ECX6239/EEX6539 - Wireless Communications
(2019/2020)
ANSWER GUIDE
Question 1
a) Explain the three basic propagation mechanisms that affect wireless mobile communication
systems and provide suitable examples. (06 marks)
ANSWER
There are three propagation mechanisms that affect wireless communication signal propagation.
These are reflection, diffraction and scattering.
When an EM wave impinges upon an object with dimensions larger than the wavelength of the
propagating wave reflection occurs.
e.g. Ground reflections from the surface of the earth and from buildings and walls.
When the transmitter and receiver are separated by a large distance and there is an obstruction in the
direct signal propagation path, diffraction occurs due to the sharp irregularities of the obstructing
surface.
e.g. Signal reception issues due to high rise buildings or foliage
When the EM waves travel in a medium where there are objects or particles with much smaller
dimensions compared to the wavelength, then, the signal is said to scatter due to the irregularities in
the channel.
e.g. Scattering due to foliage
b) In the far-field of an electromagnetic wave transmitted from a transmitting antenna the
electric field component at a distance 'r' is given as Etx = {a.cos(2πf [t – r/c])}/r. If the
receiver is fixed the received electric field component is given as Erx = {b.cos(2πf [t –
r/c])}/r where 'c' is the speed of light. (14 marks)
a) Derive an expression for the electric field component of the received signal at a distance
'r1' at time 't=t1' when the receiver antenna is moving at a velocity 'v' where the initial
location at t=0 is 'r0'.
b) When at 'r1' the receiving mobile station stops moving when it sees an obstructing wall
'd' (where d < r1) distance directly in front of it .
i. Derive an expression for the reflected wave.
ii. Compute the phase difference between the two waves.
ANSWER
a) r1 = r0 + vt1 -> Erx = {b.cos(2πf [t –( r0/c) – (vt1/c)])}/r.
b) i) There are two waves superimposed. Consider the direction and take the algebraic sum.
Erx = {b.cos(2πf [t –( r1/c)])}/r - {b.cos(2πf [t –(2d- r1/c)]}/(2d-r)
ii) Taking the difference of the phase components of the two signal you get the following
expression.
Phase difference = (d-r)[4πf/c] + π
c) A transmitter and a receiver are separated by a distance 2km. The line-of-sight path is
directly obstructed by a tall building of height 75m which is at 1km away from the
transmitter. The heights of the transmitter and the receiver are 50m and 30m
respectively. All the heights are measured from the ground. Compute the loss in dB due
to the obstacle when a signal of 1900MHz is transmitted. (06 marks)
ANSWER
Using the Knife edge diffraction model for solving this problem.
Calculate the wavelength and compute the Fresnel diffraction coefficient using the standard
formula.
V = h([2(d1+d2)]/[λ.d1.d2])​0.5
From the Gd graph shown in Figure 1, the diffraction loss for v=4.5 is approximately -25dB
d) Consider the uplink of a GSM system where the signal to noise ratio (S/N) required is 11
dB. Assume a maximum mobile transmit power of 1.0 W. Consider 0 dB antenna gain at the
mobile and 12 dB gain at the base station (BS). Path loss given by the urban area Hata
model, where frequency (f) is 850 MHz, BS antenna height is 30 meters, mobile height is 1
meter. Consider noise figure (N) to be 3 dB and the system is noise-limited. Calculate the
maximum range of the link.
ANSWER
Thermal noise = NkTB = −147.9 dBW
Gains = 16.3 dB gains.
Path loss computed using the Hata model for the urban area, distance (d) in km,
L(dB) = 69.55 + 26.16 log10(850) − 13.82 log10(30) + [44.9 − 6.55 log10(30)] log10 d
So, Using the above formulation, d = 6km
Figure 1: Relationship between the Fresnel diffraction parameter vs diffraction loss
Question 2
a) Describe the GSM architecture using a suitable diagram and explain the functions of the
main sub-systems. (12 marks)
ANSWER
Main sub-systems are,
- Base-Station Subsystem
- Network Subsystem
- Operation and Support Subsystem
- Mobile Station Subsystem
Base Transreciver Station(BTS)-BTS It has radio transreciever that define a cell and are capable of
handelling radio link protocols with MS. Functions of BTS are 1. Handelling radio link protocols 2.
Providing FD communication to MS. 3. Interliving and de- interliving. Base station controller(BSC)
IT manages radio resources for one or more BTS.It controls several hundred BTS al are connected
to single MSC. Functions of BTS include control BTS, radio resource management, handoff
management and control, radio channel setup and frequency hopping.
