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The Smith Chart

The Smith ChartContentsi.Introductionii.Waves on Ideal Transmission Linesiii.The Smith Chartiv.ReferencesIntroductionThis article deals with ideal transmission lines for electrical waves. If you would like a review of sinusoidal signals, phasors and transmission line equations, please read Backward Waves. We shall use the same notation here, except that the coordinate z that specifies location along a transmission line has z = 0 at the load instead of the source, and that we also use z as the normalized impedance Z/Z o, where Z o is the characteristic impedance of the line. To avoid confusion, -z is often represented by d.Our signals are sinusoidal waves of frequency f and wavelength λ, and fλ = v, the velocity of the waves. In free space, v = c = 2.9978 x 108 m/s, approximately. In any material medium, v = c/n, where n is the index of refraction of the medium. Velocity on a transmission line is usually expressed by the velocity factor 1/n instead. When calculating wavelength on a transmission line, the velocity factor must be taken into account. Indeed, λ = c/nf = velocity factor x c/f. For theoretical work, the angular fre quency ω = 2πf, and the propagation constant k = 2π/λ are more convenient.For clarity, we shall consider only ideal lines in this article, those with no series resistance or shunt conductance. Actual lines approximate ideal lines rather closely, so this is not a serious limitation. Energy is conserved on an ideal line; the power out is the power in. The principal parameters of a line are its capacitance C and inductance L per unit length. The wave velocity is then v = 1/√(LC) and the characteristic impedance is Z o = √(L/C). These two parameters are generally quoted for any transmission line material.Coaxial cable RG-8/U has Z o= 53Ω and 1/n = 0.66. This velocity factor is typical of polyethyene (PE) insulation. RG-59/U, with Z o= 73Ω, has the same insulati on and velocity factor. RG-141/U, with polytetrafluoroetylene (PTFE) insulation, has Z o= 50Ω and 1/n =0.70. PE foam is mainly air, so RG-8/U(foam) has a velocity factor of 0.80 but about the same Z o= 50Ω. Coaxial cable has the great advantage that the f ields are totally enclosed. The molded 300Ω "twin-lead" has a velocity factor of 0.82. Parallel-wire lines in air have even larger velocity factors, usually about 0.95.A parallel-wire line with conductors of diameter d spaced a distance s betweeen centerlines has C = πε/cosh-1(s/d) and L = (μ/π)cosh-1(s/d). The product LC = με = 1/c2, so the idealvelocity factor is 1. The characteristic impedance is Z o = (cosh-1/π)√(μ/ε). √(μ/ε) = 377Ω, the wave impedance of free space. For d = 2mmand s = 20mm, Z o= 359Ω. The inverse hyperbolic cosines are calculateddirectly by the HP-48G, but can also be expressed in terms of naturallogarithms.Waves on Ideal Transmission LinesOn an ideal transmission line, there is generally a wave moving from source to load V'e-jkz, and a wave moving from load to source V"e jkz. The coefficients V' and V" are constants independent of z, since the line is lossless. The corresponding currents are (V'/Z o)e-jkz and-(V"/Z o)e jkz. All quantities are multiplied by the time factor e jωt, which is understood. The total voltage and current at any point are the sums of the contributions of the two waves.If the line is terminated at z = 0 with an impedance Z L, then this must be the ratio of the total voltage to the total current at that point, or Z L = Z o(V' + V")/(V' - V"). This condition establishes the ratio of V" to V'. Let us define the reflection coefficientρ(z) as the complex ratio (V"e jkz)/(V'e-jkz) = (V"/V')e2jkz. We then have z(0) = Z L/Z o = [1 + ρ(0)]/[1 - ρ(0)]. This equation can be inverted to give ρ(0) = [z(0) - 1]/[z(0) + 1]. The normalized load impedance determines the reflection coefficient at the load.At any other point on the line, a similar equation holds, but the complex reflection coefficient varies in a simple way. In fact, ρ(z) = ρ(0)e2jkz. That is, only the phase changes, while the magnitude remains constant. The magnitude will be represente d by ρ. In the usual case when the termination is resistive, ρ = [r(0) - 1]/[r(0) + 1]. As we move a distance d towards the source, ρ= ρe-2jkd. Therefore, the complex number ρ, represented as a vector, rotates clockwise through an angle (4π/λ)d. When d isλ/2, a half-wavelength, the vector has rotated through a complete circle.The maximum voltage on the line will be |V'| + |V"| and the minimum will be |V'| - |V"|. The ratio of the maximum to the minmum voltage is called the voltage standing wave ratio, VSWR, and there is an analogous definition for the current standing wave ratio. Clearly, maximum voltage corresponds to minimum current, and vice versa. The VSWR S = (1 +|ρ|)/(1 - |ρ|) in terms of the magnitude of the reflection coefficient. If |ρ| is zero, then S = 1 and the maximum voltage is constant along the line. If |ρ| = 1, then S is infinite, and there are points of zero voltage which correspond to points of maximum current, called nodes.The Smith ChartIf we let ρ = u + jv, this can be plotted in the (u,v) plane in the usual way of representing complex numbers. The normalized impedance, z = Z/Z o = [1 + ρ]/[1 - ρ] is a function of ρ, and so its real and imaginary parts, z = r + jx, can be expressed in terms of the real and imaginary parts of ρ = u + jv. If lines