ANTENNAS and WAVE PROPAGATION. Solution Manual

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1 ANTENNAS and WAVE PROPAGATION Solution Manual A.R. Haish and M. Sachidananda Depatment of Electical Engineeing Indian Institute of Technolog Kanpu Kanpu , India OXFORD UNIVERSITY PRESS

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3 Contents Electomagnetic Radiation 5 2 Antenna Chaacteistics 9 3 Wie Antennas 45 4 Apetue Antennas 59 5 Antenna Aas 83 6 Special Antennas 97 7 Antenna Measuements 05 8 Radio Wave Popagation 07 3

4 4 CONTENTS

5 Chapte Electomagnetic Radiation Solution. The unit vectos in spheical coodinates, viz., a, a θ, and a φ epessed in tems of a, a, and a z as see Eample.7, can be a = a sin θ cos φ + a sin θ sin φ + a z cos θ a θ = a cos θ cos φ + a cos θ sin φ a z sin θ a φ = a sin φ + a cos φ Taking the dot poduct of each of the unit vectos with itself, a a = sin 2 θ cos 2 φ + sin 2 θ sin 2 φ + cos 2 θ = a θ a θ = cos 2 θ cos 2 φ + cos 2 θ sin 2 φ + sin 2 θ = a φ a φ = sin 2 φ + cos 2 φ = Taking the dot poducts with each othe, a a θ = sin θ cos θ cos 2 φ + sin θ cos θ sin 2 φ sin θ cos θ = 0 a θ a φ = cos θ cos φ sin φ + cos θ sin φ cos φ = 0 a a φ = sin θ cos φ sin φ + sin θ sin φ cos φ = 0 Theefoe, the spheical coodinate sstem is othogonal. Solution.2 In ectangula coodinate sstem, the gadient opeato is given b Taking the cul of φ, Epanding the deteminant, 2 φ φ = a z 2 φ z dφ φ = a d + a dφ d + a dφ z dz a a a z φ = z dφ dφ dφ d d dz 2 φ + a z 2 φ 2 φ + a z z 2 φ = 0 5

6 6 CHAPTER. ELECTROMAGNETIC RADIATION Solution.3 The cul of a vecto in ectangula coodinates is given b a a a z A = z A A A z Epanding the deteminant, Az A = a A z Taking the divegence, 2 A z A = 2 A + z A + a z A z A + a z A 2 A z 2 A z 2 A + z 2 A = 0 z Solution.4 In ectangula coodinates a a a z A = z A A A z Az A = a A A + a z z A z A + a z A Now we can epand A as A = { A a A A z z A } z { Az +a z A A z A } { A +a z z A z Az A } z { 2 } A = a + 2 A z z 2 A 2 2 A 2 z { 2 } A +a + 2 A z z 2 A 2 2 A 2 z { 2 } A +a z z + 2 A z 2 A z 2 2 A z 2 { 2 A = a + 2 A z z + 2 A 2 2 A 2 2 A 2 2 A 2 z { 2 A +a + 2 A z z + 2 A 2 2 A 2 2 A 2 2 A 2 z } }

7 7 { 2 } A +a z z + 2 A z + 2 A z 2 z 2 A z 2 2 A z 2 2 A z 2 z A = a + A + A z z A +a + A + A z z A +a z z + A + A z z a 2 A a 2 A a z 2 A z = A 2 A Solution.5 In a souce-fee egion the Mawell s equations educe to Taking the cul of the fist equation, E = jωµh H = jωɛe E = 0 H = 0 E = jωµ H Epanding the L.H.S. using the vecto identit and substituting the epession fo H on the R.H.S fom the second equation, Since E = 0 and ω 2 µɛ = k 2, we get E 2 E = jωµjωɛe 2 E + k 2 E = 0 Similal, taking and cul of the second equation, H = jωɛ E Epanding the L.H.S. using the vecto identit and substituting the epession fo E on the R.H.S fom the fist equation, H 2 H = jωɛ jωµh Since H = 0, we get 2 H + k 2 H = 0 Solution.6 Fom Eqns.3 to.34, we can conclude that V = V 0 e jk / epesents a wave with a velocit v = ω/k. Since both ω and k ae positive numbes, the

