ECE 5324/6324 NOTES ANTENNA THEORY AND DESIGN

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1 ECE 534/634 NOTES ANTENNA THEORY AND DESIGN 013 Om P. Gandhi Text: Warren L. Stutzman and Gary A. Thiele, Antenna Theory and Design, Third Edition (013), John Wiley & Sons. The identified page numbers and the equations with dashes (x-xxx) refer to the equations of the text.

2 Example: Show that for far-field region V = jβ V = jβrˆ V for any vector V such as A, H and E of the radiated fields from antennas. Solution: The radiated fields A, H and E are of the form V = K(θ,φ) r e jkr ˆ V V(r, θ, φ) ˆ V (1) where K(θ, φ) would, in general, depend upon the current distribution on the antenna. In spherical coordinates V(r, θφ, ) θˆ φˆ K(, ) ˆ θφ β j r V r e Vˆ = + + r r r sin r θ θ φ 1 ˆ 1 V ˆ ˆ 1 V = j V(rˆ V) ( V) ( ˆ V) ˆ β + θ + φ r r θ r sin θ φ () jβ V( rˆ Vˆ ) = jβ rˆ V = jβ V (3) since all terms other than the second term in Eq. are a factor of 1/βr π r smaller. For βr = >> 1, all of these terms can, therefore, be neglected. λ This is a powerful relationship which can be applied for radiated fields from any antenna. A jβ A β H= = j rˆ A µ µ µ H jβ H β µ E= = = rˆ H= rˆ H jωε jωε ωε ε (4) (5) 1

3 p. 44, 45, 50 Text General Theory of Conduction Current Antennas R P r r Formulate J ( r ). Steps 1. Calculate A. 3. µ J(r ) j( ωt βr) µ JS j( ωt βr) A = e dv = e ds 4π R 4 R V π S For volume current radiators µ I e j( ωt βr) = d 4 π R For line current radiators A β A β rˆ A E 1 H = = j = j = = rˆ E µ o µ o µ o jωµ o η E = H = β ˆ r (ˆ r A ) = jω ˆ r (ˆ r A ) jωε o jωε o µ o = ω j A ( A rˆ) rˆ = ω j ( A ˆ A ˆ θθ+ φφ ) For surface current radiators (-101) (-107) (-105) A B C = ACB ABC ( ) ( ) ( ) ( ) ( ) ( ) rˆ rˆ A = rˆ A rˆ rˆ rˆ A From Eq. (-105) we can write which is transverse to direction of propagation ˆ r. E = jω A (-104) 1 * 4. Calculate S = Re ( E H ) ππ 1 5. Total Radiated Power = ( θ φ φ θ) * * S ds = Re E H E H r sin θdθφ d 00 (-17) (-18)

4 Calculation of Magnetic Fields of Conduction Current Antennas Definition of Magnetic Vector Potential A : A Simplifying Mathematical Intermediate Step I R= RRˆ F (x,y,z) From Biot-Savart's law of electromagetism I d Rˆ J dv Rˆ B =µ o =µ o 4πR V 4π R B =µ o V J dv 4πR 1 R (1) () In going from Eq. 1 to Eq., we have used the following steps 1 Rˆ = R R J J = J(r ) R R R (3) (4) The first term in Eq. 3 is zero, since the current density J is a function of source coordinates r = (x,y,z ) whereas the curl J involves derivatives with respect to field coordinates (x,y,z). From Eq. J dv B=µ o A 4πR V (5) Thus the magnetic field at the field point can be written as curl of magnetic vector potential A where A is given by µ o J dv A = 4π R V (6) Note that calculation of B is a lot simpler if the intermediate step of first calculating A is undertaken since integral of Eq. 6 is much simpler than that of Eq. 5 or Eq. 1. a

5 Note that because of time retardation for propagating fields, Eq. 6 should be modified to µ o J(r )e A = 4 π R j( ωt βr) dv (7) Same as Eq of the text Once, the only complicated step that of integration for Eq. 7 has been done, the magnetic field B from Eq. 5 can be simplified to B= A= β j A= β j Rˆ A jβ A H = µ o (-107) text b

6 p. 45 Text A Uniform Line Source R ˆφ θ ˆr r ˆθ Fig..9 text. A uniform line source. I( z ) = I o for L < z < L (-109) R = r z cos θ (-86) L/ µ β j r jβz cosθ A = e Io e dz zˆ 4 π r L/ A z βl β j rsin cos θ µ IoLe = 4πr βl cos θ βl sin cos θ ˆ µ IL o E = jωsin θa ˆ z θ= jω sin θ θ 4πr βl cos θ From Eq. (-107) H = 1 η E θ ˆ r θ ˆ = E θ φ ˆ η Radiation pattern for a plot of normalized values of E(θ, φ) is given by (-110) (-111) E( θφ, ) F( θφ=, ) (-11) Emax For an elemental or Hertzian dipole (βl/ <<1) F(θ) = sin θ (-113) 3

7 Otherwise βl sin cos θ F( θ ) = sin θ βl cos θ (-114) Normalized Pattern factor P(θ) = F (θ) (-119) p. 34 An Infinitesimal (Hertzian) Current Dipole or An Ideal Dipole I = I o L = z βl = π z λ <<1 βl cos θ <<1 In Equations -110 to -114; sin x x 1 E = jω µi o z sin θ θ ˆ 4πr H = E θ φ ˆ = jβi o z sin θ η 4πr ˆ φ (-74a) (-74b) o λ o I z sin S= η 8 r θ rˆ (-76) * ( ) 1 Radiated Power P = Re E H ds ( θ φ φ θ) ππ 1 Re E H * E H * r sin d d = θ θ φ o 00 I z 1 I = πη = o R 3 λ r (-18) 4

8 A Linear Center-Fed Dipole (See also pp of the text) z z Rr = Ra = πη = 80 π Ω 3 λ λ S D max 1.5 So Valid only for very short length Hertzian dipoles ( z 0.0 λ ) (-169) p. 57 = = (-148) p. 54 L z = + I a 0 ẑ I m z cos Azˆ H φˆ E θˆ L z = L L I(z ) = Im sin β z z < L Note that Ia = I(z ) = I z 0 msin β = As previously assumed on p. 3 of these Notes (from Eq. -86 of the text), (6-1) (1) p. 15 From Eq R = r z cos θ 0 µ L j z cos A = Im sin β β θ + z e dz 4πr L/ L/ L + jβz cosθ β j r sin β z e dz e zˆ 0 (6-3) β j r e ˆ j60 Im β j r E = jωsin θa ˆ z θ= jη Im F( θ) θ= F( θ)e θ 4πr r (6-6) p. 154 where F(θ) is the function that gives the variation of radiated fields with angle θ. Note that this expression for the radiated θ-directed E field can also be expressed in terms of the feedpoint antenna current I a using Eq. (1) on this page. 5

9 βl βl cos cos θ cos F( θ ) = sin θ () See p. 154 of the text, Fig. 6-4, for plots of F(θ) for several values of L/λ. F(θ) is always zero for angle θ= 0 i.e. no radiated fields along the length of the dipole. * 1 * E E 15 Im S = Re ( E H θ θ θ φ) = rˆ = F ( θ) rˆ η πr 15 Ia 30 Prad F ( θ) = F ( θ )rˆ = L πr r Ra L sin π β sin β (3) ππ 15 Im Radiated Power Prad = S ds = F ( θ)r sin θdθdφ πr = Im Rrm = Ia Ra (4) π 60 Im Ra = F ( ) sin d θ θ θ (5) I a 0 Where R a is the antenna equivalent resistance at the feed point ( z = 0). The antenna equivalent resistance R a is given by Table l on p. 7. Thus the directivity D of a linear center-fed antenna of end-to-end length L is given by: Smax 10 F ( θ) max D = = (6) So Ra βl sin 6