of r = constant and x = constant are drawn on thediagram, the result is called the Smith Chart, which is shown in the figure.This is a rather complex figure, but will repay careful study.First of all, the vertical and horizontal axes are v and u, the imaginary and realparts of ρ, which is normally represented in polar form. The outer circlecorresponds to ρ = 1 as well as to r = 0. The equation of the circle for a resistance r is [u - (r/r+1)]2 + v2 = (1/1+r)2, as can be seen by expressing z in terms of u and v, and finding the real part. The circle passing through points A and the origin B, of radius 1/2, corresponds to r = 1. Finally, point A corresponds to r = ∞.The normalized reactance, x, is also constant on circles which pass through A and have their centres on the v-axis. These circles are easily found to be (u - 1)2 + (v - 2/x)2 = (2/x)2. The parts of the circles for x = +1 and x = -1 are shown. Positive, or inductive, reactance is above the u-axis, while negative, or capacitive, is below. The u-axis corresponds to x = 0, or a z that is purely resistive.We note that point C corresponds to r = 0, or to a short at the load end of the line. Point B corresponds to r = 1, so the line is terminated in its characteristic resistance. At point A, r = ∞, or the line is open. Either point A or point C makes ρ = 1. As we proceed toward the source, the vector ρ rotates clockwise from whichever point describes the particular termination. While r remains zero, x goes through the complete range of values from +∞to -∞. This will be the reactance at any point along the line, and in particular, at the source. By changing the length of the line, we can present any desired reactive impedance to the source; at two points the reactance is zero, while the resistance is zero or infinity.If the line is terminated as at point B, then ρ = 0 (and doesn't go anywhere), while the impedance presented to the source is constant at z = 1, or Z = Z0. This important case is a matched line, and there is no reflection at the load end.A point P at an arbitrary location is shown also. It happens to lie on the r = 1 circle, so if it represents the termination of a line, the termination impedance has this real part, and some reactive part as well. This point determines a reflection coe fficient ρ and an angle θ which together determine the complex ρ. Any other point on the line is represented by some point on a circle of radius ρ with centre at B.This is important enough to be represented on a separate diagram. Let us assume that a line is terminated by a resistance at point A. For concreteness, suppose ρ = 0.6 and r = 0.2 at the point A. Suppose we are dealing with a 300Ω line at 200 MHz. The termination is then 0.2 x 300 = 60Ω. Suppose B represents the source end of the line. The angle θ is 135° (say), so 135° = 2(2π/λ)d = 720°d/λ. If the velocity coefficient is 0.82, the wavelength is 0.82 x c / 200MHz = 1.23 m. Now we can find the actual length of the line: d = (135/720)(1.23) = 23 cm. This end of the line is on the r = 1 circle, so the resistive component of the impedance is 300Ω. On an actual Smith Chart, we could also read off the reactance as well. Let's suppose it is x = 2. Then X = (2)(300) = 600Ω, inductive. Therefore, at point B, the impedance looking into the line is 300 + j600 Ω.The physical line is shown in the figure at the right. It is represented as constructed from300Ω plastic twinlead. The 60Ω resistor is not a standard value, but 56Ω would do about as well. Measuring the impedance at the input is a little more difficult, unless you have the very expensive insruments that can do it directly.Lines that are exactly a quarter-wavelength long have interesting properties. It is clear that the reflection coefficients at the two ends are simply negatives of each other, at the ends of a diameter in the Smith Chart. This means that if Z/Z o= (1 + ρ)/(1 - ρ) at one end, then Z'/Z o = (1 - ρ)/(1 + ρ) = Z o/Z at the other, or ZZ' = Z o2. This is called a quarter-wave transformer. Remember that when you design such a transformer, it will work as intended only at the design frequency, for only then is it a quarter-wave long.A quarter-wave line shorted at one end, as shown in the figure, is called a quarter-wave stub, and presents a very large impedance at its open end. Such stubs can be used to support a transmission line. Although there will be a DC path to ground, signal frequencies will be isolated. They will also act as pretty good bandpass filters, too, since only the signal current will not be shorted out.In this article, I have only presented the theory of the Smith Chart with a few examples. It has been used to solve problems of many standard types that arise in transmission line design. Of course, all the calculations can be done with a pocket calculator or computer, but the chart has the great advantage of giving a graphic picture of conditions that can give a deeper understanding and help in solving unusual problems. The interested reader should certainly examine an actual Smith Chart to appreciate how easy it is to use, and how quickly it provides answers.ReferencesS. Ramo, J. R. Whinnery and T. Van Duzer, Fields and Waves in Communication Electronics (New York: John Wiley & Sons, 1965). pp. 31-41. Fig. 1.20b is a Smith Chart that can be copied if no other source is available.Return to Tech IndexComposed by J. B. CalvertCreated 16 August 2003Last revised。

