Written Examination. Antennas and Propagation (AA ) June 22, 2018.

Transcription

1 Written Examination Antennas and Propagation (AA. 7-8 June, 8. Problem ( points A circular loop of radius a = cm is positioned at a height h over a perfectly electric conductive ground plane as in figure, and radiates in free space at f = GHz. The plane of the loop belongs to a plane parallel to the xy-plane with axis oriented along the z-axis of cartesian reference system as in figure. The loop is fed with a current I s = j A, flowing along ˆφ. Find, in the half space z > :. the explicit expressions of the total radiated electric and magnetic far fields as a function of h;. the explicit expression of the radiation intensity in the case h = λ/; 3. the total radiated electric and magnetic far fields and the radiation intensity in the case h (h tends to zero. Problem ( points Eight y oriented elementary electric dipoles operate at f = 3 GHz. They are equally spaced at a distance d =.5 cm along the z axis of a cartesian reference system, as shown in figure. The length of each dipole is d e. The excitation currents of the ten dipoles satisfy I n = I e e jnα V being I e = A, and n =,,, 8.. Find the explicit expressions of the normalised array factor (ÃF when α = π/;. Determine the explicit expression of the total radiated electric and magnetic far field when α = π/; 3. Find the position of the nulls expressed in terms of θ angle of the ÃF when α = π/;. Find for what value of α the main lobe occurs at θ = π 6. 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; θ < π f max cos 6 π (θ f(θ, φ = f(θ = θ 3π ( 3π < θ π b the power density radiated at a distance r = m in the direction θ = π 3 ; c the directivity of the antenna when θ = π 3 ; 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. 7-8: Written Examination z a I s ˆ h r x PEC y Figure : Geometry of Problem. z q d y x Figure : Geometry of Problem.

3 Antennas and Propagation (AA. 7-8: 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 The loop antenna is equivalent to an elementary magnetic dipole oriented along ẑ. One can apply image theory and remove the perfectly electric conductive ground plane by introducing in the half space z <, at (,, h, another elementary magnetic dipole (or equivalently, a loop antenna, identical to the one in the z < half space, but directed along ẑ. The far field radiated by a magnetic source can be calculated using: in which E = jk e jkr πr [ˆθLφ ˆφL θ H = jk e jkr [ˆθLθ + ζ πr ˆφL φ L (θ, φ = V J m (r e jkr ˆr dv In the case of an elementary magnetic dipole placed at (,, and directed along ẑ, one gets L (θ, φ = I m d m ẑ yielding in which L θ = I m d m sin θ L φ = I m d m = jωµi s (πa Finally, carrying out the calculations, E = jk e jkr πr I md m sin θ ˆφ H = jk ζ e jkr πr I md m sin θ ˆθ The field radiated by each elementary magnetic dipole, in the far field approximation, is E i = jk e jkr i πr i I m d m sin θ ˆφ H i = jk ζ e jkri πr i I m d m sin θ ˆθ in which, for the elementary magnetic dipole located at (,, h, i = and r r h cos θ; for the elementary magnetic dipole located at (,, h, i = and r r + h cos θ. The total radiated field (E + E, H + H in the half space z >, is E tot = k e jkr πr I md m sin (kh cos θ sin θ ˆφ H tot = k ζ e jkr πr I md m sin (kh cos θ sin θ ˆθ Note that the expressions above can also be obtained by using array theory. Namely, the problem under consideration can be solved by multiplying the field radiated by a single elementary magnetic dipole 3

