网站建设与管理试题及答案,个性化网站,网站策划书模板范文,wordpress开发文档(chm)前言 CS小菜鸡控制理论入门 视频学习笔记 视频传送门#xff1a;动态系统的建模与分析】9_一阶系统的频率响应_低通滤波器_Matlab/Simulink分析 拉普拉斯变换 F(s)L{f(t)}∫0∞f(t)e−stdtF(s)\mathcal{L}\{f(t)\}\int_0^\infty f(t)e^{-st}\mathrm{d}tF(s)L{f(t)}∫0∞f(t)…前言 CS小菜鸡控制理论入门 视频学习笔记 视频传送门动态系统的建模与分析】9_一阶系统的频率响应_低通滤波器_Matlab/Simulink分析 拉普拉斯变换
F(s)L{f(t)}∫0∞f(t)e−stdtF(s)\mathcal{L}\{f(t)\}\int_0^\infty f(t)e^{-st}\mathrm{d}tF(s)L{f(t)}∫0∞f(t)e−stdt其中sσjωs\sigmaj\omegasσjω
L{f′(t)}sF(s)−f(0)\mathcal{L}\{f(t)\}sF(s)-f(0)L{f′(t)}sF(s)−f(0)
L{f′′(t)}s2F(s)−sf(0)−f′(0)\mathcal{L}\{f(t)\}s^2F(s)-sf(0)-f(0)L{f′′(t)}s2F(s)−sf(0)−f′(0)
L{∫0tf(τ)d(τ)}1sF(s)\mathcal{L}\{\int_0^t f(\tau)d(\tau)\}\frac{1}{s}F(s)L{∫0tf(τ)d(τ)}s1F(s)
一个电路例子
e′Li′′Ri′1cieLiRi\frac{1}{c}ie′Li′′Ri′c1i
sE(s)Ls2I(s)RsI(s)1cI(s)sE(s)Ls^2I(s)RsI(s)\frac{1}{c}I(s)sE(s)Ls2I(s)RsI(s)c1I(s)
sE(s)(Ls2Rs1c)I(s)sE(s)(Ls^2Rs\frac{1}{c})I(s)sE(s)(Ls2Rsc1)I(s)
I(s)sLs2Rs1cE(s)I(s)\frac{s}{Ls^2Rs\frac{1}{c}}E(s)I(s)Ls2Rsc1sE(s)
常系数微分方程⟺\iff⟺线性时不变系统
非线性化系统1.在平衡点处线性化 2.采用非线性化分析控制
拉普拉斯变换求解线性微分方程
运用L\mathcal{L}L将ttt域转化到sss域−×÷-\times \div−×÷运用L−1\mathcal{L^{-1}}L−1将sss域转化到ttt域
拉普拉斯逆变换
F(s)5−ss25s4F(s)\frac{5-s}{s^25s4}F(s)s25s45−s
F(s)−3s42s1F(s)\frac{-3}{s4}\frac{2}{s1}F(s)s4−3s12
L−1[F(s)]−3e−4t2e−t\mathcal{L^{-1}}[F(s)]-3e^{-4t}2e^{-t}L−1[F(s)]−3e−4t2e−t
s−4,−1s-4,-1s−4,−1传递函数Transfer Function的极点Poles
一阶系统的单位阶跃响应Step Response
x˙(t)gRx(t)u(t)\dot{x}(t)\frac{g}{R}x(t)u(t)x˙(t)Rgx(t)u(t)
x(t)CRg(1−e−gRt)x(t)\frac{CR}{g}(1-e^{-\frac{g}{R}t})x(t)gCR(1−e−Rgt) 时间常数tτt\tautτ满足x(τ)1−1e0.63x(\tau)1-\frac{1}{e}0.63x(τ)1−e10.63 稳定整定时间 Tss4τTss4\tauTss4τ 用于系统辨识
假设 Tss4Tss4Tss4则τ1Rg\tau1\frac{R}{g}τ1gR
CRg5⟹C5\frac{CR}{g}5\Longrightarrow C5gCR5⟹C5
u(s)⟶asa⟶x(s)u(s) \longrightarrow \frac{a}{sa} \longrightarrow x(s)u(s)⟶saa⟶x(s)本质上是一个低通滤波器
频率响应 input: Misin(ωtϕi)M_isin(\omega t\phi_i)Misin(ωtϕi) output: M0sin(ωtϕ0)M_0sin(\omega t\phi_0)M0sin(ωtϕ0) 振幅响应M0MiM\frac{M_0}{M_i}MMiM0M 幅角响应ϕ0−ϕiϕ\phi_0-\phi_i\phiϕ0−ϕiϕ (一番数学推导…)
conclusion
MG∣G(jω)∣M_G|G(j\omega)|MG∣G(jω)∣
ϕG∠G(jω)\phi_G\angle G(j\omega)ϕG∠G(jω)