史密斯圆图简介

史密斯圆图（Smith chart ）分析长线的工作状态离不开计算阻抗、反射系数等参数，会遇到大量繁琐的复数运算，在计算机技术还未广泛应用的过去，图解法就是常用的手段之一。

在天线和微波工程设计中，经常会用到各种图形曲线，它们既简便直观，又具有足够的准确度，即使计算机技术广泛应用的今天，它们仍然对天线和微波工程设计有着重要的影响作用。

Smith chart 就是其中最常用一种。

1、Smith chart 的构成在Smith chart 中反射系数和阻抗一一对应；Smith chart 包含两部分，一部分是阻抗Smith 圆图（Z-Smith chart ），它由等反射系数圆和阻抗圆图构成；另外一部分是导纳Smith 圆图（Y-Smith chart ），它由等反射系数圆和导纳圆图构成；它们共同构成YZ-Smith chart 。

阻抗圆图又由电阻和电抗两部分构成，导纳圆图由电导和电纳构成。

1.1 等反射系数圆在如图1所示的带负载的传输线电路图中，由长线理论的知识我们可以得到负载处的反射系数0Γ为：000000Lj L u v L Z Z j eZ Z θ-Γ==Γ+Γ=Γ+ 其中00arctan(/)Lv u θ=ΓΓ。

图1 带负载的传输线电路图在离负载距离为z 处的反射系数Γ为：2000L j j z in u v in Z Z j e eZ Z θβ--Γ==Γ+Γ=Γ+ 其中220u v Γ=Γ+Γ，arctan(/)L v u θ=ΓΓ。

椐此我们用极坐标当负载和传输线的特征阻抗确定下来之后，传输线上不同位置处的反射系数辐值（1Γ≤）将不再改变，而变得只是反射系数的辐角；辐角的变化为2z β-∆，传输线上的位置向负载方向移动时，辐角逆时针转动，向波源方向移动时，辐角向顺时针方向转动，如图2所示。

图2 等反射系数圆传输线上不同位置处的反射系数的辐角变化只与2z β-，其中传波常数2/p βπλ=，所以Γ是一个周期为0.5p λ的周期性函数。

smith_chart(史密斯圆图)

史密期圆图（Smith chart ）分析长线的工作状态离不开计算阻抗、反射系数等参数，会遇到大量繁琐的复数运算，在计算机技术还未广泛应用的过去，图解法就是常用的手段之一。

Smith chart 就是其中最常用一种。

阻抗圆图又由电阻和电抗两部分构成，导纳圆图由电导和电纳构成。

1.1等反射系数圆在如图1所示的带负载的传输线电路图中，由长线理论的知识我们可以得到负载处的反射系数0Γ为：000000Lj L u v L Z Z j eZ Z θ-Γ==Γ+Γ=Γ+其中00arctan(/)L v u θ=ΓΓ。

图1 带负载的传输线电路图在离负载距离为z 处的反射系数Γ为：2000Lj j z in u v in Z Z j eeZ Z θβ--Γ==Γ+Γ=Γ+其中0Γ=arctan(/)L v u θ=ΓΓ。

图2 等反射系数圆传输线上不同位置处的反射系数的辐角变化只与2z β-，其中传波常数2/p βπλ=，所以Γ是一个周期为0.5p λ的周期性函数。

史密斯圆图

史密斯圆图
史密斯圆图(Smith chart)是一款用于电机与电子工程学的图表，主要用于传输线的阻抗匹配上。

史密斯图的基本原理在于以下的算式:
反射系数Γ(reflection coefficient)和阻抗z L均为复数，z L是归一化负载值，即z L = ZL/ Z0。