8 8 CHAPTER. ELECTROMAGNETIC RADIATION velocit is positive. Theefoe, V epesents a wave tavelling in the positive diection. Solution.7 a The equiphase suface is a sphee of adius with cente at the oigin. b Equiphase suface is a plane =constant. Solution.8 Fom Eqn.53, we have Substituting into Eqn.57 A z = µ I 0dl e jk A = a φ µ I 0dl e jk sin θ θ µ I 0dl e jk cos θ Pefoming the indicated diffeentiation, µ A = a φ I 0dl jke jk sin θ + e jk Substituting in and eaanging, we get = a φ H = 0 H θ = 0 µ I 0dl e jk sin θjk + H = µ A H φ = jk I 0dl sin θ e jk + jk sin θ Solution.9 Substituting the epession fo H φ fom Eqn.6 into Eqn.63 E = jωɛ 2 sin θ a θ a θ sin θjk I 0dl sin θ sin θjk I 0dl sin θ { e jk Pefoming the indicated diffeentiation, E = a jωɛ 2 2 sin θ cos θjk I 0dl sin θ a θ jk I 0dl sin 2 { θ e jk { + jk e jk { + } jk } + } jk } jke jk jk e jk jk e jk jk 2

9 9 Substituting ωɛ = k/η and simplifing E = η a cos θjk I 0dl e jk { + } jk 2π jk { +a θ jk I 0dl sin θ jk e jk This can be witten in component fom as E = η I 0dl cos θ 2π E θ = jη ki 0dl sin θ E φ = 0 e jk e jk + jk k 2 + jk + jk k 2 } Solution.0 Fom Eqn.53, the magnetic vecto potential due to a z diected cuent element is, µ A = a z I 0dl e jk whee, = z 2 The cul of A in ectangula coodinates is given b Az A = a A A + a z z A z A + a z A Since A and A ae zeo, Diffeentiating A z with espect to, A z A = a a A z A z = µ I 0dl e jk We now compute Similal, e jk = = e jk jk e jk z 2 e jk 2 jk e jk = e jk jk z 2

10 0 CHAPTER. ELECTROMAGNETIC RADIATION and hence A z = µ I 0dl e jk jk Theefoe, the magnetic field components ae, H = A z µ = I 0dl H = A z µ = I 0dl H z = 0 e jk e jk jk jk Using the tansfomation fom ectangula to spheical coodinates see Appendi F, H = sin θ cos φh + sin θ sin φh H θ = cos θ cos φh + cos θ sin φh H φ = sin φh + cos φh Substituting the epessions fo H and H and using the elationships = sin θ cos φ and = sin θ sin φ, H = I 0dl e jk = 0 H θ = I 0dl e jk = 0 H φ = I 0dl e jk = jk I 0dl sin θ jk {sin θ cos φ sin θ sin φ sin θ sin φ sin θ cos φ} jk {cos θ cos φ sin θ sin φ cos θ sin φ sin θ cos φ} jk sin φ sin θ sin φ cos φ sin θ cos φ e jk + jk These ae the same as given b Eqns.59 to.6. The electic field is computed using, E = jωɛ H = Hz a jωɛ H H + a z z H z H + a z H Since H z = 0, E = H a jωɛ z + a H z + a H z H We now compute the patial deivates of H and H with espect to,, and z. H = I 0dl e jk + jk e jk jk + e jk jk

11 Similal, H H = I 0dl + e jk = I 0dl + e jk jk e jk + e jk 3 e jk jk + e jk 3 jk jk jk jk Using = sin θ cos φ 2 + sin θ sin φ 2 = 2 sin 2 θ, we can wite H H = I 0dl + e jk jk jk + 2 sin 2 θ 2 sin2 θ + sin2 θ 2 Futhe, we can wite the deivatives of H and H with espect to z as H z H z = I 0dl e jk z e jk 3 e jk = I 0dl e jk z 3 jk z jk jk jk + e jk z + e jk jk z 3 jk z 3 which can be witten as, H = I 0dl e jk jk cos θ sin θ sin φ jk z + e jk sin θ sin φ H z e jk = I 0dl e jk jk sin θ sin φ cos θ e jk jk cos θ sin θ cos φ sin θ cos φ cos θ jk The electic field is given b E = H a jωɛ z + a H z + a H z H jk cos θ 2 + e jk cos θ sin θ cos φ 2

