# Artículo de Friis/Texto completo

Artículo de Friis
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## Introducción

Una fórmula de transmisión una fórmula de transmisión simple para un circuito de radio es encontrada. Se hace hincapié en la utilidad de la fórmula y se discuten sus limitaciones

Ecuación 1
${\displaystyle {\frac {P_{r}}{P_{t}}}={\frac {A_{r}A_{t}}{d^{2}\lambda ^{2}}}}$
Pt =Energía de alimentación de la antena transmisora en sus terminales de entrada
Pr =Potencia disponible en los terminales de salida de la antena receptora
Ar =Área efectiva de la antena receptora
At =Área efectiva de la antena transmisora
d =Distancia entre las antenas
${\displaystyle \lambda }$ =Longitud de onda

Área efectiva

El área efectiva de cualquier antena ya sea emisora o receptora, se define para la condición en la que la antena se usa para recibir una onda electromagnética plana polarizada linealmente. El área efectiva se define como la relación entre la potencia recibida y la densidad de potencia incidente en una antena. (Ítem A )(formula 2 y 3)

Área efectiva descripción

El área efectiva está definida por:

${\displaystyle A_{eff.}={\frac {P_{r}}{P_{0}}}}$

Potencia de salida disponible:

${\displaystyle P_{r}=P_{0}A_{eff.}}$
${\displaystyle (R_{rad}=80\pi ^{2}a^{2}/\lambda ^{2})}$

Área efectiva del dipolo:

${\displaystyle A_{dip}}$=${\displaystyle {\frac {P_{r}}{P_{0}}}={\frac {3\lambda ^{2}}{8\pi }}=0.1193\lambda ^{2}}$[0.3cm] ${\displaystyle P_{o}=E^{2}/120\pi }$
${\displaystyle Pr={\frac {E^{2}a^{2}}{4(80\pi ^{2}a^{2}/\lambda ^{2})}}}$ = ${\displaystyle {\frac {\lambda ^{2}E^{2}a^{2}}{320\pi a^{2}}}}$= ${\displaystyle {\frac {\lambda ^{2}E^{2}}{320\pi ^{2}}}}$
${\displaystyle A_{dip}={\frac {\frac {\lambda ^{2}E^{2}}{320\pi ^{2}}}{\frac {E^{2}}{120\pi }}}={\frac {120\pi \lambda ^{2}E^{2}}{320\pi ^{2}E^{2}}}={\frac {3\lambda ^{2}}{8\pi }}}$

Área efectiva de una circunferencia

Área efectiva de un rectángulo

Área efectiva de un rectángulo

${\displaystyle A=0.3454\lambda *0.3612\lambda =0.1247\lambda ^{2}}$

(Ítem B)

Área efectiva Reflector Parabólico

Área reflector parabólico

${\displaystyle A=\pi r^{2}}$
${\displaystyle A_{reflector}=(\pi r^{2})*{\frac {2}{3}}}$

Ejemplo

se tiene un reflector parabólico con un diámetro de 30 mts, calcule el área del reflector y su respectiva área efectiva.
${\displaystyle A=\pi *(15mts)^{2}}$
${\displaystyle A=706.5mts^{2}}$
El área efectiva es igual a:
${\displaystyle A=706.5mts^{2}}$
${\displaystyle A=466.29mts^{2}}$
Radio igual:
${\displaystyle r={\sqrt {\frac {A}{\pi }}}}$
${\displaystyle r={\sqrt {\frac {466.29}{\pi }}}=12.1829}$

(Ítem D)

Reflector parabólico

Antena isotrópica área efectiva

${\displaystyle A={\frac {\alpha ^{2}}{4\pi }}}$

El área efectiva de una antena isotrópica la podemos tomar como si estuviéramos hablando de una antena dipolo pequeña.
Un dipolo pequeño equivale a ${\displaystyle L={\frac {\alpha }{10}}}$

ANTENAS DE BOCINA:
Las antenas de bocina fueron desarrolladas a finales del siglo XIX, para ser usadas en aplicaciones de alta frecuencia tales como las transmisiones de microondas. El primer registro de una antena de bocina Data del año 1897.
Una de las ventajas más significativas de este tipo de antenas son la ganancia el patrón de radiación y la impedancia.
Dentro de los principales tipos de antenas de bocina encontramos la de plano E plano H y piramidal.
La antena de bocina tipo E equivale a una guía de onda cuadrada. La antena de bocina tipo H equivale a una guía de onda rectangular. La antena de bocina tipo piramidal es equivalente a una guía de onda rectangular, pero con la diferencia de que inicia En punta hasta abrirse.

