ece 4370: antenna engineering solutions to test 1 (fall 2017) · 2020. 7. 31. · ece 4370: antenna...
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ECE 4370: Antenna Engineering
Solutions to TEST 1 (Fall 2017)
1. Antenna Recognition:
(b) (c)(a)
(e) (f) (g)
(d)
(h)
1. Half-Wave Dipole (b) 5. Meandered Dipole (a)
2. λ4 -Monopole (c) 6. Biconical Antenna (d)
3. Patch Antenna (h) 7. Loop Antenna (f)
4. Folded Dipole (g) 8. Broadband Monopole (e)
2. Directivity: Directivity must average to 1 over 4π steradians. If an antenna radiates into anoctant with directivity D0 and radiates into the remaining 7 octants with directivity 0, thenD0 must equal 8 in the linear scale (9 dBi).
3. Shortened Dipole:
(a) Neglecting the coupling between meandered horizontal segments, this resonating antennashould be λ/2 in length when straightened.
(b) Start with the vector magnetic potential, making the same phase approximation for theelectrically short dipole:
Ãz(r, θ, φ) =µ
4πrexp (−jkr)
+L2∫−L2
I cosz′π
Lexp(+jkz′ cosθ︸ ︷︷ ︸
≈0
) dz′ =µIL
2π2rexp (−jkr)
From this we arrive at the far-field electric and magnetic field solutions:
~̃H(r, φ, θ) ≈ jkIL sin θ2π2r
exp (−jkr) φ̂ ~̃E(r, φ, θ) ≈ jkηIL sin θ2π2r
exp (−jkr) θ̂
1
-
(c) 90 degrees (identical to the Hertzian and electrically short dipoles)
(d) We neet to calculate total radiated power for this:
PT =
2π∫0
π∫0
~Sav︷ ︸︸ ︷1
2~̃E × ~̃H
∗·r2 sin θ dθ dφ r̂
=
2π∫0
π∫0
1
2
kηIL sin θ
2π2r
kIL sin θ
2π2rr2 sin θ dθ dφ
= 2πηk2I2L2
8π4
π∫0
sin3 θ dθ
=ηk2I2L2
4π3
π∫0
sin3 θ dθ
︸ ︷︷ ︸43
=ηk2I2L2
3π3
Since PT =12I
2Ra, then radiation resistance is solved as
Ra =2ηk2L2
3π3
Interestingly, this is slightly higher than the electrically short monopole. And with noreactive impedance component, this should be easier to match than the short, straightdipole.
4. Aircraft Communications:
(a) EIRP = PT + GT (log scale) = 11 dBW
(b) From logarithmic link budget:
PR = EIRP +GR − 20 log10(
4π
λ
)− 20 log10 r
where λ = 0.43 m at 700 MHz, r is TR separation distance, GR = 6 dBi, and PR=114dBW. Solving for r gives a range of 158 kilometers.
(c) There is a mismatch factor of 0.64 on the link (1.9 dB). All else being equal, a receivedpower of -86 dBm now corresponds to a link distance of 120 kilometers, nearly a 20%decrease.
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