10 Table 1. Calculated values of the driving point resistance Ra for end-fed monopoles of different lengths Calculated h/λ. (Multiply values of by the to driving obtain point the resistance driving point R a for resistance end fed for monopoles center-fed of dipoles different of length h/λ. L (Multiply = h.) by to obtain the driving point resistance for dipoles.) h/λ = L/λ R a h/λ = L/λ R a 7

11 Fig. 1. The calculated resistance R a and reactance X a of an end-fed monopole antenna of length h (in terms of wavelength λ). Multiply by to obtain the driving point resistance R a for a center-fed dipole antenna of length L = h. 8

12 Example 1: Calculate and compare the directivities, gains, and power densities including E-fields created by dipole antennas of lengths L = 0.07 λ, 0.18 λ, 0.5 λ, and 1.1 λ. Power radiated by the antenna is 100 W and distance from the antenna to the field point r o = 10 km. Note that the radiated power and the distance r o are needed to calculate the power density and maximum electric fields. S max θ = 90 = SD o E S = η Including Ohmic losses for Prob. 5 of HW From Eq antenna efficiency From p. 7 From Eq. of Notes P rad * i = L/ λ R a ohms I a A F ( θ ) D θ = 90 D ii μw/m Emax = η S R max a e 4π r mv/m R iii r = -155 ohmic Ω Ra + Rohmic G = e r D * Note that P 1 = I R rad a a from Eq. (4) on p. 6 of Class Notes; I a = P rad R a i From Eq. () on p. 6 of Class Notes, F ( θ ) F ( θ) max = 1 cos( πl λ) θ = 90 ii From Eq. (6) on p. 6 of Notes, βl βl πl πl cos cosθ cos cos cosθ cos λ λ = =. For L/λ < ; F(θ) is max. for θ = 90. sinθ sinθ F ( θ ) max ( π λ) D 10 = R a sin L iii R ohmic is given by Eq. (9) on p. 13 of Class Notes; R = MHz ; Take a = 3.64 mm (0.185) 8 AWG wire (App. B on p. 783 of the Text). For Aluminum, from App. B.1, σ = S/m; take f MHz = 10 MHz. S f σ 9

13 Example : P rad = 1 W; f = 835 MHz; r = 1 km 30 L = h = 0.65λ= 0.65 = 3.35 cm h L 0.35 λ = λ = R a Table1 = = 181.7Ω 1 Prad = Ia Ra Ia = A Ia I I a m = = = βl πl sin sin λ βl 1 cos 60 Im 60 I m mv Emax = F( θ ) = = r θ= 90 r 1 m E rms = E peak = 7.6 mv m S max = E max η = E rms η = µw m F ( θ) 10 max D = = = (.45 dbi) R a βl sin sin (0.65 π) This is an improvement of only times (or 0.3 db) relative to a half wave dipole. S max = S o D = µw m 10

14 Radiation patterns xy plane H-plane z y.45 dbi (0.3 dbd) x Fig.. The radiation pattern of a z-directed dipole antenna for the xy plane or H-plane (normal to the orientation of the dipole). See also p. 154 (Fig. 6-4) yz plane (E-plane) θ HP HP Fig. 3. The radiation pattern of the z-directed dipole antenna for the yz plane or the E-plane. HP = (90 - θ HP ) (-16) p. 49 βl βl cos cos θhp cos 1 βl = 1 cos sin θhp (7) 11

15 pp. 57, 58 Text Ohmic Losses for a Linear Dipole dr From Eq. 6-1, p. 15 Text I( z ) = I m sin β L z = I m sin β L + z 0 z L L (1) z 0 βl Ia = I(z ) = Im sin z = 0 () Ohmic power lost in the antenna L/ 1 Pohmic = ( I dr) (3) L/ where d z dr = (πa δ)σ = R s d z πa R s 1 = = σδ ωµ σ (4) (5) (-171) p. 58 is the surface resistance which depends on the conductivity σ of the material and frequency ω (= πf). See App. B.1 of the Text for σ of various metals. 1

16 L/ 0 s ohmic m m 0 L/ R L L P = I sin β z dz + I sin β + z dz 4πa (6) L/ βl/ L 1 1 sin ( ζ) sin β + z dz = sin ζdζ= ζ β β βl sin ( βl) = β L where ζ = =β + z Rs L sin ( βl) 1 Pohmic = Im 1 = IA Rohmic 8πa ( L) β βl/ 0 (7) (8) Rs L 1 sin ( βl) Rohmic = 1 4πa βl ( βl) sin (9) P Prad R Antenna efficiency e a r = = = (-177) Pin Prad + Pohmic Ra + Rohmic For a Short Dipole (L = Δz << λ) Gain G = ed r (-155) z Ra = 0 π λ (-17) R ohmic Rs z RL s = = (-175) 6π 6πa Example 3: For the general case of a linear dipole or a monopole R ohmic is calculated from the general Eq. 9 given above. For a Short Dipole L L Ra = 0 π λ λ (-17) See e.g. Table 1 on page 7 for L/λ =0.0, R a = = Ω. 13

17 Using the conductivity of steel (see App. B.1 of the Text) σ = 10 6 S/m. From Eq or Eq. 5 s 3 R = f Ω MHz From Eq. 9, βl is small and we can expand sin x for small x 3 ( βl) L Rs L 1 β 6 Rs L Rohmic = 1 = Short dipole 4πa βl βl 6πa For L/λ = 0.0 dipole at f = 1 MHz; taking a = l/8" (-175) p. 59 λ = 300m ; L = 6m R ohmic = 6π 1 = 0.807Ω Antenna Efficiency Ra er = = = 0.19 (1.9%) Ra + Rohmic Example 4: Gain G = ed= r = Note that for short dipoles of thin wires, the ohmic resistance can be substantial and even larger than R a. Therefore, this leads to reduced efficiency of radiation. For a Half Wave Dipole L = 0.5λ; R a = 73.1 Ω; βl = πl λ = π ; βl = π; f = 10 MHz; L = 15m; a = 1/8" From Eq. 9 3 ( ) Rs L Rohmic = = = 3.33Ω 4πa 1 4π er = = (95.65%) G = D = =

18 pp Text Dipoles Versus Monopoles Above a Perfect Ground or Reflector I a I a Image antenna L L I(z ) same as on page 5 of the Notes I( z ) = I m sin β z 0 z For 0 θ 180 For 0 θ 90 (1) 60I m jβ r E = j F( θ) e ˆ θ r j60i m jβ r E F e ˆ θ r () = ( θ) (3) 15I S = m F ( θ ) rˆ P π r 1 = I R (6) rad a a 15I 15 I S F r F rˆ πr πr β L sin m a (4) = ( θ) ˆ= ( θ) F( θ ) is given as Eq. () on p. 6 of the Notes. (5) 1 P rad = I a R a (7) Since a monopole radiates in the upper half space while a dipole radiates both in the upper and lower half spaces, 1 Sdipole = S monopole for identical radiated powers (8) D = D (9) monopole dipole R a X For identical radiated powers a monopole = 1 R X a a dipole (10) I = I (11) a a I = I (1) m m 15

19 Example 4: h λ = L = 0.35 Monopole Antenna λ f = 1.5 MHz, λ = 00 m ; r = 1 km h = L = 70 m ; P rad = 10 3 W (1 KW) From Table 1 on page 7, R a =17.1 Ω (do not multiply by for monopoles) From Eq. 7, I a = 3.967A From Eq. 5, βl βl cos cos cos 15 (3.967) θ S = = 0.9 mw / m monopole 6 π 10 sin (0.7 π) sin θ θ= 90 S D = = 3.64 = D monopole Prad dipole So = 4 π r From Eq. (6) on p. 6 of Class Notes F ( θ) 10 D = max Ra βl sin Note that L = h which is the height of the monopole. 16