Antenas tipo bocina

Antena de bocinas con polarización circular
Una manera de obtener una polarización circular, es el uso de un dispositivo conocido como polarizador circular, este estos polarizadores se usan para lograr convertir señales polarizadas linealmente que se introducen la entrada de la antena, en señales con polarización circular a la salida de la misma antena.

Polarización

DE FRIIS A RAPPAPORT Y DE RAPPAPORT A FRIIS
Si se considera la antena transmisora como isotrópica, la densidad de potencia en la antena receptora: ${\displaystyle \omega ={\frac {P_{t}}{A}}={\frac {P_{t}}{4\pi d^{2}}}}$

de ganancia de la antena transmirsora es Gt. Luego, la densidad de potencia ${\displaystyle \omega ={\frac {P_{t}G_{t}}{4\pi d^{2}}}}$

En condiciones ideales, el área efectiva de la antena receptora es Ar. Luego, la potencia recibida ${\displaystyle P_{r}=\omega A_{r}=({\frac {P_{t}G_{t}}{4\pi d^{2}}})A_{r}}$

La ganancia de la antena transmisora es ${\displaystyle G_{t}={\frac {4\pi A_{t}}{\lambda ^{2}}}}$

Entonces,

${\displaystyle P_{r}=P_{t}({\frac {4\pi A_{t}}{4\pi d^{2}\lambda ^{2}}})A_{r}}$

${\displaystyle {\frac {P_{r}}{P_{t}}}=({\frac {G_{t}G_{r}\lambda ^{2}}{(4\pi )^{2}d^{2}}})}$

La ganancia de la antena receptora es: ${\displaystyle G_{r}={\frac {4\pi A_{r}}{(\lambda ^{2}}}}$ ${\displaystyle {\frac {P_{r}}{P_{t}}}=({\frac {G_{t}G_{r}\lambda ^{2}}{(4\pi )^{2}d^{2}}})}$ ${\displaystyle P_{r}=P_{t}({\frac {A_{t}A_{r}\lambda ^{2}}{(4\pi )^{2}d^{2}}})}$ ${\displaystyle G={\frac {4\pi A_{\varrho }}{\lambda ^{2}}}}$

${\displaystyle P_{r}=P_{t}({\frac {({\frac {4\pi A_{\varrho }}{\lambda ^{2}}})\lambda ^{2}}{(4\pi )^{2}d^{2}}})}$

${\displaystyle P_{r}=P_{t}({\frac {4\pi A_{\varrho }}{(4\pi )^{2}d^{2}A_{i}so}})}$

${\displaystyle P_{r}=P_{t}({\frac {4\pi A_{\varrho }}{(4\pi )^{2}d^{2}({\frac {\lambda ^{2}}{4\pi }})}})}$

${\displaystyle P_{r}=P_{t}({\frac {A_{t}A_{r}}{d^{2}\lambda ^{2}}})}$

${\displaystyle {\frac {P_{r}}{P_{t}}}=({\frac {A_{t}A_{r}}{d^{2}\lambda ^{2}}})}$

Comparación de la ecuación 3.1 con la ecuación de friis
Ecuación de Rappaport en función de las ganancias: ${\displaystyle P_{r}(d)={\frac {P_{t}G_{t}G_{r}\lambda ^{2}}{(4\pi )^{2}d^{2}L}}}$ ${\displaystyle P_{r}(d):}$ Potencia recibida en función de la separación entre T-R.
${\displaystyle P_{t}:}$ Potencia transmitida.
${\displaystyle G_{t}:}$ Ganancia de la antena transmisora.
${\displaystyle G_{r}:}$ Ganancia de la antena receptora.
${\displaystyle \lambda :}$Longitud de onda en metros.
d:Distancia de separación T-R en metros.
L:Factor de pérdida del sistema no relacionado con la propagación (L>=1). Cuando L=1, significa que no hay pérdidas en el sistema de los componentes físicos.
Utilidad de la ecuación de friis.

