Radar operates using a double path loss. The radar transmitting antenna radiates a fieldthat illuminates a target. These incident fields excite surface currents that also radiateto produce a second field. These fields propagate to the receiving antenna, where theyare collected. Most radars use the same antenna both to transmit the field and to collectthe signal returned, called a monostatic system, whereas we use separate antennas forbistatic radar. The receiving system cannot be detected in a bistatic system because itdoes not transmit and has greater survivability in a military application.

S = = or |E| = = ----(1)

We determine the power density illuminating the target at a range RT by usingEq. (1):

S_inc = ----(2)

The target’s radar cross section (RCS), the scattering area of the object, is expressed insquare meters or dBm2: 10 log(square meters). The RCS depends on both the incidentand reflected wave directions. We multiply the power collected by the target with itsreceiving pattern by the gain of the effective antenna due to the currents induced:

RCS = σ = = Ps(θr, φr, θi, φi) ----(3)

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 becomesthe transmitting source and we apply Eq. (1) to find the power density at the receivingantenna at a range RR from the target. Finally, the receiving antenna collects the powerdensity with an effective area AR.Wecombine these ideas to obtain the power deliveredto the receiver:

P_rec = S_RA_R =

We apply Eq. (4):

=

to eliminate the effective area of the receiving antenna and gatherterms to determine the bistatic radar range equation:

= (5)

We reduce Eq. (5) and collect terms for monostatic radar, where the same antennais used for both transmitting and receiving:

₌

Radar received power is proportional to 1/R⁴ and to G².

We find the approximate RCS of a flat plate by considering the plate as an antennawith an effective area. Equation (2) gives the power density incident on the platethat collects this power over an area A_R:

P_C = A_R

The power scattered by the plate is the power collected, P_C, times the gain of the plateas an antenna, G_P :

P_S = P_CG_P =  A_RG_P ( , )

This scattered power is the effective radiated power in a particular direction, whichin 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 interms of area:

P_S=

We determine the RCS σ by Eq. (3), the scattered power divided by the incidentpower density:

σ = = = (6)

The right expression of Eq. (6) divides the gain into two pieces for bistatic scattering,where the scattered direction is different from the incident direction. Monostaticscattering uses the same incident and reflected directions. We can substitute any objectfor 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 powerreceived will be delivered to the antenna terminals. If we provide a good impedancematch to this signal, it will not reradiate and the RCS is reduced. When we illuminatean antenna from an arbitrary direction, some of the incident power density will bescattered by the structure and not delivered to the antenna terminals. This leads tothe division of antenna RCS into the antenna mode of reradiated signals caused byterminal mismatch and the structural mode, the fields reflected off the structure forincident power density not delivered to the terminals.

Why Use An Antenna?

We use antennas to transfer signals when no other way is possible, such as communicationwith a missile or over rugged mountain terrain. Cables are expensive andtake a long time to install. Are there times when we would use antennas over levelground? The large path losses of antenna systems lead us to believe that cable runsare better.

Example Suppose that we must choose between using a low-loss waveguide run and apair of antennas at 3 GHz. Each antenna has 10 dB of gain. The low-loss waveguide hasonly 19.7 dB/km loss. Table 1-1 compares losses over various distances. The waveguidelink starts out with lower loss, but the antenna system soon overtakes it. When thepath length doubles, the cable link loss also doubles in decibels, but an antenna linkincreases by only 6 dB. As the distance is increased, radiating between two antennaseventually has lower losses than in any cable.

TABLE 1-1 Losses Over Distance

Example A 200-m outside antenna range was set up to operate at 2 GHz using a 2-mdiameterreflector as a source. The receiver requires a sample of the transmitter signalto phase-lock the local oscillator and signal at a 45-MHz difference. It was proposedto run an RG/U 115 cable through the power and control cable conduit, since the runwas short. The cable loss was 36 dB per 100 m, giving a total cable loss of 72 dB.A 10-dB coupler was used on the transmitter to pick off the reference signal, so thetotal loss was 82 dB. Since the source transmitted 100 mW (20 dBm), the signal was−62 dBm at the receiver, sufficient for phase lock.

A second proposed method was to place a standard-gain horn (15 dB of gain) withinthe beam of the source on a small stand out of the way of the measurement and nextto the receiver. If we assume that the source antenna had only 30% aperture efficiency,we compute gain from Eq. (1):

G =( )²η_a= ( )²η_a

(λ = 0.15 m):

G = ( )²(0.3)= 526  (27.2dB)

The path loss is found from Eq. (1-9) for a range of 0.2 km:

32.45 + 20 log[2000(0.2)] − 27.2 − 15 = 42.3dB

The power delivered out of the horn is 20 dBm − 42.3 dB = −22.3 dBm. A 20-dBattenuator must be put on the horn to prevent saturation of the receiver (−30 dBm).Even with a short run, it is sometimes better to transmit the signal between two antennasinstead of using cables.