# 9.6 Friis Transmission Formula

The two antennas shown in Fig. 9-19 are part of a free-space communication link with the separation between the antennas, R, being large enough for each to be in the far-field region of the other. The transmitting and receiving antennas have effective areas A t and A r and radiation efficiencies ξ t and ξ r, respectively. Our objective is to find a relationship between P t, which is the power supplied to the transmitting antenna, and P rec, which is the power delivered to the receiver. As always, we assume that both antennas are impedance-matched to their respective transmission lines. Initially, we consider the case where the two antennas are oriented so that the peak of the radiation pattern of each antenna points in the direction of the other.

# Figure 9-19 Transmitter–receiver configuration.

We start by treating the transmitting antenna as a lossless isotropic radiator. The power density incident upon the receiving antenna at a distance R from an isotropic transmitting antenna is simply equal to the transmitter power P t divided by the surface area of a sphere of radius R:

S iso=P t 4 π R 2.(9.65)

The real transmitting antenna is neither lossless nor isotropic. Hence, the power density S r due to the real antenna is

S r=G t S iso=ξ t D t S iso=ξ t D t P t 4 π R 2.(9.66)

Through the gain G t=ξ t D t, ξ t accounts for the fact that only part of the power P t supplied to the antenna is radiated out into space, and D t accounts for the directivity of the transmitting antenna (in the direction of the receiving antenna). Moreover, by Eq. (9.64), D t is related to A t by D t=4 π A t/λ 2. Hence, Eq. (9.66) becomes

S r=ξ t A t P t λ 2 R 2.(9.67)

On the receiving-antenna side, the power intercepted by the receiving antenna is equal to the product of the incident power density S r and the effective area A r:

P int=S r A r=ξ t A t A r P t λ 2 R 2.(9.68)

The power delivered to the receiver, P rec, is equal to the intercepted power P int multiplied by the radiation efficiency of the receiving antenna, ξ r. Hence, P rec=ξ r P int, which leads to the result

If the two antennas are not oriented in the direction of maximum power transfer, Eq. (9.69) assumes the general form

P rec P t=G t G r λ 4 π R 2 F t(θ t,ϕ t)F r(θ r,ϕ r),(9.70)

where F t(θ t,ϕ t) is the normalized radiation intensity of the transmitting antenna at angles (θ t,ϕ t) corresponding to the direction of the receiving antenna (as seen by the antenna pattern of the transmitting antenna), and a similar definition applies to F r(θ r,ϕ r) for the receiving antenna.

# Example 9-4: Satellite Communication System

A 6 GHz direct-broadcast TV satellite system transmits 100 W through a 2 m diameter parabolic dish antenna from a distance of approximately 40,000 km above Earth’s surface. Each TV channel occupies a bandwidth of 5 MHz. Due to electromagnetic noise picked up by the antenna as well as noise generated by the receiver electronics, a home TV receiver has a noise level given by

P n=K T sys B(W),(9.71)

where T sys [measured in kelvins (K)] is a figure of merit called the system noise temperature that characterizes the noise performance of the receiver–antenna combination, K is Boltzmann’s constant [1.38×10−23 (J/K)], and B is the receiver bandwidth in Hz.

The signal-to-noise ratio S n (which should not be confused with the power density S) is defined as the ratio of P rec to P n:

S n=P rec/P n(dimensionless).(9.72)

For a receiver with T sys=580 K, what minimum diameter of a parabolic dish receiving antenna is required for high-quality TV reception with S n=40 dB? The satellite and ground receiving antennas may be assumed lossless, and their effective areas may be assumed equal to their physical apertures.

Solution: The following quantities are given:

P t=100 W,f=6 GHz=6×10 9 Hz,S n=10 4,transmit antenna diameter d t=2 m,T sys=580 K,R=40,000 km=4×10 7 m,B=5 MHz=5×10 6 Hz.

The wavelength λ=c/f=5×10−2 m, and the area of the transmitting satellite antenna is A t=(π d t 2/4)=π (m 2). From Eq. (9.71), the receiver noise power is

P n=K T sys B=1.38×10−23×580×5×10 6=4×10−14 W.

Using Eq. (9.69) with ξ t=ξ r=1,

P rec=P t A t A r λ 2 R 2=100 π A r(5×10−2)2(4×10 7)2=7.85×10−11 A r.

The area of the receiving antenna, A r, can now be determined by equating the ratio P rec/P n to S n=10 4:

10 4=7.85×10−11 A r 4×10−14,

which yields the value A r=5.1 m 2. The required minimum diameter is d r=4 A r/π=2.55 m.