12 2 CHAPTER. ELECTROMAGNETIC RADIATION = I { 0dl e jk a jk cos θ sin θ cos φ jk jωɛ e jk jk sin θ cos φ cos θ + e jk cos θ sin θ cos φ 2 e jk +a jk cos θ sin θ sin φ jk e jk jk sin θ sin φ cos θ + e jk cos θ sin θ sin φ 2 e jk a z jk + 2 sin 2 θ + jk } 2 sin2 θ + sin2 θ 2 The components of the electic field in spheical coodinates can be witten as E = sin θ cos φe + sin θ sin φe + cos θe z E θ = cos θ cos φe + cos θ sin φe sin θe z E φ = sin φe + cos φe Substituting the epessions fo E, E and E z, and simplifing E = I 0dl e jk { sin θ cos φ jk jωɛ sin θ cos φ cos θ jk + sin θ sin φ jk sin θ sin φ cos θ cos θ sin θ cos φ + sin θ cos φ cos θ sin θ sin φ jk cos θ jk jk cos θ + sin θ sin φ 2 cos θ jk + 2 sin 2 θ + jk } 2 sin2 θ + sin2 θ 2 = I 0dl e jk { jk + 2 sin 2 θ cos θ + jωɛ jk sin 2 θ cos θ + jk + } 2 sin2 θ cos θ = I 0dl k e jk + 2π ωɛ jk cos θ = η I 0dl cos θ e jk + 2π jk 2 jk + sin 2 θ cos θ + 2 sin2 θ cos θ cos θ jk + sin 2 θ cos θ

13 3 E θ = I 0dl e jk { cos θ cos φ jk jωɛ sin θ cos φ cos θ jk + cos θ sin φ jk sin θ sin φ cos θ cos θ sin θ cos φ + sin θ cos φ cos θ sin θ sin φ jk cos θ jk jk cos θ + sin θ sin φ 2 + sin θ jk + 2 sin 2 θ + jk } 2 sin2 θ + sin2 θ 2 = I 0dl e jk { jk + 2 cos 2 θ sin θ + jωɛ + jk sin 3 θ jk + sin θ + + } 2 sin3 θ = I 0dl sin θ e jk jk jωɛ = I 0dl sin θ = I 0dl sin θ = jη I 0dl sin θ e jk jωɛ e jk jωɛ e jk jk jk 2 + jk jk k 2 2 jk + cos 2 θ sin θ + 2 cos2 θ sin θ jk + sin 3 θ jk + 2 jk jk + E φ = I 0dl e jk { sin φ jk jωɛ sin θ cos φ cos θ jk + sin θ cos φ + cos φ jk cos θ sin θ sin φ sin θ sin φ cos θ jk + sin θ sin φ = 0 cos θ sin θ cos φ jk cos θ 2 jk } cos θ These ae the epessions fo the components of the electic field in spheical coodinate sstem as given b Eqns

14 4 CHAPTER. ELECTROMAGNETIC RADIATION Solution. Substituting ω = 2πf and f = v/λ, Using k = 2π/λ, and v = / µɛ, we get ωµ = 2π v λ µ ωµ = k µɛ µ = k µ ɛ = kη Solution.2 The electic and magnetic fields in the fa-field egion of a Hetzian dipole can be witten as E θ = E 0 sin θ e jk H φ = E 0 η sin θ e jk whee E 0 is a comple constant. Taking the cul of the electic field, a a θ sin θa φ E = 2 sin θ θ φ 0 E 0 sin θ e jk 0 Since the electic field is independent of φ, diffeentiation with espect to φ ields a zeo. Thus, the deteminant can be epanded to get, { } E = sin θa 2 φ sin θ E 0 sin θe jk Diffeentiating with espect, Similal, H = = 2 sin θ E = a φ je 0 k sin θ e jk ωµ e jk = ja φ E 0 sin θ η E 0 = jωµ a φ η = jωµh a a θ sin θa φ θ φ 0 0 sin θ E 0 η 2 sin θ a E 0 e jk 2 sin θ cos θ η sin θ e jk sin θ e jk E 0 e jk = a 2 cos θ + a η 2 θ j E 0 e jk sin θk η 2 sin θ a E 0 θ η sin2 θ jke jk