ZL是电路的负载值，Z0是传输线的特性阻抗值，通常使用50Ω。

这是一双线性变换，属于复变函数中的保角变换。

它将z
复平面上实部r=常数和虚部x＝常数的两族正交直线变换为Γ
复平面上的正交圆族。

该图表是由菲利普·史密斯(Phillip Smith)于1939年发明的，当时他在美国的RCA公司工作。

史密斯也许不是图表的第一位发明者，一位名为Kurakawa的日本工程师声称早于其一年发明了这种图表。

史密斯曾说过，“在我能够使用计算尺的时候，我对以图表方式来表达数学上的关联很有兴趣。

”
Smith 圆图
图表中的圆形线代表阻抗的实部，即等电阻圆；中间的横线与向上和向下散出的弧线则代表阻抗的虚部，即等电抗圆。

上半圆是正值，下半圆是负值。

在图边的数字代表反射系数的角度(0-180度)和波长(由零至半个波长)。

有一些图表是以导纳值(admitt ance)来表示，把上面的阻抗圆图旋转180度即可导纳圆图。

自从有了计算机后，此种圆图的使用率随之而下，但仍常用来表示特定的资料。

对于就读电磁学及微波电子学的学生来说，在解决课本问题仍然很实用，因此史密斯图至今仍是重要的教学工具。

在学术论文里，结果也常会以史密斯图来表示。

SMITH CHART原理及应用

前言印刷電路板的ｐａｔｔｅｒｎ線路有很多必需是借助ｔｈｒｕｏｇｈ　ｈｏｌｅ完成線路路徑的佈局，對低頻電路而言ｔｈｒｕｏｇｈ　ｈｏｌｅ幾乎不會對該電路產生不良影響，不過高頻電路的阻抗（ｉｍｐｅｄａｎｃｅ）整合卻扮演關鍵性角色，換言之若將具有ｔｈｒｕｏｇｈ　ｈｏｌｅ的線路當作一般傳輸線路處理，就會面臨許多超乎預期的困擾，主要原因是在傳輸線路上如果設有ｔｈｒｕｏｇｈ　ｈｏｌｅ，該部位就會產生非連續性點阻抗，而該點或多或少會形成反射波，最後造成電路誤動作，類比電路的精度發生誤差等嚴重後果。

　該反射波的反射程度是用反射係數表示，它是用複素數處理變成複素量。

雖然電子電路經常使用複素數與ａｄｍｉｔｔａｎｃｅ等計算方式，不過實際上複素數計算相當煩瑣，其中傳輸線路與高頻電路常用的複素數計算，如果改成史密斯特性圖表（Ｓｍｉｔｈ　ｃｈａｒｔ）方式，就可輕鬆獲得相同的計算結果。

有鑑於此，本文將介紹史密斯特性圖表（Ｓｍｉｔｈ　ｃｈａｒｔ）使用上必需注意的事項。

反射係數反射係數是表示整合狀態的尺度，反射係數是負載阻抗與傳輸線路特性阻抗Z0 相異時，部份入射電力未被負載吸收，變成反射電力折返信號源時，入射電力與反射電力的比亦即反射係數可由下式求得:Γ＝反射波/入射也就是說反射係數是具有大小與位相的量，它可由上式Z R 與Z0 兩個阻抗關係求得，此外式(1)可轉換成下式:【試算例1】假設傳輸線路特性阻抗Z0 為50Ω，負載阻抗分別是0Ω、50Ω、1kΩ、j50Ω時，反射係數Г=0.5ㄥ450 ，試算負載阻抗Z R 。

①Z R=0Ω時(負載端短路)這意味著振幅大小相等，位相 1800相異的反射波折返信號源，如圖1(b)所示。

②Z R=50Ω時(整合)?=(50-50)/(50+50)=0這表示成為整合狀態，未發生反射波。

③Z R=1000O 時(不整合)?=(1000-50)/(1000+50)=0.95④Z R=∞O 時(負載端開放)這表示振幅大小相等，位相相等的反射波折返信號源，如圖1(a)所示。

史密斯圆图

例3 在Z0为50Ω 的无耗线上测得 VSWR为5，电压驻波最小点
出现在距负载λ /3处，求负
载阻抗值。解：电压驻波最小点:
rmin = K = 1/ VSWR = 1/ 5 = 0.2
在阻抗圆图实轴左半径上。以rmin点沿等 VSWR=5的
圆反时针旋转转λ /3得到 zL 0.77 j1.48 ，故得负载阻抗为 Z 38.5 j 74() L
解： / 2 0.25m
f 3 108 / 0.5 600(MHz)
20lg 20lg(| V |max / | V |min ) 0 (6)
2
电压驻波最小点距负载 0.10m 0.2λ 以| V |min 点沿ρ＝2的圆反时针（向负载）旋转0.2λ
0.028 j 0.15
yL
1.18
0.25
对应向电源波长数0.028
0.45
zL
yL 点沿等Γ线顺
时针旋转0.3λ,得
yin
j 0.6
j 0.9
0.328
？
yin 1.18 j 0.9
Gr
复平面上的反射系数圆
ZL
是一簇|G|≼1同心圆。
r圆
r 1 2 GRe GIm 1 r 1 r
2
2
上式为归一化电阻的轨迹方程，当r等于常数时，其轨迹为一簇圆；
圆心坐标 r ,0 1 r 半径
GIm
1 1 r
8
例2 已知： Z 0 50
Z L 100 j50
0.24
ZL
求：距离负载0.24波长处的Zin.
解
ZL zL 2 j Z0