La ecuación relaciona las potencias y permite calcular el balance de potencias.
Su utilidad se ve en el campo lejano.
No es útil para otro medio de propagación diferente al espacio libre.
Solo es útil en sistemas ideales que no presentan obstáculos y no tiene en cuenta perdidas.
Antiguamente, solo se podía estimar el rango de potencia del transmisor, pero no determinaba la potencia recibida.
Es útil con medidas de unidad en potencia por unidad de área.
Es útil en longitudes de onda largas.
Ecuación de Friis en función de las áreas efectivas: ${\displaystyle {\frac {P_{r}}{P_{t}}}={\frac {A_{r}A_{t}}{d^{2}\lambda ^{2}}}}$

Area efectiva: ${\displaystyle A={\frac {\lambda ^{2}G}{4\pi }}}$
${\displaystyle P_{r}:}$ Potencia de alimentación de la antena transmisora en la entrada.
${\displaystyle P_{t}:}$Potencia disponible en la terminal de salida de la antena receptora.
${\displaystyle A_{r}:}$Área efectiva de la antena receptora.
${\displaystyle A_{t}:}$Area efectiva de la antena transmisora.
${\displaystyle d:}$Distancia entre antenas.
${\displaystyle \lambda :}$ Longitud de onda.

# Introduction

Abstract—A simple transmission formula for a radio circuit is derived. The utility of the formula is emphasized and its limitations are discussed.

This note emphasizes the utility of the following simple transmission formula for a radio circuit made up of a transmitting antenna and a receiving antenna in free space:
${\displaystyle {\frac {P_{r}}{P_{t}}}={\frac {A_{r}A_{t}}{d^{2}\lambda ^{2}}}}$
where
Pt= power fed into the transmitting antenna at its input terminals.
Pr=power available at the output terminals of the receiving antenna.
Ar=effective area of the receiving antenna.
At=effective area of the transmitting antenna
d=distance between antennas.
${\displaystyle \lambda }$= wavelength
The effective areas appearing in (1) are discussed in the next section and this is followed by a derivation of the formula and a discussion of its limitations.

EFFECTIVE AREAS
The effective area of any antenna, whether transmitting or receiving, is defined for the condition in which the antenna is used to receive a linearly polarized, plane electromagnetic wave. The author suggests the adoption of the following definition:

${\displaystyle A_{eff.}={\frac {P_{r}}{P_{0}}}}$
or
${\displaystyle P_{r}=P_{0}A_{eff.}}$
Where Pr is the received power as defined above and Po is the power flow per unit area of the incident field at the antenna. In words, (3) states that the received power is equal to the power flow through an area that is equal to the effective area of the antenna. Note that the definition does not impose the condition of no heat loss in the antenna. Equation (3) shows that the effective area of an antenna is proportional to its power gain. The effective areas of antennas of special interest are given in the following:

Mobile Radio Propagation: Large-Scale Path Loss
For a small uniform current element the available output power is equal to the induced voltage squared divided by four times the radiation resistance. Thus

${\displaystyle P_{r}={\frac {E^{2}a^{2}}{4R_{rad}}}}$

where
E=effective value of the electric field of the wave.
a=length of the current element.
Rrad. = radiation resistance of the current element
${\displaystyle R_{rad}=80\pi ^{2}a^{2}/\lambda ^{2}}$ Since the power flow per unit area is equal to the electric field squared divided by the impedance of free space,
i.e., ${\displaystyle P_{o}=E^{2}/120\pi }$, we have

${\displaystyle A_{dip}={\frac {P_{r}}{P_{0}}}={\frac {3\lambda ^{2}}{8\pi }}=0.1193\lambda ^{2}}$

The effective area of a half-wavelength dipole with no heat loss is only 9.4 per cent, 0.39 decibels,2 larger than the effective area of the small dipole. Therefore

${\displaystyle A_{0.5\lambda }=0.1305\lambda ^{2}}$

The area of a rectangle with one-half wavelength and one-quarter wavelength sides is ${\displaystyle 0.125\lambda ^{2}}$ and it is, therefore, a good approximation for the effective areas of small dipoles and half-wavelength dipoles.


${\displaystyle A_{0.5}=0.1305\lambda ^{2}}$
${\displaystyle 0.1305\lambda ^{2}=\pi r^{2}}$
${\displaystyle {\frac {0.1305}{\pi }}\lambda ^{2}=r^{2}}$
${\displaystyle 0.04154\lambda ^{2}=r^{2}}$
${\displaystyle {\sqrt {0.04152\lambda ^{2}}}={\sqrt {r^{2}}}}$
${\displaystyle r=0.2038\lambda }$

B. Isotropic Antenna with No Heat Loss
The hypothetical isotropic antenna has the same radiation intensity in all directions. It has two thirds of the gain’ or effective area of the small dipole. Therefore
${\displaystyle A_{isotr.}={\frac {\lambda ^{2}}{4\pi }}.}$ .