Network subsystem( NSS) handles the switching of GSM calls between external networks and
indoor BSC. It includes three different databases for mobility management as: .HLR (Home
Location Register), VLR (Visitor Location Register), AUC (Authentication center), Mobile
switching center (MSC). It connects fixed networks like ISDN ,PSTN etc. Functions of MSC
include, call setup, supervision and relies, collection of billing information, call handling / routing,
management of signalling protocol, record of VLR and HLR. HLR (Home Location Register)
facilitates call roaming and call routing capabilities of GSM are handled. It stores all the
administrative information of subscribers registered in the networks. IT maintenance unique
international mobile subscriber identity.(IMSI). VLR (Visitor Location Register) - It is a temporary
database. It stores the IMSC number and customer information for each roaming customer visiting
specific MSC. Authentication center - It is a protected database .It maintenance authentication keys
and algorithms.It contains a register called as Equipment Identity Register. Operation
subsystem(OSS) include management of all mobile equipment in the system 1) management for
charging and billing procedure 2) maintain all hardware and network operations.
b) Compare and discuss the advantages and limitations of frequency reuse and cell splitting
techniques which aims at increasing the capacity in a cellular system. (06 marks)
ANSWER
Adjacent cells are assigned different frequencies to avoid interference or crosstalk. Main objective
is to reuse frequency in nearby cells. For example, 10 to 50 frequencies assigned to each cell and
the transmission power is controlled to limit power at that frequency escaping to adjacent cells. The
main issue is to determine how many cells must intervene between two cells using the same
frequency. As an approach to cope with the increasing capacity cell splitting is performed. Cell
splitting technique includes separation of areas in cells of high usage into smaller cells.
c) Explain the spread spectrum concept using a general model. (06 marks)
ANSWER
The general model of SS is shown below.
PN generator generates the spreading code at the transmitter. At the receiver, the same spreading
code needs to obtain the data.
d) A cellular network consists of 36 cells. Each cell has a hexagonal shaped cell area of 3 km​2​.
Total number of radio channels allocated 196. (06 marks)
i) Calculate the total area covered by the cellular network.
ii) Calculate the total channel capacity if the cell reuse is (1) N = 4 (2) N = 5 and (3)
N = 7.
ANSWER
(i) total area covered = 108 km​2
(ii)
Reuse factor
Total channel capacity
N=4
1764
N=5
1411
N=7
1008
Question 3
a) Derive an expression for the average signal-to-noise ratio (SNR) in a selection diversity
receiver with 'M' branches. (10 marks)
ANSWER
The probability that a single branch has SNR less than a threshold (r) is given by,
P= 1 – e^(-r/Γ)
The probability that all 'M' independent diversity branches receive signals which are simultaneously
less than some specific SNR threshold (r) is given by,
P​M​ = [1 – e^(-r/Γ)]^M
The average SNR of the received signal when diversity is used is given by the following expression.
Γ​avg​ = ​0 ∫​ ∞​​ (1 – P​M (r)
)
​
Substituting the expression for P​M​ will result in the following.
Γ​avg​ = ​0 ∫​ ∞​​ (1 – [1 – e^(-r/Γ)]^M dr
Simplifying the above expression will yield the following.
Γ​avg​ = Σ​M​ (-1)​i-1​ ( ​i​ M
​ ​) ​0 ​∫ ∞​​ e^(-r/Γ)]^M dr where i=1,2,3...M
Γ​avg​ / Γ = Σ​M ​( 1/ i ) where i=1,2,3...M
b) Consider a selection diversity mechanism with five branches. Each branch receives an
independent Rayleigh fading channel. Consider the average SNR value is 20dB. Calculate
the probability that SNR will drop upto 10dB in all the branches. (05 marks)
ANSWER
The probability that a single branch has SNR less than a threshold (r) is given by,
P = 1 – e^(-r/Γ)
The probability that all five (05) independent diversity branches receive signals which are
simultaneously less than some specific SNR threshold (r) is given by,
P = [1 – e^(-r/Γ)]^5
c) Describe the operation of an OFDMA transmitter and receiver in LTE system with a suitable
diagram.(05 marks)
ANSWER
Reference: Cox. C., An introduction to LTE: LTE, LTE advanced, SAE and 4G
mobile communications . John Wiley & Sons.
As explained in Dayschool 3.