20 pp Small Diameter (<< λ) Loop Antennas The loop antenna is a radiating (or receiving) coil of one or more turns of circular or rectangular form. Ferrite or air core loops are used extensively in radio receivers, direction finders, aircraft receivers, and UHF transmitters. The theory of loop antennas is derived in a manner similar to the General Theory of Conduction Current Antennas given on page 44 of Text and on page of my handout notes. We start by assuming, as seen in Fig. 1, that the current I in the loop has the same magnitude and phase. This is certainly possible for small diameter loops where πb < λ/10. z R r x Loop in the xy plane Fig. 1. A circular loop antenna of radius 'b'. From General Theory of Conduction Current Antennas, from Eq. -101, A = µ 4π π I φ ˆ e jβr bd φ (1) R 0 R = ( x x ) + ( y y ) + z () x = b cos φ ; y = b sin φ ; z = 0; x = r sin θ; y = 0; z = r cos θ (3) Note that we have defined the x-axis (the choice of which is arbitrary) such that the field point lies in the xz plane. The field point F, therefore, has coordinates (x, 0, z) in Cartesian coordinate system and (r, θ, 0) is spherical coordinate system. Substituting Eq. 3 into Eq. 1/ b R = r + b br sin θcos φ r 1 sin θcos φ r = r bsin θcos φ (4) 17

21 since, for the far-field region, r >> b. Using the far-field approximation for Eq. 1 β j r π µ Ie ˆ jβb sin θcos φ A = φ e bdφ πr 0 (5) For small radii βb = πb λ We can also write (see Fig. 1(b)) <<1, we can write jβb sin θ cos e φ =1 + jβb sin cos φ (6) φ ˆ = ˆ x sin φ + y ˆ cos φ (7) Also to be noted is that for the field point F From Eq. 5 therefore, we can write y ˆ = φ ˆ (8) A = jµi S 4πr βe jβr sin θ φ ˆ (9) (p. 86 Text Eq. 3-49) where S = πb is the area of the loop. On page 19, we compare the expressions for the radiated fields from a loop antenna to those for an ideal (infinitesimal) dipole and show duality of the two sets of fields. Ohmic resistance of a circular loop antenna can be written as follows: where R ohmic = R w = 1 Rs = = σδs πb (πaδ)σ = br s a 18 fmhz σ (10) (3-60) p. 88 Text where "b" is the mean loop radius and "a" is the wire radius; R s = 1/σδ is the surface resistance at the frequency of interest previously defined on page 1 of Class Notes. The small loop antenna is inherently inductive. For a small circular loop of N turns wound on a magnetic core 8b L= N bµ eff µ o n a (11) (Eq. 3-6 p. 88 Text)

22 Table. Field expressions for small diameter circular loop antennas and an ideal (infinitesimal) dipole antenna [see p. 4 of Class Notes]. Loop Antenna H θˆ Ideal (Infinitesimal) Dipole 19 Magnetic Vector Potential ( A ) jis jβ r βsinθe ˆ φ 4π r Magnetic Field A jβ IS jβ r H = = rˆ A β e sinθθˆ µ µ 4π r Electric Field H ηis jβ r E = = ηrˆ H β e sinθφˆ jωε 4π r (9) µ I z (-65) j r (3-48) Text e β z ˆ p. 33 Text 4π r (3-50) p. 86 Text (3-49) p. 86 Text jβ I z e jβ r sinθφˆ 4π r since rˆ zˆ = sinθφˆ (-74b) (-70) p. 33 Text jη I z (-74a) jβ r βe sinθθˆ p. 34 Text 4π r Radiated Power Density * 1 * EE S = Re( E H ) = rˆ η Radiated Power 1 P = S ds I R r ηis 3π r 4 β sin θrˆ 10I ( β S) R r (for single turn loop) ( β S ) Directivity D Rr N turn loop S 0 31,00 S λ Ω (3-5) ( z) ( π r) ηi 4 p. 86 Text ( I z) (3-53) max = S o NS 31, 00 µ eff Ω λ (3-54) β sin θrˆ ωµβ 1π z z 80π = 790 λ λ (-76) (-77) p. 33 Text (-169) p. 57 Text 19

23 For an N-turn loop, Rohmic is also higher proportional to overall length of the wire R ohmic N turn loop = N br s a (1) The effective permeability μ eff depends not only on the permeability μ r of the ferrite core material, but also on the core geometry, i.e., length to diameter ratio R, given as follows: µ eff µ r = (13) 1 + D 1 ( µ ) r where 4 D is the demagnetization factor approximately given by D [4] Example 5 (see also Ex. 3-1, p. 88, Text): D R (14) p. 87 Text Calculate the input impedance, directivity, and gain for an N = 1000 turn loop antenna wound with a AWG copper wire on a ferrite rod of diameter 3/4". This antenna is to be used at a frequency of 1.5 MHz. It is given that µeff = 50 for the ferrite that is used. Solution: From p. 783 of the Text, for AWG wire d = a = mm From p. 58, Eq ωµ f R MHz s = = = ohms σδ σ σ (15) For copper σ = S/m (p. 783, Text); R s = Ω at f = 1.5 MHz Mean loop radius b = a = mm From Eq. 1, for N = 1000-turn loop R ohmic = 9.85Ω From Eq Text (see also p. 19 of Class Notes) R r = 31,00 μ eff NS λ = 45Ω (3-54) Text 4 R. Pettengill, H. Garland, and J. Mendl, Receiving antennas for miniature receivers, IEEE Transactions on Antennas and Propagation, Vol AP-6, pp , July

24 From Eq. 11 above 6 L = 0.3H ω L = π =.18 MΩ D = 1.5; e r = R r R ohmic +R r = 0.8 (8%) G = e r D = 1.3 pp Antennas in Communication Systems P t G t P r G r Fig. 4-4 (p. 107 Text). A communication link. Voc = VA Equivalent circuit for the receiving system. Maximum available power to the receiver (for Z L = Z A ) PAm i 1 VA V E ( z) A 1 = RA = = RA 8RA 8 RA (1) For an ideal (infinitesimal) dipole V A = E i z () Maximum effective aperture area = A e,m = PAm 3 = λ = λ Sinc 8π (3) 1

25 i Sinc = E η (4) D = 1.5 for an ideal dipole (5) (4-) 4π 4π 3 D = A e,m = λ = 1.5 λ λ 8π (6) (4-3) Text For a general antenna, therefore G = 4π λ A e (7) A e = e r A em effective aperture area of an antenna (4-7) p. 108 Text Available power including also the antenna losses P A = S A e (4-6) Text or P S = G t t 4πR (4-31) GP t t GG t rλ Pr = SAer = A er = Pt 4πR (4πR) (4-33) AetA P er r = Pt R λ Friis transmission formula (4-33) We can also write Eq. (4-33) in db-form as follows: P r(dbm) = P t(dbm) + G t(db) + G r(db) 0 log R(km) 0 log f (MHz) 3.44 (4-34) Example 6: For Ground Based TV Stations Say, Channel 5 f = 76-8 MHz f 80 MHz λ = 3.75 m P rad ~ 5-10 kw say, 10 kw = 10 4 W (40 dbw)

26 G t ~ 0-50 (factor) say, G t = 30 (factor) (14.77 db ~ 15 db) EIRP = G t P rad 55 dbw W (85 dbm) say R max ~ 0-30 miles ~ 50 km; since 1 mile = 1.6 km G r 7 db 5 A e,r = λ 4π G r = 5.6 m Using the logarithmic form of the Friis communication link formula Eq. (4-34) Example 7: P r (dbm) = = 1.5 dbm = 10 mw = 56. µ W Calculate the open-circuit voltage developed across an antenna of resistance R A = 80 ohms for the above-calculated incident power density Assume R A = 80Ω V 8R oc VA A A Pr 56. µ W Sinc = = = 10 A e 5.6 m = power picked up and delivered to a matched load = SincA 8R V = 8R S A = 8 80 P = oc A inc e,r r = mv 0.19V 6 e 3

