自动控制原理考试专项复习课件.ppt

1.什么是开环控制系统和闭环控制系统，会绘制控制什么是开环控制系统和闭环控制系统，会绘制控制系统的方框图系统的方框图1.会利用电路的基本定律求解电路的微分方程会利用电路的基本定律求解电路的微分方程2.会用复阻抗法求解电路的传递函数会用复阻抗法求解电路的传递函数3.会用结构图化简的方法求系统的传递函数会用结构图化简的方法求系统的传递函数4.会绘制系统的信号流图（结点：信号，箭头上的值：增益）会绘制系统的信号流图（结点：信号，箭头上的值：增益），会用梅森公式求系统的传递函数会用梅森公式求系统的传递函数自动控制原理自动控制原理第二章第二章 控制系统的数学模型控制系统的数学模型梅森公式(Masons rule)前向通道前向通道k k的增益的增益 (The gain of the (The gain of the k kth path)th path)前向通道前向通道k k的余因式的余因式 (The cofactor of the (The cofactor of the k kth path)th path)流图的特征式流图的特征式 (The determinant of the graph)(The determinant of the graph)Figure 4 LC ladder network.An LC ladder network is shown in Figure 4.One may write the equations describing the network as follows:I1=(V1-Va)Y1,Va=(I1-Ia)Z2,Ia=(Va-V2)Y3,V2=IaZ4.Construct a flow graph from the equations and determine the transfer function V2(s)/V1(s).Figure 1 The signal flow graph is shown in Figure 1.Solution:(1)(2)The forward path is T1=Y1Z2Y3Z4.The feedback loops are L1=-Z2Y1,L2=-Y3Z2,L3=-Z4Y3,L1L3=Z2Y1Z4Y3.The determinant is =1-(-Z2Y1-Y3Z2-Z4Y3)+Z2Y1Z4Y3=1+Z2Y1+Y3Z2+Z4Y3+Z2Y1Z4Y3.The cofactor is 1=1.So the transfer function is T(s)=T1 1/=Obtain the transfer function of the differentiating circuit shown in Figure 2.Figure 2 A differentiating circuit.Solution:1.典型的输入信号典型的输入信号2.二阶系统的单位阶跃响应（二阶系统的单位阶跃响应（Y(s)y(t)），系统的终），系统的终值（值（），欠阻尼二阶系统的超调，），欠阻尼二阶系统的超调，调节时间（定义）（调节时间（定义）（）3.稳定性（充要条件）稳定性（充要条件）4.稳态误差稳态误差自动控制原理自动控制原理第三章第三章 控制系统的时域分析控制系统的时域分析 典型输入信号(Typical input signals)1.阶跃信号(the step signal)自动控制原理自动控制原理第三章第三章 控制系统的时域分析控制系统的时域分析 单位阶跃信号(Unit step signal)自动控制原理自动控制原理第三章第三章 控制系统的时域分析控制系统的时域分析 2.斜坡信号(the ramp signal)自动控制原理自动控制原理第三章第三章 控制系统的时域分析控制系统的时域分析 3.抛物线信号(the parabolic signal)A unity negative feedback control system has the plant transfer function .(a)Determine the percent overshoot and settling time(using a 2%settling criterion)due to a unit step input.(b)For what range of K is the settling time less than 1 second?Solution:A control system has the structure shown in Figure 2.Determine the gain at which the system will become stable.Figure 2 Feedforward system.Solution:10.An armature-controlled DC motor with tachometer feedback is shown in Figure 4.Assume that Km=10,J=1,and Ra=1.Determine the required gain,K,to restrict the steady-state error to a ramp input(v(t)=t for t0)to 0.1(assume that D(s)=0).Figure 4 DC motor with feedback.Solution:The closed-loop transfer function is With v(t)=t,we have V(s)=1/s2.Using the final value theorem yieldsWe desire that ess=1/10K1.自动控制原理自动控制原理第四章第四章 控制系统的复数域分析控制系统的复数域分析 根轨迹(Root locus)1.定义(Definition)根轨迹是系统某一参数从零变化到无穷时，特征方程的根在s平面上的变化轨迹。

The root locus is the The root locus is the pathpath of the roots of the of the roots of the characteristic equation characteristic equation traced outtraced out in the s-in the s-plane as a system parameter is plane as a system parameter is changedchanged.自动控制原理自动控制原理第四章第四章 控制系统的复数域分析控制系统的复数域分析根轨迹绘制的步骤(The root locus procedure)这里所说的根轨迹是指增益K从0到变化时,特征根变化的轨迹,称为180Here the root locus is the Here the root locus is the pathpath of the roots of the of the roots of the characteristic equation characteristic equation traced outtraced out in the s-plane as in the s-plane as the gainthe gain K K is is increasedincreased from zero to infinity and it is from zero to infinity and it is called called 180180。

locus.locus.自动控制原理自动控制原理第四章第四章 控制系统的复数域分析控制系统的复数域分析 Step 1 在复平面上标出开环极点和开环零点The first step is to The first step is to markmark the the open-loopopen-loop poles poles and zeros in the s-plane.and zeros in the s-plane.自动控制原理自动控制原理第四章第四章 控制系统的复数域分析控制系统的复数域分析 当K从零到无穷大变化时，根轨迹从P(s)的极点（开环极点）开始，到P(s)的零点（开环零点）结束The locus of the roots of the characteristic equation The locus of the roots of the characteristic equation 1+KP1+KP(s s)=0 0 beginsbegins at at the poles of P(s)the poles of P(s)and and ends ends at at thethe zeros of P(s)zeros of P(s)as K increases form 0 to infinity.as K increases form 0 to infinity.自动控制原理自动控制原理第四章第四章 控制系统的复数域分析控制系统的复数域分析 Step 2 确定实轴上的根轨迹。

The second step is to determine which parts The second step is to determine which parts of the of the real axisreal axis belong the root locus.belong the root locus.自动控制原理自动控制原理第四章第四章 控制系统的复数域分析控制系统的复数域分析 若实轴上某一段右边的开环零点和极点(实轴上)个数之和为奇数,则该实轴段为根轨迹段The root locus on the real axis always lies in a The root locus on the real axis always lies in a section of the real axis to the left of an section of the real axis to the left of an odd numberodd number of poles and zeros.of poles and zeros.自动控制原理自动控制原理第四章第四章 控制系统的复数域分析控制系统的复数域分析 Step 3 确定根轨迹的分支数及根轨迹的对称性。

The third step is to determine the number of The third step is to determine the number of separateseparate lociloci and the and the symmetrysymmetry of the root loci.of the root loci.自动控制原理自动控制原理第四章第四章 控制系统的复数域分析控制系统的复数域分析 根轨迹的分支数等于开环极点的个数The number of The number of separate lociseparate loci is equal to the number is equal to the number of of polespoles.根轨迹必然是连续的且关于实轴对称The root loci must be The root loci must be continuouscontinuous and and symmetricalsymmetrical with respect to the horizontal with resp。