2-2Smith Chart

12
Yin 1 1 (d ) 1 e j (d ) yin Y0 zin 1 (d ) 1 e j (d )
1 (d ) z in 1 (d )
•归一化的导纳可以由归一化阻抗在复Γ平面上旋转 180°得到，即导纳点是阻抗点关于原点的对称点。 •将Smith阻抗圆图旋转180°得到的圆图称为Smith导纳圆图。 •归一化导纳和反射系数平面点存在一一对应关系。
28
2. 归一化负载阻抗
z L (30 j 60) / 50 0.6 j1.2
v p / f 0.5*3*108 / 2*109 7.5cm
d 2cm 0.267
29
开路线变换
采用开路线可以方便的得到纯感性和纯容性电抗。传输线的特性阻抗为50Ω，工作在3GHz，相速度为光速的77%，若要实现2.12pF的电容或5.3nH的电感，所需开路短截线的长度。
2 2
1 半径： x
1 圆心： r 1, i x
x的范围为＜x＜+， x可为负（容性），也可为正（感性）。所有圆的中心都在过r =+1并垂直于实数轴r的线上。 x=时，圆的半径为零，即r =+1和i =0的一个点。 x→0时，圆的半径和圆的中心沿着垂直于实数轴（r）的线。
32
短路线变换
采用短路线可以方便的得到纯感性和纯容性电抗。传输线的特性阻抗为50Ω，工作在3GHz，相速度为光速的77%，若要实现2.12pF的电容或5.3nH的电感，所需短路短截线的长度。
33
34
在Smith圆图上找到2.12pF的电容和5.3nH的电感所对应的点，从短路点开始旋转至该两点，得到所需线的长度。实现2.12pF的电容需短路短截线0.425 λ，即 32.7mm 实现5.3nH的电感需短路短截线长0.176 λ，即 13.5mm

(完整word版)smith史密斯圆图(个人总结),推荐文档

smith chart史密斯圆图总结史密斯圆图(Smith chart)是一款用于电机与电子工程学的圆图，是最著名和最广泛的用于求解传输线问题的图解技术。

主要用于传输线的阻抗匹配上。

一条传输线(transmission line)的电阻抗力(impedance)会随其长度而改变，要设计一套匹配(matching)的线路，需要通过不少繁复的计算程序，史密斯圆图的特点便是省却一些计算程序。

Smith圆图的构成：等反射系数圆、阻抗圆图、导纳圆图。

史密斯圆图的基础在于以下的算式Γ= (Z - 1)/(Z+ 1)Γ代表其线路的反射系数(reflection coefficient)，即S-parameter里的S11，Z是归一负载值，即ZL / Z0。

当中，ZL是线路的负载值Z0是传输线的特征阻抗值，通常会使用50Ω。

圆图中的横坐标代表反射系数的实部，纵坐标代表虚部。

圆形线代表等电阻圆，每个圆的圆心为1/(R+1),半径为R/(R+1).R为该圆上的点的电阻值。

中间的横线与向上和向下散出的线则代表阻抗的虚数值，即等电抗圆，圆心为1/X,半径为1/X.由于反射系数是小于等于1的，所以在等电抗圆落在单位圆以外的部分没有意义。

当中向上发散的是正数，向下发散的是负数。

圆图最中间的点(Z=1+j0, Γ=0)代表一个已匹配(matched)的电阻数值(此ZL=Z0，即Z＝1)，同时其反射系数的值会是零。

圆图的边缘代表其反射系数的幅度是1，即100%反射。

在图边的数字代表反射系数的角度(0-180度)。

有一些圆图是以导纳值(admittance)来表示，把上述的阻抗值版本旋转180度即可。

圆图中的每一点代表在该点阻抗下的反射系数。

该电的阻抗实部可以从该电所在的等电阻圆读出，虚部可以从该点所在的等电抗圆读出。

同时，该点到原点的距离为反射系数的绝对值，到原点的角度为反射系数的相位。

由反射系数可以得到电压驻波比和回波损耗。

史密斯圆图基本原理及应用

解：先求出归一化负载阻抗0.5+j0.5，在圆图上找出与此相对应的
点P1，以圆图中心点O为中心、以OP1为半径，顺时针（向电源方向）旋转0.2到达点P2，查出P2点的归一化阻抗2j1.04，将其乘以特性阻抗即可得到z=0.2处的等效阻抗为100 j52()
微波工程基础
17
第一章均匀传输线理论之史密斯圆图及其应用
第一章均匀传输线理论之史密斯圆图及其应用
3.阻抗圆图（smith chart）
实轴右半边为电压波腹点又代表驻波比
向电源
实轴左半边为电压波节点又代表行波系数K
向负载
将反射系数圆图、归一化电阻圆图和归一化电抗圆图画在一起，为完整的阻抗圆图，也称为史密斯圆图。

微波工程基础
9
2 2
Z in 1 u jv 传输线上任意一点归一化阻抗为： Z in Z 0 1 u jv 令 Zin r jx ，则得到下列方程
微波工程基础
4
第一章均匀传输线理论之史密斯圆图及其应用
阻抗到反射系数的映射示意图---等电阻圆
微波工程基础
15
第一章均匀传输线理论之史密斯圆图及其应用
[例1-2]已知传输线的特性阻抗Z0=50Ω，终端接有下列负载阻抗，将其线上各点反射系数在圆图上表示
已知下列负载阻抗：a Z L 50 bZ L 48.5
cZ L (75 j 25) d Z L (10 j5)
Zl=(0.77+j1.48)50=38.5+j74()
微波工程基础
18
第一章均匀传输线理论之史密斯圆图及其应用
6.阻抗匹配阻抗匹配的方法1—电抗匹配