27 Chapter 8 -- Antenna Arrays (see pp Text) For a uniformly excited (UE), equally-spaced linear array (ESLA) F(x,y,z) d cos φ For N identical radiating elements (length, orientation, etc.) that are excited with identical magnitudes but progressively phase-shifted currents i.e. jα jα j( N 1) α I, Ie, Ie, Ie (1) we can write the total electric field E T as follows jβ r jβ r1 jβ r N 1 T= o N 1 (1) E E e Ee E e r = xˆ x + yˆ y + zˆ z r 1 = (x d)ˆ x + yˆ y + zˆ z β=β θ φ + θ φ + θ [ sin cos xˆ sin sin yˆ cos zˆ] () (3) From Eq. 1, we can write β j r α j jβdsinθcosφ jα jβdsinθcosφ ET = Ee o 1+ e e + e e + N 1 jβ r jnψ = Ee o e n= 0 since ( ) ( ) β r1 r = βd sin θcos φ; β r r = βd sin θcos φ (4) (3-16) (6) 4

28 From Eq. 4 where E T = E o AF N 1 jnψ jnψ 1 e Array Factor AF = e = jψ n= 0 1 e (7) AF = e j(n 1)ψ / sin(nψ / ) sin(ψ / ) Normalized AF f(ψ) = sin(n ψ / ) N sin(ψ / ) UE, ESLA (8) (8-19) p. 79 Text (9) (8- Text) where ( ) ψ= βd sin θcos φ α for an xˆ directed array x ( y y) x = βd sin θsin φ α for a yˆ directed array ( ) = βdz cos θ αz for a zˆ directed array (see Eq Text) (10) E T = NE o f(ψ) = sin(n ψ / ) * E o sin(ψ / ) (11) H T = E T = jβ ˆ r E T jωµ o jωµ o (1) * 1 * ET E S = Re T ( ET HT) = rˆ η (13) ψ * From Eq. 11, for directions of max radiation = 0, ±π, ± π, p. 80 Text A number of trends can be seen by examining the normalized array factor f(ψ). 1. As N increases the main lobe narrows. Peak for the main lobe occurs for ψ = 0 where f(ψ) =1. 5

29 For directions of zero (nulls of radiation) Zero values of f(ψ) occur for Plot of f(ψ) as a function of ψ. Fig. (8-8) p. 80 Text. i.e. Nψ = ± π, ± π, ψ = ± π N, ± 4π N, (14). More than one major lobe will exist if it is possible to get values of ψ = ± π, ± 4π. The additional lobes are called Grating Lobes. 3. The minor lobes are of width π/n in the variable ψ and the major lobes (main and grating) are twice this width i.e. 4π/N in the variable ψ. 4. The side lobe peaks decrease relative to the major lobe as 1 1 1: : 3π 5π N sin N sin N N For large N, SLL decrease as (16) i.e. or 1: 3π : 5π : (17) 0, 0 log 3π, 0 log 5π, 0, , , db (18) 5. As N increases, there are more side lobes in one period of f(ψ). See also the text, Fig. 8-8, p

30 Case A. Broadside Arrays All antennas excited in phase α = 0. From Eq. 10, for an antenna stretched along the z-axis for major lobes Subcase 1 ψ = βd cos θ = 0, ± π, ± 4π (19) π 1 λ 1 λ θ=±, cos ±, cos ± d d (0) d/λ < 1 i.e. interelement spacing less than λ Two and only two major lobes of radiation for θ = ± π/ i.e. in directions broadside to the stretch of the array For directions of first nulls ( θ FN ) 7

31 BWFN = θ θ (4) FN left FN right for Example 6: From Eq. 3 1 λ 1 λ = cos cos Nd Nd 1 sin λ λ radians λ = = Nd Nd Nd Nd >> λ d/λ = 0.5, N = 8 (5) (8-31) p. 83 Text (6) (8-33) π 1 1 θ FN =± sin =± (7) BWFN = 9 (8) Angle for first-side lobe from Eq. 11 Nψ sin = ± 1 Nψ ( ) = 8πd = 4 βd cos θ λ cos θ = ± 3π (9) 1 3 θ= cos ± =± 68 ; ± 11 8 (30) ( db down relative to major lobe) Subcase From Eq. 19, for major lobes 1 d λ ψ = βd cos θ = 0, ± π, ± 4π, (31) cos θ = 0, ± λ d, ± λ d (31a) This corresponds to six major lobes and a radiation pattern of the type shown on the next page. 8

32 Example 7: d/λ = 1.5 From Eq. 31, for major lobes For angles of maximum radiation 3π cos θ = 0, ± π, ± 4π (3) cos θ = 0, ± / 3, ± 4/ 3 (33) θ = ± 90 ; ± 48., ± (34) Six maxima of radiation θ= = θ=90 θ= d/λ > Angles for first nulls for each of these maxima are obtained from Eq. 1 3π cos θ FN = ± π N ; π ± π ± N (35) Subcase 3 d/λ = 1.0 For this case, there are four maxima of radiation (major lobes); the two fatter lobes in the above figure coalesce into single modes with directions of maximum radiation θ = 0, 180. From Eq. 31a (36) 9

33 θ = ± 90, 0, 180 (37) Four maxima of radiation From Eq. 1, directions of first nulls are: π π π ψ= d cos θ FN =± ; ± π±, λ N N (38) λ λ cos θ FN =±, ± 1 ±, (38a) Nd Nd Example 8: From Eq. 38a N = 8, d/λ = 1.0 cos θ FN = ± 1 8, ± 1 ± 1 8 = ± 1 8, 7 8, 7 8 (39) θ FN = 8.8, 97.18, 8.96, 8.96, ; (40) BWFN = for major lobes along ± 90 = 57.9 for major lobes along 0,

34 p. 315 Case B. Electronically-Scannable (Steerable) Antennas -- Phased Array Antennas The phase shift of currents (excitations) for adjacent antennas may be altered α βd cos θ o (phase delay) (41) From Eq. 11 for directions of maximum radiation (major lobes) ψ = 0, ± π, ± π, (4) ψ = 0, ± π, ± 4π, (43) ( ) βd cos θ cos θ o = 0, ± π, ± 4π, (44) cos θ = cos θ o, cos θ o ± λ d, cos θ o ± λ d, (45) For two and only two major lobes for θ = ± θ o, d/λ should be less than 0.5. θ o 1 α θ o = cos βd Example 9: N = 8, d/λ = 0.3 ; α = -30 ; θ o = 1 π /6 cos π 0.3 = ± 73.9 For α variable from -30 to -75 θ o varies from ± 73.9 to ± 46 31

35 For directions of zero radiation, from Eq. (14) on p. 5 of Class Notes, Nψ = ± π ψ = ± π N π βd ( cosθ cosθ o ) =± N cosθ FN π = cosθo ± Nβd θ FN 1 π = cos cosθo ± Nβd 1 λ = cos cosθ o ± Nd (45a) Example 9 (continued): d/λ = 0.3, N = 8 α = -30 = -π/6 θ o = ±73.9 θ FN = cos ±.4 = ; BWFN = Example 9, Part B: Let us compare the antenna array of N = 8, d = 0.3λ for the following three conditions: α 0 Broadside array -30 from Ex. 9 on this page -108 directions of max radiation θ o = ±73.9 End fire antenna array α = -βd Direction of max radiation θ o = ±90 from Eq. (0) on p. 7 of Class Notes 3 BWFN from Eq. (6) on p. 8 of Class Notes 1 λ BWFN = sin Nd = 49.5 θ o = ±73.9 BWFN = θ o = 0 1 λ BWFN = cos 1 Nd = See Eq. 5 on p. 36 of Class Notes

