论文
中北大学分校学位论文英文翻译
作者姓名：
学科、专业：通信工程
学号：062027
指导教师：副教授
完成日期：2010-6-10
目录
PROPERTIES OF ANTENNAS (3)
2.1 ANTENNA RADIATION (4)
2.2 GAIN (6)
2.3 EFFECTIVE AREA (9)
2.4 PATH LOSS (9)
2.5 RADAR RANGE EQUATION AND CROSS SECTION (11)
2.6 WHY USE AN ANTENNA? (13)
天线的功能 (13)
2.1天线辐射 (15)
2.2 天线增益 (16)
2.3天线有效面积 (19)
2.4路径损耗 (19)
2.5雷达距离方程和截面 (21)
2.6为什么使用一个天线？ (23)
英文原文：
PROPERTIES OF ANTENNAS
One approach to an antenna book starts with a discussion of how antennas radiate. Beginning with Maxwell’s equations, we derive electromagnetic waves. After that lengthy discussion, which contains a lot of mathematics, we discuss how these waves excite currents on conductors. The second half of the story is that currents radiate and produce electromagnetic waves. You may already have studied that subject, or if you wish to further your background, consult books on electromagnetics.The study of electromagnetics gives insight into the mathematics describing antenna radiation and provides the rigor to prevent mistakes. We skip the discussion of those equations and move directly to practical aspects.
It is important to realize that antennas radiate from currents. Design consists of controlling currents to produce the desired radiation distribution, called its pattern .In many situations the problem is how to prevent radiation from currents, such as in circuits. Whenever a current becomes separated in distance from its return current, it radiates. Simply stated, we design to keep the two currents close together, to reduce radiation. Some discussions will ignore the current distribution and instead, consider derived quantities, such as fields in an aperture or magnetic currents in a slot or around the edges of a microstrip patch. You will discover that we use any concept that provides insight or simplifies the mathematics.
An antenna converts bound circuit fields into propagating electromagnetic waves and, by reciprocity, collects power from passing electromagnetic waves. Maxwell’s equations predict that any time-varying electric or magnetic field produces the opposite field and forms an electromagnetic wave. The wave has its two fields oriented orthogonally, and it propagates in the direction normal to the plane defined by the perpendicular electric and magnetic fields. The electric field, the magnetic field, and the direction of propagation form a right-handed coordinate system. The propagating wave field intensity decreases by 1/R away from the source, whereas a static field drops off by 1/2R . Any circuit with time-varying fields has the capability of radiating to some extent.
We consider only time-harmonic fields and use phasor notation with time dependence jwt e . An outward-propagating wave is given by ()j kR wt e --, where k, the wave number, is
given by 2π/λ. λ is the wavelength of the wave given by c/f , where c is the velocity of light (3 ×8
10m/s in free space) and f is the frequency. Increasing the distance from the source decreases the phase of the wave.
Consider a two-wire transmission line with fields bound to it. The currents on a single wire will radiate, but as long as the ground return path is near, its radiation will nearly cancel the other line’s radiation because the two are 180°out of phase and the waves travel about the same distance. As the lines become farther and farther apart, in terms of wavelengths, the fields produced by the two currents will no longer cancel in all directions. In some directions the phase delay is different for radiation from the current on each line, and power escapes from the line. We keep circuits from radiating by providing close ground returns. Hence, high-speed logic requires ground planes to reduce radiation and its unwanted crosstalk.
2.1 ANTENNA RADIATION
Antennas radiate spherical waves that propagate in the radial direction for a coordinate system centered on the antenna. At large distances, spherical waves can be approximated by plane waves. Plane waves are useful because they simplify the problem. They are not physical, however, because they require infinite power.
The Poynting vector describes both the direction of propagation and the power density of the electromagnetic wave. It is found from the vector cross product of the electric and magnetic fields and is denoted S:
S = E ×H* W/2m
Root mean square (RMS) values are used to express the magnitude of the fields. H* is the complex conjugate of the magnetic field phasor. The magnetic field is proportional to the electric field in the far field. The constant of proportion is η, the impedance of free space (η = 376.73Ω):
2
E
==W/2m(1.1)
S S
η
Because the Poynting vector is the vector product of the two fields, it is orthogonal to both fields and the triplet defines a right-handed coordinate system: (E, H, S).