4 Antennas and Propagation (AA. 7-8: Written Examination located at (,, by the array factor of a configuration consisting of two identical antennas, fed with opposite currents and placed along the z-axis, at (,, h and (,, h, respectively. The general expression of such array factor is AF (θ, φ = C e jkˆr ˆr jkhˆr + C e ˆr Since the two sources are fed with opposite currents, one gets AF (θ, φ = e jk(hẑ ˆr e jk( hẑ ˆr = j sin (kh cos θ The expression of the radiation intensity as a function of h is When h = λ/ In the limit h, being one gets U(θ, φ = U(θ, φ = k 8π ζ I m d m sin (kh cos θ sin θ k ( 8π I m d m sin π ζ cos θ sin θ lim sin (kh cos θ = h E tot = H tot = U(θ, φ = As far as the numerical values are concerned, λ=3 cm, I m d m = jωµi s (πa k =.85, π I md m = k k 8.683, πζ I m d m =.9, 8π ζ I m d m =.97. Solution to Problem The field radiated by a y directed elementary electric dipole placed at the origin of the reference system is e jkr [ E = jkζ πr I ed e cos θ sin φ ˆθ + cos φ ˆφ H = jk e jkr [ πr I ed e cos φ ˆθ + cos θ sin φ ˆφ The normalised array factor is ÃF (θ, φ = N ej N (α+kt ˆr sin [ N (α + kt ˆr sin [ (α+kt ˆr which, in the case under consideration, being N = 8, α = π/, t = (,, λ/, one gets: ÃF (θ, φ = ÃF (θ = 8 ej 7π 8 (+ cos θ sin [ ( π/ + π cos θ [ π (π/+ cos θ sin Therefore the field radiated by the array of 8 dipoles can be written as e jkr [ E = jkζ I e d e ÃF (θ cos θ sin φ ˆθ + cos φ ˆφ πr H = jk e jkr [ I e d e ÃF (θ cos φ ˆθ + cos θ sin φ ˆφ πr To calculate the radiation intensity, one can use the formula U(θ, φ = E ζ r

5 Antennas and Propagation (AA. 7-8: Written Examination yielding k U(θ, φ = ζ π I e d e ÃF [ (cos θ sin φ + cos φ To calculate the nulls, one must study ψ = ( π/ + π cos θ = π + π cos θ. As θ spans the interval [, π, ψ spans the interval [ π, 3π. In such interval, sin ψ has the following zeroes: ψ = mπ, with m =,,,, 3. ψ/ spans the interval [ π/8, 3π/8. Therefore the value corresponding to m = must be excluded, cause it corresponds to the maximum of the array factor. To sum up, the nulls of the array factor are given by the the values of θ that satisfy: π + π cos θ = mπ θ = cos ( m with m =,,, 3. Solving the above equation gives θ =, θ =.7, θ = π and θ = π, as can be seen in Fig.. The main beam points towards θ = cos π. Note that the array factor is here = AF ψ Figure 3: Normalised array factor in Problem. plotted as a function of θ, using Matlab. This is just for the sake of better visualising the positions of the nulls and the main beam. Since you will not have Matlab during the written test, you will not be asked to plot the normalised array factor as a function of θ. To solve the fourth point, one must consider the expression of the normalised array factor for the geometry under consideration and a general α, that is when t = (,, λ/, yielding ÃF (θ, φ = ÃF (θ = 8 ej 7 (α+ π cos θ sin [ ( α + π cos θ [ π (α+ cos θ sin which attains its maximum value when α + π cos θ =, that is when α = π cos θ. In our case, θ = π 6, therefore α = π 3 radiants, as can be seen in Fig. 6. Solution to Problem 3 The total power P r radiated by the antenna can be calculated using the formula P r = ζ k 6π π dφ 5 3π π dθ f (θ, φ sin θ

6 Antennas and Propagation (AA. 7-8: Written Examination AF θ Figure : Normalised array factor in Problem Figure 5: Normalised array factor in Problem - polar representation. yielding The following holds Therefore b a P r = π f max ζ k 6π 3π π cos (θ sin θ dθ cos M (θ sin θ dθ = cosm+ (a cos M+ (b M + P r = π f max k cos 3 ( π cos 3 ( 3π ζ 6π =.35 3 The power density S r of the antenna is given by S r = k f (θ, φ ζ 6π r 6

7 Antennas and Propagation (AA. 7-8: Written Examination Figure 6: Normalised array factor in Problem, when α = π 3. Therefore, when r = m and θ = π 3, one has S r = k ( ζ 6π f max cos π = being The directivity of the antenna at θ = π 3 ( cos π 3 is given by = 96 with D(θ, φ = U(θ, φ P r π = π P r ζ k 6π f max 6 =.873 U(θ, φ = ζ k 6π f max cos (θ with π θ 3π. The directivity attains its maximum value when θ = π or θ = 3π, being D max = k ζ 6π f max π P r =