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【完结】翻译:DESIGNING CONTROL LOOPS for LINEAR and SWITCHING POWER SUPPLIES第二章

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  • 2018-3-24 14:23:48
来世纪电源网两年了,没发过什么帖子。最近一直在看这本书,就把第二章翻译出来给大家分享

三年前,第一章有人已经翻译过了http://bbs.21dianyuan.com/thread-208685-1-1.html
1919a1bcae2acc04b3d02666d717313.png
这本书不多打广告了,主要讲电源环路这块。读这本书的时候,有些地方我觉得比上学的时候学的控制原理要好得多!
我也是逐句翻译的,有些话可能会读起来比较生涩,我也会按照自己的话来表达。
有翻译得不好的地方希望大家多提建议



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  • 2018-3-24 14:26:53
 
迷人的 TL431 。。。
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  • 2018-3-24 21:59:56
 
我特意放了一张大图上来
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  • 2018-9-25 10:36:50
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  • 2018-3-24 14:27:27
 
Chapter2 Transfer Functions 第二章传递函数
We learned in Chapter 1 that a transfer function links a response signal to an excitation signal. A transfer function can be written in a lot of different ways whether it has been derived using brute-force algebra or by implementing a smarter approach (e.g., via known tools such as Thevenin/Norton transformations). What actually matters is the insight you can get just by reading the final equation. By insight, we mean your ability to immediately see where poles or zeros are located, if some gainor attenuation exists, just by reading the equation.
我们在第一章学过,传递函数将响应信号与激励信号链接起来。传递函数可以用许多不同的方式写出,不管它源自使用代数学或者通过使用更精确的方法(例如,通过诸如戴维宁/诺顿变换等已知工具)得出。真正重要的是你可以通过阅读最终方程得到的信息。通过这些信息,如果某种增益或衰减存在,只需读取方程,你能够立即看到极点或零点的位置。
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总工程师
  • 2018-3-24 15:35:18
 
什么是“强力代数学” 呢?
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版主
  • 2018-3-24 18:56:53
 
brute-force 在这里,感觉意译成 简单粗暴 也可以...
还有下面的 strictly proper, 控制论一般译成 严格真 或是 严格正则
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LV10
总工程师
  • 2018-3-25 12:24:57
 
文工的译法应该是可以的,不过我觉得还可以进一步讨论。

我们经常见到加密软件,破解(Crack)这些加密软件的方法之一称为暴力破解,那么暴力破解又是怎么个破解法呢?实际就是穷举法,把所有可能的密码组合采用排列组合的方式一一试验,直到找到正确的密码。因此,brute-force 可以译成穷举法,brute-force algebra 译成代数穷举法。


这种方法是不是很笨啊?确实笨,笨办法。那么下文提到的 Smart Approach 则与之相对可以译成聪敏方法,(聪敏这个词有点土,一时想不出有什么好的词,但译成精准似乎不妥)。
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副总工程师
  • 2018-3-24 21:58:31
 
这块我承认翻译的不好,版主的那个简单粗暴看起来不错
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  • 2018-3-24 14:32:00
 
2.1 Expressing Transfer Functions表达传递函数Linear network theory teaches us that the transfer function H of a network made of capacitors, resistors, and inductorscan be expressed the following way:
线性网络理论告诉我们,电容,电阻和电感组成的网络传递函数H,可以表示成以下形式:
2.1.png

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副总工程师
  • 2018-3-24 14:33:32
 
这图片看起来会不会很小?

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版主
  • 2018-3-25 08:36:48
 
论坛图片是按宽度控制的,你编辑图片时有意缩小左右的无效空间,即可看见稍大的图。
25 (1).png

25 (1).png
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  • 2018-3-24 14:34:55
 
In this equation, it is important that theorder of the denominator m is always greaterthan or equal to that of the numerator n. When m > nthe magnitude of H(s) goes to zero as s goes toinfinity. A transfer function satisfying this property is said to be strictly proper. The order of the denominator D(s) reflects theorder of the network. The roots that cancel this denominator are called the poles. On the other hand, the roots that cancel the numerator N(s) are called the zeros. The order of the network can be determined by the number of independent storage elements Cand L. Should you have two independent capacitors and one inductor, you have three distinct state variables making a third-order circuit: m equals 3 in (2.1) and you have three roots or polesin the denominator.
在这个等式中,最重要的是分母的阶数总是大于或等于分子的阶数。当m > n时,随著s趋向于无穷,H(s)的幅值趋向于0。传递函数符合这个性质的称“严格正则”的。分母D(s)的阶反映网络的阶。使分母为0的根叫做极点。使分子为0的根叫零点。网络的阶可以通过独立储能原件的个数确定。你有两个独立的电容和一个独立的电感,则有三个不同的状态变量构成的三阶电路,式中m=3并且在分母上有三个根或者说三个极点。
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  • 2018-3-24 22:12:58
 
这里的strictly proper  应该是“严格正则”,感谢吧主:eric.wentx 的指正网络上给出的例句也是这样翻译的
360截图1715020210490134.png


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  • 2018-3-24 14:42:46
 
For instance, if we study a first-order system featuring resistors and a capacitor, its transfer function expression may look like that:
例如,如果我们研究一阶系统特点,电阻和一个电容,它的传递函数表示可以看作如下表示: 2.2.png
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  • 2018-3-24 14:45:03
 
By reading this expression, can you immediately see if the circuit has gain (or attenuation)? Can you identify wherethe roots of the numerator (the zeros) or thedenominator (the poles) are? I cannot. To let you unveil the presence of these elements, you must rearrangethat equation into a slightly different format. Factor theterms a0 and b0 to obtain.
通过观察这个式子,如果这个线路有增益或者衰减你能立即看出来吗?你能确定分子的根(零点)或者分母的根(极点)在哪吗?我不能!为了让你揭露出这些成份的存在,你必须将这个式子调整成一个略微不同的形式。分解a0b0获得。
2.3.png


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  • 2018-3-24 14:47:48
 
Now, from this expression, you can tellthat the static gain or the attenuation of the circuit is
通过该式,你可以说线路的静态增益或者衰减是:
2.4.png

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副总工程师
  • 2018-3-24 14:52:01
 
and that a pole and a zero cohabit:
有一个极点和一个零点为: 2.5.png
If we apply this factorization to (2.1), weobtain:
如果我们对(2.1)式做因式分解,我们得到:
2.7.png
Then, the next step is to factor the polynomials so that the following form appears:

然后,下一步是分解多项式,因此得到如下形式:
2.8.png
This is the preferred factored pole-zero method as Dr. Middlebrook promoted it in his design-oriented analysis course,[1]. What is important is that a familiar structure appears in the final expression. For instance, it can be difficult to obtain a clean factored formas the previous in a second- or a third-order system. In that case, should you write something like

Middlebrook博士推荐的在他的设计向导分析课程里面的方法这是很好的分解极零点的方法。重要的是,最终表达式中会出现一个熟悉的结构。例如,有时可能难以获得一个像前面那样用一个二阶或者三阶系统的明显的分解形式。在这种情况下,你可以写一些像:
2.9.png


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  • 2018-3-24 14:53:48
 
then the reader will immediately recognizea frequency response combining a zero and the double poles of a second-ordersystem. In that case, we can show that the roots of the denominator are given by
然后你会立即意识到二阶系统的频率响应有一个零点和两个极点。在这种情况下,你可以通过下式得到分母的根
2.10.png
Depending on the quality factor value Q, these roots can be either real or complex conjugates.
根据质量因子值Q,这些根可以是实数或共轭复数。

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  • 2018-7-5 21:30:31
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quality factor :这里一般书上都称作为品质因数
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  • 2018-3-24 16:16:11
 
mark一下
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  • 2018-3-24 19:46:57
 
感谢楼主分享
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超级版主
  • 2018-3-27 09:24:19
 
拜读拜读
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  • 2018-3-27 19:32:45
 
共同学习共同进步
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  • 2018-3-27 19:36:57
 
2.1.2 The 0-db Crossover Pole    0-db交叉极点I named this factor the0-dBcrossover pole. The origin pole in an expression suchas that described by (2.11) is 0, meaning that when s = 0, thequotient goes to infinity. However, when s is combinedwith a coefficient—for instance when you have1/sRC(1 + ...)—it is advantageousto rewrite it as
我把这个因数叫做0-dB交叉极点。如式(2.11)所描述的原点极点为0,这意味着当s = 0时,商达到无穷大。但是,当s与系数结合时,例如,有1/sRC(1 + ...),这有助于改写成: 2.16.png

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  • 2018-3-27 19:41:38
 
in which I call wpo/ s the 0-dB crossover pole. Itequals 1 /RC in (2.16). It correspond to a cutoff frequency at which themagnitude of wpo/ s simply equals 1 or 0dB,hence its name. Figure 2.1 graphically represents the magnitude of wpo/ s.
其中我称wpo/ s为0dB交叉极点,在式(2.16)中它等于1/RC。它对应于一个交接频率,在这个频点上,wpo/ s的大小等于1或0dB。图(2.1)生动地描绘了wpo/ s的大小。
图2.1.png
Figure 2.1 The 0-dB crossover pole is a frequency at which themagnitude of wpo/ sis 1.
2.1 0 dB交叉极点是wpo/ s的幅值为1时的频率


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管理员
  • 2018-3-28 09:06:53
 
非常感谢支持
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  • 2018-3-27 19:47:27
 
Now, when combined with a zero, as in the G0 term from (2.14),it creates a break in the slopes and lets you purposely unveil a gain, changingwith the position of wpo. This is what Figure 2.2 shows youfor two different values of wpo.
现在,当与式(2.14)中的G0项相结合时有一个零点,它会在斜率中创造一个转折,并让你有意揭示一个增益,随着wpo.的位置而变化。这是图2.2显示了两个不同的wpo.值。
图2.2.png
Figure 2.2When combined with a zero, the gain G0 changes with the positionof wpo.
2.2当与零点结合时,增益G0 随着wpo的位置改变
The gain G0 is the mid-bandgain defined in (2.14).
增益G0就是被定义在式(2.14)的中频增益


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  • 2018-3-27 19:53:31
 
2.2 Solving for the Roots 解出根Thezeros of a transfer function are the frequency points at which the magnitude ofthe transfer function is zero: the excitation signal, the input, no longerreaches the output. For the opposite, the poles are the frequencypoints at which the transfer function goes to infinity. If we consider atransfer function as a fraction made of a numerator N(s) over adenominator D(s), then the zeros cancel the numerator while the poles cancel thedenominator. In other words, identifying the zeros and the poles of a system issimilar to respectively solving for the roots of its numerator anddenominator expressions. As all coefficients in (2.7)numerators or denominators are real, the roots can be either purely real orappear in complex conjugate pairs.
传递函数的零点是:传递函数幅值为0时的频点。激励信号,即输入,不再到达输出。相反的,极点是使传递函数幅值趋于无穷的点。如果我们认为传递函数为分数形式分子N(s)在分母D(s)的上面,零点使得分子为0,极点使得分母为0.换句话说,识别系统的零点和极点与分别解出传递函数分子和分母表达式的根相似。由于式(2.7)中分子或分母中的所有系数都是实数,因此根可以是纯实数或者为共轭复数。

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  • 2018-3-27 19:55:17
 
Let`s go through a few exampleswhere we will use SPICE notations in the upcomingequations:1k = 1000, 1Meg= 106, 1m= 0.001 and 1u= 10-6. We assume a transfer function H as follows:
让我们来看几个例子,我们将在下面的方程中使用SPICE符号:1k = 1000, 1Meg= 106, 1m= 0.001 ,1u = 10-6。我们假设传递函数为:
2.17.png


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  • 2018-3-27 19:57:49
 
The zeros of this transfer function arefound when H(s) = 0 or when N(s) = 0.On the contrary, thepoles are the roots of the denominator, D(s) and bring H(s) to infinity. Let `s rearrange (2.17) tomake it look a little friendlier:
传递函数的零点,当H(s) = 0 或者当N(s) = 0求得。相反,极点是分母D(s)的根,并使得H(s)趋于无穷。让我们重新整理式(2.17),使它看起来“亲切”一些:
2.18.png
H(s) = 0 if we have N(s) = 0:使分子为零:
2.19.png


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  • 2018-3-27 20:01:23
 
Given the factored form, we can identify thefollowing roots, our real zero:
求出零点:
2.20.png
The poles are obtained when making H(s) = ¥ or solving for D(s) = 0:
H(s) = ¥或者解出D(s) = 0获得极点:
2.21.png
Our real poles can be identified at thefollowing position:

得到极点的位置:
2.23.png


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  • 2018-3-27 20:03:57
 
When s equals one ofthese roots, we encounter either a zero or a pole. The poles or zerosfrequencies are obtained by computing the roots magnitude:
当s等于这些根中的一个时,我们遇到零点或极点。通过计算根大小来获得极点或零点频率:
2.25.png
In these simple examples, theroots are real and you do not need imaginarynotation to solve the equations.Let us have a look at a different function:
在这个例子中根都是实数没有虚数。来看个不同的函数:
2.28.png


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LV8
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  • 2018-3-27 20:06:56
 
First, we can massage the expression alittle bit to make it easier to work with. Develop the right side of D(s) and factorthe terms according to (2.9):
首先为了使表达式处理起来容易,我们稍稍地给表达式做个整形。展开分母的右侧并根据(2.9)分解式子:
2.29.png
We can identify a static gain G0:我们可以确定静态增益G0:
2.30.png
一个零点:
2.31.png
一个极点:
2.32.png

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LV8
副总工程师
  • 2018-3-27 20:11:05
 
a damped angularfrequency w0:阻尼角频率w0
2.33.png
and, finally, a quality factor Q:最后,品质因子Q
2.34.png
From (2.10), we can see that a Qgreater than 0.5 makes the polynomial within thesquare root a negative number: the roots are complex. Following the definitiongiven in (2.10), we have

2.10可以看出大于0.5Q使平方根内的多项式成为负数根是复数。根据(2.10)中的定义,解得:
2.35.png



2.34.png
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LV8
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  • 2018-3-27 20:13:00
 
These roots are said to be conjugate. Thefrequency at which these double poles appear is obtained by calculating themagnitude of either s2 or s3:
这些根是共轭的。通过计算任意一个s2或者s3的模获得这些双极点出现的频率。
2.37.png
Yes, this is the natural frequency we haveevaluated in (2.33). Should we plot(2.29), we would observe the combined actionof a pole/zero pair plus a double pole peaking at 10.25 radians per second, or510 mHz.
是的,这是我们在(2.33)中计算的固有频率。如果我们画出式(2.29)我们可以观察到一个极点/零点对加上一个以10.25弧度/秒或510 mHz的双极点峰值的组合作用。