Consider a pair of concentric spheres centered on the antenna. The fields around the antenna decrease as 1/R, 1/2R, 1/3R, and so on. Constant-order terms would require that the
power radiated grow with distance and power would not be conserved. For field terms proportional to 1/2R , 1/3R , and higher, the power density decreases with distance faster than the area increases. The energy on the inner sphere is larger than that on the outer sphere. The energies are not radiated but are instead concentrated around the antenna; they are near-field terms. Only the 1/2R term of the Poynting vector (1/R field terms) represents radiated power because the sphere area grows as 2R and gives a constant product. All the radiated power flowing through the inner sphere will propagate to the outer sphere. The sign of the input reactance depends on the near-field predominance of field type: electric (capacitive) or magnetic (inductive). At resonance (zero reactance) the stored energies due to the near fields are equal. Increasing the stored fields increases the circuit Q and narrows the impedance bandwidth.
Far from the antenna we consider only the radiated fields and power density. The power flow is the same through concentric spheres:
2211,22,44avg avg R S R S ππ=
The average power density is proportional to 1/2R . Consider differential areas on the two spheres at the same coordinate angles. The antenna radiates only in the radial direction; theref ore, no power may travel in the θ or φ direction. Power travels in flux tubes between areas, and it follows that not only the average Poynting vector but also every part of the power density is proportional to 1/2R :
221122sin sin S R d d S R d d θθφθθφ=
Since in a radiated wave S is proportional to 1/2R , E is proportional to 1/R. It is convenient to define radiation intensity to remove the 1/2R dependence:
U(θ, φ) = S(R, θ, φ) 2R W/solid angle
Radiation intensity depends only on the direction of radiation and remains the same at all distances. A probe antenna measures the relative radiation intensity (pattern) by moving in a circle (constant R) around the antenna. Often, of course, the antenna rotates and the probe is stationary.
Some patterns have established names. Patterns along constant angles of the spherical coordinates are called either conical (cons tant θ) or great circle (constant φ). The great circle cuts when φ = 0 °or φ = 90°are the principal plane patterns. Other named cuts are also used,
but their names depend on the particular measurement positioner, and it is necessary to annotate these patterns carefully to avoid confusion between people measuring patterns on different positioners. Patterns are measured by using three scales: (1) linear (power), (2) square root (field intensity), and (3) decibels (dB). The dB scale is used the most because it reveals more of the low-level responses (sidelobes).
Figure 1.1 demonstrates many characteristics of patterns. The half-power beamwidth is sometimes called just the beamwidth. The tenth-power and null beamwidths are used in some applications. This pattern comes from a parabolic reflector whose feed is moved off the axis. The vestigial lobe occurs when the first sidelobe becomes joined to the main beam and forms a shoulder. For a feed located on the axis of the parabola, the first sidelobes are equal.
2.2 GAIN
Gain is a measure of the ability of the antenna to direct the input power into radiation in a particular direction and is measured at the peak radiation intensity. Consider the power density radiated by an isotropic antenna with input power Po at a distance R: S = Po/4π2R . An isotropic antenna radiates equally in all directions, and its radiated power density S is found by dividing the radiated power by the area of the sphere 4π2R . The isotropic radiator is considered to be 100% efficient. The gain of an actual antenna increases the power density in the direction of the peak radiation:
2
024E PG S R πη== or 014PG E S R ηηπ== (1.2) Gain is achieved by directing the radiation away from other parts of the radiation sphere. In general, gain is defined as the gain-biased pattern of the antenna:
()()02
,,4P G S R θφθφπ= power density ()()0,,4P G U θφθφπ
= radiation intensity (1.3)
FIGURE 1.1 Antenna pattern characteristics. The surface integral of the radiation intensity over the radiation sphere divided by the input power Po is a measure of the relative power radiated by the antenna, or the antenna efficiency: ()2000,sin 4r e G P d d P ππθφθθφηπ
==?? efficiency where Pr is the radiated power. Material losses in the antenna or reflected power due to poor impedance match reduce the radiated power. In this book, integrals in the equation above and those that follow express concepts more than operations we perform during design. Only for theoretical simplifications of the real world can we find closed-form solutions that would call for actual integration. We solve most integrals by using numerical methods that involve breaking the integrand into small segments and performing a weighted sum. However, it is helpful that integrals using measured values reduce the random errors by averaging, which improves the result.
