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ECE 4784/6784, Wireless Communications, Assignment 1
1. (15) A half-wave dipole is situated at the top of a cellular tower, oriented vertically on the tower. The current into the dipole is a sinusoidal current at frequency 870 MHz, with peak value I = 0.1 ampere. Consider a mobile receiver at a distance of d kilometers along the horizon. Remember the dipole pattern gives field strength
E(d, θ, φ) =
60I cos( π cos(θ))
2
aθ v/m d sin(θ)
(1)
a. Assume free-space propagation laws. Find the electric field strength at the mobile location.
Also find the power flux density Φ in w/m2 for users along the horizon.
b. Repeat for the two-ray model, assuming a tower height of 40 meters, a mobile height of 2 meters, and a plane earth approximation. Assume the transmitted signal is verticallypolarized, and that the earth reflection coefficient magnitude is 1.
c. Plot both power flux densities versus d on log-log scales, over a range from 100 meters to
10 kilometers. You should see scalloping in the two-ray result, with an envelope that decays twice as fast (on log-log scale) as the free-space result.
2. (10) Given an RF pulse s(t) = 10−3 rect(
t
) cos(2π870 · 106 t + π/6) volts
.001
(2)
• find S(f ), the Fourier transform, and sketch the energy density spectrum in the frequency domain
• Find the complex envelope s(t) with respect to the reference cos(2π870 · 106 t).
˜
• find the energy delivered to a 50 Ω load
3. (Additional assignment for ECE 6784 group, building on first problem, 10 points)
The power flux density expression can be integrated over the surface of a sphere enclosing the dipole to obtain the total power radiated by the antenna. Call this number P watts. An isotropic antenna radiating P watts would produce a flux density at the same radius of
P = P/4πr2
(3)
The gain of any antenna is the ratio of the maximum flux density, which occurs at θ = π/2, to the flux density of an isotropic antenna, with the power density measured
1. (15) A half-wave dipole is situated at the top of a cellular tower, oriented vertically on the tower. The current into the dipole is a sinusoidal current at frequency 870 MHz, with peak value I = 0.1 ampere. Consider a mobile receiver at a distance of d kilometers along the horizon. Remember the dipole pattern gives field strength
E(d, θ, φ) =
60I cos( π cos(θ))
2
aθ v/m d sin(θ)
(1)
a. Assume free-space propagation laws. Find the electric field strength at the mobile location.
Also find the power flux density Φ in w/m2 for users along the horizon.
b. Repeat for the two-ray model, assuming a tower height of 40 meters, a mobile height of 2 meters, and a plane earth approximation. Assume the transmitted signal is verticallypolarized, and that the earth reflection coefficient magnitude is 1.
c. Plot both power flux densities versus d on log-log scales, over a range from 100 meters to
10 kilometers. You should see scalloping in the two-ray result, with an envelope that decays twice as fast (on log-log scale) as the free-space result.
2. (10) Given an RF pulse s(t) = 10−3 rect(
t
) cos(2π870 · 106 t + π/6) volts
.001
(2)
• find S(f ), the Fourier transform, and sketch the energy density spectrum in the frequency domain
• Find the complex envelope s(t) with respect to the reference cos(2π870 · 106 t).
˜
• find the energy delivered to a 50 Ω load
3. (Additional assignment for ECE 6784 group, building on first problem, 10 points)
The power flux density expression can be integrated over the surface of a sphere enclosing the dipole to obtain the total power radiated by the antenna. Call this number P watts. An isotropic antenna radiating P watts would produce a flux density at the same radius of
P = P/4πr2
(3)
The gain of any antenna is the ratio of the maximum flux density, which occurs at θ = π/2, to the flux density of an isotropic antenna, with the power density measured