36 For a one-dimensional antenna array The array factor of a one-dimensional antenna array from Eq. (8) of Class Notes p. 5 is as follows: AF ( Nψ ) sin = (1) sinψ Where ψ is given by Eq. (1) on p. 5 of Class Notes. From Eq. (1) here, for directions of max radiation ψ = 0, ±π, For directions of zero radiation or nulls of radiation Nψ =± π, ± π, or ψ = ±π/n for first nulls of radiation. Table of general relationships for one-dimensional z-directed ẑ phased array antennas α 0 α α = -βd directions of max. radiation principal lobe/s ψ = 0 θ o = ±90 broadside array 1 α θo = cos βd see p. 31 of Class Notes θ o = cos -1 (1) = 0 End fire array directions of first nulls derived on p. 3 1 λ cos cosθ o ± Nd 1 λ = cos ± Nd see Eq. () on p. 7 of Class Notes 1 λ cos cosθ o ± Nd 1 α λ = cos ± βd Nd see Eq. 45a on p. 3 of Class Notes λ 1 cos 1 Nd BWFN 1 λ sin Nd see Eq. (6) on p. 8 of Class Notes calculate θ FN1, θ FN BWFN = θ FN - θ FN1 1 λ cos 1 Nd 1 λ = 4sin Nd see Eq. 5 on p. 36 of Class Notes 33

37 Case C. End Fire Arrays From the previous section, we can see that in order to get a single major lobe for θ o = 0 i.e. along the line or stretch of the array, we need α = - βd and d/λ < 0.5 (47) For this case, the two major lobes on the previous page coalesce into one major lobe in the end fire direction. Example 10: N = 0, d/λ = 0.4 α = βd = πd λ = 144 (48) For directions of first nulls from Eq. 14 π ψ=βd ( cos θfn 1) =± (49) N (49a) 1 7 FN cos θ =± =± BWFN = θ FN = 57.9 (50) (50a) Example 11: It is interesting to note that for a given stretch of the array (N-1)d or approximately Nd, BWFN is smallest for broadside arrays, intermediate for phased arrays and broadest (largest) for end fire arrays. For N = 0, d = 0.4 broadside array ( α= 0 ) 1 λ 1 1 BWFN = sin = sin = Nd 8 as compared to 57.9 for an end fire array. 34

38 Example 10 (continued): N = 0 ; d/λ = 0.4 For Hanson-Woodyard end fire array (p. 85 Text) π α = βd + N (8-37) ; λ 1 d < 1 0 (8-38a) 180 = d < λ = 153 For directions of first nulls (from Eqs. 10, 14 on pp. 5, 6 of Class Notes) π π ψ = βd z cosθ α z = 144 cosθ FN 153 =± =± =± 18 N 0 cos θ FN 153 ± = = , 135 = rather than 7/8 or in Eq. 49a θ FN = ± ( ) 1 cos = ±0.36 BWFN = 40.7 rather than 57.9 for an ordinary end fire array (see Eq. 50a on p. 34 of Class Notes) 35

39 Nd/λ BWFN for antenna arrays Example 11-1: 1 λ = cos 1 Nd 1 λ = 4sin Nd Nd = 5.0 λ BWFN = 3.07 for a broadside array BWFN = for an end fire array p. 93 Directivity of a Uniformally Excited, Equally Spaced Antenna Array From Eq. 13 we can write AF E S = T E * T ˆ r = E o sin Nψ η η sin ψ ˆ r (53) S max = S o max AF max N S o (54) max 1 I A R A0 R A1 R N 1 N 1 A Ai i= 0 Power radiated by the antenna array = ( + + ) 1 = I R (55) 36

40 In general where D o and RA0 R D = N D A0 array o N 1 RAi i= 0 pertain to an isolated element of the antenna array. (55a) Ignoring Mutual Impedance Effects R A0 = R A1, = R N-1 Power radiated by the antenna array = 1 I A RA0 N = N P o (56) where P o is the power radiated by the zeroth element max S AF D= max = Do = DoN NPo N 4πr (57) Example 1: where D o is the directivity of each of the antenna elements. Calculate the directivity of an antenna array of 0 half wavelength (L = λ/) dipoles that are fed in phase and consequently radiate in broadside directions. Neglect the mutual impedance effects for this problem. Solution: Example 13: AF D = max = NDo = = 3.8 N a. Calculate the directivity/gain of an array of 30 vertical monopoles above ground each of length H = L/ = 0.35 λ that are spaced a distance d = 0.λ from each other. b. Calculate the relative phase difference between monopoles if the major lobe of radiation is to be in the end fire direction assuming an ordinary end fire array. c. Calculate the BWFN for this array. Solution: a. D = N D o = = b. From Eq. 47 α = βd = πd λ = 7 37

41 Each of the successive elements should be fed with a current that is lagging in phase by 7 from the previous element. c. From Eq. 49a, cos θ FN = 1 λ Nd =1 λ 6λ = = 5 6 BWFN = cos -1 (5/6) = D and 3-D Uniformly Excited, Equally-Spaced Antenna Arrays N x : No. of antennas in x-direction N y : No. of antennas in y-direction N z : No. of antennas in z-direction A -D array of identical elements Neglecting phase terms E T = E 1 AF x AF y AF z 38

42 AF x N sin x dx sin cos = 1 sin dx sin cos x x ( β θ φ+α ) ( β θ φ+α ) AF y Ny sin d sin sin = 1 sin d sin sin ( β y θ φ+αy) ( β y θ φ+αy) As always N sin z AF = z 1 sin ( βd cos θ+α ) z ( βd cos θ+α ) H T = E T = jβˆ jωµ o jωµ o z r E T z z = ˆ r E T η T = 1 AF x AF y AF S S z where S S 1 1 is the radiated power density due to one of the elements. These arrays are also called mattress Arrays. Example 14: A Unidirectional Broadside Array In order to obtain a unidirectional broadside array, we can use a -D antenna array of N z = 1, N y =, N x which can be an arbitrary number. By using a back row of antennas that are placed with d y = λ/4 and α y = 90, we can obtain an antenna pattern as shown. ŷ λ /4 ˆx 39

43 Universal field-pattern chart for arrays of various numbers n of isotropic point sources of equal amplitude and spacing Array factor or f( ψ ) 40

44 41

45 0 90 An enlarged version of Fig. (b) from previous page. α = 90 ; d = λ/4 Note a broad unidirectional (cardiod type) pattern possible with this arrangement. 4

46 Reactance of Linear Dipoles We have previously calculated R in = R a + R ohmic for linear dipole or monopole antennas. We need to know the input reactance X in or X a in order to design matching networks to match power in or out of the antenna. Like the current distribution on a linear dipole, the input reactance can be written as though a two-wire line of length L/ had been opened up as shown in the following: D d E E L/ S D E L/ E z a. L/ D D g/ z = 0 -g/ z D D S(z) z E E b. c. For a two-wire line of Fig. a, each of diameter d, Zo 10 S = n εr d (1) For the opened-up line of Fig. b, we can define an average characteristic impedance Z o Z L/ 1 o = Z o(z) dz L/ () 0 43

47 For the completely opened-up transmission line of Fig. c, we can define g/+ L/ 10 4z Zoa = n dz L ε d r g/ 10 L = n 1 ε d r (3) The reactance Z D D of an open-circuited transmission line of length L/ can be written from Transmission Line Theory L ZDD = Zin = jzoa cot β (4) Combining Eq. 3 and 4, we can write βl ZDD = jxin = j Zoa cot (5) where L ( )L is the effective "electrical" length of the antenna. Reactance of Linear Monopoles Above Ground We have previously shown that Similarly, R in monopole = 1 R in dipole (6) 1 L Zoa = Zoa = 60 n 1 (7) monopole dipole d From Eq. 5 Xin monopole 1 Xin = (8) dipole Example 15: Calculate the feed point impedances R in + jx in for linear dipoles of length (a) L = 0.5λ (half wave dipole) and (b) L = 0.3λ. Assume that the antenna wire is No. 19 AWG (d = m from Table B., p. 63) and frequency f = 30 MHz. Take copper as the material for the antenna. 44