In a system the transmitter output impedance or the receiver input impedance may not match the antenna input impedance. Peak gain occurs for a receiver impedance conjugate matched to the antenna, which means that the resistive parts are the same and the reactive
parts are the same magnitude but have opposite signs. Precision gain measurements require a tuner between the antenna and receiver to conjugate-match the two. Alternatively, the mismatch loss must be removed by calculation after the measurement. Either the effect of mismatches is considered separately for a given system, or the antennas are measured into the system impedance and mismatch loss is considered to be part of the efficiency.
Example Compute the peak power density at 10 km of an antenna with an input power of 3 W and a gain of 15 dB.
First convert dB gain to a ratio: G = 151010 = 31.62. The power spreads over the sphere area with radius 10 km or an area of 4π42(10) 2m . The power density is
()282
3(31.62)75.5/410W S nW m m π==? We calculate the electric field intensity using Eq. (1-2):
()()975.510376.75333/
E S V m ημ-==?= Although gain is usually relative to an isotropic antenna, some antenna gains are referred to a λ/2 dipole with an isotropic gain of 2.14 dB.
If we approximate the antenna as a point source, we compute the electric field radiated by using Eq. (1.2):
()()0,,4j k R PG e E R θφηθφπ
-= (1.4) This requires only that the antenna be small compared to the radial distance R. Equation (1.4) ignores the direction of the electric field, which we define as polarization. The units of the electric field are volts/meter. We determine the far-field pattern by multiplying Eq. (1.4) by R and removing the phase term jkR e - since phase has meaning only when referred to another point in the far field. The far-field electric field ff E unit is volts:
()()0,,4ff PG E θφηθφπ= or ()()2
014,,ff G E P πθφθφη??=???? (1.5) During analysis, we often normalize input power to 1 W and can compute gain easily from the electric field by multiplying by a constant 4πη = 0.1826374.
1.3 EFFECTIVE AREA
Antennas capture power from passing waves and deliver some of it to the terminals. Given the power density of the incident wave and the effective area of the antenna, the power delivered to the terminals is the product.
d e f f P S A = (1.6)
For an aperture antenna such as a horn, parabolic reflector, or flat-plate array, effective area is physical area multiplied by aperture efficiency. In general, losses due to material, distribution, and mismatch reduce the ratio of the effective area to the physical area. Typical estimated aperture efficiency for a parabolic reflector is 55%. Even antennas with infinitesimal physical areas, such as dipoles, have effective areas because they remove power from passing waves.
2.4 PATH LOSS
We combine the gain of the transmitting antenna with the effective area of the receiving antenna to determine delivered power and path loss. The power density at the receiving antenna is given by Eq. (1.3), and the received power is given by Eq. (1.6). By combining the two, we obtain the path loss:
()212,4d t
A G P P R θφπ= Antenna 1 transmits, and antenna 2 receives. If the materials in the antennas are linear and isotropic, the transmitting and receiving patterns are identical (reciprocal) [2, p. 116]. When we consider antenna 2 as the transmitting antenna and antenna 1 as the receiving antenna, the path loss is
()12
2,4d t
AG P P R θφπ= Since the responses are reciprocal, the path losses are equal and we can gather and eliminate terms:
1212
G G A A == constant
Because the antennas were arbitrary, this quotient must equal a constant. This constant was found by considering the radiation between two large apertures [3]:
24G A πλ
= (1.7) We substitute this equation into path loss to express it in terms of the gains or effective areas:
2
1212224d t P A A G G P R R
λπλ??== ??? (1.8) We make quick evaluations of path loss for various units of distance R and for frequency f in megahertz using the formula.
path loss(dB)=()()()1220log U K fR G dB G dB +-- (1.9) where Ku depends on the length units:
Example Compute the gain of a 3-m-diameter parabolic reflector at 4 GHz assuming 55% aperture efficiency.
Gain is related to effective area by Eq. (1.7):
24A
G πλ=
We calculate the area of a circular aperture by ()2/2A D π=. By combining these equations,
we have
22a a D Df G c ππηηλ????== ? ?????
