AntennasAntennasTransmitterITransmission LineTransmitting AntennaElectromagnetic WaveIReceiverITransmission LineReceivingAntennaElectromagnetic WaveIAntennas are transducers that transfer electromagnetic energybetween a transmission line and free space. Amanogawa, 2006 Digital Maestro Series1AntennasHere are a few examples of common antennas:Linear dipole fedby a two-wire lineLinear monopolefed by a single wireover a ground planeCoaxial groundplane antennaGround planeLinear elements connectedto outer conductor of thecoaxial cable simulate theground planeLoop antennaUda-Yagi dipole arrayLoop dipoleMultiple loop antenna wound around a ferrite coreParabolic (dish) antennaLogperiodic arrayPassive elements Amanogawa, 2006 Digital Maestro Series2AntennasFrom a circuit point of view, a transmitting antenna behaves like anequivalent impedance that dissipates the power transmittedTransmitting AntennaTransmitterITransmission LineElectromagnetic WaveITransmitterZgVgTransmission LineI 1 2P( t ) = Req I 2ZeqI= Req + jXeqThe transmitter is equivalent to a generator. Amanogawa, 2006 Digital Maestro Series3AntennasA receiving antenna behaves like a generator with an internalimpedance corresponding to the antenna equivalent impedance.ReceiverITransmission LineReceivingAntennaElectromagnetic WaveIIZeqZin 1 2P( t ) = Rin I 2VeqZRTransmission LineIThe receiver represents the load impedance that dissipates the timeaverage power generated by the receiving antenna. Amanogawa, 2006 Digital Maestro Series4AntennasAntennas are in general reciprocal devices, which can be used bothas transmitting and as receiving elements. This is how theantennas on cellular phones and walkietalkies operate.The basic principle of operation of an antenna is easily understoodstarting from a twowire transmission line, terminated by an opencircuit.|I|ZgVgZR Open circuit|I|Note: This is the returncurrent on the second wire,not the reflected currentalready included in thestanding wave pattern. Amanogawa, 2006 Digital Maestro Series5AntennasImagine to bend the end of the transmission line, forming a dipoleantenna. Because of the change in geometry, there is now anabrupt change in the characteristic impedance at the transitionpoint, where the current is still continuous. The dipole leakselectromagnetic energy into the surrounding space, therefore itreflects less power than the original open circuit the standingwave pattern on the transmission line is modified| I0 |Zg|I|VgZ0|I|| I0 | Amanogawa, 2006 Digital Maestro Series6AntennasIn the space surrounding the dipole we have an electric field. Atzero frequency (d.c. bias), fixed electrostatic field lines connect themetal elements of the antenna, with circular symmetry.EE Amanogawa, 2006 Digital Maestro Series7AntennasAt higher frequency, the current oscillates in the wires and the fieldemanating from the dipole changes periodically. The field linespropagate away from the dipole and form closed loops. Amanogawa, 2006 Digital Maestro Series8AntennasThe electromagnetic field emitted by an antenna obeys Maxwellsequations E = j H H = J + j EUnder the assumption of uniform isotropic medium we have thewave equation: E = j H = j J + E H = J + j E= J + HNote that in the regions with electrical charges 22 E = E E = ( ) E22 Amanogawa, 2006 Digital Maestro Series9AntennasIn general, these wave equations are difficult to solve, because ofthe presence of the terms with current and charge. It is easier touse the magnetic vector potential and the electric scalar potential.The definition of the magnetic vector potential isB=ANote that since the divergence of the curl of a vector is equal tozero we always satisfy the zero divergence condition B = ( A) = 0We have also E = j H = j A ( E + j A) = 0 Amanogawa, 2006 Digital Maestro Series10AntennasWe define the scalar potential first noticing that ( ) = 0and then choosing (with sign convention as in electrostatics) ( E + j A) = ( ) E = j A Note that the magnetic vector potential is not uniquely defined,since for any arbitrary scalar field B = A = ( A + )In order to uniquely define the magnetic vector potential, thestandard approach is to use the Lorenz gauge A + j = 0 Amanogawa, 2006 Digital Maestro Series11AntennasFrom Maxwells equationsH=1 