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LV6
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  • 2018-3-28 09:38:44
 
楼主加油
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LV6
高级工程师
  • 2018-3-28 09:56:19
 
楼主是那么牛X
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  • 2018-3-28 11:32:32
 
经常看英文资料,其实你也可以的
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  • 2018-3-30 16:52:27
 
大牛
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版主
  • 2018-3-30 17:56:36
 
等最后可以分享下翻译的第二章中文文档
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LV8
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  • 2018-4-1 11:50:39
 
2.2.1 Poles and Zeros Found by Inspection通过观察找出极点和零点In the previous paragraphs, we have learnedthat zeros and poles could be found by solving for the roots of the consideredtransfer function. To make our life easier, we have seen that rearranging the equation in a factored form could help usidentify the poles/zeros positions in a quicker way. Unfortunately, despite thesimplicity brought by rearranging the expression, the starting point stillremains the transfer function equation that you must derive from node and meshanalysis of the studied network. Rather than deriving the transfer functionthrough classical analysis techniques, is there a way to actually detect wherethe poles and zeros are hidden and write the transfer function just by lookingat the network arrangement—in other words, by inspection? After all, we know that the final result should fit the formatgiven by(2.8). Let us see how we could do that.
上一段,我们学习过可以通过解出目标传递函数的根找到零点和极点。为了让我们的工作更容易一些,我们看到,用一种较快的方法,因式分解等式的形式可以帮助我们鉴别出零点和极点。不幸地,尽管通过改写表达式简单产生,起点仍然是必须从所研究网络的节点和网孔分析中得到传递函数方程。而不是通过传统的分析技术推导出传递函数,有办法检测到极点和零点的隐藏位置,并通过观察网络布置来写出传递函数 - 换句话说,通过观察?毕竟我们知道最终结果应该符合(2.8)的样式。让我们来看看我们可以怎么做。


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LV8
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  • 2018-4-1 11:54:18
 
If we understand thata transfer function links an output signal (the response)to an input signal(the excitation), then a zero at a certain frequency prevents the excitationfrom reaching the output. Let us try to apply this theory to the passive filterappearing in Figure 2.3. We can see an ac source delivering a signal through a resistorR1 to a network made of a series-parallel combination of two resistorsand a capacitor. There is one distinct storage element, the capacitor C1;this is a first-order network. As such, without knowing whether there are polesor zeros, it must fit the format given by (2.3):

如果我们知道,传递函数从输出(响应)链接到输出(激励),然后零点在某频率阻止激励到达输出。让我们来应用这条定理,无源滤波器如图(2.3)所示。交流源释放信号通过R1到两个电阻和一个电容组成的网络。这里有一个明显的储能元件,C1,是一阶网络。因此,不知道是否有极点或零点,他必须符合式(2.3)的形式。 2.38.png
t2.3.png
Figure 2.3 A simple first-order system featuring azero and a pole
.2.3,一个简单有零点和极点的一阶系统。



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LV8
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  • 2018-4-1 11:58:08
 
The brute-force algebra would be tocalculate the expression of impedance Z1and apply
用代数学可以计算出阻抗Z1的表达式
2.39.png
If you go ahead anddevelop the equation, you will end up with a moderately complicated expression, but chances to make mistakes during the expansions are real.Furthermore, without additional work on the final result, it is unlikely that polesand zeros pop up at the end.
如果继续展开式子,你会得到一个一般复杂的表达式,但是在展开的过程中可能会犯错。而且,如果不对最终结果进行额外的处理,最终不会出现极点和零点。


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LV8
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  • 2018-4-1 12:00:06
 
To start the derivation, let us observe thesystem at dc, when s = 0. This is exactlywhat SPICE does when it starts the simulation: to calculate the dc operating pointof the circuit under study, also called the bias point, SPICE opens all capacitorsand shorts all inductors. With the equivalent network, it calculates all dccurrents and voltages that the simulator will use for the rest of the simulation.In our network, we can do the same. When C1 is open, we areleft with R1 and R3.Therefore, the dcattenuation G0 is simply
来观察系统在直流时,当s=0.这正是SPICE开始模拟时所要做的:计算被研究电路的直流工作点(也称为偏置点),SPICE开路所有电容器并短路所有电感器。通过等效网络,计算所有直流电流和电压,模拟器会使用仿真的剩余部分。在我们的网络里,我们可以做同样的事情。当C1开路,留下R1和R3。因此直流衰减G0是:
2.40.png
Now, if you recall the definition of azero, it is a frequency point at which the excitation no longer reaches theoutput. In other words, in Figure 2.3, what element in the input signal pathcan stop its propagation?
现在,回顾零点的定义,它是激励不再接近输出的频点。换言之,在图(2.3),什么成分在输入信号路径上可以阻止它的传播


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LV8
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  • 2018-4-1 12:08:40
 
Either an element in series with the signaloffers an infinite impedance at a certain frequency or an element linking thesignal path to the ground becomes a short circuit, again at a certain frequencypoint. In our example, the only element that can stop the signal from reachingthe output is the series combination of R2 and C1.When its resultingimpedance is null (short circuit), we have a zero in the transfer function:
在某一频率与信号串联提供一个无穷的阻抗,或者在某一频率,将信号短路至地。在该例中只有一种要素可以阻止信号接近输出R2和C1的串联结合。当它的阻抗变为无效时(即短路),我们就有了在传递函数上的一个零点。
2.41.png
You can immediately see the zero positionis the root of the numerator:你可以立即看出零点的位置是分子的根:
2.42.png
We can now write the partial transferfunction of our network by combining(2.40) and (2.42):现在我们可以写出该网络的部分传递函数通过式(2.40)和式(2.42)的结合
2.43.png
We are almost there but we lack thedenominator expression D(s). This is where

the poles hide.我们还缺少分母D(s)的表达式。这是极点的隐藏位置。


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LV8
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  • 2018-4-1 12:12:13
 
2.2.2 Poles, Zeros, and TimeConstants 极点,零点和时间常数
By definition, a gain is a dimensionlessexpression. When you say the voltage gain of a system is 20 dB, it is anothermeans to say that the gain is 10 V/V or 10. In other words, if we go back to(2.7), and only consider second-order terms (for the sake of simplicity), wecan write the transfer equation of second-order networkas follows:
通过定义,增益是无量纲的。当说系统的电压增益是20dB时,其实是在说增益是10v/v或者是10。换言之,如果我们回到式(2.7),并且仅考虑二阶系统(为了简单起见),我们可以写出二阶网络的传递方程:
2.44.png




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LV8
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  • 2018-4-1 12:15:12
 
A term multiplied by shas the dimension of a frequency, Hz. A term multipliedby s2has a dimension of a squaredfrequency, Hz2. To make sure that all s-terms and s2-terms lose theirdimension when multiplied by a coefficient, these coefficients must have theinverse dimension. Therefore, the terms b1 /b0 and a1 /a0 must have adimension of Hz-1, whereas the terms b2/b0 anda2/a0 must have a dimension ofHz-2. Whatoffers a dimension of Hz-1oractually seconds? A time constant does. What has a dimension of Hz-2 or squared seconds? A product of time constants. And this makessense when you consider the expression of the zero found in (2.42) its denominatorhas indeed the dimension of a time constant. How to get the poles then? We need toidentify the time constants of our system but in a different manner than whatwe have shown for the zeros. As we stated, the transfer function denominator D(s) of a linearnetwork does not depend on its excitation or response signals. It only dependson the network structure alone. If you look at transfer functions describing a givennetwork, its output impedance, its input admittance, and so on, then you will seethat all these equations share a common denominator D(s).
一项乘以s有频率的量纲Hz。一项乘以s2有平方的量纲Hz2。为了确定所有的s项和s2项当乘以一个系数失去它们的量纲,这些系数必须有反量纲。因此,b1/b0 和 a1/a0项必须有Hz-1的量纲,b2 /b0 a2/ a0必须有Hz-2的量纲。是什么提供了Hz-1的量纲或者实际上的秒呢?是时间常数提供的。什么提供了Hz-2或者说是秒的平方呢?时间常数的产物。这个意义是:当考虑被发现的零点的表达式(2.42),它的分母确有时间常数的量纲。怎么获得极点呢?我们需要确定我们系统的时间常数,但是采用与我们已经显示的零点不同的方式。正如我们所述,线性网络的传递函数分母D(s)不依赖于他的激励或响应信号。他仅依赖网络的结构。如果观察给出的网络传递函数描述,他的输出阻抗,输入导纳等...,然后会看到所有的等式都共用了一个分母D(s)。

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LV8
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  • 2018-4-1 12:17:41
 
To study the network alone, we are going tobring its excitation signal to zero. How do we do that? Well, if the excitationsignal is a voltage source, we set it to zero: replace it by a short circuit.If the excitation signal is a current source, then open circuit it. Let`s applythis technique to Figure 2.3 by shorting to ground the left terminal of R1:
为了只研究网络本身,我们将把它的激励信号归零。我们怎么做呢?如果激励信号是电压源,我们设置它为0用短路代替电压源。如果激励是电流源,那就开路。让我们应用此方法将图(2.3)R1左端短接至地:
The time constant is easily calculated byevaluating the resistance seen from the capacitor terminals. The first one is obviously R2, in series withthe parallel combination of R1 and R3:
时间常数很容易通过从电容末端看到的阻抗计算。R2,与R1R3并联值串联。
2.45.png


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  • 2018-4-1 12:19:21
 
The pole definition, for this simplefirst-order system, is simply the inverse ofthe equivalent time constant:对于这个一阶系统极点的定义为时间常数的倒数
2.46.png
t2.4.png
Figure 2.4 Byshorting the input voltage, we can reveal the time constants of the system.通过将输入电压短路,我们可以得到系统的时间常数

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LV8
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  • 2018-4-1 12:22:40
 
The expression of our denominator D(s) is therefore因此我们的分母D(s)的表达式为
2.47.png
The complete transfer function then becomes完整的传递函数变为:

2.48.png
This is what is called a low-entropyequation by analogy to thermodynamic laws. Simplyput, the entropy of a system qualifies its degree of internal disorder: toproduce the work the system has been designed for, you need to bring less externalenergy when its entropy is low (elements are well organized and well ordered)than when it is high (elements are in disorder; this is a chaoticorganization). For our equations, a low-entropy expression gives you immediateinsight, without further work, on the transfer function it realizes. For theopposite, a high-entropy equationdoes not reveal anything and requires further energy through factorizations orexpansions before it tells you where poles and zeros are. Theanalytical technique we just described lets you write low-entropy equations by inspection, justby looking at the network schematic and identifying its time constants. Let us check this again via another simple example that appears inFigure 2.5 with an inductive first-order network. Let us try to derive thetransfer function applying what we learned.
通过类比于热力学定律这被称作是低熵等式。简单地说,系统的熵是他的内部混乱程度:当它的熵很低时难以产生系统设计的工作,你需要带来一些外部能量(分子是系统的,整齐的)比当熵高时(分子在无序状态下,这是一个混乱的系统)。对于我们的等式,低熵等式给你的信息,在传递函数上它的实现无需进一步工作。相反的,在它告诉你零点和极点位置之前,通过因式分解或者展开因式一个高熵等式不显示任何事情并且不需要进一步的能量。我们刚刚描述的分析技术可以让你通过观察来写低熵方程,只需查看网络原理图并确定其时间常数即可。让我们经过另一个简单的例子检查这一理论,如图(2.5)有一个电感的一阶系统。让我们尝试着应用所学过的东西得出传递函数。
t2.5.png
Figure 2.5 The capacitor has beenreplaced by an inductor.2.5用电感代替电容


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LV8
副总工程师
  • 2018-4-1 12:26:33
 
First, we start from s= 0, the dc transfer function. If capacitors areopen circuited at dc, inductors, for the opposite, are considered as shortcircuits. When shorting L1, the attenuation G0 is immediatelywritten as
首先我们从s=0开始即直流传递函数。如果电容在直流时时开路的电感正好相反被认为是短路。当短路L1,衰减增益G0立刻被写出:
2.49.png
The zero is found by identifying a networkthat prevents the excitation from reaching the output of the circuit understudy. A series element admittance can become zero at a frequency point or anetwork connecting the signal path to ground can have an impedance that dropsto zero. In our circuit, the series path is R1 and offers a fixedvalue. On the other hand, a network that can potentially shunt the excitationsignal to the ground is R2 and L1.This is the placewhere the zero hides. To unveil it, simply solve
通过识别网络阻止激励接近当下研究线路的输出找出零点。串联元件导纳在某频率点可以变为零,或者连接信号路径对地有零阻抗。在该线路中串联路径是R1并且提供一个固定的值。另一方面,一个网络R2L1,可能会分流激励信号至地。这就是零点隐匿的位置。为了揭示它,简单地解:
2.50.png



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LV8
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  • 2018-4-1 12:32:11
 
We have our zero position:我们有了零点的位置:
2.51.png
If the time constant of the resistor Rdriving the capacitor C is RC, then the time constant of an inductor L driven by a resistor R is L/ R. Whatimpedance drives the inductor L1 in Figure 2.5? Todiscover it, we set the excitation signal to zero. The impedance is the seriescombination of R2 and R1 in parallel with R3:
如果电阻R“驱动”电容C的时间常数是RC,则有电阻R驱动电感L的时间常数是L/R.图(2.5)中,什么阻抗驱动电感L1?为了找到它,我们令激励信号为0。阻抗是R1R3的并联的值再与R2串联。
2.52.png
The pole definition, for this first-ordersystem, again, is the inverse of the equivalent time constant:对于该一阶系统极点的定义是时间常数的倒数:
2.53.png
The expression of our denominator D(s) is therefore我们的分母Ds)表达式:
2.54.png
The complete transfer function then becomes完整的传递函数变为:

2.55.png
You can see how efficient this fastanalytical method can be to express the transfer function of a simple network. The appendix at the end of this chapter shows it at work in abridge impedance determination. Try to derive it yourself using the classicalalgebra technique, and you will quickly adopt these fast analytical techniques!Of course, in a small chapter portion, we have just scratched the surface, and asyou complicate the network under analysis with more storage elements, you needto apply different techniques such as the extra-element theorem (EET). I encourageyou to check [26] at the end of thischapter, as they will offer you a means to learn and make this techniqueefficiently work for you.

