中北大学分校学位论文英文翻译
10 2
99
34100.5586850.310G
(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
212122221dtPAAAAfBfPRRc
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.
1.5 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 TR by using Eq. (1.2):
2
,4TTinc
T
PGSR
(1.11)
The target’s radar cross section (RCS), the scattering area of the object, is expressed in square
中北大学分校学位论文英文翻译
11 meters or dB2m: 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,,,4reflected
srriiTTTpowerPRCSpowerdensityincidentPGR
(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 RR from the target. Finally, the receiving antenna collects the power density with an effective areaRA. We combine these ideas to obtain the power delivered to the receiver:
2
2,,,44RTTrriirecRRT
R
APGPSARR
We apply Eq. (1.7) to eliminate the effective area of the receiving antenna and gather terms to determine the bistatic radar range equation: 2
322,,,4TRrriirecTTR
GGPPRR
(1.13) We reduce Eq. (1.13) and collect terms for monostatic radar, where the same antenna is used for both transmitting and receiving:
22344recTPGPR
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 areaRA: 2
,4TTCRT
PGPAR
The power scattered by the plate is the power collected,CP, times the gain of the plate as an antenna,PG:
中北大学分校学位论文英文翻译
12 2
,,4TTii
sCPRPrr
T
PGPPGAGR
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:
2
22
44TTR
sTPGAPR
We determine the RCS σ by Eq. (1.12), the scattered power divided by the incident power den
sity: 22
22
,,444RiiRrrsR
TTT
GGPAPGR
(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.
1.6 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.
中文译文:
中北大学分校学位论文英文翻译
18 0,,4ffPGE 或 2 014,,ffGEP (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4= 0.1826374来可以计算增益。
1.3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。 deffPSA (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。
1.4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
AGPPR 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 122,4dt
AGPPR
中北大学分校学位论文英文翻译
19 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:
12
12 GGAA =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:
24GA
(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
121222
4dtPAAGGPRR
(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)=1220logUKfRGdBGdB (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
2
/2AD
22
aaDDfGc
(1.10)
其中D 是直径和
a
中北大学分校学位论文英文翻译
20 2
99
34100.5586850.310G
(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
在两个小孔之间当频率在增加时传输会发生什么变化?如果我们假定有效面积依然保持为常数不变,作为在一个抛物反射面上,传输的增加是频率的平方:
2
212122221dtPAAAAfBfPRRc
其中B 在一个固定的范围内是一个常数。无论频率是多少,接收孔径捕获同样的功率,但发射天线增益的增加是频率的平方。因此,接收功率的增加也是频率的平方。只有当频率变化时其增益是一个固定值的天线,它的路径损耗的增加是频率的平方。
1.5雷达距离方程和截面
雷达使用双路径损耗来运作。雷达发射天线辐射一个照亮一目标的场。这些事件场激发表面电流,也辐射产生出第二个场。这些场传播到接收天线,他们在那里收集。大部分雷达使用相同的天线来传递场并收集返回的信号,称为单系统,而我们在双基地雷达中使用单独的天线。在一个双系统中无法侦测到接收系统,因为它不传输和在军事上的应用具有更大的生存能力。
我们在一个TR 范围内通过使用方程式(1.2)来确定有启发性目标的功率密度:
2
,4TTincT
PGSR
(1.11) 目标的雷达散射截面(RCS ),对象的散射面积是表示在平方米或dB2 m :10 log(平方米)。RCS 取决于事件和反射波的方向。我们乘以通过目标与它的由于电流诱导的有效天线的增益接的收方向来收集的功率:
中北大学分校学位论文英文翻译
21 RCS/
2
,,,4srriiTTT
PPGR
(1.12) 在一个通信系统,我们叫Ps 为等效各向同性辐射功率(EIRP ),这相当于输入功率和天线增益的积。目标成为传递的来源和我们应用方程式(1.2)来找到在接收天线上在一个从目标的RR 范围内的功率密度。最后,接收天线收集功率密度与有效面积RA 。我们把这些想法结合起来,以获取传递给接收器的功率:
22 ,,,44RTTrriirecRRTRAPGPSARR 我们应用方程式(1.7)来消除接收天线的有效面积和收集的条件来确定双基地雷达距离方程:
2
322,,,4TRrriirecT TRGGPPRR (1.13) 我们减少方程式(1.13),并为单雷达收集条件,这里的同一天线是用于发射和接收的:
22344recTPGPR
雷达的接收功率是成正比与的1/4R和2G 的。
我们通过考虑板作为一有有效面积的天线来找到一平板的近似RCS 。方程式(1.11)在板上给出了功率密度事件,该板在一面积RA 上收集这种功率:
2,4TTCRTPGPAR
功率由板分散是收集的功率,CP 乘以作为天线的板的增益PG : 2
,,4TTiisCPRPrrT PGPPGAGR 这个分散的功率是在一个特定的方向的有效辐射功率,这在天线是输入功率和在某一特定方向上增益的积。我们通过使用有效面积来计算板的增益,并找到就面积而言的分散的功率:
2
22
44TTR
sTPGAPR
中北大学分校学位论文英文翻译
22 我们由方程式(1.12)来确定RCS σ,分散的权力除以事件的功率密度:
22
22
,,444RiiRrrsR TTTGGPAPGR (1.14) 为了双站散射,方程式(1.14)的正确数式把增益划分为两件,在这分散的方向是不同于这次事件的方向的。单散射使用相同的事件和反射的方向。我们可以把任何对象代替为平板和使用一个有效的面积的观点及其相关的天线增益。 天线是一个有着独特RCS 特征的对象,因为部分的接收到的功率将传递给天线终端。如果我们提供了一个良好的阻抗匹配到这个信号,它不会重新辐射和RCS 是减少的。当我们从一个任意方向照亮天线,一些事件的功率密度将由结构分散和不会传递给天线终端。这导致天线RCS 在重新辐射信号的天线模式上分工所造成的终端错配和结构模式,为了事件的功率密度不传递给终端场反映了结构。
1.6为什么要使用一个天线?
当没有其他办法是可行的时候,我们使用天线来传输信号,如与导弹或在崎岖的山区地形上的通信。电缆昂贵,而且需要很长的时间来安装。当我们在水平地面上会使用天线的时候就这些?天线系统的大的路径损失会令我们相信,电缆运行更好
中北大学分校学位论文英文翻译
10 2
99
34100.5586850.310G
(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
212122221dtPAAAAfBfPRRc
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.
1.5 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 TR by using Eq. (1.2):
2
,4TTinc
T
PGSR
(1.11)
The target’s radar cross section (RCS), the scattering area of the object, is expressed in square
中北大学分校学位论文英文翻译
11 meters or dB2m: 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,,,4reflected
srriiTTTpowerPRCSpowerdensityincidentPGR
(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 RR from the target. Finally, the receiving antenna collects the power density with an effective areaRA. We combine these ideas to obtain the power delivered to the receiver:
2
2,,,44RTTrriirecRRT
R
APGPSARR
We apply Eq. (1.7) to eliminate the effective area of the receiving antenna and gather terms to determine the bistatic radar range equation: 2
322,,,4TRrriirecTTR
GGPPRR
(1.13) We reduce Eq. (1.13) and collect terms for monostatic radar, where the same antenna is used for both transmitting and receiving:
22344recTPGPR
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 areaRA: 2
,4TTCRT
PGPAR
The power scattered by the plate is the power collected,CP, times the gain of the plate as an antenna,PG:
中北大学分校学位论文英文翻译
12 2
,,4TTii
sCPRPrr
T
PGPPGAGR
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:
2
22
44TTR
sTPGAPR
We determine the RCS σ by Eq. (1.12), the scattered power divided by the incident power den
sity: 22
22
,,444RiiRrrsR
TTT
GGPAPGR
(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.
1.6 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.
