4^2} \sqrt{\omega^2 + 8^2} } \Bigg\rvert_{\omega = 1} = \frac{ \sqrt{2} }{ \sqrt{17} \sqrt{65} } = 0.04254 \end{aligned}
W(jω)
=
(s+4)(s+8)s+1
s=jω=ω2+42ω2+82ω2+1
ω=1=17652=0.04254对于后者:
∣ W d ( e j ω T ) ∣ = ∣ z 2 − 0.98904 z z 2 − 1.88391 z + 0.88692 ∣ z = e j ω T = ∣ e 2 j ω T − 0.98904 e j ω T e 2 j ω T − 1.88391 e j ω T + 0.88692 ∣ T = 0.01 , ω = 1 = ∣ e 0.02 j − 0.98904 e 0.01 j e 0.02 j − 1.88391 e 0.01 j + 0.88692 ∣ = ∣ ( cos 0.02 + j sin 0.02 ) − 0.98904 ( cos 0.01 + j sin 0.01 ) ( cos 0.02 + j sin 0.02 ) − 1.88391 ( cos 0.01 + j sin 0.01 ) + 0.88692 ∣ = ∣ 0.01081 + 0.01011 j 0.002904 + 0.0011599 j ∣ = ∣ 4.40952 + 1.72018 j ∣ = 4.73317 \begin{aligned} \Big\lvert W_d \left( e^{j \omega T} \right) \Big\rvert &= \Bigg\lvert \frac{z^2-0.98904z}{z^2-1.88391z+0.88692} \Bigg\rvert_{z = e^{j \omega T} } \\ &= \Bigg\lvert \frac{e^{2 j \omega T}-0.98904e^{j \omega T}}{e^{2 j \omega T}-1.88391e^{j \omega T}+0.88692} \Bigg\rvert_{T=0.01, \\\ \omega = 1} \\ &= \Bigg\lvert \frac{e^{0.02j}-0.98904e^{0.01j}}{e^{0.02j}-1.88391e^{0.01j}+0.88692} \Bigg\rvert \\ &= \Bigg\lvert \frac{ \left( \cos 0.02+ j \sin 0.02 \right)-0.98904 \left( \cos 0.01+ j \sin 0.01 \right)}{\left( \cos 0.02+ j \sin 0.02 \right)-1.88391\left( \cos 0.01+ j \sin 0.01 \right)+0.88692} \Bigg\rvert \\ &= \Bigg\lvert \frac{0.01081+0.01011j}{0.002904+0.0011599j} \Bigg\rvert \\ &= \Big\lvert 4.40952+1.72018j \Big\rvert = 4.73317 \end{aligned}
Wd(ejωT)
=
z2−1.88391z+0.88692z2−0.98904z
z=ejωT=
e2jωT−1.88391ejωT+0.88692e2jωT−0.98904ejωT
T=0.01, ω=1=
e0.02j−1.88391e0.01j+0.88692e0.02j−0.98904e0.01j
=
(cos0.02+jsin0.02)−1.88391(cos0.01+jsin0.01)+0.88692(cos0.02+jsin0.02)−0.98904(cos0.01+jsin0.01)
=
0.002904+0.0011599j0.01081+0.01011j
=
4.40952+1.72018j
=4.73317代入(3),即可得到离散系统的传函:
W ( z ) = ∣ W ( j ω ) ∣ ∣ W d ( e j ω T ) ∣ W d ( z ) = 0.04254 4.73317 W d ( z ) = 0.008988 z 2 − 0.98904 z z 2 − 1.88391 z + 0.88692 = 0.008988 z 2 − 0.008889 z z 2 − 1.88391 z + 0.88692 \begin{aligned} W(z) &= \frac{ \big\lvert W \left( j \omega \right) \big\rvert}{ \big\lvert W_d \left( e^{j \omega T} \right)\big\rvert} W_d (z) = \frac{0.04254}{4.73317} W_d (z) \\ &= 0.008988 \frac{z^2-0.98904z}{z^2-1.88391z+0.88692} \\ &= \frac{0.008988z^2-0.008889z}{z^2-1.88391z+0.88692} \end{aligned} W(z)=
Wd(ejωT)
W(jω)
Wd(z)=4.733170.04254Wd(z)=0.008988z2−1.88391z+0.88692z2−0.98904z=z2−1.88391z+0.886920.008988z2−0.008889z