(1.10) where D is the diameter and a η is the aperture efficiency. On substituting the values above, we obtain the gain:
()()29934100.5586850.310G π?????==????? (39.4dB)
Example Calculate the path loss of a 50-km communication link at 2.2 GHz using a transmitter antenna with a gain of 25 dB and a receiver antenna with a gain of 20 dB. Path loss = 32.45 + 20 log[2200(50)] - 25 - 20 = 88.3 dB
What happens to transmission between two apertures as the frequency is increased? If we assume that the effective area remains constant, as in a parabolic reflector, the transmission increases as the square of frequency:
2
21
2122221d t P A A A A f Bf P R R c λ??=== ???
where B is a constant for a fixed range. The receiving aperture captures the same power regardless of frequency, but the gain of the transmitting antenna increases as the square of frequency. Hence, the received power also increases as frequency squared. Only for antennas, whose gain is a fixed value when frequency changes, does the path loss increase as the square of frequency. 2.5 RADAR RANGE EQUATION AND CROSS SECTION
Radar operates using a double path loss. The radar transmitting antenna radiates a field that illuminates a target. These incident fields excite surface currents that also radiate to produce a second field. These fields propagate to the receiving antenna, where they are collected. Most radars use the same antenna both to transmit the field and to collect the signal returned, called a monostatic system, whereas we use separate antennas for bistatic radar. The receiving system cannot be detected in a bistatic system because it does not transmit and has greater survivability in a military application.
We determine the power density illuminating the target at a range T R by using Eq. (1.2):
()2,4T T inc T
P G S R θφπ=
(1.11) The target’s radar cross section (RCS), the scattering area of the object, is e xpressed in square meters or dB 2m : 10 log(square meters). The RCS depends on both the incident and reflected wave directions. We multiply the power collected by the target with its receiving pattern by
the gain of the effective antenna due to the currents induced:
()2,,,4r e f l e c t e d
s r r i
i T T T p o w e r P R C S p o w e r d e n s i t y i n c i d e n t P G R θφθφσπ=== (1.12) In a communication system we call Ps the equivalent isotropic radiated power (EIRP), which equals the product of the input power and the antenna gain. The target becomes the transmitting source and we apply Eq. (1.2) to find the power density at the receiving antenna at a range R R from the target. Finally, the receiving antenna collects the power density with
an effective area R A . We combine these ideas to obtain the power delivered to the receiver:
()()()2
2,,,44R T T r r i i rec R R T R A P G P S A R R σθφθφππ==
We apply Eq. (1.7) to eliminate the effective area of the receiving antenna and gather terms to determine the bistatic radar range equation:
()()2322,,,4T R r r i i rec T
T R G G P P R R λσθφθφπ= (1.13) We reduce Eq. (1.13) and collect terms for monostatic radar, where the same antenna is used for both transmitting and receiving:
()22344rec T P G P R
λσπ= Radar received power is proportional to 1/4R and to 2G .
We find the approximate RCS of a flat plate by considering the plate as an antenna with an effective area. Equation (1.11) gives the power density incident on the plate that collects this power over an area R A :
()2,4T T C R T
P G P A R σθφπ= The power scattered by the plate is the power collected,C P , times the gain of the plate as an antenna,P G :
()()2,,4T T i i s C P R P r r
T P G P P G A G R θφθφπ==
This scattered power is the effective radiated power in a particular direction, which in an antenna is the product of the input power and the gain in a particular direction. We calculate the plate gain by using the effective area and find the scattered power in terms of area:
22244T T R s T P G A P R ππλ
= We determine the RCS σ by Eq. (1.12), the scattered power divided by the incident power density:
()()2
222,,444R i i R r r s R T T T G G P A P G R θφθφλπσπλπ=== (1.14) The right expression of Eq. (1.14) divides the gain into two pieces for bistatic scattering, where the scattered direction is different from the incident direction. Monostatic scattering uses the same incident and reflected directions. We can substitute any object for the flat plate and use the idea of an effective area and its associated antenna gain. An antenna is an object with a unique RCS characteristic because part of the power received will be delivered to the antenna terminals. If we provide a good impedance match to this signal, it will not reradiate and the RCS is reduced. When we illuminate an antenna from an arbitrary direction, some of the incident power density will be scattered by the structure and not delivered to the antenna terminals. This leads to the division of antenna RCS into the antenna mode of reradiated signals caused by terminal mismatch and the structural mode, the fields reflected off the structure for incident power density not delivered to the terminals.
