英语翻译Low-pass FiltersAn integrator is the simplest filter mat

英语翻译
Low-pass Filters
An integrator is the simplest filter mathematically,and it forms the building block for most modern integrated filters.Consider what we konow intuitively about an integrator.If you apply a DC signal at the input (i.e.,zero frequency),the output will describe a linear ramp that grows in amplitude until limited by the power supplies.Ignoring that limitation,the response of an integrator at zero frequency is infinite,which means that it has a pole at zero frequency.(A pole exists at any frequency for which the transfer funtion’s value becomes infinite.)
We also know that the integrator’s gain diminishes with increasing frequency and that at high frequencies the output voltage becomes virtually zero.Gain is inversely proportional to frequency ,so it has a slope of –1 when plotted on log/log coordinates (i.e.,-20dB/decade on a Bode plot).
You can easily derive the transfer function as Vout/VIN = Xc/R = 1/sCR =W0/s .Where s is the complex- frequency Variable σ+jw and W0 is 1/RC.If we think of s as frequency,this formula confirms the intuitive feeling that gain is inversely proportional to frequency.
The next most complex filter is the simple low-pass RC type (Figure 1).Its characteristic (transfer funtion) is Vout/VIN = 1/1 + sCR= W0/s + W0 .When s = W0,the function reduces to w0/ w0 ,i.E.,1.When s tends to infinity,the function tends to zero ,so that is a low-pass filter.When s =- W0 ,the denominator is zero and the function’s value is infinte,indicating a pole in the complex frequency plane.The magnitude of the transfer function is plotted against s in Figure 2 ,where the real component of s(σ) is toward us and the positive imaginary part (jw) is toward the right .The pole at - w0 is evident .Amplitude is shown logarithmically to emphasize the function’s form .For both the integrator and the RC low-pass filter,frequency response tends to zero at infinite frequency ;that is ,there is a zero at s = ∞ .This single zero surrounds the complex plane.
But how does the complex function in s relate to the circuit’s response to actual frequencies When analyzing the response of a circuit to AC signals ,we use the expression jwL for impendance of an inductor and 1/jwC for that of a capacitor .When analyzing transient response using Laplace transforms ,we use sL and 1/sC for the impedance of these elements .The similarity is apparent immediately .The jw in AC analysis is in fact the imaginary part of s ,which ,as mentioned earlier ,is composed of a real part s and an imaginary part jw .

低通滤波器积分器数学上是最简单的过滤器,并且它形成积木为多数现代联合过滤器.考虑什么我们直觉konow关于积分器.低通滤波器,如果您运用一个DC信号在输入(即.零的频率),产品将描述在高度增长,直到由电源限制的一架线性舷梯.忽略那个局限,积分器的反应以零的频率是无限的,因此它意味着它有一根杆以零的频率.(A杆存在以调动funtion的价值变得无限.)的任何频率.
们也知道积分器的获取减少随着频率的增加,并且在高频率产品电压变得实际上零.获取与频率相反地是比例,因此它有?（我不知道!）倾斜?当密谋在日志或日志时协调(即.- 20dB/decade在预示剧情).您能容易地获得传递函数作为Vout/VIN = Xc/R = 1/sCR =W0/s.那里s复杂频率易变的σ jw和W0是1/RC.如果我们认为s作为频率,这个惯例证实直觉的感觉获取与频率相反地是比例.
下多数复杂过滤器是简单的低通RC类型(图1).它的特征(调动funtion)是Vout/VIN = 1/1 sCR= W0/s W0.当s = W0,作用减少到w0/w0,i.E.,1.当s趋向到无限时,作用趋向到零,因此是一台低通滤波器.当s = - W0,分母是零时,并且作用的价值是infinte,表明一根杆在复杂频率飞机.传递函数的巨大在表2被密谋反对s,真正的组分s (σ)的地方往我们和正面虚构部分(jw)朝右.杆在- w0是显然的.高度对数显示强调作用的形式.为积分器和RC低通滤波器,频率响应趋向到零以无限频率; 那,有没有零在s = （我不知道抱歉 啊!） 20.这唯一零围拢复平面.
但复函数在s与对实际频率的电路的反应怎么关连?当分析电路的反应对AC发信号时,我们为感应器的impendance和1/jwC使用表示jwL为那电容器.当分析瞬变响应使用Laplace变换时,我们为这些元素阻抗使用sL和1/sC.相似性立刻是明显的.jw在AC分析实际上是s的虚构部分,如前面提到,由一真正的部s和一虚构部分jw组成.
有2处不知道.就这样了抱歉啊!