你可以看到这种快速分析方法可以有效地表达简单网络的传递函数。这一章结尾的附录介绍了它用在桥阻抗计算上。尝试使用经典代数学自己推导它,并且你将很快采纳这些快速分析技术!当然,在小章节中,我们只是抓住了表面,随着分析中使用更多存储元素使网络复杂化,你需要应用不同的技术,例如额外元素定理(EET)。我鼓励你在本章结尾处查阅[2-6],因为它们将为你提供一种学习方法,使这种技术为你效力。


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  • 2018-4-8 21:27:06
 
2.3 Transient Response and Roots暂态响应和根

The stability of aclosed-loop system can be assessed in different ways. If an ac sweep teaches usabout phase margin at a given crossover frequency, it does not readily tell ushow the system will react to an incoming perturbation or a sudden change in theinput setpoint. A common test consists of exciting the control system inputwith a given stimulus. As detailed in Chapter 1, there are numerous types ofavailable stimuli: a step, a Dirac pulse, a linear ramp, and so on. Generally,the response to a step is the most popular choice, in particular for regulatorssuch as linear or switching converters. If stepping an electronic load in thelaboratory does not require a particular care, applying the technique to atransfer function implies that some precautions will be observed. First, ourtransfer function is expressed in the Laplace domain, whereas the step belongsto the time domain. A transformation is needed to transit from one domain tothe other. Once the step is converted into the Laplace domain, it becomes theexcitation signal U(s) to the transfer function H(s) understudy. Then, as its output signal Y(s) is also expressed using Laplace notation, an inverseLaplace-transform is necessary to return to the time-domain and see theresulting waveform. Figure 2.6 shows this process.
闭环系统的稳定性可以用不同的方式讨论。如果交流扫描告诉我们在给定交接频率的相位裕度,它不会告诉我们系统将如何对输入扰动或输入给定点的突变做出反应。通常测试由激励控制系统输入与给定的激励构成。第一章提到过的有多种可用的激励:阶跃,脉冲,斜坡等等...。通常选择对阶跃的响应,尤其对于:线性转换器或者开关转换器。如果在实验室中用电子负载加载并不需要特别小心,将该技术应用于传递函数意味着可以观察到一些预防措施。首先传递函数是用拉氏域表达的,然而阶跃是属于时域。需要转换从一个域到另一个域。一旦阶跃信号转化到拉氏域,它就成为正在研究传递函数H(s)的激励信号U(s)。然后,由于其输出信号Y(s)也使用拉普拉斯符号来表示,所以需要拉普拉斯反变换来返回到时域并且看到所得到的波形。图2.6显示了此过程。

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  • 2018-4-8 21:28:37
 
t2.6.png
Figure 2.6 You can assess the time-domain response of a Laplace transfer functionby exciting itsinput with a unity step.你可以通过单位激励它的输入来评估传递函数的时域响应。
What is the Laplace-transform of a unitystep? Let`s have a look at the waveform that appears in Figure 2.7: it is 0 forall negative time values and equals 1 at t = 0and all values beyond. This is atime-domain waveform and, according to Figure 2.6,we must transpose it into theLaplace domain before driving the transfer function of interest. In Chapter 1,we learned the definition of the Laplace transform:

阶跃的拉氏变换是什么?让我们看看2.7图所示的波形,时间为负时它的值为0,从t=0开始,值为1这是时域波形,根据2.6图所示,我们必须把它转换为拉氏域。第一章,我们学过拉氏变换的定义:
2.56】.png


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  • 2018-4-8 21:30:23
 
As u(t) equals 1 from 0 to ¥, the equation becomes:随着u(t)0¥都为1,等式变为:
2.57.png
This is the classical definition of a unitystep in the Laplace domain for s > 0.  这是单位阶跃当s > 0在拉氏域的定义。
t 2.7.png
Figure 2.7 A unity step function is 0 for t <0 and jumps to 1 for t 0.单位阶跃是t < 0时为0t 0时跳为1.


2.58.png
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  • 2018-4-8 21:32:34
 
If an input signal 1/s enters atransfer function H(s), the resulting output signal isnothing else than
如果输入信号1/s进入传递函数H(s),输出结果为:
2.58.png
From this Laplace-domain expression, weneed to extract its time-domain correspondence through a reverseLaplace-transform:

从这个拉氏域表达式,我们需要通过反拉氏变换提取出它对应的时域表达式:
2.59.png
Is it as simple as that? Well, itdepends if you use a mathematical solver or try to derive the equation yourself. Inthe second case, you have to realize that the answer isnot the product of the inverse Laplace transform of each individual term: youmust rewrite the expression as a sum of individual terms, each having its own inverseLaplace-transform equivalent. Since the Laplace-transformis a linearoperator, the inverse-Laplacetransform of this sum will be the sum of all individual inverseLaplace-transform terms. However, as most of our transfer functions are oftenin the form of a rational function N(s) D(s), you need to split them into a sum of ratios of small polynomials.This technique is called partial fractionexpansion, and articles on the subject canbe found on mathematical textbooks or on the Web. Theinvolved algebra looks simple but you need to be careful when expanding complextransfer functions.

反拉氏变换和拉氏变换一样简单?它依靠是否使用数学求解器或者你自己尝试得出的等式。第二种情况,你必须意识到,答案不是每个单项的逆拉普拉斯变换的结果,你必须重写表达式为独立项相加的形式,每一项有它自己的拉氏反变换。由于拉氏变换是线性算子,因此这个拉氏反变换会是所有独立拉氏反变换项的和。但是,我们的传递函数通常是用有理数N(s)/D(s)的形式,你需要拆开他们为多项式的比的和。这叫做部分分式展开式,相关文章可以在数学教材或者网页上找到。涉及的代数学看起来简单,但是当展开复杂的传递函数的时候你需要当心。

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副总工程师
  • 2018-4-8 21:41:17
 
Let`s try to apply the technique to ourfirst transfer function (2.17) and see its response to a unity-step stimulus:
让我们尝试将该方法应用于我们的第一个传递函数(2.17),并看到它对单位阶跃激励的响应:
2.60.png
The technique we are going to use is calledthe Heaviside cover-upmethod, named after the English electrical engineer Oliver Heaviside. In the case of (2.60), theory tells us that it can berewritten or expanded into the following terms:

我们将要用到的方法叫做Heaviside遮盖法,得名于英国电气工程师OliverHeaviside。在式(2.60)情况下,该方法告诉我们它可以被改写或者被展开为如下形式:
2.61.png


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  • 2018-4-8 21:42:33
 
As you can see, eachdenominator cancels for the roots we already found in (2.23) and (2.24), but itnow also includes s = 0. To determine the value of coefficient a1, a2, and a3, the idea is to make s equal the selected root (0 for a1, 1k for a2, and so on) while multiplying(2.60) by the denominator including that root (by s for a1, by s + 1k for a2, and so on). Therefore, the concerned denominator on bothsides of the fraction naturally disappears, a bit like if you were covering itup with your finger. It then leaves a simple equationin s, where s takes the value of the selected root. Sounds complicated? Not really as the following details show:
正如你所看到的,我们已经在(2.23)和(2.24)中找到的根使得每个分母都为0,但现在也包括了s = 0。为了确定系数a1, a2, a3的值,办法是使s等于所选的根(a1的为0a2的为–1k,以此类推)同时用(2.60)乘以包含该根的分母(a1乘以sa2的乘以s+1,以此类推)。因此,分数两边的相关分母自然消失,有点像用手指遮盖它。然后在s中留下一个简单的等式,其中s取所选根的值。听起来很复杂?不,如下所示:
2.62.png
2.63.png
2.64.png

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  • 2018-4-8 21:44:32
 
This is it, if we rewrite (2.61) with the right coefficients, we have
就是这样,如果我们重写有正确系数的式(2.61)
2.65.png
The inverse Laplace-transform of theprevious equation is thus

前面等式的反拉氏变换为:
2.66.png
We can look at theinverse Laplace-transform tables to get each individual term:
我们可以查看反拉普拉斯变换表以获得每一项:
2.67.png
2.68.png
2.69.png

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  • 2018-4-8 21:45:47
 
Assembling all these terms according to (2.66), we now have our time-domain expression:
根据(2.66)集合所有项,有时域的表达式: 2.70.png
The first immediateremark is that the exponents on the exponential terms are the roots of thecharacteristic equation of H(s). The secondremark concerns the zeros. They do influence thetime-domain response, but only the poles directly impact the decaying timeconstants in the response as we just saw. The third remark concerns theirsigns, negative. It means that when t goes to infinity, since allexponential terms go to zero, the output signal reaches a steady-state valuegiven by the first term: with a 1-V step, the output should reach 166 μV. By theway, we knew it from the start, after we rearranged (2.17) into (2.18) as thedc gain, G0, is 1/6k or 166u.This is confirmed if we plot (2.70) in Figure 2.8.
第一点是在指数项的指数是Hs的特征方程的根。其次涉及到零点。它们确实会影响时域响应,但只有极点直接影响我们刚刚看到响应中的衰减时间常数。最后他们的符号是负的。这也意味着当t趋于无穷,由于所有指数项趋于0,输出信号接近通过第一项给出的稳态值:随着1-V阶跃,输出应该接近166 μV。顺便说一下,我们从一开始就知道,当我们把(2.17)重新排列成(2.18)时,直流增益G0就是1 / 6k166u。如果我们在图2.8中绘制(2.70),则证实了这一点。
t 2.8.png
Figure 2.8 Time-domain response of the transferfunction H(s) described by (2.17) when subject to a 1 Vinput step.
2.8,在1-v输入阶跃下传递函数H(s)的时域响应


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  • 2018-4-8 21:46:10
 
From these derivation lines, if we try toobtain the step response from a transfer function that contains positive rootsin its characteristic equation, the time-domain response will containexponential terms featuring positive exponents. As t goes to infinity, these exponentialterms do not die out to zero but keep increasing, making the output signalseverely diverging: the system response is not bounded; you do not controlanything. The question that immediately arises is, if myanalysis shows that all roots are negative at the considered operatingpoint, can they move and suddenly become positive?
如果我们尝试从一个在特征方程里有正根的传递函数中获得阶跃响应,则时域响应将包含具有正指数的指数项。当t趋于无穷大时,这些指数项不会消失,而是继续增加,使得输出信号严重发散:系统响应不受限制; 你什么都控制不了。立即出现的问题是,如果我的分析表明,所有的根都是负的,在所考虑的工作点,它们是否会移动并突然变为正?

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  • 2018-4-9 09:25:20
 
楼主加油
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  • 2018-4-9 11:44:23
 
谢谢分享啊   
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  • 2018-4-14 22:17:27
 
2.3.1 When the Roots Are Moving
In the previous examples, we havefixed-values roots (e.g., –1k or –30k). They do not depend on other variables, and the transient responseto a step won`t change as long as the roots remain constant. Now, let usconsider a more practical case such as the unity-return regulator appearing inFigure 2.9.
在前面的例子中,我们有固定的根(如–1k或–30k)。他们不依靠其他变量,并且只要根是常数,对于阶跃的暂态响应就不会改变。现在让我们思考一个更实际的情况,如图2.9所示的单位反馈控制器
t2.9.png
Figure 2.9 This circuit mimics a first-order converter H stabilized by a 60-dB gain compensator G  featuring a single pole response.

图2.9  这个电路模拟一个由单极响应的,60-dB增益补偿器G,稳定的一阶变换器H.



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  • 2018-4-14 22:21:57
 
We can see a transfer function H(s) that, forinstance, could be the simplified expression of a voltage-mode switchingconverter operated in the discontinuous conduction mode (DCM), a buck-boost. Wecan show that the zero wz1 is created by the outputcapacitor Cout  and its equivalentseries resistance (ESR):
我们可以看到一个传递函数H(s),例如,它可以是工作在间断导通模式下电压模式开关转换器(升降压转换器)的简化表达式。我们可以写出它的零点wz1由输出电容Cout和它的等效电阻产生:
2.71.png
The pole wp1 depends on the load Rload and Cout: 极点wp1由负载电阻R loadCout产生。
2.72.png
To stabilize the power supply, we havedesigned a compensator G(s) that places a pole wp2  while offering some gain G2. This is obviously not the best-known compensation, but we keep itsimple for the sake of the example. We will learn in Chapter 3 that a unityfeedback system closed-loop response TCL(s) obeys the following expressionwhere the term 1 +G(s)H(s) represents the characteristic equation of the closed-loop transferfunction:

为了使得电源稳定,我们要设计补偿器G(s),它在提供一些增益G2的同时放置一个极点wp2。这显然不是最有名的补偿,但这个例子为了简单起见。我们将在第3章学习一个整合的反馈系统闭环响应TCLs),表达式如下,这里1+G(s)H(s)项代表闭环传递函数的特征方程:
2.73.png
In this equation, TOL(s) represents the open-loop gaindefined as:在这个方程中,TOL(s)是开环增益,定义为:
2.74.png


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  • 2018-4-14 22:25:53
 
Let us develop (2.73)and see if we can make it fit a familiar format. If we develop andrearrange all the terms, we can show that (2.73) can be rewritten as
展开式(2.73)并重写每一项,使它成为一个我们熟悉的形式,为:
2.75.png
This format fits a second-order equation:这个形式是一个二阶方程:
2.76.png
In which we canidentify the following terms: 我们可以在其中找出以下项:
2.77.png


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  • 2018-4-14 22:41:35
 
The closed-loop poles of (2.76) are givenby the roots of its denominator, D(s). What is worth noting is that the open-loop zero of H(s), wz1, now appears in thecharacteristic equation and affects the closed-loop poles. This is always true:the zeros that appear in the plant transfer function or the ones you will placein the compensator G always show up in the characteristic equation. If you place lowfrequency zeros to improve the phase margin in a system to be compensated,these zeros will turn into low-frequency poles once the loop is closed.Low-frequency poles imply a slowly reacting system. This is something to keepin mind when evaluating the compensation strategy to follow: is it worthselecting a high crossover frequency requiring the placement of one or twozeros at low frequency, or is it better to adopt a different crossover pointand avoid assigning zeros at low frequency?
式(2.76)的闭环极点由它的分母D(s)给出。值得注意的是H(s)的开环零点,wz1,现在出现在特征方程里并且影响着闭环极点。这是事实,出现在被控对象传递函数中的零点或者将要放置在补偿器G中的零点总是出现在特征方程中。如果将低频零点置于待补偿系统的相位裕度上,一旦闭合,这些零点将变为低频极点。低频极点意味着一个响应缓慢的系统。在评估补偿策略时,需要注意以下几点:是否值得选择一个高交接频率,需要在低频处放置一个或两个零点,还是采用不同的交接点,避免在低频处出现零点?
For stability purposes, what matters now isthe expression of (2.76) denominator roots. Since we put our transfer equationunder a known second-order form, the roots or the poles of the denominatorfollow the definition of (2.10):
出于稳定的目的,现在重要的是(2.76)分母根的表达。由于我们把特征方程置于一个已知的二阶形式中,分母的根或极点遵循(2.10)的定义:
2.80.png