2.6 WHY USE AN ANTENNA?
We use antennas to transfer signals when no other way is possible, such as
communication with a missile or over rugged mountain terrain. Cables are expensive and take a long time to install. Are there times when we would use antennas over level ground? The large path losses of antenna systems lead us to believe that cable runs are better.
中文翻译：
天线的性能
一个方法是一本有关天线的书是从讨论天线如何辐射开始的。从麦克斯韦方程组，我们得出了电磁波。之后，进行冗长的讨论，其中包含了很多数学，我们在讨论这些波
如何在导体上激发电流。这个情节的下半部分，就是电流的辐射和产生的电磁波。你可能已经研究过这个问题，或如果你想增进你的阅历，参考在电磁学方面的书籍。研究电磁学就有了一些数学方面的见识来描述天线的辐射，并提供严谨，以防止错误。我们跳过讨论这些方程，直接转到实际的问题。
这是很重要的认识到，天线在电流中产生辐射。设计构成的控制电流产生预期的辐射分布，其所谓的方向图。在许多情况下，问题是如何防止电流中的辐射，如在集成电路。每当一电流变为在距离上从其返回电流中分隔，它发热着。简单地说，我们的设计就是为了保持两个电流靠近在一起，以减少辐射。一些讨论会忽略电流的分布，而作为代替则会考虑得出的数量，如在各个领域的光圈或磁流在一个缝隙或靠近边缘的一微带斑纹。你会发现我们使用任何的概念来提供深入的或简化的数学。
天线1转换电路领域，必将成为传播的电磁波，并通过互惠，从通过电磁波中收集动力。麦克斯韦方程组预测，任何时间变电气或磁场产生相反的领域和形成一电磁波。波有其两个领域的面向正交，而且它在正常的方向上传播给由垂直的电场和磁场所定的飞机。电场、磁场和传播的方向形成一个右手坐标系统。传播波场的强度从源端上降低1/R，而静电磁场下降到1/2R。任何电路随时间变化的场在一定程度上有能力辐射。
我们只考虑时间谐波领域和使用相符号与时间依赖性jwt
e。一个外向型传播的波给出了()
j kR wt
e--，其中k是波数，给出了2π/λ。λ是波所给予的c/f的波长，其中c 是光速（在自由空间中为3×8
10米/秒）和f是频率。从源头上减少了波的相位来增大距离。
考虑一个两线传输的线路与场，必将给它。在一个单一线路上的电流将会辐射，但只要地面返回路径是很近的，其辐射几乎会取消其他线的辐射，因为两者有180°的相位差和波程有差不多一样的远。除了线路变得越走越远外，在波长方面，场所产生的两个电流在所有的方向上将不再取消。在一些方向的相位延迟是不同的辐射从电流在每条线上，和电力从该路线上逃逸。我们保持电路从辐射提供了密切地面的回赠。因此，高速逻辑要求地球平面去减少辐射及其不受欢迎的串扰。
2.1天线辐射
天线辐射球面波在以天线为核心的坐标系统的径向方向上传播。在大的距离上，球面波可以近似平面波。平面波是有用的，因为他们把问题简化了。他们不是自然的，然而，因为它们需要无限的功率。
该玻印廷矢量描述两个方向的传播和功率密度的电磁波。这是从矢量穿过产生的电场和磁场中发现的，并标注为S ：
S = E ×H* W/2m
均方根（RMS ）值是用来表达场的重要性。H*是复杂的共轭的磁场相。磁场在远区场上是与电场成正比的。比例常数是η，自由空间中的阻抗(η = 376.73Ω)： 2
E S S η== W /2m (1.1)
因为玻印廷矢量是两个场的矢量的产物，这是正交的两个场以及三重定义了一个右手坐标系统：(E, H, S)。
考虑一对以天线为核心的同心球形。靠近天线的场减少为1/R, 1/2R , 1/3R 等等。恒指定的条件将要求功率辐射与辐距离和将不会被保存的功率一起增长。场方面的比例1/2R , 1/3R ，和更高，功率密度随距离减少，比面积增加的速度快。在球形里面的能源大于在球形外部的能量。这些能量不辐射，但是代替集中在天线周围，它们是近区场的条件。只有1/2R 条件的玻印廷矢量（1/R 场的条件）所代表的辐射功率，因为该球形的面积的增长为2R ，并给出了一个常数的积。所有辐射功率流经内部球体将传播到球形的外部。符号的输入抗依赖于近区场的场类型的优势：电气（电容式）或磁场（电感）。在共振（零抗）上储存的能量是平等的，因为是近区场。存储场的增多增加了电路的Q 和缩小阻抗带宽。
