1.3 Stability of Closed-Loop Control Systems

? An important consequence of feedback control is that it can cause oscillatory

responses (振荡响应). 反馈控制的一个重要的结果是会产生振荡响应。

? If the oscillation has a small amplitude and damps out (衰减) quickly, then the

control system performance is generally considered to be satisfactory.如果振荡的幅值很小并且衰减很快，那么一般认为控制系统的运行状态是令人满意的

? However, under certain circumstances the oscillations may be undamped (无衰减的)

or even have an amplitude that increases with time until a physical limit is reached,

such as a control valve being fully open or completely shut. 然而，在某些情况下，振荡有可能是无阻尼的，甚至幅值会随时间而增大，直到达到了物理极限，比如一个被完全打开或关闭的控制阀。

In these situations, the closed-loop system is said to be unstable.在这些情况下，闭环系统是不稳定的

In this unit we analyze the stability characteristics of closed-loop systems and present several useful criteria for determining whether a system will be stable. 在本节中，我们对闭环系统的稳定性特性做出分析，并提出几个用于判断系统是否稳定的判据 Additional stability criteria based on frequency response analysis are not discussed here. 另外的基于频率响应分析的稳定性判据在这里不做讨论

But first we consider an illustrative (用作说明的) example of a closed-loop system that can become unstable.首先，我们考虑一个闭环系统的例子，这个系统可以变得不稳定

To determine the effect of Kc on the closed-loop response c(t), we consider a unit step (单位阶跃) change in set point, R(s) = 1/s. We have derived the closed-loop transfer function (闭环传递函数) for set-point changes:为了判断KC对闭环响应c(t)的影响, 我们考虑对设定值施加一单位阶跃变化，R(s)=1/s。可以得到随设定值变化的闭环传递函数：

After Kc is specified, c(t) can be determined from the inverse Laplace transform (拉普拉斯逆变换) of Eq. (4). But first the roots of the cubic (立方的) polynomial (多项式) in s must be determined before performing the partial fraction expansion (部分分式展开). These roots can be calculated using standard root-finding techniques.KC确定之后，c(t)可以通过对方程（4）进行拉普拉斯反变换得到。但是在运算部分分式展开式之前，首先要得到s的三阶多项式的根。这可以通过标准的求根方法来得到.

The unstable response for Example is an oscillation where the amplitude grows in each successive (连续的) cycle. 本例中的不稳定响应是幅值在每一次循环中不断增大而产生的振荡

In contrast, for an actual physical system the amplitudes would increase until a physical limit is reached or an equipment failure occurs. 相反，在实际物理系统中，幅值增大到物理极限或导致设备故障为止

Since the final control element usually has saturation (饱和) limits, the unstable response would manifest (表明，显露) itself as a sustained (持续的)oscillation with a constant amplitude instead of a continually increasing amplitude.因为终端控制元件通常都有饱和限制，所以不稳定响应最终会表现为幅值不变地持续振荡，而不是不断增大。

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? Clearly, a feedback control system must be stable as a prerequisite (先决条件) for

satisfactory control. 很明显，一个反馈控制系统能够可靠控制的先决条件是稳定 ? Consequently, it is of considerable (应注意的，考虑的) practical importance to be able

to determine under what conditions a control system becomes unstable. 因此，考虑系统在什么情况下变得不稳定是非常重要的

? For example, what values of the PID controller parameters Kc, τI (τ, tau), and τD keep

the controlled process stable?例如，PID控制器的参数取什么值时能够保持控制过程 ? 稳定？

disturbance (one that is not sustained) occurs.在发生暂态扰动之后，一个开环稳定过程将会返回到初始的稳定状态下。

? By contrast there are a few processes, such as exothermic chemical reactors (放热化学

? These processes are extremely difficult to operate without feedback control.他们在没

? Before presenting various stability criteria, we introduce the following definition for

unconstrained linear systems (无约束线性系统). 在介绍各种稳定性判据之前，我们先介绍关于无约束线性系统的定义

? We use the term \to refer to the ideal situation where there are no

physical limits on the output variable.我们使用术语“无约束”，来特指对输出变量无任何物理约束的理想状况。

? Definition of stability.:An unconstrained linear system is said to be stable if the output

response is bounded (有界的) for all bounded inputs. Otherwise it is said to be unstable.稳定性的定义：对于一个无约束线性系统，如果对所有的有界输入，输出响应都是有界的，那么该系统是稳定的，否则就是不稳定的。

? By a bounded input, we mean an input variable that stays within upper and lower

limits for all values of time. 所谓有界输入，是指输入变量值在任何时刻都保持在上、下界范围之内。

? For example, consider a variable x(t) that varies with time. If x(t) is a step or sinusoidal

(正弦) function, then it is bounded. 比如，考虑变量x(t)，随时间t变化。如果x(t)是阶跃或正弦函数，则它是有界的。

? However, the functions x(t) = t and x(t) =e3t are not bounded.而函数x(t) = t和x(t)