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  • 2018-4-14 22:48:12
 
From this expression, what matters is thequality factor value:
1. Q < 0.5: the expression under the square root is strictly positive;the roots are separate and real.
2. Q = 0.5: the expressionunder the square root is zero; the roots are coincident and real.
3. Q > 0.5: the expressionunder the square root is negative; the roots are imaginary with a real part.
从这个表达式看出,重要的是品质因数的值:
1.Q < 0.5:根号下的表达式是正数,根为两个不相等的正实数。
2.Q = 0.5: 根号下的表达式是0,根为两个相等的实数。
3.Q > 0.5: 根号下的表达式是负数,根为有实部的虚数。
In the expression of Q given by(2.78), there are parameters that can change, while some are less likely tomove. This is part of the design discussion to see whether these parameters canbe considered real threats or are unlikely to affect the final result despite smallvariations. It is your responsibility, as a design engineer, to cover the caseswhere large variations of parameters are likely to happen. Whether they are dueto operating conditions (e.g., temperature, bias points) or production spreads(e.g., tolerance, change of component), you must check their impact on the stability.Here, for the sake of the example, we will consider only the power stage zerobrought by the output capacitor equivalent series resistance (ESR) described by(2.71). This parasitic element not only moves in relation to the capacitortemperature (the resistance increases as the temperature drops), it is alsoaffected by a wide spread in production. Figure 2.10 shows the equivalentcircuit of a typical capacitor. This model can be supplemented with other  parasitics, such as a leakage resistance oran equivalent series inductance (ESL), but this simple approach will do for ourexample.

在式 (2.78)给出Q的表达式中,有些参数可以改变,而有些则不太可能移动。这是设计讨论的一部分,看看这些参数是否可以被认为是真正的威胁,或者尽管有小的变化,也不太可能影响最终结果。作为设计工程师,你有责任在设计时去涵盖可能发生参数变化的情况。无论它们是由于工作条件(例如温度,偏置点)还是材料一致性问题(例如,容差,部件的变化),你都必须检查它们对稳定性的影响。在这里,为了举例,我们只考虑(2.71)所描述的输出电容等效串联电阻(ESR)带来的功率级零点。这种寄生元件不仅相对于电容器温度移动(电阻随着温度下降而增加),而且还受到材料一致性问题的影响。图2.10显示了典型电容的等效电路。这个模型可以补充其他寄生效应,例如泄漏电阻或等效串联电感(ESL),但是这个简单的近似可以作为我们的例子。
t2.10.png
Figure 2.10 A typical capacitor always exhibits stray elements such as an ESR. This parasitic element introduces a zero in thetransfer function.
2.10 一个典型电容总是有杂散的元素像等效电阻。在传递函数中这个寄生参数引入一个零点。


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  • 2018-4-16 12:22:21
 
学习轻松多了。谢谢
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  • 2018-7-14 06:02:25
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通俗易懂,果然好东东
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  • 2018-4-15 12:29:39
 
辛苦了     
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LV8
副总工程师
  • 2018-4-15 12:40:56
 
谢谢楼主的翻译 啊
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  • 2018-4-16 21:03:35
 
Let`s assume a 470-μF output capacitor. Looking into themanufacturer datasheet, we found that its typical ESR is 90mW at 25°C. However, it can vary from 50 mW to 200 mW if we consider atemperature range from –40°C to +105°C and production spreads. The zerogiven by (2.71) will thus move between
假设一个470-μF的输出电容。查看参数表,我们找到了ESR典型值为90 mW(25°C)。如果我们只考虑温度范围为–40°C ~+105°C和材料差异,ESR可以从50 mW 变化到200 mW。通过式(2.71),零点会移动在此区间:
2.81.png
2.82.png
For the sake of the example, we will assumethat the rest of elements constitutive of H and G have the following values:
为了举例,我们假定构成HG的其他元素具有下列值:
z.png



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LV8
副总工程师
  • 2018-4-16 21:08:38
 
Based on these numbers, the quality factor Q describedby (2.78) will change as the zero modifies its position inrelations to ESR variations. For the lowest zero position, we have
基于这些数值由式2.78描述的品质因子Q将随着零点的变化而改变。较低的零点位置:
2.83.png
when the ESR reduces athigher temperature, the quality factor becomes:当在高温时ESR减小,品质因数变为:
2.84.png
The ESR value for which Qequals 0.5 is found by solving for wz1                 Q = 0.5时的ESR值可以通过求解Wz1找到
2.85.png
It happens when the zero reaches:零点接近:
2.86.png
With a 470-μF capacitor, it corresponds to anESR of 116 mΩ

对应的ESR116 mΩ
As the ESR varies, it affects the qualityfactor and changes the nature of the roots. These roots can therefore be purelyreal; in that case, Q = 0.312, and the step response of the closed-loop transfer function isnonringing. According to (2.80), when we have a 1.7-kHz zero, these roots havethe following values:
ESR改变,它影响品质因数并且改变根的性质。这些根可以是纯实数的,在这种情况下,Q = 0.312,并且闭环传递函数的阶跃响应是没有震荡的。根据(2.80)1.7-kHz的零点,根有如下值:
2.87.png


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  • 2018-4-16 21:10:52
 
If we multiply (2.75)by 1/s and extract the time-domainresponse, we obtain, for these conditions, a signal plotted in Figure 2.11. Please note the presence of a light overshoot though, despite a Q less than0.5. This overshoot is brought by the zero presence thataffects the transient response but not the steady-state value. The finalvalue is not 1 V as expected. This is normal; the closed-loop gain defined by (2.77)is less than 1 (0.98 exactly): we have a permanent static error of 20 mV.
如果我们将(2.75)乘以1 / s并提取时域响应,我们得到如图2.11所示的信号。尽管Q值小于0.5,但请注意小过冲的存在。这个过冲是通过零点的存在产生的,影响暂态响应而不是稳态值。最终的值不是期望的1v。这是正常的闭环增益被定义通过(2.77),是小于1(实际上是0.98)有固定的稳态误差20mv
t2.11.png
Figure 2.11 The step response for real roots is anonringing signal with a very minor overshoot.

实数根的阶跃响应是无震荡的,有非常小的过冲。

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LV8
副总工程师
  • 2018-4-16 21:30:31
 
When the ESR exactly equals 116 mW, both roots are coincident and equal:
ESR116mΩ,两个根相等:
2.88.png
The step responseappears in Figure 2.12. Theory tells us that the step response of asecond-order system featuring coincident poles should not bring overshoot, butwe can see it in the picture. This is because thestudy considers only a system having only poles, not an extra zero as we havehere. Its presence changes the transient response.

阶跃响应出现在图2.12。理论告诉我们具有重合极点的二阶系统阶跃响应,不应带来过冲但我们可以在图片中看到它。这是因为被研究的系统仅有极点,没有零点的存在。它的存在改变了暂态响应。
t2.12.png
Figure 2.12 With coincident poles, the response is fast,and the overshoot is still weak. 有一致的极点响应快并且过冲小


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  • 2018-4-16 21:33:26
 
Now, when the ESR is really low, the rootsbecome imaginary conjugates: 现在,当ESR非常低,根变为虚数对时:
2.89.png
In that case, a more pronounced overshootstarts to appear, as shown in Figure2.13.在此情况下,一个明显的过冲出现在图2.13中。
t2.13.png
Figure 2.13 As the imaginary coefficients now appear, theclosed-loop ac response starts to peak, inducing some ringing in thetime-domain response. 当虚数对出现时闭环交流响应开始出现峰值在时域响应中引起一些振荡

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LV8
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  • 2018-4-16 21:38:28
 
If for any reason the ESR would becomenegligible (i.e., you chose a multilayer type of capacitor or parallel a lot oflow-ESR capacitors), the roots would simply become
如果由于某种原因ESR变得可以忽略(如,你选择了一个多层电容器或者并联了很多低ESR的电容器)根会变为:
2.90.png
The corresponding step response exhibitsmore overshoot than the previous signal but is still stable. Figure 2.14 detailsthe results:

相应的阶跃响应展现出比前面的信号更大的过冲但是仍是稳定的。图2.14显示了结果:
t2.14.png
Figure 2.14 Despite a pronounced overshoot, the response is stillstable. 尽管过冲显著响应仍是稳定的
From (2.90), we can compute the naturalangular frequency w0 as

(2.90)我们可以计算出固有角频率w0
2.91.png
A similar value is obtained if you use(2.79) or (2.89).
如果使用(2.79)或(2.89),则获得类似的值。
As can be observed, by changing the ESRvalues, the roots exhibit different real and imaginary coefficients. Actually,we can see that a decreasing ESR makes the quality factor grow while the realpart of the roots decreases. This makes sense: the real part of the roots expresses losses that damp the second-order system. The Q  increase is simply the result of these damping effects going down as the ESR vanishes.

通过ESR值的改变,可以观察到,根展现出不同的实数和虚数系数。实际上,我们可以看到ESR的下降使品质因子增加,而根的实数部分减少。这是有道理的:二阶系统根的实部表示阻尼损耗这里多说一下:这个二阶系统特征根实部是衰减系数,由阻尼比和自然频率相乘得到,虚部系数叫做阻尼振荡频率。)Q增加的结果仅仅是这些阻尼效果随着ESR的下降而下降。

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  • 2018-4-16 21:59:31
 
我的理解“This makes sense: the realpart of the roots expresses losses that damp the second-order system.”这句话时我翻译出来的意思,大家有没有不同的见解?
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LV10
总工程师
  • 2018-4-17 11:53:31
 
有点不同见解:

对于复数 a+bi ,a 、b 都是实数,a 通常称为实部,而 bi 称为虚部,imaginary coefficients 我认为应该译成 虚部系数,这里的系数是 b。

quality factor 通常称为 品质因子,而不是质量因子。

实部表示对二阶系统的阻尼损耗,不是振荡,damp = 阻尼。

最后一句,我觉得应该这样翻译:Q增加的结果仅仅是这些阻尼效果随着ESR的下降而下降。(文中没有讲ESR消失,实际也不可能消失)

仅供参考。
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LV8
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  • 2018-4-17 21:22:23
 
1,2同意,而且1是很低级的错误。3,突然发现,把 the second-order system,拿到前面去,就翻译通顺了“二阶系统根的实部表示阻尼损耗”,因为之前我觉得那个damp是做动词的。可是阻尼的确有阻碍震荡的作用啊,毕竟是损耗嘛。

4,vanishes 就是消失,成为0的意思,不过我觉得你说的ESR不可能消失也是个事实,所以第4点,参考你的翻译。

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LV10
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  • 2018-4-17 22:40:49
 
说的很有道理,另外 vanishe 确实是消失或者说为 0,因此你译成ESR为零也没有错。
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  • 2018-4-30 10:32:38
 
2.4 S-Plane and Transient Response  S平面和暂态响应
In stability analysis, it is interesting toplot the trajectory of these roots and see how they move in relation to aconsidered parameter. In the present case, we have chosen the  ESR of the output capacitor, but in most ofthe textbooks the selected parameter is a gain coefficient k. A typicalexample shown in the literature is that of Figure 2.15.
在稳定性分析中,感兴趣的是绘制这些根的轨迹并查看它们如何相对于所考虑的参数移动。在当前情况下,我们选择了输出电容的ESR,但是在大多数教科书中,选择的参数是增益系数k2.15是文献中显示的典型例子
t2.15.png

Figure 2.15 In this typical example, the parameter k is a gain inserted in the control loop and subjectto wide variations.
图2.15 在这个典型的例子中,参数k是在控制回路中插入的一个增益并且受到很大的变化。
This is a unity-gain return control system,and it is easy to derive the transfer function from input to output:

这是一个单位反馈系统,传递函数很容易从输入到输出推导:
2.92.png
G and H are individual transfer functions,made of a numerator and a denominator:

GH是独立的传递函数,由分子和分母组成。
2.93.png
If we now reinject these definitions in(2.92), we obtain:

现在将这些代入到2.92),得到:
2.95.png




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LV8
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  • 2018-4-30 10:59:27
 
This equation shows that the poles andzeros put in G(s) to shape the open-loop response for adequate crossover frequencyand phase margin now appear in the closed-loop transfer equation with the terms NG(s) and DG(s). The characteristic equation oftennoted X(s) (Chi, pronounced like key) can be rewritten as:
这个式子表明,极点和零点放进G(s)为了适当的交接频率和相位裕度以形成开环响应,现在出现在具有NG(s) DG(s)项的闭环传递方程中。特征方程X(s)可以被写成:
2.96.png
This is an interesting equation because itshows that the zeros you will put in the compensator G (e.g., toboost the phase margin at low frequency) will turn into poles when the loop isclosed. A low-frequency open-loop zero turning into a low frequency closed-looppole will slow down the response to an incoming perturbation or a setpointchange. If we now look at the parameter k, we can see several cases,depending on whether k is small or high:
这是一个有趣的方程,因为它显示了你将放入补偿器G中的零点(例如,在低频率时提高相位裕度)将在闭合回路时变成极点。低频开环零点转变为低频闭环极点将减缓对输入扰动或给定点变化的响应。如果我们现在看看参数k,我们可以看到几种情况,取决于k(的值)是小还是大:
k is small,then the characteristic equation can simplify to X(s) (s) = DG(s)DH(s): the closed-loop poles are those already present in the open-loopgain equation.

K小,传递函数可以被简单的看成X(s) (s) = DG(s)DH(s)闭环极点是已经存在于开环增益方程中。
k nowincreases, and as the roots are a continuous function of k they move away from their open-loopdefinition to fully satisfy (2.96).

k现在增加,根是k的连续函数,k移动远离他们的开环定义式满足(2.96)
If k increases furtherand becomes really high, then the left side of (2.96) can be neglected and thedefinition becomes: X(s) (s) = DG(s)DH(s). The closedloop poles are now given by the open-loop zeros!