48 a. From the table on driving point resistance, p. 7 of Class Notes R = = 73.1Ω (9) ri L / λ= 0.5 From p.13 of Class Notes, Eq. 9 Rs 1 L sin ( βl) Rohmic = πa π 4 4β sin (10) σ copper = S/m 1 4 Rs = = fmhz σδ for copper 4 4 Rs = = f = 30 MHz (11) L = λ = 5 m R ohmic = π =1.411Ω (1) 4 From Eq. 3 on p. 44 of Class Notes Taking L 1.0 L = 0.51λ R in = R ri + R ohmic = 74.53Ω (13) 10 Z 10 n oa = = Ω 4 (14) From Eq. 5 on p. 44 of Class Notes π 0.510λ cot = λ βl jxin = j cot =+ j 31.3Ω (15) Z in = R in + jx in = j31.3ω (16) 45

49 Note that if we had constructed a slightly shorter, say L = 0.49λ dipole L = λ = 0.5λ jx in = j cot β L 0 Z in = R ri + R ohmic + j0 = j j0ω (17) b. You can solve for the numbers for part b of the problem following the procedure indicated above. Example 16: Solution: Feedpoint impedance for a linear monopole of length L/ = 0.5λ. From Eq. 16 Z in monopole = 1 Z in dipole = j15.65ω Examples on Calculation of Im (Z A ) or Reactance of Antennas Example 17: (See also Fig. 6-6, p. 157 Text) Example 18: L/λ = 0.4; wire radius a = λ (same as in Fig. 6-6, p. 157 Text). Assume L = 1.04 L. From Eq. 3 on p. 44 of Class Notes 0.8λ Zoa = 10 n 1 = 68.15Ω 0.001λ From Eq. 5 on p. 44 of Class Notes πl ZDD = j Zoa cot = j68.15 cot (0.4π 1.04) = j184.3ω λ taking L = 1.04L (from Table 6- on p. 159 Text). From the graph in Fig. 6-6, p. 157 Text Im ( ZA ) = j180ω L/λ = 0.3; wire radius a = λ (one of the wire radii on p. 8 of Class Notes). 46

50 From Eq. 3, p. 44 of Class Notes 0.6λ Zoa = 10 n 1 = 54.08Ω 0.008λ π L ZDD = j54.08 cot L = j54.08 cot 0.3π = j351.4ω λ L taking L / L = From graph on p. 8 of Class Notes which is close. Reactance X a = 150 = 300Ω Examples on Mutual Impedance Effects Example 19: A half-wave dipole above ground x Distance to ground d g = λ /4 d g = distance to image antenna = λ / From Fig. 8-5a, b, for d/λ = 0.5 d g d g 1 z Image antenna Z1 = 1.5 j30; I = I1 180 = I1 47 [ ] V = I Z + I Z = I Z Z V Z 1 1 = = (73+ j4.5) ( 1.5 j30) I1 = j7.5 Feedpoint impedance of the half-wave dipole placed at a distance of λ /4 from the ground = j7.5ω rather than 73 + j4.5ω. Radiation Pattern We can consider the above situation as a -element (N = ) antenna array in the x direction and write E = E AF T 1

51 Figure 8-5 The mutual impedance between two resonant parallel dipoles as a function of their spacing relative to a wavelength. (a) The real part. (b) The imaginary part. 48

52 Figure 8-6 The mutual impedance between two resonant collinear dipoles as a function of spacing relative to a wavelength. (a) The real part. (b) The imaginary part. 49

53 ( ψ ) ( ψ ) sin N / AF = sin / where ψ=βd sin θcos φ+α = βd sin θcos φ+π x x g =πsin θcos φ+π From pp of Class Notes, Eqs. (54)-(57) R 73 D = G = Do AF = 1.64 N 4 = 5.60 R 85.5 A,isolated max A,with ground effect Without ground effect D = G = 1.64 Example 0: A broadside array of five λ/4 monopoles (α = 0) d = λ / Ant. #1 # #3 #4 #5 d 1 = λ / d 13 = λ d15 = λ d14 = 3 λ / I1 = I = I3 = I4 = I5 because it is a broadside array V Z 1 1 = = Z11 + Z1 + Z13 + Z14 + Z15 I j7.5 = [(73+ j4.5) + ( 1.5 j30) + (4 + j18) + ( 1.8 j1) + (1 + j9) ] = = j13.75ω monopoles Z5 = Z1 by symmetry 50

54 Same as Z 1 d5 = 3 λ / Z = Z1 + Z + Z3 + Z4 + Z5 d 4 = λ 1 = ( 1.5 j30) + (73 + j4.5) + (4 + j18) + ( 1.8 j1) monopoles 1 = [ 50. j11.5 ] = 5.1 j5.75 Ω [ ] Z = Z4 by symmetry Same as Z 3 Z3 = Z13 + Z3 + Z33 + Z34 + Z35 = Z13 + Z3 + Z33 Same as Z 13 1 = (4 + j18) + ( 1.5 j30) + (73 + j4.5) monopoles 56 + j18.5 = = 8 + j9.5 [ ] Note that for each of the antennas, the input impedances are slightly different and each of these values are different than 73+ j4.5 or j1.5ω Directivity for an isolated λ/4 monopole. From Eq. (55a) on p. 37 of the Class Notes, including mutual impedance effects D = D AF R A isolated o max 5 RAi i= = 3.8 N 5 = Re Z Z Re Z Z Re Z ( + ) ( + ) ( ) Do = for a single isolated λ/4 monopole above ground Note that a directivity of 1.1 is higher than ND o of = 16.4 which would be obtained for this antenna array neglecting mutual impedance effects. 51

55 Inclusion of mutual impedance effects can often lead to an increased gain relative to the value had the mutual impedance effects been neglected. Example 1: Two monopole antennas separated by λ/4 antenna is grounded.). (Note that the second V1 = Z11I1 + Z1I 0= Z1I1 + ZI λ/4 I 1 λ/4 I Z 1 (36 j5) / I = I1 = I1 Z (73+ j4) / j e j64.78 j180 = I1 = 0.5 e e I 1 + j e + j115. = 0.5 e I 1 For the driven antenna 1 V1 I Z 1 = Z11 Z1 I = + 1 I (1) 1 From p. 307, Fig. 8-5 of the Text (see also p. 48 of the Class Notes) From Eq. (1) 36 j5 j34.8 Z1 = = 1.91 e d =λ /4 monopole 73+ j4 j115. j34.87 Z1 0.5 e = e = j3.ω Calculate current I 1 for a transmitter power of 100 W Antenna 1 is the only antenna that is driven and is to be fed (current in antenna is created by induction) 1 1 Prad = 100W = I1 RA1 = I I1 =.8A Because of induced current (by mutual impedance effect) 5

56 j115. j115. I = 0.5 I1 e = 1.19 e A Example : Calculate the feedpoint impedances of two parallel antennae separated by a distance of λ/4 and fed with a phase shift α = -90º. Each of the antennas is a λ/ dipole. Rad. pattern 1 From Fig. 8-5, p. 307 Text (p. 48 of Class Notes) Z11 = 73 + j4.5 ; Z1 = 36 j j4.5 V V 1 1 = I1Z11 + IZ1 Z1 = = Z11 j(36 j5) I1 Z1 = 48 + j6.5ω V V = I1Z1 + IZ Z = = j(36 j5) + (73 + j4.5) I Z = 98 + j78.5ω Power fed to Ant. 1 = Power fed to Ant. = 1 I KW 1 I KW 1 I 1 R A1 + R A = 10 KW Total power = ( ) I1 = 11.7 A RA,isolated 73 G = N G 1 = RAi i= 1 =

