ZI I sin eI I sin 0-jke-jikrH.=OE.=j2r24r(d) The radiation fields are differentin different directions, andthis propertyis called the directivity of theantenna.The portion of the field intensity expression that describes theelevation and the azimuthal dependence is called the directivityfactor, and is denoted by f (0, Φ)For the z-directed electric current element, f(O, Φ) = sin 0.(e) The directions of the electric and the magnetic fields areindependent of time, and the radiation fields are linearly polarizedThe above properties (a), (b), (c), and (d) are common for allantennas with finite-sizes.UV
(d) The radiation fields are different in different directions, and this property is called the directivity of the antenna. For the z-directed electric current element, . f (,) = sin (e) The directions of the electric and the magnetic fields are independent of time, and the radiation fields are linearlypolarized. kr r I l H j e 2 sin j − = kr r ZI l E j e 2 sin j − = The portion of the field intensity expression that describes the elevation and the azimuthal dependence is called the directivity factor, and is denoted by f (, ) . The above properties (a), (b), (c), and (d) are common for all antennas with finite-sizes
Strictly speaking, there is energy exchange also in the far-fieldregion. However, the amplitudes of the field intensities accountingfor energy exchange is at leastinversely proportional to the squareofthedistance,while the amplitudes ofthefieldintensities forenergy radiation are inverselyproportionalto the distance.Consequently, the exchanged energy is much less than theradiated energy in the far-field region, while the converseis truein the near-fieldregion.ENear-zonefieldsFar-zonefields
E r Strictly speaking, there is energy exchange also in the far-field region. However, the amplitudes of the field intensities accounting for energy exchange is at least inversely proportional to the square of the distance, while the amplitudes of the field intensities for energy radiation are inverselyproportional to the distance. Near-zone fields Far-zone fields Consequently, the exchanged energy is much less than the radiated energy in the far-field region, while the converse is true in the near-field region
Differentantenna types produceradiationfields of differingpolarizations. Antennas can produce linearly, circularly, or ellip.ticallypolarized electromagnetic waves.The polarization properties of the receiving antenna mustmatch that ofthe receivedelectromagneticwaveTo calculate the radiation power Pr, we take the integrationof the real part of the complex energy flow density vector over thesphericalsurface ofradius r in the far-field region, as given byP =f Re(S.)dsWhere S. is the complex energy flow density vector in the far-fieldregion,i.e[E.Pe.HPZS.=E。xH=e, IE。lH.=eZu7
Different antenna types produce radiation fields of differing polarizations. Antennas can produce linearly, circularly, or ellipticallypolarized electromagnetic waves. To calculate the radiation power Pr , we take the integration of the real part of the complex energy flow density vector over the sphericalsurface of radius r in the far-field region, as given by = S Pr Re(Sc ) dS Where Sc is the complex energy flow density vector in the farfield region, i.e. H Z Z E r E H r r 2 2 * c | | | | | || | S = E H = e = e = e The polarization properties of the receiving antenna must match that of the received electromagnetic wave
ZI?1 sin ? We find= Re(S.)S. =e42r2If the medium is vacuum, Z= Z, = 120元 Q, the radiated power isobtainedasTP,= 80元r21where Iis the effectivevalue of the electric currentThe resistive portion accountingfor the radiation process may bedefined as the radiationresistance, and is given byPRFor the electric current element.R =80元AThe greater the radiation resistanceis, the higher will be the powerradiated for a given electric current.U7
We find Re( ) 4 sin 2 2 c 2 2 2 Sc = e = S r ZI l r If the medium is vacuum, Z = Z0 = 120 , the radiated power is obtained as 2 2 2 r 80π = l P I where I is the effective value of the electric current. The resistive portion accounting for the radiation process may be defined as the radiation resistance , and is given by 2 r r I P R = For the electric current element, 2 2 r 80π = l R The greater the radiation resistance is, the higher will be the power radiated for a given electric current
Example. If an electric current element is placed at the origin alongthe x-axis, find the far-zone fields.Solution: Sincelllul, we fnd4元rA = A, sin 0cosdA。 = A cospcos 0Aβ=-A sin Φg,uP(x, y, z)Forthefar-zone fields, consideringonly the parts inversely proportional tothe distancer we find1OH =-(e sin p+e, coso cos p)e-ikrSince the far-zone fields are TEM wave, the electric field intensityisZI1(ee cos cos p-e, sin p)e-ibE- ZHxe, =2元rU7
Example. If an electric current element is placed at the origin along the x-axis, find the far-zone fields. Since I l = ex I l ， ， , we find x Ax A = e kr x r I l A j e 4π − = sin cos cos sin cos x x r x A A A A A A = − = = For the far-zone fields, considering only the parts inversely proportional to the distance r we find kr r I l j ( sin cos cos )e 2 j − = − + H e e Since the far-zone fields are TEM wave, the electric field intensity is kr r r ZI l Z j ( cos cos sin )e 2 j − = = − − E H e e e Solution: r Il z y x , P(x, y, z) O