B = J + j E ( A) = J + j ( j A )From vector calculus B = J + j E ( ) = ( ) 2 ( A) = ( A) A = J + A j 22Lorenz Gauge A = j Amanogawa, 2006 Digital Maestro Series ( A) = j 12AntennasFinally, the wave equation for the magnetic vector potential is A + A = A + A = JFor the electric field we have2222 D = E = ( j A ) = 22 + j A = + j ( j ) = The wave equation for the electric scalar potential is + = + = 2222 Amanogawa, 2006 Digital Maestro Series13AntennasThe wave equations are inhomogenoeous Helmholtz equations,which apply to regions where currents and charges are not zero.We use the following system of coordinates for an antenna bodyzJ( r ') ( r ')dV 'r'Observation pointr r'ryxRadiating antenna body Amanogawa, 2006 Digital Maestro Series14AntennasThe generals solutions for the wave equations are J ( r ') eA( r ) = V 4r r' j r r 'dV ' ( r) = V41 ( r ') e j r r 'r r'dV 'The integrals are extended to all points over the antenna bodywhere the sources (current density, charge) are not zero. The effectof each volume element of the antenna is to radiate a radial wavee j r r 'r r' Amanogawa, 2006 Digital Maestro Series15AntennasInfinitesimal AntennazJ(0)Observation point (0)Szxr' = 0r = r r'dV 'yInfinitesimal antenna bodyI = constant phasor z >1j r>>1( j r )2 r=2r >> 1 Amanogawa, 2006 Digital Maestro Series21AntennasAt large distance we have the expressions for the Far Field j I z eH i 4 rE i j rsin 2 r >> sin j I z e 4 r j r At sufficient distance from the antenna, the radiated fields are perpendicular to each other and to the direction of propagation. The magnetic field and electric field are in phase and E =H = H These are also properties of uniform plane waves. Amanogawa, 2006 Digital Maestro Series22AntennasHowever, there are significant differences with respect to a uniformplane wave: The surfaces of constant phase are spherical instead of planar, and the wave travels in the radial direction. The intensities of the fields are inversely proportional to the distance, therefore the field intensities decay while they are constant for a uniform plane wave. The field intensities are not constant on a given surface of constant phase. The intensity depends on the sine of the polar angle .The radiated power density is 2 11 *P( t ) = Re E H = i rH 22 {} I z 2= ir sin 2 4 r Amanogawa, 2006 Digital Maestro Series232AntennaszP( t )HJrEThe spherical wave resembles a plane wave locally in a smallneighborhood of the point ( r, , ). Amanogawa, 2006 Digital Maestro Series24AntennasRadiation PatternsElectric Field and Magnetic FieldzE or HyxFixed rxPlane containing the antennaPlane perpendicular to the antennaproportional to sinomnidirectional or isotropic Amanogawa, 2006 Digital Maestro Series25AntennasTimeaverage Power Flow (Poynting Vector)zP( t )yxFixed rxPlane containing the antennaPlane perpendicular to the antennaproportional to sin2omnidirectional or isotropic Amanogawa, 2006 Digital Maestro Series26AntennasTotal Radiated PowerThe timeaverage power flow is not uniform on the spherical wavefront. In order to obtain the total power radiated by the infinitesimalantenna, it is necessary to integrate over the sphere 2Ptot =d 0 = 20 d r22sin P (t ) 4 I z = 3 4 2 I z 2 3= 2 r 0 d sin 2 4 r =4 3Note: the total radiated power is independent of distance. Althoughthe power density decreases with distance, the integral of thepower over concentric spherical wave fronts remains constant. Amanogawa, 2006 Digital Maestro Series27AntennasPtot1 = Ptot 2Ptot1Ptot 2 Amanogawa, 2006 Digital Maestro Series28AntennasThe total radiated power is also the power delivered by thetransmission line to the real part of the equivalent impedance seenat the input of the antenna 2 z 2 1 24 2 I z 1 2Ptot = I Req == I 4 223 3 ReqThe equivalent resistance of the antenna is usually called radiationresistance. In free space2 o2 z = o == 120 [ ] Req = 80 o Amanogawa, 2006 Digital Maestro Series2[ ]29AntennasThe total radiated power is also used to define the average powerdensity emitted by the antenna. The average power densitycorresponds to the radiation of a hypothetical omnidirectional(isotropic) antenna, which is used as a reference to understand thedirective properties of any antenna. Power radiation pattern of