Wireless Systems: Comprehensive Assessment with Detailed Solutions

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Wireless
Assessment 1

Table of Contents
Question 1..................................................................................................................................2
Question 2..................................................................................................................................3
Question 3..................................................................................................................................4
Question 4..................................................................................................................................7
Question 5..................................................................................................................................8
Question 6..................................................................................................................................9
Question 7..................................................................................................................................9
Question 8................................................................................................................................11
Question 9................................................................................................................................11
Question 10..............................................................................................................................12

Question 1
Ordering Pizza Scenario:
A. The guest will first ask for the pizza ordering menu. Then, the host will give the menu
and the guest will choose the pizza and then will place an order.
B. The guest finds the telephone number and then dial the telephone number. Then, a
connection is established and the tone is sent to the dialled number.
C. The guest waits as the order is being confirmed. The host first talks with the order
clerk and then allocation of the particular order is done to the guest and then the order
is confirmed with the guest.
D. Order clerk then finally receives the order information and then asks for the payment
and gets an approval for the payment and finally, the pizza order is placed.
Figure 1: ordering Pizza
Delivering Pizza Scenario:
A. Pizza cook cooks pizza with all the specified information given by the guest and
finally gives that pizza order to the order clerk.
B. Firstly, order clerk packs the pizza and then prepare it for the delivery and also makes
a delivery receipt which he then puts in the delivery van.
C. The delivery van is used to transfer all the pizzas that were ordered by the guests form
the order clerk. In the ordering of the pizza the telephone line was considered as the
transmitting line but in this case, the road will be considered as the transmitting line.
D. At last, the host gives a confirmation about the payment information and then
confirms delivery by returning to the order clerk.
E. In the end, the guest is given the pizza by the host.

Figure 2: Delivering Pizza
Question 2
In the case of the French PM, for instance, he directly addresses his comment to the Chinese
PM. However, two translators transmit the message via the telephone system. The translator
of the French PM translates his words into English, which is then translated into Chinese by
his Chinese PM interpreter. In the case of the French PM, for instance, he directly addresses
his comment to the Chinese PM. However, two translators transmit the message via the
telephone system.
As previously, the Prime Ministers continue to contact one another directly. An intermediate
network now translates the signal before it is transmitted. The intermediate translates each
link's messages. The intermediate converts each link's messages.

Question 3
A.
 Maximum Amplitude i.e. Am = 15
 Time period i.e. T = 3 seconds
 Frequency = 1/T = 1/3 = 0.3333 Hz
 f (t) = Am Sin(wt + ɸ)
ɸ represents the phase
Here, wt = 0 so, f (t) = 0
Am Sin (ɸ) = 0
So, ɸ=0
Phase i.e. ɸ = 0o

B.
 Maximum Amplitude i.e. Am = 4
 Time period i.e. T = 6.5 seconds
 Frequency = 1/T = 1/6.5 = 0.1538 Hz
 f (t) = Am Sin(wt + ɸ)
ɸ represents the phase
Here, wt = 0 so, f (t) = 0
Am Sin (ɸ) = 0
So, ɸ=0
Phase i.e. ɸ = 0o

C.
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 Maximum Amplitude i.e. Am = 7.5
 Time period i.e. T = 2.25 seconds
 Frequency = 1/T = 1/2.25 = 0.444 Hz
 f (t) = Am Sin(wt + ɸ)
ɸ represents the phase
Here, wt = 0 so, f (t) = 0
Am Sin (ɸ) = Am
So, Sin (ɸ) =1
Phase i.e. ɸ = 90o

Question 4
The general form of the equation – A Sin (2π f t)
A is amplitude
f is frequency
Time period = 1/f
a) 10𝑆𝑖𝑛(2𝜋(100)𝑡)
From the above equation, we get
A = 10
f = 100 Hz
T = 1/100 sec = 0.01 sec
ɸ = 0
b) 20𝑆𝑖𝑛(2𝜋(30)𝑡 + 90)
From the above equation, we get
A = 20
f = 30 Hz
T = 1/30 sec = 0.33 sec
ɸ = 0
c) 5𝑆𝑖𝑛(500𝜋𝑡 + 180)
From the above equation, we get
A = 5
f = 500/2 = 250 Hz
T = 1/250 sec = 0.04 sec
ɸ = 180;