现在k进一步增加,并且到很高,式(2.96)左边项可以被忽略并且定义式变为X(s) (s) = DG(s)DH(s)。现在闭环极点就是开环零点。

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  • 2018-5-1 18:42:54
 
不错。
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LV6
高级工程师
  • 2018-5-2 08:47:09
 
楼主辛苦了
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  • 2018-5-5 21:03:54
 
It is important to check the values takenby the roots as k changes. For instance, are there some conditions where the realparts of the roots dangerously diminish, contributing to eliminate the damping?Can these roots suddenly reverse their negative sign and turn positive, makingthe system output diverging with all consequences behind? This makes sense,since as we discovered with (2.70) the poles directly influence the time-domainresponse of our system.
检查根随着k的值发生变化是很重要的。例如,是否有一些情况下,根的实部减少,有助于消除阻尼?这些根能否突然由负变正,使系统输出偏离所有结果?这是有道理的,因为我们发现(2.70)极点直接影响我们系统的时域响应。
Network theory teaches us that the complete or total responseof a system is generally the sum of its natural or free response rn(t) and its forced response rf (t). The free response is obtained bysetting the input u(t) to zero and considering nonzero initial conditions. On thecontrary, the forced response is obtained by solely considering the input andsetting all initial conditions to zero. The forced response in (2.70) was 166 μV, whereasthe rest of the terms composed the natural response. The response of a SISOsystem is thus given by
网络理论告诉我们一个系统的完全响应通常是它的固有响应自由响应rn(t)和强制响应rf (t)之和。通过设置输入u(t)0并认为非0初始条件得到自由响应。相反地,强制响应是通过单独地考虑输入并且设置所有初态为0获得。在(2.70)中,强制响应是166 μV,其余项为固有响应。一个单输入单输出的系统响应为:
2.97.png
In this expression, pi are thepoles of the characteristic equation, and Ci are the coefficients of theexponential terms. The number of poles depends on thepolynomial degree of the denominator: two poles for a second-order system,three for a third-order system, and so on. The easiest way to study these rootsis to place them on a dedicated map called an Argand diagram, commonly denotedas the s-plane. The variable sis classically defined by

在这个表达式里pi是特征方程的极点Ci是指数项的系数。极点的数量决于分母的多项式次数:两个极点为二阶系统,三个极点为三阶系统,等等...。研究这些根最简单的方法是将它们放在一个称为Argand图的专用图上,通常称为S平面。变量s定义:
2.98.png


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  • 2018-5-5 21:05:08
 
The s-plane is thus a two-dimensionalgraph where the vertical axis represents the imaginary portion of s, jw, and thehorizontal axis, its real part, s. The area of the planecorresponding to negative roots is called the left half plane (LHP), whereasthe area situated to the right of the vertical axis is simply called the right halfplane (RHP); the roots placed in this area have positive real parts. The polesare illustrated by a cross, ´, whereas the zeros aredesignated via a circle, . The location ofthe poles in the s-plane will affect the signal delivered by (2.98) as follows:
因此,s平面是二维图,其中,纵轴代表s的虚部,即jw,横轴为s的实部。与负根对应的平面的区域称为左半平面(LHP),而位于纵轴右侧的区域简称为右半平面(RHP); 放在这个区域的根有正实部。极点用×来表示,零点用○表示。s平面中的极点位置将影响信号传递通过(2.98)式如下:
1.If the pole is real, pi = -s, it is placed in the LHP; it is called a LHP pole. Its contributionto the response is in the form Ce-st, a decaying exponential component of duration 1/s. Therefore, if the polelocation is far from the origin of the s-plane, then the response is aquickly decaying signal. For the opposite, as the pole`s location moves closerto the 0 point, the response will be a slowly decaying signal. Suppose we havein the time-domain expression a term looking like 5e-0.1t; the signal starting with anamplitude of 5 V will decay and lasts 10 s before dying out to zero.
1.如果极点是实数,pi = -s,它在左半平面,称为左半平面极点。它对响应的贡献是Ce-st,的形式,它是一个持续时间为1 /s的衰减指数分量。因此,如果极点位置远离s平面的原点,则响应是快速衰减的信号。相反,当极点的位置靠近原点时,响应将是缓慢衰减的信号。假设我们在时域表达式中有一项为5e-0.1t; 5 V开始的信号将衰减并持续10s,然后消失为零。
2.A pole appearing at the origin, pi = 0, defines a term of constant amplitude, defined by the initialconditions: 5e-pit = 5e0 = 5 V
极点出现在原点上,pi=0,定义为常值项,由初始条件定义:5e-pit = 5e0 = 5 V
3. If the pole is real but this time appears in the RHP, it is called a RHPpole and is of the form pi =s. It is a positive root. Itscontribution to the output signal is of the form Cest ,a continuously growing component. If you deal with a systemfeaturing a closed-loop RHP pole, it is unstable.
如果极点是实数,但是它出现在右半平面,称为右半平面极点并且形式如pi =s。它是正根。输出信号形如Cest,是一个持续增长的分量。如果你处理一个闭环右半平面极点系统,它是不稳定的。
4. When solving the characteristic equationroots, you can find conjugate pairs of roots, leading to a form pi = -s± jω. In this case, the contribution to the time-domain signal is adecaying sinusoidal signal of frequency w obeying theform Ae-st sin(ωt +φ). A and φ are imposed by the initialconditions. Again, the real part s representsthe losses that damp the response. It is important to note that if the realpart is 0, we have imaginary pole pairs of the form pi =±jω. This is an undampedoscillatory component of frequency ω.
当求解特征方程的根时可能找到共轭的根形如pi = -s± jω。在这种情况下,作用到时域信号是一个形如Ae-st sin(ωt +φ)频率为ω的衰减周期信号。Aφ是由初始条件强加的。再次,实部s代表响应的阻尼损失。重要的是要注意,如果实部是0,我们有形式为pi =±jω的虚部极点对。这是频率ω的无衰减振荡成分。
5. If the complex pole pair is found to bein the RHP, pi = s± jω, the response is an exponentially growing sinusoidal signal.
如果发现复数形式极点对在有右半平面中,即pi = s± jω,那么响应是呈指数增长的正弦信号。

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  • 2018-5-5 21:06:48
 
A graphical representation of these variouscases appears in Figure 2.16. This is a system featuring two poles, hence asecond-order type. These poles could be that of a LC filter in which the quality factoris purposely adjusted for instance by an added resistance. In the first part(a), the poles are separate and purely real. The quality factor is very low;the system is overdamped. The response is similar to that of two cascaded RC filters. Inthe second plot (b), the poles are coincident; the quality factor is equal to0.5. The response is faster but there is no overshoot as the roots do notinclude imaginary parts. In (c), the poles are split and represented byconjugate roots. The quality factor is beyond 0.5. An imaginary portion is now present,and you see oscillations. However, the real parts provide the damping and representthe losses in the network that calm down the oscillations. This is a decaying signal.In (d), all losses have disappeared, and the system response is purely oscillatory.The roots are imaginary, and the real parts (the damping) have gone: we havesustained oscillations. In (e), the quality factor is negative and the poles havejumped in the RHP. The exponential exponent is now positive and oscillations grow,the system diverges. In (f), the imaginary portions have disappeared, thesystem still diverges but without oscillations.
2.16显示了这些不同情况。这是一个有两个极点的系统,因此是一个二阶系统。这些极点可能是一个LC滤波器的极点,其中品质因数可以通过增加电阻调整。在(a)中,两个极点是不相等的并且是纯实数的。品质因数很低;系统过阻尼。其响应类似于两个串联的RC滤波器。在(b)中,两极点是相等的;品质因数等于0.5。响应速度更快,但不存在超调,因为根不包含虚部。在(c)中,极点由共轭根表示。品质因数大于0.5。现在有虚部,你可以看到振荡。然而,实部提供了阻尼,并在网络中表现损耗平稳震荡。这是一个衰减的信号。在(d)中,所有的损失都消失了,系统响应是纯粹的振荡。根是虚数,实部(阻尼)已经消失:我们有持续的振荡。在(e)中,品质因数是负的,极点在右半平面中出现。幂为正,振荡增长,系统发散。在(f)中,虚部消失,系统仍然发散但没有振荡。
t2.16.png
Figure 2.16 Dependingon the closed-loop denominator roots position, the output response of a systemconverges if there are poles occurring in the LHP. If RHP poles appear, thesystem is unstable.

图2.16依照闭环分子根的位置,若有极点出现在左半平面,系统的输出响应向一点集中,若出现在有半平面,系统不稳定。
Figure 2.17 shows another example whereseveral poles and zeros are represented. They correspond to the followingequation roots:

2.17显示了另一个例子其中有几个极点和零点。它们对应于以下等式根:
2.99.png
t2.17.png
Figure 2.17 The s-planeallows the placement of poles and zeros and helps to check how they move inrelationship to a selected parameter.

s平面允许放置极点和零点,并有助于检查它们如何移动与选定参数的关系。

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  • 2018-7-9 21:45:32
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图2.16里的所有Sz 都应该为Sp!
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  • 2018-5-5 21:08:58
 
Please note than one of the zeros, sz3, lies inthe RHP; it is a positive root.
请特别注意一点,零点sz3,在右半平面而且是正根
Based on what (2.97) taught us, thetime-domain response of such a transfer function to a 1-V step is made of threeterms (three poles) plus the forced response. The forced response is found bycalculating the dc gain of (2.99):
根据(2.97),这种传递函数对1V阶跃的时域响应是由三项(三个极点)加上强制响应组成的。强制响应是通过计算(2.99)的直流增益得到的:
2.100.png
The natural response is made of one puredecaying term (one real pole) and two other terms that are damped sinusoidalsignals (conjugate poles pair):

固有响应是由一个纯阻尼项(一个实数极点)和两个其他项组成,两个其他项是抑制周期信号(共轭极点)。
2.101.png
When all terms have died out to zero, theoutput is 1.33 V. As you can see, thezeros do not explicitly appear in the various terms, but they play a role inthe coefficientsand also in the polarity:the output is negative. Why? We have three zeros and three poles. The two LHPpoles and zeros phase lag/lead compensate for each other. The third pole lagsby 90°, so what about the RHP zero (RHPZ)? It alsolags by 90°, making a total of 180° or a polarity reversal,bringing a negative output for a positive 1-V step! The RHP zero, rather thanadding a phase lead as a LHP zero would, further degrades the phase by lagging,just as a pole does. We will see that in more details in a few paragraphs.

当所有项都消失为零时,输出为-1.33 V。正如你所看到的,这些零点不是出现在各个项中,而是它们在系数和极性中起作用,即输出是负的。为什么?我们有三个零点和三个极点。两个左半平面极点和零点相位滞后/超前相互补偿。第三个极点使相角滞后90°,右半平面零点呢?它同样使相角滞后90°,使得总滞后180°或者使得极性相反,为1v阶跃带来负的输出。右半平面零点,而不是像左半平面零点那样使得相位超前,将会进一步通过滞后降低相位,就像极点一样。我们将在下几节中更详细地看到。

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  • 2018-5-5 21:10:42
 
Suppose now that sz3 becomes a LHP zero, theresponse would transform into
现在假设sz3变为左半平面零点,相应会变为:
2.102.png
The sign is now positive since we have aneven number of zeros and poles, making the total phase change to 0°: a positive step gives a positive voltage. Pleasenote that shifting of the zero into the LHP has affected the C coefficients but not the exponential terms that remain exclusivelylinked to the locations of the poles.
现在符号是正的,因为我们有偶数个零点和极点,使得总相位变为0°:正阶跃产生正电压。请注意,将零点移动到左半平面中会影响系数C,而不会影响与极点位置相关的指数项。
Now, assume the third pole becomes a RHPpole, sp 3 = 3 in (2.99) and the third zero is still in the LHPas in (2.102). The time-domain response can be expressed as follows:

现在假设第三个极点变为右半平面极点sp 3 = 32.99并且第三个零点仍在左半平面2.102.时域响应可以被表示成
2.103.png
The first term is enough to make the systemunstable: the exponent is positive and forces the expression to diverge as t increases.This is the RHP pole effect. If you look at the last term, it is negativeagain: the two LHP pole/zero pairs compensate each other with a total 0° phase. However, the third LHP zero adds a 90° phase lead that is normally compensated by 90° lag brought by a LHP pole. Here, as this pole is in the RHP, itbrings another 90° phase lead, whereas itshould be a lag, as with a LHP pole: the total phase lead is now 180°, or another phase reversal. However, as the first positiveexponential term immediately dominates the response, you will not see thenegative output.
第一项足以使系统不稳定:指数是正的,迫使表达式随着t的增加而发散。这是右半平面极点的影响。如果观察最后一项,它是负的,两个左半平面零极点对儿总相角为0°相互抵消。但是,第三个左半平面零点增加了一个90°的相角超前,通常由左半平面极点带来的90°滞后补偿。在这里,由于这个极点在右半平面中,它带来了另一个90°的相位超前,而它应该是一个滞后的,如同一个左半平面极点:总相位超前180°,或者是另一个相位反转。然而,由于第一个正指数项直接主导了反应,所以你不会看到负输出。
We have graphed all three responses in Figure 2.18. As expected, theRHP pole brings a diverging output. The RHP zero, despite bringing anadditional phase lag, does not make the system diverging. As a preliminaryconclusion, if a RHP zero forces you to be more careful when compensating theconverter, a single RHP pole in the characteristic equation (thus a closed-looppole) is absolutely insurmountable.

我们在图2.18上绘出三个响应。正如期望的那样,右半平面极点带来了偏差。右半平面零点带来了额外的相位滞后,但不会使系统发散。作为一个初步的结论,如果一个右半平面零点迫使你更加小心在补偿转换器的时候,那么特征方程中的单个右半平面极点(是一个闭环极点)绝对是无法克服的。
t2.18.png
Figure 2.18 Various time-domain outputs depending where thethird poles and zeros are in the s-plane.