从天线到目前为止，我们只考虑辐射的场和功率密度。功率流是相同的通过同心的球形：
221
1,22,44avg avg R S R S ππ= 平均功率密度是成正比于1/2R 的。考虑在同一坐标的角度上的两个球形的面积的差异。
天线的辐射，只有在径向方向;因此，没有功率可能在θ或φ方向上游走的。功率在面
积中的通量管上游走，并如下，不仅平均坡印亭矢量，而且功率密度的每个部分都是与1/2R 成正比的：
22
1122sin sin S R d d S R d d θθφθθφ=
自从在一个辐射波S 是成正比于1/2R 的之后，E 是成正比于1/R 。界定辐射强度以此来消除1/2R 的依赖是很方便的：
U(θ, φ) = S(R, θ, φ) 2R W/solid angle
辐射强度，只取决于辐射的方向和在所有的距离上保持不变。一探针天线测量相对辐射强度（方向图）是通过在天线的周围移动轨迹在一个圆圈（常数 R ）上。当然很多时候天线在旋转而且探头是固定的。
一些方向图已经建立了名称。方向图随着球面坐标系的常数角度就叫做锥形（常数θ）或大圈（常数φ）。大圈削减当φ=0°或φ= 90°是主要的平面方向图。其他命名削减也使用，但他们的名字取决于特定的测量定位，而且它是必要的注释，这些方向图小心地在人们对不同定位器的测量方式之间去避免造成混乱。方向图通过采用3个规模来衡量的：（1）线性（功率），（2）平方根（磁场强度），及（3）分贝（dB ）。该分贝的规模是最常用的，因为它揭示了更多的低层次的反应（旁瓣）。
图1 .1显示出了方向图许多特点的。半功率波束有时也只是被称为波束。第十功率和零的波束也被使用在一些应用中。这方向图是来自一个抛物反射面，该反射面的装置是移下来的轴。残余的瓣当第一副瓣变为加入到主波束时发生，并形成了一个肩膀。为了让一个装置坐落于抛物线的轴上，第一旁瓣都是平等的。
2.2 增益
增益是天线在一个特定的方向上的指导输入功率到辐射能力的一个衡量，也是衡量顶峰的辐射强度。通过各向同性天线的输入功率Po 在距离R 来考虑功率密度辐射：S = Po/4π2R 。一各向同性天线在所有的方向上的辐射是相同的，其辐射功率密度S 是由辐射功率除以一个球形的面积4π2R 得到的。各向同性散热器被认为是100％的效率。一个实际天线的增益增加了在顶峰辐射方向上的功率密度： 2
024E PG S R πη== or 014P G E S R ηηπ== (1.2)
增益是通过指导辐射远离其他部分的辐射范围来取得的。在一般情况下，增益定义为天线的增益偏颇的规模：
()()02,,4P G S R θφθφπ=
功率密度 ()()0,,4P G U θφθφπ
= 辐射强度 (1.3)
图1.1 天线方向图的特点 在辐射面积上的辐射强度的表面积分除以输入功率Po ，这是天线或天线效率的相对功率辐射的一个衡量： ()2000,sin 4r e G P d d P ππθφθθφηπ
==?? 效率 这里的Pr 是辐射功率。在天线上的物质损失或者由于弱阻抗匹配的反射功率减少辐射功率。在这本书，在设计的过程中上述方程中的积分和那些后续表达的概念多于我们执行的操作。只有现实世界中的理论简化，我们才能找到封闭形式的解决方案，该方案将要求实际一体化。我们通过用数值方法来解决大部分积分，该方法涉及把积分分成小片
段和做一加权和。不过，使用实测值的积分减少平均随机误差，提高了结果，这是有帮助的。
在一个系统中发送器的输出阻抗或接收器的输入阻抗可能与天线的输入阻抗不匹配。最大增益出现上一个接收器的阻抗共轭匹配天线，这意味着电阻部分是相同的和无功部分是相同的幅度，但有相反的迹象。增益测量的精度要求在天线和接收机之间的调谐器使得它俩共轭匹配。此外，错配的损失必须通过计算后的测量来消除。无论错配的影响对于一个给定系统来说是分开考虑，或天线测量到系统阻抗和错配的损失被认为是效率的部分。
例子 计算峰值功率密度在10公里长的输入功率为3瓦特的和增益为15分贝的天线。
先把dB 增益转换成一个比例：G = 151010 = 31.62。功率分布在 半径为10公里或面积为4π42(10)平方米的球面上。功率密度是：
()2
823(31.62)75.5/410W S nW m m π==?
