Written Examination. Antennas and Propagation (AA ) April 26, 2017.

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1 Written Examination Antennas and Propagation (AA. 6-7) April 6, 7. Problem ( points) Let us consider a wire antenna as in Fig. characterized by a z-oriented linear filamentary current I(z) = I cos(kz)ẑ with λ z λ. Starting with the general expression of the far field radiation vector N(θ, φ), for wire antennas, derive the explicit expression of the electric and magnetic radiated far field by the antenna under discussion. Find the expression of the radiation intensity and discuss the properties of the radiation pattern. Problem ( points) Two Huygens sources, operating at f = 3 GHz, are positioned at P = (λ/8, λ/8, ) and P = ( λ/8, λ/8, ), respectively, as in Fig.. Each Huygens source consists of Find: a x directed electric dipole of length d e = λ/5, excited by a current I e = A; a y directed magnetic dipole of length d m = d e, excited by a magnetic current I m = ζ I e V, being ζ the characteristic impedance of free space. a) the explicit expressions of the total radiated electric and magnetic far fields; b) the explicit expressions of the radiation intensity; c) the direction along which the radiation intensity attains its maximum value; d) the direction along which the radiation intensity attains its maximum value in the x y plane. Problem 3 ( points) The far-field pattern f(θ, φ) of a certain antenna radiating in free space at f =.3 GHz is independent of φ and is given by Assuming f max = [V m, find: a) the total radiated power; f(θ, φ) = f(θ) = { fmax cos 5 (θ) θ π π < θ π () b) the power density radiated at a distance r = m in the direction θ = π ; c) the directivity of the antenna when θ = π ; d) the maximum value of the directivity. Remember: you may not work on the exam with anyone else, ask anyone questions, or consult textbooks, notes or sites on the Web for answers. You can consult only notes provided by the instructor during the assignment.

2 Antennas and Propagation (AA. 6-7): Mid Term Written Examination z l = λ y l = λ x Figure : Geometry of Problem. y I m I e I m z x I e Figure : Geometry of Problem.

3 Antennas and Propagation (AA. 6-7): Mid Term Written Examination Notation In the following pages you can find the solutions of the various problems included in this test. Throughout this document, vectors are written in bold type. ˆx, ŷ, ẑ represent the unit vectors in a cartesian reference system; ˆr, ˆθ, ˆφ are the unit vectors in a spherical reference system. Solution to Problem See slide n.ro -9 of file 9 A&P 7 W ireantenna.pdf on toccafondi/wordpress/wpmyfiles/antennaspropagation Solution to Problem The far field radiated by an electric source can be calculated using in which E = jk e jkr πr [ˆθ (ζ N θ ) + ˆφ (ζ N φ ) H = jk [ ˆθ (ζ N φ ) + ζ πr ˆφ (ζ N θ ) N (θ, φ) = V J(r )e jkr ˆr dv In the case of a d e long elementary electric dipole placed at (,, ), directed along the x-axis, and fed with the current I e, one gets N θ (θ, φ) = I e d e cos θ cos φ Therefore, carrying out the calculations, N φ (θ, φ) = I e d e sin φ [ E = jkζ πr I ed e cos θ cos φ ˆθ sin φ ˆφ H = jk e jkr [ πr I ed e sin φ ˆθ + cos θ cos φ ˆφ The fields radiated by magnetic sources can be written in the following form: in which E = jk e jkr πr [ L φ ˆθ L θ ˆφ H = jk [ L θ ˆφ + Lφ ˆφ ζ πr L (θ, φ) = V J m (r )e jkr ˆr dv In the case of a d m long elementary magnetic dipole placed at (,, ), directed along the y-axis, and fed with the current I m, one gets Therefore, carrying out the calculations, L θ (θ, φ) = I m d m cos θ sin φ L φ (θ, φ) = I m d m cos φ E = jk e jkr [ πr I md m cos φ ˆθ cos θ sin φ ˆφ 3

4 Antennas and Propagation (AA. 6-7): Mid Term Written Examination H = jk [ ζ πr I md m cos θ sin φ ˆθ + cos φ ˆφ Being d e = dm and I m = ζ I e one can write that the field radiated by a Huygens s source with the geometrical configuration as the one in the exercise and placed in (,, ) is E = jkζ πr I ed e ( + cos θ) [ cos φ ˆθ sin φ ˆφ H = jk e jkr [ πr I ed e ( + cos θ) sin φ ˆθ + cos φ ˆφ Taking the displacement into account one gets [ E = jkζ πr I ed e ( + cos θ) e j π sin θ cos φ e j π sin θ sin φ cos φ ˆθ sin φ ˆφ H = jk e jkr [ πr I ed e ( + cos θ) e j π sin θ cos φ e j π sin θ sin φ sin φ ˆθ + cos φ ˆφ and [ E = jkζ πr I ed e ( + cos θ) e j π sin θ cos φ e j π sin θ sin φ cos φ ˆθ sin φ ˆφ H = jk e jkr [ πr I ed e ( + cos θ) e j π sin θ cos φ e j π sin θ sin φ sin φ ˆθ + cos φ ˆφ Therefore the total field can be written as E + = jkζ πr I ed e ( + cos θ) cos [ π [ sin θ (sin φ + cos φ) cos φ ˆθ sin φ ˆφ H + = jk e jkr [ π [ πr I ed e ( + cos θ) cos sin θ (sin φ + cos φ) sin φ ˆθ + cos φ ˆφ The radiation intensity is U (θ, φ) = k ζ 8π I e d e ( + cos θ) cos [ π sin θ (sin φ + cos φ) The direction along which the radiation intensity attains its maximum value is θ =. In the x y plane, θ = π. To maximixe the radiation intensity in this plane, one has to maximize the factor [ cos π (sin φ + cos φ) Being the square of a cosine, it attains its maximum value when its argument equals or π. Therefore (sin φ + cos φ) = φ = 35 and φ = 35 On the other hand π (sin φ + cos φ) = π (sin φ + cos φ) = which is clearly impossible. Therefore, in the x y plane the radiation intensity attains its maximum value when φ = 35 and φ = 35, as can be seen in Fig. 3. Solution to Problem 3 The total power radiated by the antenna can be calculated using the formula = π ζ 6π dφ π dθ f (θ, φ) sin θ

5 Antennas and Propagation (AA. 6-7): Mid Term Written Examination yielding The following holds Therefore b a Figure 3: Plot of cos [ π (sin φ + cos φ) = π f max π ζ 6π cos (θ) sin θ dθ cos N (θ) sin θ dθ = cosn+ (a) cos N+ (b) N + = π f max ζ 6π =.8939 The power density S r of the antenna is given by S r = f (θ, φ) ζ 6π r Therefore, when r = m and θ = π, one has S r = ( ζ 6π f max cos π ) =.6 8 being ( cos π ) = 3 The directivity of the antenna at θ = π is given by with D(θ, φ) = U(θ, φ) π = π ζ 6π f max 3 =.6875 U(θ, φ) = ζ 6π f max cos (θ) with θ π. The directivity attains its maximum value when θ =, being D max = ζ 6π f max π = 5

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