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.

2