Question 4
a) Explain the power control mechanisms used in CDMA IS-95 system and their limitations
under fast fading conditions. (04 marks)
ANSWER
In CDMA IS-95 both open loop power control and closed loop power control are used.
For uplink, open-loop sets the transmit power of the mobile user based on the measurements of the
downlink ​channel strength via a pilot signal.
Closed-loop power control operates at 800 Hz and involves 1 bit feedback from the base station to
the mobile, based on measured SINR values.
SINR is estimated based on the output of the Rake receiver.
SINR threshold depends on the multipath channel statistics which cannot be predicted in advance.
Therefore, an outer loop adjusts the SINR threshold as a function of frame error rates.
Accuracy of power control is also limited by the 1-bit quantization.
Although the estimations arrive on the feedback path at a higher rate, due to the limitation on the
1bit quantization, the fast fading events cannot be tracked.
b) Explain the operation of a 'M' branch Rake receiver. (06 marks)
ANSWER
A Rake receiver utilizes multiple correlators to separately detect the M strongest multipath
components. The outputs of each correlator are weighted and summed to get a better estimate.
Combining the time delayed versions of the original signal transmission improves the signal to
noise ratio.
Weighted aggregate of the correlator outputs is given by S' = ∑​i α​
coefficients are
​ m​ S​m weighted
​
denoted as 'α​m​'.
c) In spread spectrum terminology, 'chip' is used to refer to the sample period of a symbol. If
there are two symbols, each as 'n' chips, transmitted over two pseudo random sequences x1
and x2. Explain how the signal detection is performed when the Rake receiver operation is
implemented using matched filters. In your answer consider additive noise component at the
receiver. State any assumptions you make. (10 marks)
ANSWER
In a Rake receiver the inner products of the received signal with shifted versions of the transmitted
binary sequences are taken. Each output is then weighted by the channel tap gain of the appropriate
delay and summed. The filter is matched to the channel response and then sampled at time n + L
(delayed tap in the PN sequence). Let the channel impulse response be 'h' and the additive noise at
the channel be denoted as 'w​n​'. Assume that the coherence time is much greater than the delay in
each multipath component, 'h' is known to the receiver and 'h' doesn't vary over the symbol period.
The output can then be written as the channel response 'h' convolved with the input signal x added
noise component w​n​.
Question 5
a) Discuss the concept of linear modulation and state the main advantages as well as
limitations. (03 marks)
ANSWER
Power efficiency describes the ability of a modulation technique to preserve the bit error rate of the
digital message at low power levels.
It is often expressed as the ratio of the signal energy per bit to noise power spectral density required
at the receiver input for a certain probability of error.
Depending on the type of digital modulation, the amount of signal power should be increased to
obtain an acceptable bit error probability.
Bandwidth efficiency describes the ability of a modulation scheme to accommodate data within a
limited bandwidth. Bandwidth efficiency reflects how efficiently the allocated bandwidth is utilized
and is defined as the ratio of the throughput data rate per Hertz in a given bandwidth.
b) Define the terms (i) power efficiency and (ii) bandwidth efficiency related to digital
modulation. (05 marks)
ANSWER
In linear modulation techniques, the amplitude of the transmitted signal varies linearly with the
modulating digital signal. Linear modulation techniques are bandwidth efficient so allows to
accommodate greater traffic but do not have a constant envelope and the linearity depends on the
circuitry used in the signal generation.
c) Draw the constellation diagrams for the quadrature phase shift keying (QPSK) digital
modulation scheme. (04 marks)
ANSWER
d) (i) Explain how the bits are encoded in binary phase shift keying (BPSK). (04 marks)
ANSWER
In BPSK, the phase of a constant amplitude carrier signal is switched between two values according
to the two possible signals m1 and m2, corresponding to binary 1 and 0, respectively, which are
phase separated by 180​0​.
Consider the amplitude of the carrier signal as 'A' and the energy per bit is 'E​b​'.
Then, considering the phase difference between the two signals corresponding to the binary
symbols, the transmitted BPSK signal s(t) is given as follows.
For binary '1'
s(t) = √(2 E​b​ /T) cos (2π f​c​ t + θ​c​)
For binary '0'
s(t) = -√(2 E​b​ /T) cos (2π f​c​ t + θ​c​)
Then, s(t) = m(t) √(2 E​b​ /T) cos (2π f​c​ t + θ​c​)
(ii) A dense constellation diagram reveals that its modulation scheme is less energy efficient
compared to another modulation scheme which has a sparse constellation diagram.