15 At lage distances, we can ignoe the tem containing / 2 and using the elationship k/η = ωɛ, we can wite H = jωɛ { = jωɛe a θ E 0 sin θ e jk } 5 Solution.3 Assuming fa field condition, fom Eqn.69, E θ = jη ki 0dl sin θ E θ = η ki 0dl sin θ e jk In the - plane θ = 90. Substituting dl= m, I 0 = 0 A, f= MHz which coesponds to λ = 300 m, and k = 2π/λ = 2π/300 and η = Ω we get Theefoe, we get E θ = π = =6279 m = 0 3 Solution.4 The magnetic vecto potential fo a diected Hetzian dipole is given b Epessing a is spheical coodinates, The magnetic field is given b µ 0 A = a I 0dl e jk A = µ 0 I 0dl e jk a sin θ sin φ + a θ cos θ sin φ + a φ cos φ H = µ A = a a θ sin θa φ µ 2 sin θ θ φ A A θ sin θa φ { I 0 dl = a 2 sin θ cos φ e jk cos θ sin φ e jk sin θ θ φ a θ sin θ cos φ e jk sin θ sin φ e jk φ + sin θa φ cos θ sin φ e jk } sin θ sin φ e jk θ I 0 dl = {a 2 cos θ cos φ e jk cos θ cos φ e jk sin θ

16 6 CHAPTER. ELECTROMAGNETIC RADIATION a θ jk sin θ cos φ e jk + sin θa φ = I 0dl e jk = a θ jki 0 dl jk cos θ sin φ e jk sin θ cos φ e jk cos θ sin φ e jk } {a θ cos φjk + a φ cos θ sin φjk + } cos φe jk + jki 0 dl a φ cos θ sin φe jk jk + jk The electic field is given b, E = jωɛ H = a a θ sin θa φ jωɛ 2 sin θ θ φ 0 H θ sin θh φ = jki 0dl jωɛ 2 sin θ { a sin θ cos θ sin φ e jk + cos φ e jk + θ jk φ jk a θ sin θ cos θ sin φ e jk + jk + sin θa φ cos φ e jk + } jk = k I 0 dl ωɛ 2 sin θ {a cos 2 θ sin φ e jk + + sin φ e jk + jk jk +a θ sin θ cos θ sin φ + sin θa φ jke jk { jke jk + jk The components of the electic field ae, E = η I 0dl + jk + e jk jk 2 2 sin θ sin φ sin2 θ e jk = η I 0dl sin θ sin φe jk E θ = η I 0dl jk 2 sin θ = η jki 0dl E φ = η jki 0dl + jk sin θ cos θ sin φe jk cos θ sin φe jk cos φe jk } + e jk jk 2 } + jk + jk k 2 + jk k 2 + jk k 2

17 7 Solution.5 The electic field, E z at 0, 00, 0 is the same as E θ = 00, θ = π, φ = 2, and fom Eqn.65 π 2 E θ = jη ki 0dl e jk + jk k 2 Compaing it = 2 cos6π 0 6 t with it = I 0 cosωt we get, I 0 = 2 A, ω = 6π 0 6 and k = ω = 6π 06 = ad/m. Fom the data given in the poblem, we have v dl = 0.5 m and = 00 m. Theefoe, Since a z E z θ=90 = a θ E θ, e j E θ = j j = j0.088e j j = j j0.592 = j E θ = V/m E z = = V/m If the dipole is oiented along diection, the field also gets oiented along the diection. Theefoe, E = V/m

18 8 CHAPTER. ELECTROMAGNETIC RADIATION