图2.18 不同时域输出取决于第三极点和零点在s平面中的位置。

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  • 2018-5-7 21:07:07
 
2.4.1 Roots Trajectories in theComplex Plane在复平面上的根轨迹



The analysis of the characteristic equation(the closed-loop transfer function denominator) is usually carried at a certainoperating point. However, we have seen that variables affecting thecharacteristic equation can move. This is often the case for various gainvalues, as k in the previous example. At the time scientists studied powerelectronics in the 1950s, valve-based amplifierscould have wide gain dispersions at warm-up or at various ambient temperatures,and stability could be affected. A significant benefit of feedback was toprogram the gain via an external set of resistors, making the final chain gaininsensitive to the amplifier gain variations. Besides gain variations, therecan be perturbations effects such input voltage or load changes but alsoopen-loop poles or zeros that can slide along the frequency axis whenproduction spreads or loading conditions are involved. In our example fromFigure 2.9, sweeping the output capacitor ESR value degraded the transient response.Mathematically, it meant that the roots evaluated at different ESR cases havechanged their values. If we would plot these roots in the s-plane and linkall points together, we would obtain a so-called Evans root-locus representation,named after the work of W. R. Evans in the 1950s. By looking at the path takenby the poles, we can check if certain values of the swept parameter bring theroots close to the vertical axis (pure imaginary roots, no damping) or, evenworse, make them jump on the right side of the s-plane.
特征方程(闭环传递函数的分母)的分析通常在某一工作点进行。但是,我们已经看到影响特征方程的变量可以移动。对于不同的增益值,通常是这种情况,在前面的例子中是k。当科学家在20世纪50年代研究功率电子学时,基于阀门的放大器在预热时或在各种环境温度下可能具有广泛的增益变化,并且稳定性可能受到影响。反馈的一个重要好处是通过一组外部电阻对增益进行编程,使最终链增益对放大器增益变化不敏感。除了增益变化之外,还可能存在扰动效应,例如输入电压或负载变化,而且包括材料差异或负载条件在内,开环极点或零点可能会沿着频率轴滑动。在图2.9的例子中,扫描输出电容ESR值会降低瞬态响应。在数学上,这意味着在不同ESR情况下已经改变了根的值。如果我们将这些根部绘制在s平面上,并将所有点连接在一起,我们将获得所谓的Evans根轨迹表示法,这个表示法是在二十世纪50年代以W. R. Evans的作品命名的。通过查看极点的路径,我们可以检查扫描参数的某些值是否使根靠近垂直轴(纯虚根,没有阻尼),或者更糟糕的是,让它们跳到s右半平面。
Several mathematical programs can computeand plot the poles or zeros on the s-plane when a given parameter is swept. Mathcad is one of them withwhich root locus graphs are very simple to obtain. For instance, Figure 2.19depicts a root locus plot for (2.80) as the zero sz1 is swept from 1.5 kHz to more than 100 kHz. The appendix at the endof this chapter shows how we plotted this picture.
当一个给定的参数被扫描时,一些数学软件可以计算并绘制s平面上的极点或零点。 Mathcad是其中之一,根轨迹图很容易获得。 例如,图2.19描绘了当零点sz11.5kHz扫描到超过100kHz时(2.80)的根轨迹图。 本章末尾的附录告诉我们如何绘制这张图。
t2.19.png
Figure 2.19 In this picture, we can clearly see how the roots moveas the ESR zero is swept from 1.5 kHz up to 150 kHz.

图 2.19 图中,当ESR零点被扫描从1.5 kHz150 kHz时,我们可以看到根的移动。


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  • 2018-5-7 21:07:41
 
Should the roots approach the imaginaryaxis or worse, move to the right-half place section, stability would be atstake, as exemplified in Figure 2.16. Fortunately, on the drawing, we see thatdespite an ESR zero going to almost infinity (the ESR vanishes to 0), the rootsalways remain in the left plane, guaranteeing some damping. Calculations showthat in these conditions, Q does not exceed 3.5 and the step response is stable despite apronounced overshoot. The trajectory describing the loci of the poles canreveal a lot of information, but the study of the technique is outside thescope of this book. The reader interested in an in-depth analysis of this methodwill find a lot of information in Chapters 13 and 14 of [6].
如果根接近虚轴或者更糟,移动到右半部分,稳定性将受到威胁,如图2.16所示。 幸运的是,在图上,我们看到,尽管ESR零点几乎无穷大(ESR消失为0),但根始终保持在左侧面,从而保证了一定的阻尼。 计算表明,在这些条件下,Q不超过3.5,尽管出现明显的超调,但阶跃响应仍然稳定。描述极点轨迹的根轨迹可以揭示大量的信息,但是这项技术的研究不在本书的范围之内。有兴趣深入分析这种方法的读者可以在[6]的第1314章找到大量的信息。

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  • 2018-5-7 21:12:04
 
英文厉害了。
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  • 2018-5-21 22:01:27
 
2.5 Zeros in the Right Half Plane右半平面零点

Right half plane zeros are usually found inswitching converters transferring the energy to the load in two steps: theenergy is first stored in the inductor during the on-time and then dumped intothe load during the off-time. During the on-time, the load is isolated from theinput source while the inductor is energized. During the off-time, the energystored in the inductor is released to the load: boost, buck-boost, or flybackconverters operate this way. These three converters are called indirect energy-transferconverters. They all obey the two-step conversion process we described. For theopposite, the buck converter is a direct energy-transfer converter. You do not need an intermediate step to store energy before transmitting it to the load. This intermediate storing step actually creates a delay because the controller must always go through an energy-storing stepbefore answering the increase in the delivered power need. Should it take timeto store more energy while the power demand is fast, the converter cannotmomentarily keep up its power delivery and the output voltage falls. This eventlasts until the stored energy in the inductor has increased.
通常在开关转换器分两步传递能量给负载时发现右半平面零点:在导通时间内,能量首先存储在电感器中,然后在关断期间转储到负载中。在导通期间,电感器通电时负载与输入源隔离。在关断期间,存储在电感中的能量释放到负载:升压,升降压或反激式变换器以这种方式工作。这三个转换器被称为间接能量转换器。他们都服从我们描述的两步转换过程。相反,降压转换器是一个直接的能量转换器。在传输到负载之前,不需要中间步骤来储存能量。这个中间存储步骤实际上会产生一个延迟,因为控制器必须在应对供能需求增加之前总是经过一个能量存储步骤。在电力需求快的情况下,如果需要时间来存储更多的能量,转换器不能暂时保持其功率输出,并且输出电压下降。这种情况一直持续到感应器储存的能量增加。
The output variable momentarily going inthe opposite direction compared to what the control expects is the typicalsignature of a transfer function featuring zero located in the right-halfportion of the s-plane, also called a RHP zero or RHPZ.
与控制所期望的相比,输出变量暂时趋向向相反方向是传递函数特征零点位于s平面的右侧的典型标志,也称为右半平面零点或RHPZ


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  • 2018-5-21 22:06:28
 
2.5.1 A Two-Step ConversionProcess两步变换过程

Figure 2.20(a) represents a classical boostconverter where two switches appear: a power switch SW, usually a MOSFET, and a diode,sometimes called the catch diode. In the continuous conduction mode (CCM) ofoperation, the inductor current iL flows inthe power switch SW during the on-time or dTsw, where d is theinstantaneous duty ratio and Tsw theswitching period. During the off-time, or (1 - d)Tsw, the powerswitch is open and the output diode routes the current to an output network madeof the capacitor and the load. Regardless of the control method, voltage or current-mode,this configuration assumes that energy is first stored in the inductor duringthe on-time and then transferred to the output during the off-time.
2.20a)代表一个典型的升压转换器,其中出现两个开关管:一个功率开关SW(通常是一个MOSFET)和一个二极管,有时也被称为钳位二极管。 工作在连续导通模式(CCM)时,电感电流iL在导通时间或dTsw期间流入功率开关SW,其中d是瞬时占空比,Ts是开关周期。 在关断期间或(1 - dTsw期间,功率开关断开,并且输出二极管流出电流到电容和负载所组成的网络。 不管控制方法如何,电压或电流模式,这种结构都假定能量在导通期间首先存储在电感中,然后在关断时间内传递到输出。
t20.png
Figure 2.20 A boost converter features two power switches. Theycan be replaced by a single-pole double-throw switch that represents theoperations of the diode and the transistor.

图 2.20 升压转换器的特点是有两个功率开关。他们可以用一个单刀双掷开关代替,以体现二极管和晶体管运行特点
Figure 2.20(b) shows an equivalentrepresentation of the boost converter, where the switch/diode network has beenreplaced by a single pole double throw switch that alternatively routes theinductor current in the two different branches: the power switch or the outputdiode. If a designer would observe the currents circulating in the outputdiode, he or she would see a typical waveform shown in bold in Figure 2.21. Ourboost converter is designed to deliver power to a given load. The variable ofinterest, in our case, is thus the available output current Iout. Thiscurrent is actually made of a dc portion on which is superimposed a switchingripple. In theory, the ripple goes into the capacitor and the dc currentcirculates in the load. The dc current delivered by the boost converter isnothing other than the diode average current Id. Mathematically, this current canbe expressed by

2.20b显示了升压转换器的等效表示形式其中开关/二极管网络已被单刀双掷开关代替电感电流由二择一地流入到两个不同的路径功率开关或输出二极管。如果设计人员观察输出二极管中流动的电流,他或她会看到图2.21中粗体显示的典型波形。我们的升压转换器旨在为给定的负载提供能量。我们感兴趣的变量是输出电流Iout。这个电流实际上是由一个直流部分叠加了开关纹波组成的。理论上,纹波进入电容器,直流电流在负载中循环。升压转换器传递的直流电流是二极管的平均电流。在数学上,这个电流可以表示为:
2.104.png
t2.21.png
Figure 2.21 The currentobserved in the output diode at the beginning of the event. 在二极管中观察到的电流


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  • 2018-7-14 06:04:29
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楼主的英文功底果然深厚
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  • 2018-7-14 11:36:44
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经常看英文资料你也可以的,都是被逼出来的
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  • 2018-5-21 22:08:50
 
where Id is the average diode current also equal to the dc output current Iout. D represents the averaged duty ratio.
这里Id是平均二极管电流同样也是输出的直流Iout,D代表着占空比
If we graph the circulating currents in the diode at the switch opening, we obtain the drawing presented in Figure 2.21. When the switch opens, the current no longer circulates in the power switch but is routed, via the diode, to the output. The average value—or the dc current—circulating in the load isthe area of A0, averaged over a switching cycle Tsw.
如果我们在开关断开时画出二极管中流过的电流,我们可以得到图2.21所示的图。 当开关断开时,电流不再流过功率开关,而是通过二极管流向输出。 流入负载的平均或直流电流是在开关周期Tsw上A0的面积。 2.105.png
Now, if a sudden current demand occurs onthe output, the controller senses the transient and immediately increases theduty ratio by a small value ˆdto build moreenergy in the inductor. This is what Figure 2.22 shows.
现在如果输出端出现突然的电流需求控制器会检测到该瞬态并立即将占空比增加一个很小的值ˆd以在电感中增加更多能量。如图2.22所示。
t2.22.png
Figure 2.22Asan answer to the output current demand, the controller asks the inductor tostore more energy.
为了满足输出电流需求,控制器要求电感储存跟多的能量


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  • 2018-5-21 22:09:30
 
In theory, the new area, A1, should be larger than A0 to cope with the output powerdemand increase. However, as the switching period is fixed, the increase in theon-time duration simply shortens the off-time interval. As this is the timeduring which the current circulates in the diode to satisfy A1 > A0, the onlycondition is that the new peak current Ipeak1 is larger than the first one, Ipeak0. This is what is sketched in Figure 2.23.
理论上新的曲面A1应该大于A0以应对输出功率需求的增加。然而,由于开关周期是固定的,导通时间的增加简单地缩短了关断时间间隔。由于电流在二极管中循环以满足A1> A0的时间,唯一的条件是新的峰值电流Ipeak1大于第一个Ipeak0。这就是图2.23所描绘的内容。
t2.23.png
Figure 2.23 As the on-time reduces theoff-time duration, the only way to pass more power is to make sure the new peakcurrent is larger than the first one. Unfortunately, the inductor opposescurrent changes. 当导通时间减小了关断时间只有一种办法可以通过更多的能量是新的峰值电流大于之前的。不幸的是电感拒绝电流的改变。

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  • 2018-5-21 22:21:21
 
2.5.2 The Inductor CurrentSlew-Rate Is the Limit被限制的电感电流摆率
What is the pace at which the averageinductor current can change? Lenzs law instructs usthat the instantaneous current change rate in an inductor obeys the followingformula:
平均电感电流的变化率是多少?楞次定律告诉我们:在电感中瞬间电流变化率服从下面的公式:
2.106.png
On average, over a switching cycle, itsimply follows 平均在一个周期它为
2.107.png
现在计算另一边电感平均值的组成。通过考虑VinVout-VinL上施加时间段的加权,我们有:
2.108.png
Lets assume the following boost operating parameters:假定用以下运行参数:
t.png



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LV8
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  • 2018-5-21 22:27:29
 
With a 58.3 percent duty ratio, theconverter delivers 24 V. We are in steady state and (2.108) gives 0. Nowsuppose that the duty ratio jumps to D1 = 59 percent or a difference of 0.7 percent. What is the inductoraverage current slope in this case? Considering a large output capacitor, theoutput voltage stays constant during the duty ratio change. Applying (2.108)gives a transient average inductor voltage of
58.3%占空比下转换器输出24v。在稳态下VL0.现在假设占空比跳到59%或者70%。在这种情况下电感平均电流怎样变化?考虑一个大的输出电容,在占空比改变时输出电压保持稳定。应用式(2.108)给出的瞬态平均电感电压:
2.109.png
Back to (2.107), themaximum average current slope authorized by the inductor is therefore a rather modest value.回到式(2.107)最大电流斜率:
2.110.png
When the duty ratio changes from 58.3percent to 59 percent, it implies an output voltage change of 当占空比由58.3%变为59%,那意味着输出电压改变:
2.111.png
With a constant 240-Ω load, the output current will increase to 240Ω的负载,输出电流增至:
2.112.png
Brought back to the inductor change, theoutput current variation given by (2.112) must be accompanied by an averageinductor current variation of

回到电感变化,(2.112)给出的输出电流变化必须伴随平均电感电流变化
2.113.png
Given an average inductor slope 160 mA/ms, thiscurrent variation will only be possible within a timeframe of  给定一个平均电感电流斜率160mA/ms,这个电流变化将只可能在一个时间范围内
2.114.png


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LV8
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  • 2018-5-21 22:31:50
 
