3.4-7 First check to see if the loop is indeed electromagnetically small. Ie sinθ ˆφ H* = 2. ˆrr 2 sinθ dθ dφ =

Transcription

1 ECE 54/4 Spring 17 Assignment.4-7 First check to see if the loop is indeed electromagnetically small f 1 MHz c 1 8 m/s b.5 m λ = c f m b m Yup. (a) You are welcome to use equation (-5), but I don t like it, because it is one of these unit-specific formulas that do not appeal to my sense of generality. The fields, given by (-49) and (.5), are used to find the Poynting vector The average real power is given by E = ηβ S Ie 4r sinθ ˆφ H = β Ie S sinθ ˆθ 4r S = 1 E 1 H* = ηβ 4 S I 1 r sin θ ˆr P = Re ( S ) ds = sphere 1 ηβ 4 S I 1 1 r sin θ ˆr ˆrr sinθ dθ dφ = 1 ηβ 4 S I sin θ dθ = 1 1 ηβ 4 S I Then the radiation resistance is η = 7.7 Ω f = 1 MHz c = m/s β= f c b =.5 m S = b = P I = 1 ηβ 4 S.8 mω

2 (b) If we also assume that wire radius is small compared to the loop radius, we can assume the current is distributed around the wire uniformly as we assumed for the straight, round wire. Then we can use the straight, round wire results L R o = R eff s a Since the loop is small, the current is approximately uniform around the loop, and the effective length is equal to the wire length, i.e. the circumference of the loop. f 1 MHz ( ) S/m µ = H/m R s = f µ σ copper σ.85 mω a( No. 8 AWG) 1.4 mm b =.5 m b R o = R s a = R b s a.5 Ω (c) The radiation efficiency is not great, as expected for a small antenna e r = + R o.8 (d) The input reactance is inductive (and for a circular loop with radius large compared to the wire radius) is given by (-) L = µbln 8b a.5 µh X = ω L = fl 9 Ω which is much larger than the real part of the impedance.

3 .5-1 Example - θ pattern plot Note that equation (-7) is correct, but equation (-7) is incorrect. Example -4 θ pattern plot

4 4.- Several students were concerned about what β should be used in the formula h = jβµ eff NSsinθ ˆφ Should one use the free-space β, or the β for a wave in the ferrite material, or some effective β. If we look at the derivation of this formula, we can sort this out. First, we note that we can re-write the expression for the electric field of a small square loop (.49), in terms of the magnetic moment, m E = ηβ e m 4r sinθ ˆφ where m = IS Note that here the η and β refer to free space, as seen in the derivation. Does this formula generalize to magnetic moments generated by something other than a single square loop in free-space? Yes it does, though you would need to study multipole expansions do get a rigorous proof. Also, without proof, we note that the magnetic moment of a multi-turn loop around a ferrite core is given by m = µ eff NIS The ferrite is only within the coil, in the near field. It affects the strength of the magnetic moment, but not how the fields generated by that magnetic moment propagate into the far field. The η and β in the electric field expression are still those of free space. We plug this expression for magnetic moment into the expression for the electric field above and compare with equation (4-4), which we take to be the definition of effective length (though the antenna terminal current, IA, is incorrectly left out of that equation) E = ηβ µ eff NIS e 4r sinθ ˆφ = jωµi A 4 e r h Equation (4-4) is a general relation for effective length of an antenna in a uniform medium, and the µ in that equation is that of the surrounding medium, not any magnetic materials that might be part of the antenna. Solving for the effective length, we have h = j ηβ ωµ I I A µ eff NSsinθ ˆφ Since we assume uniform current, I, around the loops, this current is also the current at the antenna terminals, IA. Also using the usual relation We have finally ωµ = ηβ h = jβµ eff NSsinθ ˆφ and we are now convinced that the β that appears in this equation is that of free space.

5 Now for solving the assigned problem. First check the wavelength. c 1 8 m/s f 1 MHz λ = c f m It s not obvious how the ferrite might effect the current variation along a wire, but here the wavelength is several orders of magnitude larger than the loop, so it s probably OK to assume small loop behavior, even with the multiple turns. The maximum effective length occurs when sinθ = 1. c m/s f 1 MHz β = f c d 1.17 cm S = d µ eff 8 N = m.1 m 1 h max = βµ eff NS 1.88 mm 4.-1 For a small, isolated, hot object the integral in (4-18) reduces to the product of the object solid angle times the object temperature, and equation (4-19) can be used. One can approximate solid angles as the product of two orthogonal planar angles. For narrow, circularly symmetric solid angles, it is somewhat more accurate to use the following Δθ / Δθ Ω = sinθ dθ dφ = sinθ dθ = [ cosθ ] / = cos Δθ ( 1) Δθ = 1 cos = 4 Δθ sin 4 4 Δθ Δθ / which is a factor of about /4 smaller, and represents the error of approximating the area of a circle by the are of the circumscribing square. However, in this problem, we have the ratio of two solid angles, so any constant factor in your approximation will cancel out, even the angle units won t matter, as long as you are consistent. T A.8K HP.79 Δθ s.5 T s = Ω A Ω s T A HP Δθ s T A K

6 4.4-9 Starting from (4-) using G = P t G r λ t ( 4 R) λ = c f and adding some reference quantity factors, we obtain = P t G t G r P P f f R R c 4 R f Now, take the base 1 log and multiple by 1 1log P = 1log P t P +1logG +1logG log f t r f log R R + log c 4 R f Now we choose the reference values and find the value of the trailing constant P = 1 mw f = 1 MHz R = 1 km c log 4 R f.45 Then we can write the unit specific equation (though I am not a fan) ( dbm) = P t ( dbm) + G t ( db) + G r ( db) log f ( MHz) log R( km).45

7 4.5-4 First check the wavelength. c 1 8 m/s f 15.5 MHz λ = c f m which is much larger than the loop dimensions, so we can use the small loop results. (a) Though the approximation is more dubious, we will use the analogous results of the straight, round wire, ideal dipole, as we did in problem.4. Generalized, this result is R o = R s circumference of loop circumference of wire = R s ( a + b) ( w + t) = R s a + b w + t where R s = f µ σ where a and b are the loop dimensions, and w and t are the trace width and trace thickness of the loop wire cross-section. Plugging in the numbers we have f 15.5 MHz µ H/m σ ( copper) S/m R s. mω a 41. mm b 1 mm w.5 mm t.7 mm R o 4.4 mω The radiation resistance is given by f = 15.5 MHz c = m/s β= f c η = 7.7 Ω S = ab = 1 ηβ 4 S.1 mω Again the radiation efficiency is not great, as expected for a small antenna (b) For the small loop we have the same directivity as an ideal dipole. Since we are given no information regarding impedance or polarization mismatch, we assume those factors are unity A e = λ 4 G.8 1 m (c) For the power received we then have e r = + R o.149 D = 1.5 G = e r D.4 E 1 µv/m S = 1 E η = SA e W A very small power