Smith Chart 史密斯圆图

Smith Chart Tutorial Part1To begin with we start with the definition of VSWR, which is the ratio of the reflected voltage over the incident voltage. The Reflection coefficient Γ is simply the complex (ie has phase) version of VSWR:-Define voltage standing wave ratio (VSWR)V V Voltage reflection coefficient - Complexmax min =+−V V V V 1212ΓV 1 e+j. β.lΓΓΓ =V V = V V At the load (l =0) ; =V V 212121e ee j j j −+−⇒.....βββlll2but this may be complex number if there is an instantaneous phase change which we’ll call (φ) on reflection.Phase Diagram0=L (a) ..... ∅()()DiagramCrank on d represente = V V=constant is L ith not vary w do V & V line lossless a For 0>L At .21221212.)0((L)l l βφβφφ−−−ΓΓΓ∴=Γ=Γj j j e e V V eCrank DiagramWe use a crack diagram as a way of representing the reflection coefficient phasor.()V = V VV V V V 12121e e e e e j j j j j +−+−−+∴=+=+........ββββφβl l ll l 1122ΓOALPAt the origin of argand diagram.OP = magnitude of total voltage/incidentvoltageβ.LOV Γ|As we saw previously the crack diagram with a circle drawn between points A & C is the beginnings of a Smith chart less the constant resistance and reactance circles/lines.VSWR =V V = 1+1- or =VSWR -1VSWR +1max min ΓΓΓ| V |V maxV minLLength OP’ -crank diagramvoltageincidentvAt B - 2. = -= 2.-= 4. = min mingφβπφβππλπφl l l min −∴From standing wave pattern measure VSWR ⇒ | Γ | @ l min ⇒ φ at load.SmithChart - Impedance (Z) or Admittance Y chart(1) Crank diagram + constant resistance & constant reactance circles.(2) Graphical solution to the equationcomplex= 11).2()(l βφ−ΓΓΓ−Γ+=j oin e Z Z(3) Smith Chart is a reflection coefficient diagramθ = φ-2.β.LSmith ChartOpen circuit x ⇒ ∞Short circuit v=0,x=0X = 0 ∴A is the matched point no reflectionOImpedance is plotted on the smith chart by first normalising to the characteristic impedance of the system (usually 50 ohms). In a 50 ohm system the centre of the smith chart is a pure 50 ohms.For example say we wanted to plot an impedance of 150 + j75ΩFirst normalise ie 150/50 = 3Ω ; 75/50 = 1.5Ω normalised impedance = 3+ j1.5ΩSo the real part of the impedance will lie somewhere along the r = 3 constant resistance circle ie:-Next we follow the constant reactance line at 0.75 to find the intersection of the r = 3 circle to get to our impedance point.3+j1.5Using the Smith Chart(1) Moving along the T.L = rotating around the Smith Chart.BACKWARDSFORWARDSFORWARDS (TO LOAD)BACKWARDS (FROM LOAD)(2) Constant |Γ| or VSWR circlesFor a lossless line |Γ| & VSWR do not vary with L.Γ| or VSWR|Γ| = 0 VSWR = 1|Γ| = 1∞1+1-Vmax = Z(max)Zo (real VSWR)Vmin =Z(min)ZoΓΓSS S ==1Vmin Vmax(3) Measure Lmin/λg ..... determines φ (at load).|V|LFORWARD by Lmin/λg takes us theload.λπlming−(4) Reading Z from chart also can get |Γ| & φ(5)Z Z o =+−11ΓΓAdmittance = Y/Y 0 =φY/Yo-|Γ|1 On a Smith chart point diametrically oppositeZ Z gives Y Y Note Y Y = G + j. Conductance Susceptance On admittance chart r circles g circles & x circles b circles.Note g =G Y and b = B Y o o0o o=→→1Z oβ(6) To transform an impedance along a T.L, rotate around the VSWR circle:-BACKWARDS by Lmin/λg takes us toZin.ZL/Zol/λg(7) Represent a series inductance on a smith chart.0.5Read values off the reactance scale0.20Therefore, assuming a frequency of say 1GHz the value of series inductance represented on the above Smith Chart is given by:-2.38nH E 1*2π0.3*50 2πN.XL 50 wrt 0.3 0.2 - 0.5 chart Smith from read )(X Reactance 9L L ===ΩΩ=Ω=fSimilarly for a series capacitor(8) Represent a series capacitance on a smith chart.Read values off the reactance scaleTherefore, assuming a frequency of say 1GHz the value of series capacitance represented on the above Smith Chart is given by:-ohms)50(usually factor g normalisin the is N Where 6.36pF 5.0*50*E 1*2π1.N.X 2π1 C 50 w.r.t 0.5 0.5 - 1.0 chart Smith from read )(X Reactance 9C C ===ΩΩ=Ω=fTo represent shunt reactance we need to plot admittance onto the Smith Chart. It is easiest to use a Smith chart with both impedance (usually in black) lines and admittance lines (usually in red) on the same chart. Or you can rotate the Smith chart 180 degrees.(9) Represent a shunt inductance on a smith chart.0.8Read values off the admittance scale0.2Therefore, assuming a frequency of say 1GHz the value of shunt inductance represented on the above Smith Chart is given by:-()ohms)50(usually factor on normaisati N 13.26nH6.0*E 1*2π50Y *2πNL 50 w.r.t 0.6mhos 0.2 - 0.8 chart Smith from read )(Y Admittance 9L L ====Ω=Ω=f(10) Represent a shunt capacitance on a smith chart.Read values off the admittance scale0.2Therefore, assuming a frequency of say 1GHz the value of shunt inductance represented on the above Smith Chart is given by:-()ohms)50(usually factor on normaisati N 2.5pF 50*E 1*2π0.8N *2πYC 50 w.r.t 0.8mhos 0.2 - 1.0 chart Smith from read )(Y Admittance 9C C ====Ω=Ω=f。

