Fundamentals of Wireless Communication: Assignment 1 Solutions

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WIRELESS - ASSIGNMENT -1

Table of Contents
Ques 1:.............................................................................................................................................3
Ques. 2:............................................................................................................................................3
Ques. 3:............................................................................................................................................4
Ques. 4:............................................................................................................................................6
Ques. 5:............................................................................................................................................8
Ques. 6:............................................................................................................................................8
Ques. 7:............................................................................................................................................9
Ques. 8:..........................................................................................................................................10
Ques. 9:..........................................................................................................................................10
Ques. 10:........................................................................................................................................10

Ques 1:
Description of Pizza Ordering:
- The pizza menu was recommended by the guests, and the host will provide the menu
items to guests.
- After that, the guests will select the food for placing the order.
- Guest will check the telephone number or check the live status of the order. It will
connect the telephone lines and also send the dialed number to the receiver.
- The host will remind the status of ordered pizza, acknowledged with the guests and the
guest have to wait.
- Clerk of the order will receive the order, then ask for the payment, and also place the
order to the cook of the pizza.
Description of Pizza Delivery:
- The cook of the pizza will prepare according to the order given by guests to the clerk.
- The clerk will pack the pizza in the box of pizza, the prepare the receipt of the delivery
and insert into the delivery van.
- Van will deliver pizza to the host where the road line of transmitting between ordering &
delivery of the system of pizza.
- The details of payment were confirmed by the host, return to the clerk and confirm the
pizza delivery.
- At last, the pizza was successfully delivered to the guest.
Ques. 2:
(a)

(b)
Ques. 3:
(a)
- AM (Maximum Amplitude) = 15---------------------------- (1)
- T (Time Period) = 3 Seconds
 Frequency = (1/T) = (1/3) = 0.333 Hertz--------------- (2)
- X(t) = Am sin (ɷt + ϕ) (Therefore, ϕ = Phase)
 Here now (ɷt = 0)
 X(t) = 0
 Am sin (ϕ) = 0
 ϕ = 0
 Phase (ϕ) = 0o --------------------------------------------- (3)
(b)

- Am = 4 -------------------------------------------------- (Answer)
- T = 6.5 Seconds
 f = (1/T) = (1/6.5) = 0.1538 Hertz--------------- (Answer)
- Here now (ɷt = 0)
 X(t) = 0
 Phase (ϕ) = 0o -------------------------------------- (Answer)
(c)
- AM (Maximum Amplitude) = 7.5 ---------------------- (1)
- T (Time Period) = 2.25 Seconds
 Frequency = (1/T) = (1/2.25) = 0.444 Hertz ----------- (2)
 Here at (ɷt = 0): Am is max.

 Am sin (ɷt + ϕ) = Am
 sin (ϕ) = 1
 ϕ = 90
 Phase (ϕ) = 90o -------------------------------- (3)
Ques. 4:
- Sin A (2П(f)t)
- Time (t) = 1/f
- A is Amplitude
- f is frequency
(a) 10 sin(2П(100) t)
frequency = 100 hertz
Amplitude = 10
Time (t) = 1 / 100 seconds = 0.01 seconds
Phase (ϕ) = 0
(b) 20 sin(2П(30) t)

frequency = 30 hertz
Amplitude = 20
Time (t) = 1 / 30 seconds = 0.333 seconds
Phase (ϕ) = 0
(c) 5 sin(500Пt + 180)
frequency = 500 / 2 = 250 hertz
Amplitude = 5
Time (t) = 1 / 250 seconds = 0.04 seconds
Phase (ϕ) = 180
Plot shift is 180o
Shift of phase = 180o

(d) 8 sin(400Пt + 270)
frequency = 400 / 2 = 200 hertz
Amplitude = 8
Time (t) = 1 / 200 seconds = 0.005 seconds
Phase (ϕ) = 270
Ques. 5:
(a) Find the value of R:
- Pixel / Seconds = 30 * (480 * 500) = 72,00,000
- Every pixel will take almost 32 values
Therefore, log2 (32) = 5
- 5 Bit / Pixel values
- Source Rate (R) = 5 * 72,00,000
= 36 * 106 (3,60,00,000) mbps
(b) Find the channel capacity:
- B = 4.5 * 106
SNRdb = 35
SNRdb = 10 log10 = 31632
- C = 4.5 * 106 (1 + 31632) log2 (Therefore, C = 8 log2 (1 + SNR))
= 4.5 * 106 * log2 (3163)
= 4.5 * 106 *11.6272
= 52.32 * 106 bps
Ques. 6:
- 4 GHz = 4 * 109 Hertz