If the duty ratio is swept from 58.3percent to 59 percent in less than 51.6 μs, the inductor current will not build up at a sufficient pace tomake the output current rise at the same speed. As an immediate result, theoutput current drops rather than increases. On the contrary, if the duty ratiosweep is slow enough, the current can increase in the inductor at sufficientspeed to compensate the reduction in (1 - d): the outputvoltage goes up. How do we make sure the inductor will always have time tobuild up enough current in case a fast transient occurs? By simply rolling offthe crossover frequency or, in other terms, severely limiting the converterbandwidth so that fast transient demands never translates into a fast dutyratio change. If you fail to limit the bandwidth, in a fast output powertransient demand, the output voltage will go down despite an increase of dutyratio. In control theory, you have reversed the control loop and oscillationsoccur. This situation lasts until the current in the inductor builds up to theright value. To prevent this from happening, the RHP zero effect naturallylimits the available bandwidth for a given converter. Intuitively, a converterfeaturing a large inductor, hence operating in a deep CCM, will have a low-frequencyRHP zero, severely hampering its possible response time.
如果占空比从58.3%上升到59%的时间小于51.6μs,电感电流不会以足够的速度增加,使输出电流以相同的速度上升。作为结果,输出电流下降而不是增加。相反,如果占空比变化足够慢,则电感器中的电流可以以足够的速度增加,以补偿(1-d)中的减少:输出电压升高。我们如何确保电感在发生快速瞬变时总是有足够的时间增加足够的电流?通过简单地转变交叉频率,或者换句话说,严格限制转换器带宽,以便快速瞬态要求不会转化为快速占空比变化。如果不能限制带宽,在快速输出功率瞬态需求下,随着占空比增加,输出电压下降。在控制理论中,你已经颠倒了控制回路并发生了振荡。这种情况将一直持续到电感中的电流达到正确值。为防止这种情况发生,右半平面零点效应自然地限制给定转换器的可用带宽。直观地说,具有大电感量的转换器在CCM工作下,将具有低频右半平面零点,严重阻碍其响应时间。

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  • 2018-5-22 10:51:25
 
楼主又更新了
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  • 2018-5-24 10:17:58
 
支持,期待据续~
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  • 2018-5-24 10:12:35
 
楼主看到后回复一下我的邮件谢谢wjf@21dianyuan.com
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  • 2018-6-4 20:08:53
 
2.5.3 An Average Model toVisualize RHP Zero Effects可视化右半平面零点效应平均模型

Average models lend themselves very well to illustrating the effects of a RHP zero.We have built an open-loop boost converter around an auto-toggling modeldescribed in [7]. Figure 2.24 portrays the adopted schematic. In this testfixture, a 1-mH inductor driven at a 100-kHz switching period delivers 100 mAto a load (Vout = 24V). The duty ratio is first slowly swept between 58.3 percent and 59 percent.As shown in Figure 2.25, the inductor current nicely follows the demand, andthe output voltage slope is always positive. If we now sweep the duty ratio ata higher speed, Figure 2.26 indicates that despite a regular slope in theinductor current, it does not build up at a sufficient pace to answer theoutput current demand. As a result, both vout(t) and iout(t) drop. If the system would operatein closed loop, oscillations would occur because the control law is reversed:the duty ratio increases but the output voltage goes down.
平均模型借助它自身,非常适合说明有半平面零点的影响。 我们已经在[7]中描述模型的基础上构建了一个开环升压转换器。 2.24描绘了该示意图。在该测试设备下,用100kHz开关频率驱动的1mH电感为负载提供100mA电流(Vout = 24V)。 首先缓慢地在58.3%59%之间扫描占空比。 如图2.25所示,电感电流能满足需求,并且输出电压斜率总是正值。 如果我们现在以更高的速度扫描占空比,图2.26显示,尽管电感电流有规律的斜率,但它不会以足够的速度上升以满足输出电流需求。 结果,voutt)和ioutt)下降。 如果系统工作在闭环状态,则会发生振荡,因为控制规律相反:占空比增加,但输出电压下降。
2.24.png
Figure 2.24 A voltage-mode average model whose control input isswept at two different paces can demonstrate the existence of a RHPZ in theCCM-operated boost converter. 输入为两个不同的速度扫描一个电压模式平均模型,可以展示在CCM模式运行的升压转换器右半平面零点的存在。
t2.25.png
Figure 2.25When the duty ratio slowly changes and gives timeto the inductor current to build up, the output voltage variation is positive, as it should be. 当占空比缓慢的增加并且给予电流(足够的)上升时间,输出电压变量为正。


2.26.png
Figure 2.26In this example, the inductor current builds up tooslowly with regard to the output current demand. As a result, the outputcurrent drops until the current in the inductor builds up to the right value. 关于输出电流需求,电感电流的建立速度太慢。 结果,输出电流下降,直到电感中的电流达到正确值。
How do we prevent this problem fromhappening? Well, a solution is to clamp the maximum slew rate on the duty ratiocontrol input. In that way, even if a sudden variation is detected on theoutput, the error voltage will always rise at a speed where the inductorvolt-second limit is never reached, giving sufficient time for the inductorcurrent to build up. How do we limit the slew rate? By rolling off the crossoverfrequency fc at aposition usually well below the worst-case RHP zero position.

我们如何防止这个问题发生? 解决办法是限制占空比的转换速度。 这样,即使在输出端检测到突变,误差电压也会始终以电感伏秒未到极限速度上升,从而为电感电流上升提供足够的时间。 我们如何限制变化速度? 通过衰减位于低于最差情况右半平面零点交接频率fc


2.25.png
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  • 2018-6-4 20:11:00
 
2.5.4 The Right Half Plane Zero inthe Boost Converter在升压转换器中的右半平面零点

We have shown the consequences of a RHPZ ina boost converter. It is now interesting to analytically derive its position inthe transfer function of the boost converter. To simplify the analysis, we willconsider a voltage-mode control. The reader interested in current mode controland compensation examples can find more information in [8].To derive the transfer function, we can start from the output current expressiongiven in (2.104). This is a large-signal (nonlinear) equation that we must transforminto a small-signal expression. The fastest way to do that is to find partial derivatives coefficients for each of thevariables, the duty ratio D and the inductor current IL:
我们已经看到了在升压转换器中右半平面的影响。现在在升压转换器传递函数中
去分析它的来源。为了简化分析,我们认为在电压模式下。读者对电流模式和补偿感兴趣可以在[8]里找到更多的信息。为了得到传递函数,我们可以从输出电流表达式(2.104)开始。这是一个(非线性)大信号等式,必须转换到小信号表达式。最快的方式是找出每个变量偏导数系数,占空比D和电感电流IL
2.115.png
In this equation, the ac inductor current ˆiLappears. What is the expression of an ac inductor current? Simply, its the ac inductor voltage divided by the inductor impedance. Let us find the expression of the ac inductor voltage byfirst deriving its average large signal expression, already found in (2.109):

在等式中交流电感电流ˆiL出现了。交流电感电流的表达式是什么?是交流电感电压比上感抗。让我们通过推导出它在(2.109)中的平均大信号表达式,找到交流电感电压表达式。
2.116.png



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  • 2018-6-11 17:09:45
 
辛苦辛苦
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  • 2018-6-4 20:13:51
 
On average, when the converter is at the equilibrium, this equationgives zero. However, under an ac excitation, the averageinductor voltage is also ac modulated around zero. By using a partialderivative, we can see that the ac inductor voltage, in this case, is expressedby
当转换器处于平衡状态时,等式为0。然而,在交流激励下,平均电感电压也在0附近被交流调制。通过使用偏导数,我们可以看到交流电感电压为:
2.117.png
In this equation, the input term Vin has disappeared since theinput voltage is considered constant during the ac analysis. Furthermore, if we consider a large output capacitor, itsimpedance at the ac excitation can be considered close to zero. In this case,if we consider vˆout ≈ 0, we can further simplify theexpression:
在这个等式中,输入项Vin已经消失了,因为在交流分析期间输入电压被认为是恒定的。此外,如果我们考虑有一个大的输出电容,它的容抗在交流激励下可以被认为接近于0。所以如果我们认为vˆout ≈ 0,我们可以更进一步简化表达式:
2.118.png
With the ac inductor voltage on hand, it is easy to obtain the acinductor current we are looking for:
有交流电感电压在手,很轻易的就得到了我们正在寻找的交流电感电流:
2.119.png
Substituting (2.119) in(2.115) gives the final ac output current expression:
将式(2.119)带入到式(2.115)得到最终的输出交流电流表达式:
2.120.png
The average inductor current IL is the source current Iin. Considering a 100 percent efficiency power conversion, we canwrite
平均电感电流IL是源电流Iin考虑效率为100%,我们有:
2.121.png

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  • 2018-6-4 20:15:49
 
Substituting (2.122) in(2.120), we obtain 将式(2.122)带入到式(2.120)
2.123.png
Now factoring the firstterm and rearranging, we have 现在分解第一项并重写:
2.124.png
这里:
2.125.png
This expression linksthe output current to the duty ratio input. In thisequation, we can see a pole at the origin given by theinductor L and a zero featuring a positiveroot: this is the RHPZ ωz2 we are looking for. Please note that both roots depend on the duty ratio and aremoving in relation to the input/output conditions.
该表达式将输出电流链接到占空比输入。在这个等式中,我们可以看到极点在原点由电感L给出,以零为特征的正根,这个右半平面零点ωz2是我们寻找的。请注意,这两个根都取决于占空比,并且与输入/输出条件相关。
If we apply the boost converter numericalvalues from Figure 2.24, we have the following positions:
应用图2.24升压转换器的参数,我们有:
2.127.png


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  • 2018-6-4 20:17:58
 
Our interest now liesin the phase lag brought by this transfer function. The argument of a quotientis the numerator argument minus that of the denominator:
现在我们关注的是传递函数带来的相角的滞后。商的相角是其分子的相角减去分母的相角。
2.129.png
In dc, for w = 0, this expression becomes 直流下,ω0,表达式变为:
2.130.png
[url=]The presence of theorigin pole justifies this permanent phase lag of 90°. Now, with a normal zero, asthe frequency increases, we would expect its contributing phase to reach 90°, canceling theorigin pole action. Unfortunately, because this is a RHP zero, the total phaselag at w = ¥ becomes
原点极点的存在表明固定的相角滞后90°,现在有一个零点,随着频率增加,我们期望着它贡献90°的相角以抵消原点极点的作用。不幸地是,这是一个右半平面零点,当w = ¥ 时总相角滞后:
2.131.png [/url]
[url=]This is the effect ofthe RHPZ: its phase lag is 90° compared to the 90° phase lead brought by a LHP zero. To obtain a complete picture, we can obtain the transfer functionusing the average model of Figure 2.24 but also using (2.124) and a calculation software such as Mathcad. Figure 2.27 shows the plots we have obtained. The superposition of both curves confirms the validity ofthe analytical equation we have derived. Also, as expected, the total phase lagreaches 180°. This is the effect of the origin pole andthe RHP zero, which brings an additional –90° rather than 90°as it should for a normal zero. Please note that this RHPZalso exists, at the same location, in a current-mode boost converter.
这就是右半平面零点的影响,相较于左半平面零点带来的相角超前90°,而右半平面零点却带来了相角滞后90°。为了获得完整的图像,我们可以使用图2.24的平均模型获得传递函数,也可以使用(2.124)和Mathcad等计算软件。图2.27显示了我们获得的图。两条曲线的叠加证实了我们得出的分析方程的有效性。 另外,如预期的那样,总相位滞后达到180°。这是原点极点和右半平面零点的影响,它会带来额外的-90°而不是90°。请注意,此右半平面零点同样存在,在电流模式升压转换器中同样位置
2.27.png
Figure 2.27 The total phase lagreaches 180° whereas it should be 0° ifthe zero of the numerator would lie in the left-half portion of the s-plane.
2.27总相位滞后达到180°,而如果零点位于s平面的左半部分,则它应该为0°
[/url]
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LV8
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  • 2018-6-4 20:23:53
 
In this example,should we need to stabilize the converter, the RHPZ position clearly bounds themaximum crossover frequency. To avoid any stability issue, it is recommended tolimit the crossover frequency fc to less than 30 percent of the minimum RHPZ position. In ourexample, it would imply a crossover frequency of
在这个例子中,如果我们需要稳定转换器,右半平面零点位置清楚地界定了最大交接频率。为避免任何稳定性问题,建议将交叉频率fc限制在最小RHPZ位置的30%以内。 在该例中,这意味着交叉频率:
2.132.png
The compensationblock G must thus betailored to force crossover below this value.
补偿框G必须是量身定做的,以强制交叉低于该值。
One final note onRHPZ: The previous example assumes CCM to illustrate the presence of the RHPZ.It is little known that a RHPZ can also exist in the discontinuous conductionmode (DCM) of operation. However, as it is located in the higher portion of thefrequency spectrum, its influence is usually neglected at lower frequencieswhere crossover takes place.
关于右半平面零点的最后一个注意事项:前面的例子假定CCM说明右半平面零点的存在。很少人知道右半平面零点也可能存在于非连续导通模式(DCM)中。 但是,由于它位于频谱的较高部分,因此在发生交叉的较低频率时通常忽略其影响。


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LV8
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  • 2018-6-4 20:36:40
 
2.6 Conclusion 结论



Understandingtransfer functions is key for the design of fast and stable closed-loop systems.Even if you will never do root locus calculations, it is important to realize thatpoles positions can move in relation to certain operating parameters. Once theseparameters are identified (e.g., our output capacitor ESR), you will know how toefficiently compensate these variations over the power supply lifespan,ensuring a robust design. The RHP zero presence can sometimes hamper theavailable bandwidth in converters like boost or flyback architectures. Again,being able to analytically locate its worst-case position and visualize itseffects on a Bode plot is an important point to let you safely pick a crossoverfrequency value. Finally, we have seen how fast analytical techniques candramatically improve your analysis speed and unveil poles and zeros in a fewminutes. It requires dexterity and practice but once you master the technique,going back to classical brute force algebra will be difficult!
理解传递函数是设计快速、稳定闭环系统的关键。即使你永远不会做根轨迹计算,但重要的是要认识到极点位置可以相对于某些参数移动。一旦确定了这些参数(例如,输出电容ESR),你就会知道如何在电源寿命期间有效地补偿这些变化,从而确保可靠的设计。 右半平面零点存在有时会阻碍转换器中的可用带宽,如升压或反激式架构。再者,在波德图上能够分析找出其最坏情况的位置并可视化它的影响,是重要的一点以让你安全选择交叉频率值。最后,我们已经看到分析技术,能够在几分钟内快速提高分析速度并揭示极点和零点。它需要熟练和练习,但是一旦你掌握了技巧,回到经典的代数将会很困难!