=e3t则是无界的。

? If GOLis a ratio of polynomials in s (i.e., a rational function 有理函数), then the

closed-loop transfer function in Eq. (6) is also a rational function. After a rearrangement it can be factored (因式分解) into poles (pi) and zeroes (zi) as如果GOL是s多项式的比（即有理数），那么方程（6）中的闭环传函也是有理函数。通过整理，它可以表示为如公式（7）所示的被因式分解为极点和零点的表达形式

? where K‘ (it is called “K prime”) is a multiplicative (乘子) constant selected to give the

? To have a physically realizable system, the number of poles must be greater than or

equal to the number of zeroes, that is, n≥m, 为了使系统能够物理实现，极点的个数必须大于或等于零点的个数，即n≥ m。

? Note that a pole-zero cancellation (零极点对消) occurs if a zero and a pole have the

same numerical value.若零、极点有相同数值，注意零极点对消

? By contrast, if all of the poles are negative (or have negative real parts), then the

system is stable. 相反，如果所有的极点都是负数（或实部都为负），那么系统是稳定的。

? These considerations can be summarized in the following stability criterion:这可以用

? General Stability Criterion.:The feedback control system in Fig. 1.3.1 is stable if and

only if all roots of the characteristic equation are negative or have negative real parts.

Otherwise, the system is unstable.通用稳定性判据：图1.3.1所示的反馈控制系统是稳定的，当且仅当所有的特征方程的根都是负的或其实部是负的。否则，系统是不稳定的。

In 1905 Routh published an analytical technique for determining whether any roots of a polynomial have positive real parts. 1905年，劳思发表了用于判断多项式的根是否存在正实部的解析方法。

According to the General Stability Criterion, a closed-loop system will be stable only if all of the roots of the characteristic equation have negative real parts. 根据通用稳定判据，仅当所有的特征方程的根都具有负实部时，一个闭环系统才是稳定的。 Thus, by applying Routh' s technique to analyze the coefficients of the characteristic equation, we can determine whether the closed-loop system is stable. 因而，通过劳思的方法来分析特征方程的系数，我们就可以判断出闭环系统是否稳定。

This approach is referred to as the Routh Stability Criterion.It can be applied only to systems whose characteristic equations are polynomials in s.它仅能用于特征方程在s平面上为多项式的情况

Thus, the Routh Stability Criterion is not directly applicable to systems containing time delays, since an e-θs term appears in the characteristic equation where θ is the time delay. 因此，劳思稳定判据不能被直接应用于带有时延的系统中，因为特征方程中含有e-θs项，这里θ是时间延迟。

However, if e-θs is replaced by a Pade approximation (帕德近似) then an approximate stability analysis can be performed. 然而，如果用帕德近似代替e-θs项，那么也可以（对含有时延的系统）做出近似的稳定性分析。

An exact stability analysis of systems containing time delays can be performed by direct root-finding or by using a frequency response analysis.对含有时间延迟的系统，可以直接采用直接求根法或频域响应分析法来进行精确的稳定性分析

We arbitrarily assume that an> 0. If an < 0, we simply multiply Eq.(12) by -1 to generate a new equation that satisfies this condition.我们可以任意地假设an>0。如果an<0，则只要把方程（12）两边乘以负1得到新的方程，仍能够满足假设条件。 A necessary (but not sufficient) condition for stability is that all of the coefficients (a0, a1, . ,an) in the characteristic equation must be positive.稳定的必要（而非充分）条件是，特征方程的所有的系数(a0, a1, . ,an)均为正数。

If any coefficient is negative or zero, then at least one root of the characteristic equation lies to the right of, or on, the imaginary axis (虚轴), and the system is unstable. 如果有一个系数为负或零，则至少有一个特征方程的根位于虚轴的右方或虚轴上，这样系统就是不稳定的。

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? If all of the coefficients are positive, we next construct the following Routh array :如

? The Routh array has n + 1 rows where n is the order of the characteristic equation,

Eq.(12). 劳思阵含有n+1行，n为特征方程（12）式的阶数。

? The Routh array has a roughly triangular (三角形的) structure with only a single

element in the last row. 劳思阵具有大致的三角形状，最后一行仅有一个单元

? The first two (rows) are merely the coefficients in the characteristic equation, arranged

according to odd and even powers (奇偶次幂) of s. 前两行仅仅是特征方程的系数，根据s的奇、偶次幂排列。

? The elements in the remaining rows are calculated from the formulas:其它行的元素由

? …Note that the expressions in the numerators (分子) of Eqs. (13) to (16) are similar to

the calculation of a determinant (行列式) for a 2×2 matrix except that the order of subtraction is reversed. 注意，式（13）到（16）的分子表达式类似于计算一个2×2阶的行列式，但减法的次序是颠倒的。

? Having constructed the Routh array, we can now state the Routh Stability Criterion:

? A necessary and sufficient condition for all roots of the characteristic equation in Eq.

(12) to have negative real parts is that : all of the elements in left column of the Routh array are positive.式（12）特征方程的所有根均具有负实部的充要条件是，劳思阵的左列的所有元素均为正值。 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?

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