# Analysis of the link budget for a FM Analog telephone link

Not what you're looking for?

The frequency band 870 - 890 MHz is used for analog (FM) cellular telephone systems. A base station transmits a signal with a power of 1 watt to an antenna with a gain of 10 dB. The signal is received by a cell phone 2 km away with an antenna gain of 0 dB.

a. For the path length of 2.0 km, calculate the free space path loss in dB at a frequency of 880 MHz, assuming line of sight (LOS) conditions between the base station and the cell phone.

b. Calculate the power at the input to the cell phone receiver, in dBW. Set your answer out as a link budget and include 3 dB miscellaneous losses.

c. The cell phone receiver has a system noise temperature of 450 K and a noise bandwidth of 30 kHz. Calculate the noise power and carrier to noise ratio (CNR) in the receiver.

d. The threshold carrier to noise ratio (CNR) for successful reception of the FM signal is 20.0 dB. What is the link margin under line of sight (LOS) conditions? Under what conditions might the margin be reduced to 0 dB, or the CNR at the receiver fall below 20 dB?

##### Purchase this Solution

##### Solution Summary

An Analog (FM) cellular telephone system is analysed to find the Free space path loss, Noise power in the receiver and threshld link CNR. The link margin is also determined for LOS conditions and factors that may reduce this margin to below 0 dB explored

##### Solution Preview

(a) Path loss in this case is just the free space loss FSL.

The free space loss is given in terms of the operating wavelength of the transmission (Lamda) and the distance of the transmission (r) by (1), where λ and r are measured in m.

FSL = 20*log10{4pi*r/Lamda} (1)

Putting in values:

r = 2 km = 2 x 10^3 m

Lamda = c/f where c is the speed of light, c = 3 x 10^8 m/s, f is the frequency f = 880 MHz

Lamda = (3 x 10^8)/(880 x 10^6) = 0.34 m

Therefore using (1) the free space loss is determined as

FSL = 20*log10{(4п x 2 x 10^3)/0.34} = 97.4 dB

(b) Power received (Pr) at input to the cell phone can be deduced by performing the power link budget and thus the power received is given by (2) as the dB sum of the power transmitted{Pt + Gt}, the gain of the receive site antenna Gr minus the total path loss LP. Note that Pt is the actual dB power transmitted and Gt is the gain of the transmit ...

##### Purchase this Solution

##### Free BrainMass Quizzes

##### Basic Physics

This quiz will test your knowledge about basic Physics.

##### Variables in Science Experiments

How well do you understand variables? Test your knowledge of independent (manipulated), dependent (responding), and controlled variables with this 10 question quiz.

##### Introduction to Nanotechnology/Nanomaterials

This quiz is for any area of science. Test yourself to see what knowledge of nanotechnology you have. This content will also make you familiar with basic concepts of nanotechnology.

##### Classical Mechanics

This quiz is designed to test and improve your knowledge on Classical Mechanics.

##### Intro to the Physics Waves

Some short-answer questions involving the basic vocabulary of string, sound, and water waves.