急求电力专业英语文章及翻译 ！！明天就要交了 求路过大神帮小弟一个忙

已经发给你了，这里是个备份。
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Often a differential equation is Fig.3.10 solved by integration. The integration may be accomplished by analytical methods or by numerical methods on a digital computer. Integration may also be performed electronically with an op-amp circuit. Indeed, op-amps were developed initially for electronic integration of differential equations.
(1) An Integrator. The op-amp circuit in Fig.3.11 uses negative feedback through a capacitor to perform integration.
We have charged the capacitor in the feedback path to an initial value of U1, and then removed this prebias voltage at t=0. Let us examine the initial state of the circuit before investigating what will happen after the switch is opened . Since u+ is approximately zero, so will be u_, and hence the output voltage is fixed at –U1. The input current to amplifier, Ui/R, will flow through the U1 voltage source and into the output of the op-amp. Thus the output voltage will remain at – U1 until the switch is opened.
After the switch is opened at t=0, the input current will flow through the capacitor and hence the Uc will be ( 公式 ) Thus the output voltage of the circuit is ( 公式 )(3.10) Except for the minus sign, the output is the integral of Ui scaled by I/RC, which may be made equal to any value we wish by proper choice of R and C.
(2) Scaling and Summing. We need two other circuits to solve simple differential equations by analog computer methods. Scaling refers to multiplication by a constant, such as ( 公式 ) where K is a constant. This is the equation of an amplifier, and hence we would use the inverting amplifier in Fig.3.3 for the – sign or the noninverting amplifier in Fig.3.5 for the + sign.
A summer produces the weighted sum of two or more signals.Fig.3.12 shows a summer with two inputs. We may understand the operation of the circuit by applying the same reasoning we used earlier to understand the inverting amplifier. Since u-=0, the sum of the currents through R1 and R2 is ( 公式 )(3.11)
The output voltage will adjust itself to draw this current through RF, and hence the output will thus be the sum of U1 and U2, weighted by the gain factors, RF/R1 and RF/R2 , respectively. If the inversion produced by the summer is unwanted, the summer can be followed by an inverted, a scalier with a gain of – 1. Clearly, we could add other inputs in parallel with R1 and R2. In the example to follow, we shall sum three signals to solve a second order differential equation.

时常一个微分方程式是根据整合解决的 Fig.3.10 。 整合可能是完成的藉着分析方法或在一部数传计算机上的数字方法。 整合也可能电子地被一个 op-安培的线路所运行。 的确， op-安培最初为微分方程式的电子的整合发展。
(1) 一个整合之人。 在 Fig.3.11 的 op-安培的线路使用否定回应完成的一个电容器运行整合。
我们有充满感情的电容器在那回应路径到一起始 U1 的价值, 然后离开的这一 prebias 电压在 t=0. 让我们调查那起始状态线路在调查什么将会发生之前在那之后开关是打开。 自从 u 以后+ 大约是零, 因此意志是 u_, 和因此输出电压是固定的在 -U1.那输入对喇叭筒,Ui/R, 的涌流将会流过 U1 电压来源和进入 op-安培的输出之内。 如此输出电压将会保持在 - U1 直到开关是打开。
在那之后开关是打开在 t=0, 那输入现在的意志流过电容器和因此 Uc 将会如此是 ( 公式 ) 线路的输出电压是 ( 公式 )(3.10) 除为那负号告示, 输出是那整体 Ui 是依比例决定被我/RC, 可能是制造与任何的相同价值我们希望被适当的选择 R 和 C。
(2) 剥落而且总计。 我们需要其他二个线路解决类比计算机方法的简单微分方程式。 剥落提及一个常数的乘法, 像是 K 是哪里( 公式 ) 一常数。这是喇叭筒的相等, 和因此我们会在 Fig.3.3 使用反转喇叭筒为那 - 告示或非反转喇叭筒在 Fig.3.5 为那 + 告示。
夏天农产品那重量总数二或更多 signals.Fig.3.12 表演夏天与二输入。 我们可能藉由应用相同的推论我们了解线路的操作二手的早了解反转喇叭筒。 自从 u 以后-=0, 涌流完成的 R1 和 R2 的总数是 ( 公式 )(3.11)
输出电压将会经过射频对平局这涌流调整它本身, 和因此输出将会如此是那总数 U1 和 U2, 重量被那增益因素、射频/R 1 和射频/R 2, 分别地。 如果倒转是生产在夏天之前是不必要的, 夏天能是跟随被一反转, 一有鳞的与一增益 -1. 清楚地, 我们可以把其他的输入加入平行与 R1 和 R2. 在例子中跟随, 我们将总数三信号解决一第二次序微分方程式。

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