Shifting the whole plot to the left by 180o. Overall phase shift of 180 o
d) 8𝑆𝑖𝑛(400𝜋𝑡 + 270)
A = 8
f = 400/2=200 Hz
T = 1/200 sec = 0.005 sec
ɸ = 270
Question 5
A. Pixel rate = pixel/day =(480 * 500) * 30= 7200000
Every pixel can take around 32 values
So, log2 32 = 5
5 bits / pixel
Source Rate i.e. R = 7200000 * 5 =36000000 = 36 * 106 Mbps
B. B = 4.5 * 106
Using Shannon’s criteria,
SNR dB = 35 = 10log10 (SNR) = 31632
Capacity of the channel = 4.5 *106 * log2 (1+3162)
Capacity of the channel = 4.5 *106 * log2 (3163)
Capacity of the channel = 52.32 * 106 bps
Question 6
4 Giga Hertz = 4 * 109 Hertz
35863 Km = 35.863x106 m
PL = 20 log10 (4 * 109) +20log10 (35.863x106) - 147.56Db
PL = 195.6 dB
Question 7
𝑠 (𝑡) = 5 sin (200𝜋𝑡) + sin (600𝜋𝑡)
𝑠 (𝑡) = 5 sin (2𝜋f1 𝑡) + sin (2𝜋f2 𝑡)

Let’s suppose that we are having three frequencies namely f1, f2 & f3.
If there is a periodic signal then every frequency will be considered as an integer.
Multiple of f0
f1 = n1 * f0
f2 = n2 * f0
f3 = n3 * f0
f0 = gcd (f1, f2, f3)
Therefore, the given signal is f0 = gcd (100,300) = 100
Hence, fundamental frequency = 100
𝑠 (𝑡) = 5 sin (200𝜋𝑡) + sin (600𝜋𝑡)
Performing Fourier transform on sin (2𝜋A𝑡)
15 = 1/2i [∫ (f-100) - ∫ (f+100)) + 1/2i (∫ (f-300) - ∫ (f+300)]
Hence, spectrum of s (t) is
s (f) = 1/2i [5(∫ (f-100) - ∫ (f+100)) + (∫ (f-300) - ∫ (f+300))]
Bandwidth = fmax – fmin = (f+300) - (f-100) = 400 Hertz
Bitrate i.e. Capacity = 2 * Bandwidth * log2 L
L is the number of levels & Bandwidth was calculated on the previous page.
Now, for m=2
Channel Capacity = 2 * 400 * log2 2 = 800 bits per sec
Now, for m=4
Channel Capacity = 2 * 400 * log2 4 = 1600 bits per sec

Now, for m=8
Channel Capacity = 2 * 400 * log2 8 = 2400 bits per sec
Question 8
For a particular noiseless channel,
Bitrate = 2 * Bandwidth * log2 L
Where L is the number of signal level
Bitrate is bits/second
Bandwidth is a quantity which remains fixed. The bit rate is directly proportional to L i.e.
number of signal levels
If there is an increase in the signal level i.e. L then there may be a reduction in the overall
system reliability.
Question 9
The technique used to switch essentially allows the physical path to be passed from one node
to another. Mainly used in a physical layer is a circuit switch. We supply information
physically to pass on. For the transmission of data, a certain bandwidth is assigned. Two
types of circuit switches exist-space and time division.
During the virtual circuit switching packet, the address is used. The principal difference is
that the circuit switch works on the physical layer, while in the virtual layer data link layer
works. The virtual approach has one major advantage: although the allocation of resources is
on demand, the source can verify the resource availability without truly reserving them. And
even when switching packets, it is very quick because the packet is used and we can use all
bandwidth simultaneously.
Question 10
The LOS is a type of transmission that can only transmit and receive data when stations can
be seen and transmitted without any obstacle.

The maximum LOS distance between two antennas is, dm = √ 2Rht + √ 2Rhr
R ≈ 64 * 105 m
Height of one antenna is double of the other, this is given.
dm = √ 2Rh + √ 2R (2h) = √ 2Rh + 2√ Rh = √ Rh (2 + √ 2)
√ h = dm / (2 + √ 2) * √ R
h = (dm) 2 / (2 + √ 2) * R
It is already given in the question that d= 40 Km = 40 * 1000 = 40000 m
h = (dm) 2 / (2 + √ 2) * R
h = (40000) 2 / (2 + √ 2)2 * 64 * 105
h = 16 * 103 / (2 + √ 2) 2 * 64 = 1000/ 46.627
h = 21.446 which is approximately 21.45 m
Hence,
Antenna 1 height = 21.45 m
Antenna 2 height = 42.90 m