q. A 100 MHz FM carrier is modulated by an audio tone causing 20 KHz frequency deviation. Determine the carrier siwng and highest and lowest carrier frequencies

What is the modulation index (or deviation ratio) of an FM signal with carrier swing of 150 KHz when the modulating signal is 15 KHz?

An FM receiver provides a voltage gain of 200,000(106dB) prior to its limiter. The limiters quieting voltage is 200 mV. What is the receivers sensitivity?200 mV/200,000= 1V->sensitivity Determine the maximum sideband power if the carrier output is 1 kW and calculate the total maximum transmitted power.Max sideband power occurs when k = 1. At this percentage modulation each side frequency is of the carrier amplitude. Since power is proportional to the square of the voltage, each has of the carrier power. x 1kW = 250W Total sideband power = 2 x 250 = 500W. Total transmitted power = 1kW + 500W = 1.5kWQ .consider the case of a 2 cm long copper dipole that has a wire diameter of 1 mm being employed to receive a narrowband pager signal centered at a frequency of 150 MHz What is antennas Ohmic efficiency?Ans: Rr=202(/)2

=202(0.02/2)2

=0.0197 Ohms

To calculate the Ohmic resistance, we begin by evaluating the surface resistance:Rs=(fo/cu)1/2

=[ 1.5 108 4 10-7/(5.8 107)]1/2

=0.00319 Ohms

Then,Ro=(1/3) Rs(2rw)-1

=(1/3) 0.00319 0.02 /(2 0.0005)

=0.00677 Ohms

The antenna Ohmic efficiency then is=Rr/(Rr+ Ro)

=74%

Q,Loop circumference 2rl= 0.2, wire is copper of radiusrw= 0.001, andf= 3 MHz Calculate the following parameters: Radiation resistance,Rr Ohmmic resistance,Ro Efficiency, Reactive impedance,X=L

radiation resistance Rr= 31,200 (rl2/2)2= 31,200 (0.01/)2= 0.316 OhmsFromEquation Ohmic resistance Ro= Rs(rl/rw) = [fo/(2cu)] [100/] = [4.52 10-4][31.8] = 1.44 10-2OhmsThen the antenna efficiency is = Rr/(Rr+ Ro) = 0.316/0.330 = 95.8%Finally usingwe calculate the reactive impedance as X = 2f L = 2 3 106 rlo[ln(8 rl/rw) - 1.75)] = 858 Ohms

Q. Consider a 3 GHz transmitter separated from a receiver by a distance of 5 km. An earthen ridge with a straight, sharp top is located 2 km from the transmitter and juts 25 m above the direct path between the transmitter and receiver.Estimate the path loss by modeling the ridge as an absorbing screen with a knife-edge top.

The relevant dimensions ared1= 2000 m,d2= 3000 m, he= 25 m, and= 0.1 m., the diffraction parameter is = 25 [(2 5000)/(0.1 2000 3000)]1/2= 3.23The excess path loss is foundas Lke(dB) = 20 log [3.14 3.23 1.41] = 23.1 dBThe free space path loss is found as LF(dB) = 32.4 + 20 log 5 + 20 log 3000 = 115.8 dBFinally, the total path loss is L (dB) = 115.8 + 23.1 = 138.9 dB