斯密斯原图的原理及应用

斯密斯原图的原理及应用1. 前言斯密斯原图（Smith Chart）是一种用于电磁波传输线的复阻抗和反射系数分析的图形工具。

它由贝尔实验室的菲利普·斯密斯（Philip H. Smith）于1939年发明，至今仍广泛应用于射频电路设计和天线工程中。

本文将介绍斯密斯原图的原理和其在工程中的应用。

2. 原理斯密斯原图是通过将电路中的复阻抗映射到复平面上来表示的。

复平面的圆心表示纯电阻，半径为1的单位圆表示纯电抗，斯密斯原图上的每个点表示一个复阻抗。

斯密斯原图中的等电阻附近的等压线呈放射状分布，而等电抗的等压线则相互平行。

通过斯密斯原图，我们可以直观地了解电路中复阻抗的特性，方便进行参数的计算和设计。

3. 应用下面列举了斯密斯原图在射频电路设计和天线工程中的常见应用：3.1 驻波比分析斯密斯原图可以用于分析和计算电路的驻波比。

驻波比是衡量信号在传输线上反射程度的一个指标。

在斯密斯原图上，驻波比可以通过观察负载点与单位圆的交点来判断。

当负载点位于单位圆上时，表示负载是纯电阻，无反射；当负载点位于单位圆外部时，表示存在反射，其离圆心的距离与驻波比成正比。

3.2 阻抗匹配为了最大限度地将信号传输到负载端，常常需要进行阻抗匹配。

阻抗匹配意味着使源阻抗和负载阻抗之间的阻抗值相等。

斯密斯原图可以帮助我们直观地找到合适的匹配点，并计算出匹配时所需的传输线长度和阻抗变换器。

3.3 天线设计斯密斯原图在天线工程中也有着重要的应用。

通过斯密斯原图，我们可以确定天线的输入阻抗，以及选择合适的天线匹配网络。

此外，还可以预测和优化天线的辐射模式和增益。

3.4 参数调整与优化当我们需要调整电路的参数以达到某种设计要求时，斯密斯原图可以作为一个强大的工具。

通过斯密斯原图，我们可以直观地观察到将某个元件加入电路后会对整体的复阻抗产生怎样的影响，并优化设计。

3.5 反射系数计算通过斯密斯原图，我们可以直观地计算出反射系数的值。

第2.5章史密斯圆图

5) 距离最近的为电压最大点，lmax 0.25
dmax lmax lLz 0.25 0.208 0.042 dmin dmax 0.25 0.25 0.042 0.292
d/ 0.35 0.29 0.042 0
例 2.5-5
已知：Z0 = 250W; ZL = 500- j150W; l = 4.8l
VSWR- 1
GL
=
= VSWR + 1
0.518
4) 由z L点沿等圆向电源方向旋转0.35λ ，至zin点，
则可得 zin 0.36 j0.342 lin 0.35 0.208 0.5 0.058
其输入阻抗为 Zin 18 j17.1()
其输入反射系数为
Gin 0.52 in 138 0 2.41rad
圆心坐标 r ,0
1 r
半径
1
1 r
r =∞；圆心（1，0）半径=0 r =1；圆心（0.5，0）半径=0.5 r =0；圆心（0，0）半径=1
GIm GRe
x圆
GRe
12

GIm

1 x
2

1 x
2
GIm
第二式为归一化电抗的轨
迹方程，当x等于常数时，
0

lm in

2

其电长度
lmin =
lmin l
=
1 4p
(F
L
?
p)
对于一般位置： f (z) = f L - 2b z
对于相距/2的两点:
lmin
=
l (f