(Ítem C Arrays )

Dimensiones en Lambda
Vista Frontal
Orden de los dipolos

${\displaystyle A_{pinetree}=n*0.5\lambda *0.5\lambda }$

C. Broadside Arrays (Pine- Tree Antennas)
The effective area of an antenna array made up of a curtain of rows of half-wave dipoles spaced half a wavelength was calculated several years ago by the method of Pistolkors. Equal amplitude and phase of the currents in all the dipoles and no heat loss were assumed. The effective area of such an array with a reflector that doubled the gain was found to be approximately equal to the actual area occupied by the array; thus
${\displaystyle A_{pire-tree}\approx n\times 0.5\lambda \times 0.5\lambda }$
where n is the total number of half-wave dipoles in the front curtain. Formula (7) is a good approximation for large antennas. For example, an antenna of 6 rows of 17 dipoles each gave a calculated effective area only 3 per cent below the value obtained by (7). It should be pointed out that the heat loss in the connecting transmission lines will reduce the effective areas in actual antennas.

n=102 dipolos Ares(pineetree) = n ∗ 0, 5λ ∗ 0, 5λ

${\displaystyle Apt=102*0.5\lambda *0.5\lambda ={\frac {51\lambda ^{2}}{2}}}$
${\displaystyle A_{circulo}=\pi r^{2}----A_{pinetree}=25.5\lambda ^{2}}$

Radio

${\displaystyle \pi r^{2}=25.5\lambda }$
${\displaystyle r^{2}={\frac {25.5\lambda }{\pi }}}$
${\displaystyle r^{2}=8.117\lambda ^{2}}$
${\displaystyle r={\sqrt {8.117\lambda ^{2}}}}$
${\displaystyle r=2.849\lambda }$

Dipolo

.
D. Parabolic Reflectors
The effective area of the parabolic type of antenna with a proper feed has been found experimentally to be approximately two thirds of the projected area of the reflector.
.

___________________________________________________________________________________

(Ítem E)

E. Electric Horns-Aperture Sides ${\displaystyle >>\lambda }$
The effective area of a very long horn with small aperture dimensions is 81 per cent of the area of the aperture. For an optimum horn, where the aperture is dimensioned to give maximum gain for a given length of the horn, the effective area is approximately 50 per cent of the area of the aperture.
.

DERIVATION OF TRANSMISSION FORMULA (1)
Having defined the effective area of an antenna, it is a simple matter to derive (1). As shown in Fig. 1, consider a radio circuit made up of an isotropic transmitting

antenna and a receiving antenna with effective area Ar. The power flow per unit area at the distance d from the transmitter is
${\displaystyle P_{0}={\frac {P_{t}}{4\pi d^{2}}}.}$
Assuming a plane wave front at the distance d, definition (2) for the effective area and formula (8) give
${\displaystyle {\frac {P_{r}}{P_{t}}}={\frac {A_{r}}{4\pi d^{2}}}}$
Replacing the isotropic transmitting antenna in the illustration with a transmitting antenna with effective area At will increase the received power by the ratio ${\displaystyle At/A_{i}sotr}$, and we obtain
${\displaystyle {\frac {P_{r}}{P_{t}}}={\frac {A_{r}}{4\pi d^{2}A_{isotr}}}}$
Introducing the effective area (6) for the isotropic antenna, we have (1).

LIMITATIONS OF TRANSMISSION FORMULA (1)
In deriving (1), a plane wave front was assumed at the distance d. Formula (1), therefore, should not be used when d is small. W. D. Lewis, of these Laboratories, has made a theoretical study of transmission between large antennas of equal areas with plane phase fronts at their apertures and he finds that (1) is correct to within a few per cent when
${\displaystyle d\geq {\frac {2a^{2}}{\lambda }}}$
where a is the largest linear dimension of either of the antennas.
Formula (1) applies to free space only, a condition which designers of microwave circuits seek to approximate. Application of the formula to other conditions may require corrections for the effect of the “ground,” and for absorption in the transmission medium, which are beyond the scope of this note.
he advantage of (1) over other formulations is that, fortunately, it has no numerical coefficients. It is so simple that it may be memorized easily. Almost 7 years of intensive use has proved its utility in transmission calculations involving wavelengths up to several meters, and it may become useful also at longer wavelengths. It is suggested that radio engineers hereafter give the radiation from a transmitting antenna in terms of the power flow per unit area which is equal to ${\displaystyle P_{t}A_{t}/\lambda ^{2}d^{2}}$, instead of giving the field strength in volts per meter. It is also suggested that an antenna be characterized by its effective area, instead of by its power gain or radiation resistance. The ratio of the effective area to the actual area of the aperture of an antenna is also of importance in antenna design, since it gives an indication of how efficiently the antenna is utilizing the physical space it occupies. *The directional pattern, which has not been discussed in this note,is, of course, always an important characteristic of an antenna.

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