我们使用方程式（1-2）来计算电场强度： ()()9
75.510376.75333/E S V m ημ-==?= 虽然增益通常是与各向同性天线有关的，一些天线的增益涉及到一个λ/2的各向同性增益为2.14分贝的偶极子。
如果我们把天线近似作为一个点源，我们通过使用方程式（1.2）来计算电场辐射：
()()0,,4jkR
P G e E R θφηθφπ
-= (1.4) 这仅要求天线相比径向距离R 要小.方程式（1.4）忽略了电场的方向，这我们界定为两极分化。电场的单位是伏特/米。我们通过用R 乘以方程式（1.4）及消除相位jkR e -来确定远区场的方向图，只有当提到在远区场中的另一点时，相位才是有意义的。远区场电场ff E 的单位是伏特：
()()0,,4ff PG E θφηθφπ
= 或 ()()2
014,,ff G E P πθφθφη??=???? (1.5) 在分析的过程中，我们往往把输入功率看成为1瓦特，就很容易地从电场中乘以一个常数4πη= 0.1826374来可以计算增益。
2.3有效面积
天线通过波来获得功率并传递一些到终端。给出入射波的功率密度和天线的有效面积，功率传送到终端是成果。
d eff P SA = (1.6)
为了一口径天线，如1角，抛物反射面，或平板阵列，有效面积是物理面积乘以口径效率。在一般的情况下，损耗归因于材料，分布和错配减少了有效面积到物理面积的比例。为抛物反射面估计的典型口径效率是55％。甚至有极小的物理面积的天线，如偶极子，有有效面积，因为他们通过的波消除了功率。
2.4路径损耗
我们把发射天线的增益与接收天线的有效面积结合起来，以确定传递的权力和路径损耗。在接收天线的功率密度是在方程式（1.3）中给出，以及接收功率是在方程式（1.6）中给出。把两者结合，我们就得出路径损耗： ()212,4d t
A G P P R θφπ= 天线1发射和天线2接收。如果天线中的材料是线性和各向同性的，发射和接收方向图是相同的（交互）[2，第116页]。当我们考虑到天线2作为发射天线和天线1作为接收天线，路径损耗是 ()122,4d t
AG P P R θφπ= 由于反应是交互的，路径的损失都是平等的，我们可以收集和消除条款：
1212
G G A A ==常数 由于天线是任意的，这系数必须等于一个常数。这个常数是通过考虑在两个大光圈[3]之间的辐射来得出的：
2
4G A πλ= (1.7) 我们把这个方程代入到路径损耗，以表达它在增益或有效面积中的条件：
2
1212224d t P A A G G P R R
λπλ??== ??? (1.8) 我们对路径损耗作出快速评价，以距离R 的各种单位和在公式中使用百兆的频率f 。
路径损耗(dB )=()()()1220log U K fR G dB G dB +-- (1.9)
其中Ku 取决于长度单位：
例子 计算直径为3米，频率为4 GHz ，假设口径效率为55％的抛物反射面的增益。
增益是在方程式（1.7）中与有效面积有关： 24A G πλ
= 我们通过()2
/2A D π=来计算一圆孔的面积。结合这些方程，我们有： 22a a D Df G c ππηηλ????== ? ?????
(1.10) 其中D 是直径和a η是口径效率。把值代入上式，我们得到的增益为：
()()29934100.5586850.310G π?????==????? (39.4dB )
例子 计算一频率为2.2 GHz ，发射天线增益为25dB ，接收天线增益为20dB 的50公里的通讯线路的路径损耗。
路径损耗= 32.45 + 20 log[2200(50)] ? 25 ? 20 = 88.3dB