57 Methods of Matching Power to the Antennas A. Transmission Line Matching Method Example 3: Match an antenna of impedance Z a = 10 - j300ω to a twin-wire line of characteristic impedance Z o = 300Ω using (a) series elements and (b) a shunt element. Take f = 30 MHz z a = Z a 10 j300 = = j1 Z o 300 This normalized impedance is shown as point A on the Smith Chart on page 55 of Class Notes. If the antenna is not matched Voltage reflection coefficient ρ = Z a Z o Z a Zo = (1) 1+ρ VSWR = = ρ () Power reflection coefficient = P r P inc = ρ = (3) i.e % of the input power P inc is reflected and only 6.46% of the transmitter power is radiated -- a truly poor situation! Approach A: Use of series elements to match the antenna From point A, we move on the transmission line circle C to point B on p Smith Chart, which corresponds to the intersection with real part z B of 1.0 circle. Length AB = ( )λ = 0.356λ 3.56m (4) Z B B = z B Z o = (1 + j8)300 = j400ω (5) As shown in Fig., we can compensate for j400ω by using two capacitors as shown each of reactance jx se = j 400 = j100 This gives the values of series capacitances C se = 4.4 pf (6) 54

58 Example of the Transmission Line Matching Method f = 30 MHz ; λ o = 10 m ; Z a = 10 - j300ω 55

59 Approach B: Fig.. An alternative design using a shunt element to match the antenna An undesirable feature of the above design Approach A is that it takes a fairly long length AB = 0.356λ over which the transmission line is not matched. For the alternative Approach B, we work in terms of admittances. Y A = 1 10 j300 ; y A = j1 (7) j1 This is shown by point a on the Smith Chart on page 55. Now, we need to move only a distance ab = ( )λ = 0.106λ = 1.06 m (8) and use (as sketched in Fig. 3) a shunt element to match the line. Fig j Y sh = j 300 mho j (9) ω L sh 300 L sh = 6 = 0. µh (10) 8 π

60 USE OF LUMPED ELEMENTS FOR MATCHING AN ANTENNA Example 4: A Matching Circuit for an Antenna of a Cellular Telephone Topology 1 The antenna impedance is given to be 50 - j0ω. The solid-state source to which this impedance is to be matched has an internal impedance, say 15 + j130ω. A possible matching circuit is sketched as follows: V s Fig. 1 For maximum power transfer to the antenna Z AB = Z S = 15 j130ω (1) jx sh (50 j0) ZAB = + jx se 50 + j( Xsh 0) ( 0 Xsh + j50 Xsh ) 50 j( Xsh 0) (50) + ( X 0) = + jxse sh = 15 j130ω () Equating real parts on both sides of Eq. ( ) 1000 X + 50 X X 0 = X 40 X sh sh sh sh sh 35 X sh Xsh 43,500 = 0 (3) X sh 600 ± (600) , ± 540 = = = ; Ω Taking the capacitive shunt reactance -j44.86ω and equating the imaginary parts on both sides of Eq., we get 57

61 1 Xse = j104.6ω= j ω Cse 1 Xsh = = 44.86Ω ωcsh For f = 900 MHz Csh = 3.94 pf 1 Xse = = 104.6Ω ωcse Cse = 1.69 pf The matching circuit for Topology 1 is as follows: A B Topology A Fig. B Fig. 3 This problem may be easier to solve in terms of admittances 58

62 j YAB = = = + 15 j130 (15) + (130) 50 + j(xse 0) jxsh (4) Equating real parts = 17,15 (50) + (Xse 0) (5) Taking the sines inductance (Xse 0) = = 57, Xe = 53.6; 13.6Ω Xse = 53.6Ω Lse = nh Xsh = 85.59Ω Csh =.06 pf Implications for Power Transfer a. Without conjugate matching, for an oscillator voltage Vs = V RMS power Power transferred to the load = () IrmsRA = 50 = 1.5 mw ( ) + (130 0) b. With conjugate matching Power transferred to the load = () I rms Re Zs = 15 = 66.7 mw ( ) Needed for 600 mw power transferred to the load Vs = 6V RMS 59

63 A REACTIVE THREE-ELEMENT CIRCUIT FOR ANTENNA MATCHING Example 5: A reactive three-element network is a versatile circuit for matching power onto the antenna. To illustrate the procedure, let us look at the circuit of Fig. 1. The antenna equivalent impedance is R A + jx A. R A jx A D A 3-reactance matching network. Figure 1 In order to match power into the antenna, it is necessary that the impedance of the network between points A and B be purely resistive and have the same value as Z o, the characteristic impedance of the transmission line. From Fig. 1, the expression for the impedance Z AB can be written as: ( + + ) RA jxa jx1 jx ZAB = + jx3 RA + jxa + jx1+ jx (1) We select X 1 and X such that the reactance in the denominator of the first term is zero, i.e., Equation 1 can then be rewritten as: XA + X1+ X 0 () ( ) RA jx jx ZAB = + jx3 (3) RA We select X such that X Zo R = (4) A 60

64 and X 3 such that This would then give X 3 = -X (5) Z AB = Z o + j0 and the antenna would then be matched onto the transmission line. To illustrate the procedure by a numerical example, let us say that the antenna is a monopole and its impedance ZA has been calculated and found to be j460ω. Let us take Z o = 300 ohms (we must, of course, make sure that the diameter of the feeder line is not overly thin for the current-carrying requirement). From Eq. 4, X =± =± 1.Ω (6) The upper sign corresponds to an inductance L = 1.1/ω and the lower sign corresponds to a capacitance We can use either type. C = 1 ω 1.. Case 1: For inductive element X jx = jω L = j1.ω If ω is prescribed, L can be calculated. From Eq., X 1 = - X - X A = This implies an inductor for jx 1. From Eq. 5, = 438.8Ω X 3 = - 1.Ω One possible 3-reactance matching network is, therefore, shown in Fig.. 61

65 Ω Ω Ω Figure Case : For capacitive element X 1 jx = = j1.ω jωc From Eq. on p. 60 of Class Notes, X 1 = -X - X A = = 481.Ω jx1 = jω L1 = j481.ω (an inductor) From Eq. 5 on p. 60 of Class Notes, jx 3 = - jx = + j1.ω (also an inductor) and a second possible 3-reactance matching network is shown in Fig. 3. Z o = 300Ω feeder line j1.ω j481.ω -j1. Ω Figure 3 6

66 Gain of a dipole antenna placed in a corner reflector of corner angles α = 180 (flat reflector) and α = 90 (90 corner reflector). H. Jasik, Antenna Engineering Handbook, McGraw Hill & Co. α = 90 α =

67 From: K. F. Lee, Principles of Antenna Theory, John Wiley & Sons, 1984 α = 30 α = 30 64

68 H. V. Cottany and A. C. Wilson, "Gains of Finite Size Corner Reflector Antennas," IEEE Transactions on Antennas and Propagation, Vol. AP-6, 1958, pp Contours of constant gain for a 90º corner reflector = 41. Width of the reflecting planes in wavelengths, W/λ Maximum for infinite reflectors = 1.9 db (19.5) 65

69 Some Commonly Used Feeder Lines for Antennas 1. Twin Wire Transmission line 10 S Zo = n εr d (Replace εr by ε eff for relatively thin dielectric sheathing) Example: S d =1. for Z o = 300Ω (air-filled line). Wire Above Ground Transmission Line 60 S 60 4h Zo = n = n εr d εr d Example: h d = 3.05 for Z o = 150Ω (air-filled line) 3. Coaxial Line Z o = 60 n b ε r a Example: 66

