ANT Notes 7 Example Solution

1. Diners in the ground-floor dining room of a 52-story (160 meter) building in the center of a metropolis are not receiving reliable cell-phone service. To improve the service, you install a passive link to the dining room: a �/2 dipole antenna (�/4 vertical with �/4 skirt) on the roof of the building connected to a coaxial cable (having 2 dB/100 m of loss) extending down to the dining room, where it is terminated with a �/4 stub.

Assume a user in the dining room has a cell phone 10 meters below the �/4 stub in the room, operating at 1 GHz, and with an equivalent effective area as the �/4 stub in the room. The top of the building is at a 1.2 km line-of-sight distance from the cell tower antenna, with 10 dB gain. What is the total dB path loss? Note that the directivity of a �/2 dipole is 1.64 and �/4 stub is 1.5, and both are lossless.

Solution:

Let’s make a list of what we know: Gtower = 10 dBi = 10 (dimensionless, gain of the tower antenna) Rtower to building = 1200 m (radial distance from the tower to the building) Rstub to phone = 10 m (radial distance from the stub antenna to the phone) Lcoax = 160 m (length of the coax) coax loss = 2 dB / 100 m f = 1E9 (frequency of operation) Ddipole = 1.64 (directivity of the dipole antenna) Dstub = 1.5 (directivity of the stub) �dipole = �stub = 1 (the efficiency of the dipole and stub is 1 since both are lossless)

Cell phone tower to dipole at top of the building:

We can use the Friis Transmission formula for the cell phone tower to the top of the building: dipole received dipoletower dipoledipoletower = = = 6.5 × 10 or -82 dB () () (see Equation 9.69, Equation 12 on Equation Sheet)

Now for the coaxial cable loss:

160 meters (-2 dB / 100 meters) = -3.2 dB

And for the inside link:

Using again the Friis Transmission formula:

rec stubstubstubstub = = 1.28 × 10 or -49 dB ()

Total loss:

We add up all of the losses from the three segments: −82 − 3.2 − 49 = −134 dB