Question

QUESTION 12, 13, 14????

In 3.14, can write Gt and Gr directly in the answer, or assume Gt and Gr =1 (specify hypothesis)

P3.8. Path loss, fading, and delay spread are the three most important radio propagation issues. Explain why those issues are important in a cellular system. What is the difference between fast fading and slow fading? Р3.9. How is radio propagation on land different from that in free space? What does a small delay spread indicate about the characteristics of a fading channel? If the delay spread is 1 microsecond, will two different frequencies that are 5 MHz apart experience correlated fading? A wireless receiver with an effective diameter of 250 cm is receiving signals at 20 GHz from a transmitter that transmits at a power of 30 mW and a gain of 30 dB. P3.10. P3.11. P3.12. 2sl (a) What is the gain of the receiver antenna? (b) What is the received power if the receiver is 5 km away from the transmitter? AO P3.13. Consider an antenna transmitting at 900 MHz. The receiver is traveling at a speed of 40 km/h. Calculate its Doppler shift.2. H Consider an antenna transmittinga power of 5 W at 900 MHz. Calculate the received power at a distance of 2 km if propagation is taking place in free space. 25 P3.14. P3.15. Repeat Problem P3.11. Calculate the average fading duration if p 0.1.

Answer 1

Solution to P3.12:

Given Information :

Diameter of wireless receiver, d_e = 250 cm
Carrier Frequency, f_c = 20 Ghz
Transmitted power, P_t = 30 mW
Transmitter Gain = 30 dB= 1000
Distance of the receiver = 5km

From the above information,

(i) The effective area can be calculated as : $A = \frac{\pi (d_e)^2}{4} = \frac{\pi * (250 * 10^{-2})^{2}}{4} = 4.91 m^2$

(ii) The wavelength is calculated as : $\lambda = \frac{c}{f_c} = \frac{3*10^8}{20*10^6} = 0.015m$
(where c= velocity of light)

(a) GAIN OF RECEIVER ANTENNA, G_r :
The gain of receivers antenna is given by the formula $G_r = \frac{4\pi *A}{\lambda ^2} = \frac{4\pi * 4.91}{(0.015)^2} = 2.74*10^5=54.38dB$

(b) RECEIVED POWER. P_r
The received power is given by the expression-
$P_r = \frac{A* G_t* P_t}{4\pi *d^2 } = \frac{4.91*1000*30*10^{-3}}{4\pi *(5*10^{3})^{2}} = 4.69*10^{-7} W$

Solution to P 3.13:

The doppler shift is calculated as follows:

$\Delta f_{doppler}= \frac{v}{\lambda}= 33.33 Hz$

Solution to P. 314:

Transmitted Power, P_t = 5W
Transmitted Frequency= 900 MHz
Distance of received power = 2 km = 2 X 10³ m

Gt = Gr = 1 as propagation is occurring in free space.

Therefore, received power, $P_{r}= \frac{900*10^6*5}{4\pi 2k^{2}} = 89.52kW$