- 35863 kilometer = 35.863 * 106 meters
 (20 log10 (4 * 10^9)) + 20 log10 (35.863 * 10^60 – 147.56)
 195.574 dB
Ques. 7:
(a) Calculate Bandwidth
- S (t) = 5 sin (200Пt) + sin (600 Пt)
- S (t) = 5 sin (2 Пf1t) + sin (2 Пf2t)
Suppose three frequency of signals were f1, f2, f3.
Every frequency were periodic as an integer with the multiple of frequency.
 f1 = n1f0
 f2 = n2f0
 f3 = n3f0
 f0 = gcd (f1, f2, f3)
Therefore, signal given f0 = gcd (100, 300) = 100
Therefore, fundamental frequency is 100.
- S (t) = 5 sin (200Пt) + sin (600 Пt)
Transformation = sin (2ПAt)
- Is = 1/2i [ (f - A) – (f +A)]ժ ժ
Therefore, S(f) = 5 * 1/2i [ (f - 100) - (f + 100)] + 1/2i ( (f - 300) - (f + 300))ժ ժ ժ ժ
Therefore, S (f) = 1/2i [5 (f - 100) - (f + 100)] + ( (f - 300) - (f + 300))ժ ժ ժ ժ
Bandwidth (B) = fmaximum - fminimum
 (f + 300) – (f + 100)
 Bandwidth (B) = 400 Hertz
(b) Channel Capacity
It is the capacity of the noiseless signals having maximum rate of data. The Nyquist theorem
is used for the bit rate calculation that defines the maximum rate of bits.
- (Capacity) Bit Rate = bandwidth * 2 * log2L
- L is total count of levels
- Calculated Bandwidth is 400 Hertz
- For M is 2
Capacity of channel = 400 * 2 * log22 = 800 bits per second
- For M is 4

Capacity of channel = 400 * 2 * log24 = 1600 bits per second
- For M is 8
Capacity of channel = 400 * 2 * log28 = 2400 bits per second
Ques. 8:
The Nyquist theorem was used for the noiseless channel to calculate the bit rate:
- Bit Rate = Bandwidth * 2 * log2(L)
With the help of above equation,
- Bandwidth = channel bandwidth
- L = Level of signals
- Bit Rate = bits/seconds
The bandwidth cannot be changed because its quantity was fixed. Therefore, rates were
proportional to the levels of signal counts.
The main disadvantage is the increased signal levels that reduce system reliability.
Ques. 9:
The technique of virtual switching was used for providing the physical paths so that information
will be passed to one point to another.
Circuit switching was used in the physical layer where it provides the information that can be
passed. The bandwidth was allocated for data transferring. Circuit switch is of two types – time
division and space division.
The main advantage of the virtual approach was the allocation of resources on demands. The
source of the approach will check the resources availability without reserving it.
Ques. 10:
- LOS is Line of sight communication
- ht = transmitter antenna height
- hr = receiver antenna height
the maximum distance of the line of sight in the communication of the height of two antennas
as hr and ht,
- dm = √2 R h t + √ 2 R hr (R = Earth radius)
- R ~ 64 * 105 meters
- So, dm = √ 2 R h t + √ 2 R(2h)
- So, dm = √2 R h+ √2 Rh = √ R h (2 + √2)

 √h = dm / √ R (2 + √ 2)
 h=(dm)2 / (2 + √ 2) * R = (40 * 103)2meters / (2 + √ 2) * 64 * 105
16∗108meters / (2 + √ 2) * 64 * 105
 h = 16 * 103 / (2 + √ 2)2 * 64 meters = 1000 meters / 11.6568 * 4 = 1000 / 46.267
meters.
 h = 21.446 ~ 21.45 meters
 Antenna 1 height = 21.45 meters
 Antenna 2 height = 42.90 meters