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Appendix 2A: Determining a BridgeInput Impedance确定一个桥输入阻抗



We are going to use an example given by Dr.Vatché Vorpérian on page 12 of [1]. The circuit diagram appears in Figure 2.28:
我们将用到Vatché Vorpérian博士在【1】的第12页给出的例子。如图2.28所示的线路。
t2.28.png
Figure 2.28 The input impedance of this bridge can bederived in a few steps when using analytical techniques. 2.28当使用分析技术时,几步内得出这个桥的阻抗。
The exercise consists of determining theimpedance Zin(s) seen from theleft side of Figure 2.28:

确定从左边看进去的阻抗
2.133.png
If Dr. Vorpérian used the extra element theorem (EET) to obtain the result, weare going to apply the same technique we used in this chapter to find the inputimpedance. There is one storage element, the capacitor C; this is thusa first-order system. Its general expression can be put under the followingform. Analysis will further tell us if a pole or a zero are present:

如果使用有限元分析获得结果,我们将应用我们在本章寻找输入阻抗同样的技术。有一个储能原件电容C,因此为一阶系统。表达式如下,进一步分析可得到极点和零点
2.134.png
First, let us derive the input resistanceseen at dc, R0: open-circuit thecapacitors and short-circuit the inductors, if any. Our circuit simplifies tothat of Figure 2.29.

首先我们得到直流时的输入阻抗R0将电容开路电感短路。将线路简化为图2.29那样
t2.29.png
Figure 2.29Thedc input resistance, when the capacitor is removed, is really easy to find.将电容移除直流输入阻抗很容易被找到。



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The input resistance is simply R1 in series with the series-parallelcombination of the remaining elements: R1与其余的原件并联结合串联。
2.135.png
Now, let`s see if a zero exists. A zero inthis circuit would prevent the excitation signal from reaching the output. Aswe deal with an impedance expression, the excitation signal is the inputcurrent Iin and theresponse is the input voltage Vin. What inFigure 2.28 would nullify Vin? Ashort circuit involving the branch where C lies.
现在看,如果有零点存在。在线路中的零点会阻止激励信号接近输出。处理阻抗表达式,激励信号是输入电流Iin并且响应是输入电压Vin。什么使得Vin无效了呢?短路C所在的支路。
If we have a short circuit, then Vin = 0. IfVin = 0,then node 2 is grounded and R1 comes in parallelwith R2. The circuitappears in Figure 2.30.

如果短路Vin=0.如果Vin=0则节点2为地R1变为与R2并联。如2.30图所示。
t2.30.png
Figure 2.30Thezero expression is found by nullifying the response signal, which is Vin. 2.30通过使响应信号,Vin无效,找到零点表达式。
The expression seenfrom the input impedance port is simply从输入阻抗端口看到的表达很简单
2.136.png
To cancel this expression, we set itsnumerator to zero and solve for the root. In our case, this is simply.为了使该式为0,我们设分母为0并解出根。有:
2.137.png
Thus,因此,
2.138.png

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Now that the zero has been found, let`sgive a look to the pole. To derive the denominator expression D(s), we need toset the excitation to zero and find the time constant of the resulting network.The excitation, in our case, is the input current Iin.This is a current generator. When set to zero, it transforms into an opencircuit, leading to the above updated schematic.
现在,零点已经找到了。现在来找极点。为了得到分母得表达式,我们需要设置激励为0,并且找出网络得时间常数。在该例中激励为输入电流Iin这是一个电流发生器。当设置电流为0,进入开路状态,更新原理图为:
t2.31.png
Figure 2.31Thetime constant is found by setting the excitation, Iin, to zero. 2.31,通过设置激励为0得到时间常数。
The next step is to find the resistor R driving thecapacitor. Again, looking at the open port in Figure 2.31, a simple resistive arrangement isobtained:

下一步去寻找驱动电容C的电阻R。再看一下图2.31中的开口端,可以得到一个简单的电阻排布:
2.139.png
The time constant istherefore 时间常数为:
2.140.png
Leading to a poleexpression of 极点的表达式:
2.141.png
Here we are: we havederived the input impedance expression in less than 10 steps! When gatheringequations (2.135), (2.138), and (2.141), we have:

在这里,我们在不到10步中得到输入阻抗表达式!汇集等式(2.135),(2.138)和(2.141)时,有:
2.142.png
From this expression,it is easy to identify the dc term and the associated pole and zero.

从这个表达式中,很容易识别直流项和相关的极点和零点。
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  • 2018-6-18 10:36:00
 
Appendix 2B: Plotting Evans Lociwith Mathcad使用Mathcad绘制根轨迹图

There are several waysto plot Evans roots loci on a computer. One of them uses the popularmathematical tool Mathcad. Let`s assume we want to plot the roots loci of asecond-order transfer function defined as
有几种方法可以在计算机上绘制根轨迹。其中一个使用流行的数学工具Mathcad。让我们假设我们绘制一个二阶传递函数的根轨迹
2.142.png
The denominator of this expression includesthe poles of the transfer function. These poles are the roots of the equation

分母包含传递函数的极点。极点是(特征)方程的根。
2.143.png
To use Mathcad for the resolution, let`s open a new sheet and enter thefollowing equations: 要使用Mathcad,让我们打开一个新工作表并输入以下公式:
2.143b.png
The Q and theresonant frequency w0 are arbitrarilyselected for this example. As the expression of the denominator follows theform of a second-order polynomial f (s) = as2 + bs + c, we ask the software to identify the individual coefficientsby using the function denom where Q is the variable. Should we need toalso plot the zeros of the transfer function, we would replace denom by thekeyword numer in the following expression:

该例中Q和自然频率都是任意的。分母的表达式形如二阶多项式f (s) = as2 + bs+ c我们要求软件通过关键字“denom”来识别各项系数,其中Q是变量。如果我们需要画出传递函数的零点,我们将关键字“denom”替换成“numer”。
2.143c.png
According to thisresult, the denominator function to solve is in the form of根据这个结果,解出分母函数的为:
2.144.png


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Mathcad can solve this function via the keyword polyroots whose resultsare passed to a two-dimensional vector, X: Mathcad可以通过关键字“polyroots解出方程,结果为一个二维向量X
2.144b.png
We can extract the real and imaginary partsvia two dedicated keywords, Im and Re:
我们可以通过专用的关键字 Re和“Im”提出实部和虚部
2.144c.png
These two-dimensionalvectors include the conjugate roots s1, s2 and will be accessed via subscripted notations. Please note that thesubscripted notation for a vector manipulation in Mathcad is obtained bypressing [ right after the vector name:
这些二维向量包括共轭根s1s2,并且将通过下标符号使用。 请注意,Mathcad中矢量操作的下标表示法是通过在矢量名称后面按“[”来获得的:
2.144d.png
We now haveeverything to vary Q from 0.1 to 10and plot the points individually made of Re(X(Q))0 or1 as the vertical coordinate and Im(X(Q))0 or1 as the horizontal coordinate. The completecalculation sheet including the roots loci graph appears in Figure 2.32.
我们现在有了将Q0.1变为10的所有内容,并将由ReXQ))01独立的点绘制为垂直坐标,ImXQ))01绘制为水平坐标。根轨迹图的完整计算表如图2.32所示。
t2.32.png
Figure 2.32This figure gathers all notations and the plotshowing how the roots move as a function of Q. 这张图收集了所有符号以及显示了根随着Q的变化。
To display ´ for thepoles, right click on the graph and select traces/symbols. Pick a cross forboth variables and you are done. Should it be zeros instead, pick the as the availablesymbol, affected by the color of your choice.
要显示“极点”,请右键单击图形并选择迹线-符号。为这两个变量选择x,然后就完成了。如果它是零点,选择○作为可用符号,受你选择的颜色影响。
Wecan see the poles moving in the same direction as Q increases. When Q equals0.5, both roots are real and coincident. As Q continues to grow, the poles split,giving birth to an imaginary portion. As Q continues to increase, the realportion (the damping) vanishes and the roots eventually reach the imaginaryaxis when Q reaches infinity.
我们可以看到随着Q的增加,极点向相同的方向移动。当Q等于0.5时,两个根是实数并且相等。随着Q增长,极点分开,产生了虚部。随着Q持续增加,当Q达到无限时,实部(阻尼)消失并且根部最终达到虚轴。
Thereare plenty of solutions to plot roots loci in Mathcad. The presented solution issimple and quick to implement. As a drawback, you will find difficult to associatea given set of roots in the plot with the corresponding Q value. You can alwaysuse the available cursor, but it is not really practical. Searching on the Web willgive more comprehensive solutions where Q is displayed when a root isidentified in the plot but at the expense of an increased sheet complexity.

Mathcad中有很多方法可以绘制根轨迹。这个方法实施起来简单快捷。作为一个缺点,你会发现难以将图中的一组给定的根与相应的Q值相关联。你可以使用可用的光标,但它并不实用。在Web上搜索将提供更全面的方法,其中当在图中标识根时显示Q,但代价是增加的图表复杂性。



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  • 2018-6-18 10:40:48
 
Evans Loci.rar (8.19 KB, 下载次数: 29)
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  • 2018-6-18 10:46:20
 
Appendix 2C: Heaviside Expansion Formulas  Heaviside展开公式
We have seen how to obtain the time-domainresponse of a system whose transfer function is known when its input issubjected to a step function. You have to decompose the quotient into partialfractions and then sum up the inverse Laplace expression of each individualfraction. Despite its simplicity, the process requires two steps: decompose thequotient into partial fractions, and then find the inverse Laplace function ofeach partial fraction. Heaviside proposed another method that can give you thetime-domain response in one single step. It is not really well known, and Icould not find a lot of information on the Web. This is the Heaviside developmentformula described in [1]. It is expressed the following way:
我们已经看到,当输入受到阶跃函数的影响时,如何获得已知传递函数系统的时域响应。 必须将商分解为部分分式,然后对每个独立的分式的拉氏反变换相加。尽管简单,但该过程需要两个步骤:将商分解成部分分式,然后找到每个部分分式拉氏反变换。Heaviside提出了另一种方法,可以在一个步骤中提供时域响应。 它并不是很出名,我在网上找不到更多信息。 这是[1]中描述的Heaviside开发公式。表示如下:
2.145.png
The exercise consistsof identifying the denominator roots s1, s2 to sn if you have n roots. Then, you have to take the derivative of the denominator.Once you have these elements on hand, simply apply the following formula andthe resulting time-domain equation shows up:
2.146.png
Let`s take the transfer function alreadygiven in (2.60). It represents the system transferfunction multiplied by the step function, 1/s:
(2.60)为例,系统的传递函数乘以1/s
2.147.png
We have three roots for D(s). These roots are: D(s) 有三个根为:
2.148.png
We now develop thedenominator: 展开分母:
2.149.png
and derive it: 对其求导得:
2.150.png
We now calculate 现在计算:
2.151.png
and 并且:
2.152.png
By applying (2.146), wehave our time-domain response immediately: 应用(2.146)得到时域响应:
2.153.png
Developing, we obtain 展开:
2.154.png
This is exactly whatwe found in (2.70) without going through the pain of partial decomposition.Simple, yet elegant, isn`t it?

这正是我们在(2.70)中发现的,没有经过部分(分式)分解的痛苦。简单而优雅,不是吗?


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  • 2018-6-18 10:49:12
 
Ok, let`s go throughanother small example. Assume the following transfer function: 另外一个例子,假设传递函数为:
2.155.png
We have two roots for D(s).These roots are 分母的根为:
2.156.png
Let`s derive D(s): D(s)求导
2.157.png
We now calculate 现在我们计算:
2.158.png
and 并且:
2.159.png
By applying (2.146), wehave our time-domain response almost immediately: 应用(2.146)得到时域响应:
2.160.png
Rearranging and factoring,we have  整理带入:
2.161.png
We can identify complex sine and cosine functions. The final expressionis :我们可以识别复杂的正弦和余弦函数(我们可以利用欧拉公式将等式化简为正弦和余弦函数)。 最后的表达是
2.162.png
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Appendix 2D: Plotting a Right HalfPlane Zero with SPICE SPICE画出右半平面零点

We have seen that a RHPZ is a numeratorroot featuring a positive real part. A RHPZ can be put under the following form:
我们已经看到右半平面零点是带有正实部的分子根。右半平面零点可以采用以下形式:
2.163.png
The negative signindicates the presence of a positive root, located in the right half portion inthe s-plane. To show the effect of a RHPZin a Bode plot, we can try to artificially create one with SPICE. A possiblecircuit appears in Figure 2.33. The transfer function of such a configurationis quickly obtained:

负号表示存在正根位于s平面右半部分。为了在波特图中显示右半平面零点的影响,我们可以用SPICE创建一个。如图2.33所示的电路中。 快速获得这种结构的传递函数:
2.164.png
t2.33.png
Figure2.33 A RHPZ is artificially created with an op amp-baseddifferentiator and an adder. 右半平面零点是用一个基于运算放大器的微分器和
Where 这里 2.164b.png
The ac plot of such atransfer function is given in Figure 2.34. As expected, the magnitude isactually that of normal zero: it increases as the frequency increases. But thephase, rather than also heading towards +90°, it actually lags by 90°, as a pole woulddo.

这种传递函数的交流(扫描)如图2.34所示。正如预期的那样,这个幅值实际上是正常的零点,随频率的增加而增加。但是相角,而不是朝着(超前)90°方向发展,它实际上滞后了90°,就像一个极点那样。
t2.34.png
Figure 2.34Asexpected, the simulation results reveal a magnitude similar to that ofclassical zero, but the phase, rather than going up, is actually going down to –90°. 正如预期的那样,仿真结果揭示幅值是普通的零点(表现的那样),但是相角事实上是滞后90°


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  • 2018-7-4 11:57:57
 
请问楼主:现在这本书有中文版的吗?在作者主页显示有了
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  • 2018-7-5 10:23:28
 
你看到的是那个中文的广告吧?
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  • 2018-8-2 12:13:58
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是的,后面的章节还会继续吗?
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  • 2018-11-1 11:51:33
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外国的教材都是讲的很细致、很透彻,国内的很多都是跳跃性大,建议分享一下,谢谢
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  • 2018-11-2 10:19:43
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感谢翻译,翻译挺费时间的
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  • 2018-11-16 17:04:05
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这个必须要赞一个啊 有心人哪
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