斯密斯圆图_Smith_chart

阻抗匹配与史密斯(Smith)圆图: 基本原理本文利用史密斯圆图作为RF阻抗匹配的设计指南。

文中给出了反射系数、阻抗和导纳的作图范例，并用作图法设计了一个频率为60MHz的匹配网络。

实践证明：史密斯圆图仍然是计算传输线阻抗的基本工具。

在处理RF系统的实际应用问题时，总会遇到一些非常困难的工作，对各部分级联电路的不同阻抗进行匹配就是其中之一。

一般情况下，需要进行匹配的电路包括天线与低噪声放大器(LNA)之间的匹配、功率放大器输出(RFOUT)与天线之间的匹配、LNA/VCO输出与混频器输入之间的匹配。

匹配的目的是为了保证信号或能量有效地从“信号源”传送到“负载”。

在高频端，寄生元件(比如连线上的电感、板层之间的电容和导体的电阻)对匹配网络具有明显的、不可预知的影响。

频率在数十兆赫兹以上时，理论计算和仿真已经远远不能满足要求，为了得到适当的最终结果，还必须考虑在实验室中进行的RF测试、并进行适当调谐。

需要用计算值确定电路的结构类型和相应的目标元件值。

有很多种阻抗匹配的方法，包括:•计算机仿真: 由于这类软件是为不同功能设计的而不只是用于阻抗匹配，所以使用起来比较复杂。

设计者必须熟悉用正确的格式输入众多的数据。

设计人员还需要具有从大量的输出结果中找到有用数据的技能。

另外，除非计算机是专门为这个用途制造的，否则电路仿真软件不可能预装在计算机上。

•手工计算: 这是一种极其繁琐的方法，因为需要用到较长(“几公里”)的计算公式、并且被处理的数据多为复数。

•经验: 只有在RF领域工作过多年的人才能使用这种方法。

总之，它只适合于资深的专家。

•史密斯圆图: 本文要重点讨论的内容。

本文的主要目的是复习史密斯圆图的结构和背景知识，并且总结它在实际中的应用方法。

讨论的主题包括参数的实际范例，比如找出匹配网络元件的数值。

当然，史密斯圆图不仅能够为我们找出最大功率传输的匹配网络，还能帮助设计者优化噪声系数，确定品质因数的影响以及进行稳定性分析。

SMITH原图

一、Smith图圆的基本思想
( z') l e
j 2 z '
| l | e
j ( l 2 )
| l | e
j
θ的周期是 1/2λg。这种以| Γ|圆为基底的图形称为 Smith圆图。 3. 把阻抗(或导纳)，驻波比关系套覆在|Γ|圆上。这样，Smith圆图的基本思想可描述为：消去特征参数Z０，把β归于Γ相位；工作参数Γ为基底，套覆 Z(Y)和ρ。
±1
1
±1
1
二、Smith圆图的基本构成
i
x 1 = 感抗 x 1 = /2
i
r= 0
1 r=
2 r=
0
x 0 = sh rte .c or d 容抗图
0
图 7-2 等电阻图
7-3 等电抗图
x -1 = /2 3. 标定电压驻波比实轴表示阻抗纯阻点。因此，可 x =1 由电阻r 对应出电压驻波比。 4. 导纳情况
电阻圆始终和直线
r 1
相切。
二、Smith圆图的基本构成
园心坐标
r
r r 1 r
半径
i 0
1 1 r
0 1 2
2 3
0
1 2
0 0 0
1
1 2
1 3
二、Smith圆图的基本构成
虚部又可得到方程
2 (r 1) i 0 x
2 2 i
Y
Y 0.011 j Z0
三、Smith圆图的基本功能
[例2] 已知阻抗
Z 1 j ，求反射系数
i 0.088 1+j
和
0
2.60
r

Smith Chart 使用说明

史密斯原图的简单说明。
阻抗圆图快速记忆技巧 1.三个特殊点匹配点中心点（0，0） Γ=0 开路点右边端点（1，0） Γ=1 短路点左边端点（-1，0） Γ=-1
Z =1
Ρ=1 2.三条特殊线
Z =∞
r = ∞, x = ∞
Z =0 r = 0, x = 0
Hale Waihona Puke （1）实轴为纯电阻线；（2）左半实轴上的点为电压波节点，该直线段是电压波节线、电流波腹线。该直线段上某点归一化电阻 r 的值为该点的 K 值；（3）右半实轴上的点为电压波节点，该直线段是电压波腹线，电流波节线。该直线段上某点归一化电阻 r 的值为该点的 ρ 值。 3.两个特殊面（1）上半圆，归一化电抗值 x>0，上半圆平面为感性区；（2）下半圆，归一化电抗值 x<0，下半圆平面为容性区 4.两个旋转方向因为已经规定负载端为坐标原点，当观察点向电源方向移动时，在圆图上要顺时针方向旋转；反之，观察点向负载方向移动时，在圆图上要逆时针旋转。 5.四个参数在圆图上任何一点都对应有四个参量：Γ、x、ρ（或 Γ ）、φ