70 b a = for ε r =.1 (Teflon) coaxial line of Z o = 50Ω Some of the other transmission lines useful for printed antennas are: a. Miscrostripline b. Slot line etc. Ground Effect on Radiation Pattern of an Antenna We have previously considered the effect of ground for the radiation from a vertical monopole antenna. The net effect was that the monopole antenna of length L/ radiates electromagnetic fields much like a dipole of length L albeit for the upper half plane i.e. for field points above ground. For a horizontal dipole antenna placed at a distance h from the ground as sketched in Fig. 1, an image antenna 1 is created, which has a current excitation that is equal in magnitude (for high conductivity ground) but 180º out of phase with that in the installed antenna #1. 90 φ Fig. 1. A horizontal dipole antenna above ground. From Eq. 10 on p. 4 of the Class Notes, this can be considered as a two-element array (N y = ) with a phase difference α y = π or 180º. 67

71 ψ sin 1 ET E1 AF E = = y 1 = E1 cos ( βh sin θsin φ+π) ψ sin = E1 sin ( βh sin θsin φ) neglecting the phase factors both in writing AF y and E 1. Note that Eq. 1 could also have been written by following a procedure similar to that for Eq. 4 on page 4 of the Class Notes. (1) β j ( r1 r1) β j ( h sin θsin φ E ) T = E1+ Ei = E 1 1 e = E 1 1 e β j h sin θsin φ β j h sin θsin φ = E 1 e e = E1sin βh sin θsin φ ( ) ignoring the phase factors, as also done in writing Eq. 1. From Eqs. 1 and () for θ = π/ i.e. xy plane. For maxima of radiation AF = sin ( βh sin θsin φ) sin ( βh sin φ ) (3) Example 6 For first nulls of radiation βh sin φ o = ± π, ± 3 π, (4) βh sin φ FN = 0, ± π, ± π, (5) a. Calculate the spacing h to ground for a half-wave dipole antenna if the maximum of radiation is desired for angle φ o = 30º off the horizon. b. Calculate the directions of maximum and zero radiation for the selected h. c. Calculate the gain of the antenna, without and with mutual impedance effects. Solution: From Eq. 4 for φ o = 30º, sin φ o = 0.5 a. h = λ 4 sin φ o, 3λ 4 sin φ o, = λ, 3λ, 5λ, (6) 68

72 h = λ / φ ο h sin φ o In order to keep the number of principal maxima to a minimum number, we select the smallest spacing to the ground plane i.e. h = λ/ (7) b. For this spacing itself, we note from Eq. 4 that the directions of maximum radiation are: βh sin φ o = π sin φ o = + π (8) φ o = 30 (wanted), φ o = 150 (unwanted) Negative sign is ignored in Eq. 8 since that gives angles φ o = -30º, -150º (both into the ground). We will see later how to eliminate the unwanted radiation for φ o = 150º. If we had taken a larger h of say 3 λ/ from Eq. 6, we would have had many more directions of maximum radiation. For directions of first null, from Eq. 5, φ FN = 0 and sin -1 (1) or 0 and 90º for the principal maximum of φ o = 30º, and φ FN = 180º and sin -1 (1) or 180º and 90º for the principal maximum at φ o = 150º. The radiation pattern is sketched in Fig.. φ o = 150º φ o = 30º Fig.. 69

73 c. Ignoring mutual impedance effects R a1 = 73Ω (same as for an isolated half wave dipole). From Eq. 1 E max = E 1 ; S max = E max η = 4S 1 Gain = 4G 1 = = 6.56 Height above ground, h/λ Fig. 3. Angles φ of maximum and zero radiation for a horizontal dipole antenna above ground (From Eq. 1,, or 3). Elimination of Unwanted Principal Lobes of Radiation As seen in Example 4, there is an unwanted principal lobe of radiation for φ o = 150º that we would like to eliminate leaving thereby one and only one principal lobe of radiation for the desired direction φ o = 30º. A possible solution for this problem is as sketched in Fig. 4. jo e Fig. 4. An arrangement of two horizontal dipoles above ground. 70

74 We take two horizontal dipoles 1, above ground. Distance to ground h is the same for both dipoles 1 and. Shown in Fig. 4 also are the two image antennas 1,. Assuming that antenna #1 is leading in phase by α (i.e. antenna is lagging in phase by α). α - β d 1 cos φ o = 0 (9) for addition of signals along φ o = 30º principal lobe. α + β d 1 cos φ o = π (10) for complete cancellation of radiation in the back direction. Note in both Eqs. 9 and 10, φ o = 30º and d 1 cos φ o = d 1. From Eqs. 9 and 10, both the unknown α and d 1 can now be found: α = π = 90 (phase lead angle for antenna #1) (11) βd 1 cos φ o = π d 1 = λ 4 cos φ o = 0.89λ (1) This arrangement would cancel the principal lobe for φ o = 150º (in Fig. ) while reinforcing the principal lobe for the φ o = 30º angle of radiation. 71

75 p. 349 Text General Theory of Aperture Antennas (or Displacement Current Antennas). For comparison, see also the General Theory of Conduction Current Antennas on p. 44 of the Text or p. of Class Notes. aperture R J = nˆ H s a Source of Fields H a Define Equivalent surface currents Source of Fields a (9-10) Ms = Ea nˆ E (9-11) 7 µ J(r ) j( ωt βr) ε M A = e ds R = r r s(r ) j( ωt βr) 4π F = e ds R 4 π R Sa Sa (9-1) β j r µ e β j r jβ rr ˆ ε e ˆ = ˆn Hae ds jβ rr 4πr = ˆn Eae ds 4πr Sa Sa Q P F jβ rˆ F jβ F rˆ A jβ A E = = = H1 = = (1) ε ε ε µ µ = jωη F rˆ H1 jβ H E 1 1 = = = ω j A () jωεo jωεo ET = E1+ E = jωa jωη F rˆ β j r jβ e j rr ˆ ET E1 E rˆ nˆ Ea rˆ ( nˆ β = + = η Ha) e ds 4πr Sa ET jβ rˆ ET rˆ E H T T = = = jωµ o jωµ o η 1 ET E S = Re T ( E H ) = rˆ η Total radiated power = S ds sphere (9-16) (9-17) (9-13) (3) Chapter 9 Aperture Antennas 7

76 p. 466 Text A Rectangular Microstrip Patch Antenna λ / y W E E 1 Notes 1. Note that the E-fields have no variation in the y direction; hence are identical for the front and back edges separated by width W. E z. Because of a separation distance of λ/, the E-fields at edges E 1 and E are 180º out of phase; also no variation in y- direction. Fig. 1. Distributions and variations of electric fields at the four edges of the patch antenna. (9-b) p. 346 Text z y x uniform magnitude; - -directed half cycle cosinusoidal variation Fig.. Equivalent surface currents at the four edges. Maximum radiation in z-direction (normal to the patch) due to two uniform magnitude "dipoles" corresponding to edges and ; radiation from edges F and B i.e. the front and back edges of Fig. 1 cancels out. 73

77 The array factor for the equivalent current dipoles at edges E 1,E can be written from Eqs. 8 and 9 on page 5 of the Class Notes. For an x-directed array of two elements N = ; α x = 0 ; 0 ψ=βl sin θcos φ+α x Normalized AF = Nψ sin ψ sin ψ ψ sin cos βl = = cos sin θcos φ ψ sin (1) For a two-element (two-edge "currents") antenna array E = E AF T o For a uniformly-excited rectangular aperture of length W (e.g. edges E 1,E ), from Eqs. 9-36a, b (note that equivalent current M s yˆ here rather than parallel to ˆx on page 354 of the text) Eθ = Eo cos φf ( θφ, ) (11-5a) where E E cos sin f (, ) φ = o θ φ θφ (11-5b) L y βw sin sin θsin φ f ( θφ, ) = AF βw sin θ sin φ βw sin sin θsin φ βl = cos sin θcos φ) βw sin θsin φ (11-5c) In Eqs. 9-36a, b thickness t βl sin z u 1 since t << λ βlz u 74