anomnidirectional average antennaPower radiation pattern of the actual antennazxPaveP( t, ) Amanogawa, 2006 Digital Maestro Series30AntennasThe timeaverage power density is given byPave =Total Radiated PowerSurface of wave front=2 I z2 1=== ( I z) 2 122 3 4 r 4 r4 rPtotThe directive gain of the infinitesimal antenna is defined asD( , ) =P( t , r, )Pave I z I z 2= sin 3 4 r 2 4 r 22 1 3 2= sin 2 Amanogawa, 2006 Digital Maestro Series31AntennasThe maximum value of the directive gain is called directivity of theantenna. For the infinitesimal antenna, the maximum of thedirective gain occurs when the polar angle is 90 3 2 = 1.5Directivity = max { D( , )} = sin 22The directivity gives a measure of how the actual antenna performsin the direction of maximum radiation, with respect to the idealisotropic antenna which emits the average power in all directions.zPavePmax90x Amanogawa, 2006 Digital Maestro Series32AntennasThe infinitesimal antenna is a suitable model to study the behaviorof the elementary radiating element called Hertzian dipole.Consider two small charge reservoirs, separated by a distance z,which exchange mobile charge in the form of an oscillatory curentI( t)t+++z+++ Amanogawa, 2006 Digital Maestro Series33AntennasThe Hertzian dipole can be used as an elementary model for manynatural charge oscillation phenomena. The radiated fields can bedescribed by using the results of the infinitesimal antenna.Assuming a sinusoidally varying charge flow between thereservoirs, the oscillating current isI( t)current flowingout of reservoir dd= q( t) = qo cos( t) dtdt charge onreference reservoirphasor I o = j qoRadiation patternIoqo Amanogawa, 2006 Digital Maestro Series34AntennasA short wire antenna has a triangular current distribution, since thecurrent itself has to reach a null at the end the wires. The currentcan be made approximately uniform by adding capacitor plates.ImaxzIoIoThe small capacitor plate antenna is equivalent to a Hertzian dipoleand the radiated fields can also be described by using the results ofthe infinitesimal antenna. The short wire antenna can be describedby the same results, if one uses an average current value giving thesame integral of the currentIo = Imax 2 Amanogawa, 2006 Digital Maestro Series35AntennasExample A Hertzian dipole is 1.0 mm long and it operates at thefrequency of 1.0 GHz, with feeding current Io = 1.0 Ampres. Findthe total radiated power. = c f 3 10 10 = 0.3 m = 300 mm z = 1 mm Hertzian dipole 4 2 Io z 1 2 23 2 =120 () ( 1 10 )Ptot = 3 4 12 0.3o892Ioz= 4.39 mWFor a short dipole with triangular current distribution and maximumcurrent Imax = 1.0 AmpreIo = Imax 2 Amanogawa, 2006 Digital Maestro SeriesPtot = 4.39 / 4 1.09 mW36AntennasTimedependent fields - Consider the farfield approximation j I z sin j ( t r ) H ( t ) = Re H e e i Re 4 r I z sin 2 i Re jcos( t r )+ j sin( t r ) 4 r I z sin isin( t r ) 4 rj t{}()E ( t ) = Re E e{j t}37 I z sin i sin( t r ) 4 r Amanogawa, 2006 Digital Maestro SeriesAntennasLinear AntennasConsider a dipole with wires of length comparable to thewavelength.zL2'zz'r'ri 'i L1 Amanogawa, 2006 Digital Maestro Series38AntennasBecause of its length, the current flowing in the antenna wire is afunction of the coordinate z. To evaluate the farfield at anobservation point, we divide the antenna into segments which canbe considered as elementary infinitesimal antennas.The electric field radiated by each element , in the farfieldapproximation, is E' = i j I z e 4 r ' j r 'sin 'In farfield conditions we can use these additional approximations 'r ' r z 'cos Amanogawa, 2006 Digital Maestro Series39AntennasThe lines r and r are nearly parallel under these assumptions.zL2r'r'zz'z 'cos This length is neglected ifr ' r z 'cos Amanogawa, 2006 Digital Maestro Series40AntennasThe electric field contributions due to each infinitesimal segmentbecomes you cannotneglect here E' = i e j I z e4 r 4 z 'cos you canneglect here j rj z 'cossin The total fields are obtained by integration of all the contributionsE = i j e 4 r j r j r L2j z cos dzsin I( z) e L1 j eH = i 4 r Amanogawa, 2006 Digital Maestro Series L2j z cos dzsin I( z) e L141AntennasShort DipoleConsider a short symmetric dipole comprising two wires, each oflength L