Path Loss at 4 GHz

Calculate the path loss at 4 GHz for a distance of 35863 Km.

Q1.
Guest to Host : Asks for Pizza
Host Picks Telephone
Telephone signals Telephone
Clerk Picks Telephone
Hosts sends Message to Telephone
Telephone transfers message to Telephone
Clerk receives message through telephone
Clerk disconnects telephone
Host disconnects Telephone
Clerk places order to Cook
Cook makes the Pizza
Cook informs Clerk Pizza made
Clerk Calls Delivery Man
Pizza is parceled to Host
Host received Pizza

Q2.
French Prime Minister passes message to Translator in French
Translator converts message to English
Message in English is forwarded to Telephone from Translator
Telephone signals other Telephone about message
Other Telephone picks up message and forwards to other translator
Translator now receives message and converts to Chinese
Chinese Message forwarded to Chinese Prime Minister
Chinese Prime Minister replies in Chinese to translator
Translator converts message to English
Translator forwards message over Telephone
Other Telephone receives message and forwards to other translator
Translator converts message to French
French Prime Minister receives reply in French from Chinese Prime Minister

Q3 .
a. Amplitude : 15 ; Time period = 3
Frequency = .0.33 Hz ; Phase =0
b. Amplitude : 4; Time period =6.5
Frequency = 0.154Hz ; Phase =0
c. Amplitude : 7.5; Time period = 2.5
Frequency = 0.4Hz ; Phase =90
Q4. Equation for waveform A Sin (2ℿ (f)t + θ)
10 Sin(2 ℿ(100) t)
Amplitude = 10. ; Frequency = 100 Hz ; Time Period = 1/100 =10msec
Phase = 0◦

20 Sin(2 ℿ(30) t + 90)
Amplitude = 20. ; Frequency = 30 Hz ; Time Period = 1/30 =
33.33msec; Phase = 90◦
5 Sin(500 ℿ t + 180)

Amplitude = 5; Frequency = 250 Hz ; Time Period = 1/250 = 4msec; Phase
= 180◦
8 Sin(400 ℿ t + 270)
Amplitude = 8; Frequency = 200 Hz; Time Period = 1/200 = 5msec ; Phase =
270◦

Q5.
Count of pixels = 480 * 500. = 240000 pixel
Count of Bits for each pixel = 5 bits (Log232)
Bits per Frame = 5 * 240000 = 1200000 bits
Bits transferred in 1 sec = 1200000 x 30 = 36 x 106 bits
Bits transferred in one second is the source data rate.
Bandwidth of Channel = 4.5 MHz = 45 x 105
Signal to Noise Ratio (SNR) = 35 decibels
Capacity of channel = B * log2 *(1+SNR)
= 45 x 105 x log2 (1 + 103.5)

= 45 x 105 x log2 (1 + 3162)
= 45 x 105 x log2 (3163)
= 45 x 105 x 11.6272 = 52.32 x 106 Bps
Q6. Path Loss at 4 GHz for distance of 35863 Km is.
20Log10 (4 x 109) + 20log10 (35.863x106) – 147.56dB
192.04 + 151.092 – 147.56dB = 195.57dB
Q7. Fundamental Frequency = 100Hz.
Spectrum = ( ∆ ( f −100 ) −∆ ( f +100 ) ) +1 ¿
Bandwidth = (f + 300) – (f -100) = 400Hz.
Channel Capacity M =2; 2 * 400 * Log22 = 800 bits per second
Channel Capacity M = 4; 2 * 400 * Log24 = 1600 bits per second
Channel Capacity M = 8; 2 * 400 * Log28 = 2400 bits per second
Q8. Data Rate of channel can be increased by increasing the number of
levels of signal. More number of levels will allow more bits per signal sample.
However the number of levels must by a power of 2 else the full utilization of
increased number of levels is not possible.
Q9. Switching refers to connecting two disjoint mediums to each other by
method of switches. Initially unconnected source and destination become
connected. The method of this connection can be done using circuit

switching where the physical medium from two points is connected to form a
dedicated path. In Packet switching Virtual circuit, the physical medium is
not connected but set of mediums is so arranged and reserved so that data
can flow between two points without intervention of any other device. But
since, physically the circuit is not switched, it is called virtual circuit. Further
data transfers happen in packets thus Packet switching. In case of circuit
switching, the data is exchanges as stream of bits.
Packet switching is advantageous is most applications as it allows merging of
traffic from various sources into single channel and be forwarded to required
destination. This saves lot of infrastructure cost. Further having large
capacity medium allows for accommodation of burst traffic flows. Bandwidth
can be scaled dynamically and shared as on need basis.
Q10.
If Antenna has height a and b, such that b = 2 x a.
Distance between the two is given by dm =√2 Ra+ √2 R 2 a
where R is Radius of Earth.
Calculation for height of Antenna.
40 x 103 =√2 Ra+ √4 Ra
40 x 103 = √ 2 a( √ R+ √ 2 R)

√2 a = 40 x 103 / (( √ 2+1) √ R)
2a = 1600 x 106 / (2.414)2 x 64 x 105
a = 1600 x 106 / 2 x 11.6568 x 64 x 105
= 21.445 m
Therefore, antenna 1 height = 21.445m and antenna 2 height = 42.90
m
Bibliography
Forouzan’s, B. A. (n.d.). Data Communication & Networking. Mcgraw Hill.