中文译文:
中北大学分校学位论文英文翻译
18 0,,4ffPGE 或 2 014,,ffGEP (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4= 0.1826374来可以计算增益。
1.3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。 deffPSA (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。
1.4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
AGPPR 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 122,4dt
AGPPR
中北大学分校学位论文英文翻译
19 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:
12
12 GGAA =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:
24GA
(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
121222
4dtPAAGGPRR
(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)=1220logUKfRGdBGdB (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
2
/2AD
22
aaDDfGc
(1.10)
其中D 是直径和
a
中北大学分校学位论文英文翻译
20 2
99
34100.5586850.310G
(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
在两个小孔之间当频率在增加时传输会发生什么变化?如果我们假定有效面积依然保持为常数不变,作为在一个抛物反射面上,传输的增加是频率的平方:
2
212122221dtPAAAAfBfPRRc
其中B 在一个固定的范围内是一个常数。无论频率是多少,接收孔径捕获同样的功率,但发射天线增益的增加是频率的平方。因此,接收功率的增加也是频率的平方。只有当频率变化时其增益是一个固定值的天线,它的路径损耗的增加是频率的平方。
1.5雷达距离方程和截面
雷达使用双路径损耗来运作。雷达发射天线辐射一个照亮一目标的场。这些事件场激发表面电流,也辐射产生出第二个场。这些场传播到接收天线,他们在那里收集。大部分雷达使用相同的天线来传递场并收集返回的信号,称为单系统,而我们在双基地雷达中使用单独的天线。在一个双系统中无法侦测到接收系统,因为它不传输和在军事上的应用具有更大的生存能力。
我们在一个TR 范围内通过使用方程式(1.2)来确定有启发性目标的功率密度:
2
,4TTincT
PGSR
(1.11) 目标的雷达散射截面(RCS ),对象的散射面积是表示在平方米或dB2 m :10 log(平方米)。RCS 取决于事件和反射波的方向。我们乘以通过目标与它的由于电流诱导的有效天线的增益接的收方向来收集的功率:
中北大学分校学位论文英文翻译
21 RCS/
2
,,,4srriiTTT
PPGR
(1.12) 在一个通信系统,我们叫Ps 为等效各向同性辐射功率(EIRP ),这相当于输入功率和天线增益的积。目标成为传递的来源和我们应用方程式(1.2)来找到在接收天线上在一个从目标的RR 范围内的功率密度。最后,接收天线收集功率密度与有效面积RA 。我们把这些想法结合起来,以获取传递给接收器的功率:
22 ,,,44RTTrriirecRRTRAPGPSARR 我们应用方程式(1.7)来消除接收天线的有效面积和收集的条件来确定双基地雷达距离方程:
2
322,,,4TRrriirecT TRGGPPRR (1.13) 我们减少方程式(1.13),并为单雷达收集条件,这里的同一天线是用于发射和接收的:
22344recTPGPR
雷达的接收功率是成正比与的1/4R和2G 的。
我们通过考虑板作为一有有效面积的天线来找到一平板的近似RCS 。方程式(1.11)在板上给出了功率密度事件,该板在一面积RA 上收集这种功率:
2,4TTCRTPGPAR
功率由板分散是收集的功率,CP 乘以作为天线的板的增益PG : 2
,,4TTiisCPRPrrT PGPPGAGR 这个分散的功率是在一个特定的方向的有效辐射功率,这在天线是输入功率和在某一特定方向上增益的积。我们通过使用有效面积来计算板的增益,并找到就面积而言的分散的功率:
2
22
44TTR
sTPGAPR
中北大学分校学位论文英文翻译
22 我们由方程式(1.12)来确定RCS σ,分散的权力除以事件的功率密度:
22
22
,,444RiiRrrsR TTTGGPAPGR (1.14) 为了双站散射,方程式(1.14)的正确数式把增益划分为两件,在这分散的方向是不同于这次事件的方向的。单散射使用相同的事件和反射的方向。我们可以把任何对象代替为平板和使用一个有效的面积的观点及其相关的天线增益。 天线是一个有着独特RCS 特征的对象,因为部分的接收到的功率将传递给天线终端。如果我们提供了一个良好的阻抗匹配到这个信号,它不会重新辐射和RCS 是减少的。当我们从一个任意方向照亮天线,一些事件的功率密度将由结构分散和不会传递给天线终端。这导致天线RCS 在重新辐射信号的天线模式上分工所造成的终端错配和结构模式,为了事件的功率密度不传递给终端场反映了结构。
1.6为什么要使用一个天线?
当没有其他办法是可行的时候,我们使用天线来传输信号,如与导弹或在崎岖的山区地形上的通信。电缆昂贵,而且需要很长的时间来安装。当我们在水平地面上会使用天线的时候就这些?天线系统的大的路径损失会令我们相信,电缆运行更好