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Parallel-Search - Spread Spectrum Communication - Exam, Exams of Communication

Unity Degree College - LucknowCommunication

Main points of this exam paper are: Parallel-Search, Transmitter, Receiver Block, Spectrum System, Explain, Spread Spectrum Communication, System Employs, Spectrum, Hop Spread, Frequency

Typology: Exams

2012/2013

Uploaded on 04/15/2013

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Cork Institute of Technology
Bachelor of Engineering (Honours) in Electronic Engineering – Award
(Bachelor of Engineering in Electronic Engineering – Award)
(NFQ – Level 8)
Summer 2005
Spread Spectrum Communications
(Time: 2 Hours)
Examiner: Dr R.A. Guinee
Answer 3 questions Prof. C. Burkley
Mr. J. Ryan
Q1. (a) Draw the transmitter and receiver block diagram for a Frequency Hop spread
spectrum system and explain its operation. Discuss the advantages and
disadvantages of this form of spread spectrum communication. (14 %)
(b) A Frequency Hop spread spectrum system employs a fast hop with
k = 10 hops per message bit
M = 1024 frequencies
Message bit rate =2.5 kBPS
Final RF multiplication =10
Determine (i) the RF signal bandwidth, (ii) Processing gain in dB, (iii) pN
Code generator clock rate and (iv) the frequency separation in kHz. (12 %)
(c) Explain the parallel method of receiver code search used in a frequency
hopping acquisition scheme and its advantages. (7.33 %)
[33.33 %]
Q2 (a) A pseudonoise (pN) code generator produces a 63-bit sequence at a clock rate of
20MHz. What is the equation and graphical form of the autocorrelation function
(ACF) of the sequence assuming that the pulses have values ±2 volt? Comment on
the shape of the ACF and the line density of the PSD in terms of the pN sequence
length and clock rate (10 %)
(b) The data sequence x(t), given by 100110001, is fed to the DS/BPSK spread
spectrum transmitter in Fig Q2-(a) at a rate of 75 BPS with the leftmost bit first.
Data Signal
x(t)
pN Code
g(t)
Carrier
20
Ptcos ω
BPSK o/p
ttgtxPts
0
cos)()(2)(
ω
=
BPS
K
Modulator
Fig. Q2_(a)
Fig. Q2_(b)
1 0 1 0
g(t)
pf3

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Cork Institute of Technology

Bachelor of Engineering (Honours) in Electronic Engineering – Award

(Bachelor of Engineering in Electronic Engineering – Award)

(NFQ – Level 8)

Summer 2005

Spread Spectrum Communications

(Time: 2 Hours)

Examiner: Dr R.A. Guinee

Answer 3 questions Prof. C. Burkley

Mr. J. Ryan

Q1. (a) Draw the transmitter and receiver block diagram for a Frequency Hop spread

spectrum system and explain its operation. Discuss the advantages and

disadvantages of this form of spread spectrum communication. (14 %)

(b) A Frequency Hop spread spectrum system employs a fast hop with

k = 10 hops per message bit

M = 1024 frequencies

Message bit rate =2.5 kBPS

Final RF multiplication =

Determine (i) the RF signal bandwidth, (ii) Processing gain in dB, (iii) pN

Code generator clock rate and (iv) the frequency separation in kHz. (12 %)

(c) Explain the parallel method of receiver code search used in a frequency

hopping acquisition scheme and its advantages. (7.33 %)

[33.33 %]

Q2 (a) A pseudonoise (pN) code generator produces a 63-bit sequence at a clock rate of

20MHz. What is the equation and graphical form of the autocorrelation function

(ACF) of the sequence assuming that the pulses have values ±2 volt? Comment on

the shape of the ACF and the line density of the PSD in terms of the pN sequence

length and clock rate (10 %)

(b) The data sequence x(t), given by 100110001, is fed to the DS/BPSK spread

spectrum transmitter in Fig Q2-(a) at a rate of 75 BPS with the leftmost bit first.

Data Signal x(t)

pN Code g(t)

Carrier 2 P cos ω 0 t

BPSK o/p

s ( t )= 2 Px ( t ) g ( t )cos ω 0 t

BPSK

Modulator

Fig. Q2_(a)

Fig. Q2_(b)

g(t) 1 0 1 0

The pN code generator in Fig Q2-(b) emits the PRBS stream g(t) at a rate of 225 chips per sec from an initial seed 1010. (i) Sketch the final transmitted sequence x(t)g(t). (ii) What is the bandwidth of the transmitted signal? (iii) What is the processing gain? (iv) If the received signal delay T (^) d is a single chip with respect to the receiver PRBS g(t) sketch the resultant despread sequence. (14 %)

Correlator

pN Code g ( t − T ˆ d )

A *^2 P ⋅ x ˆ( t − T ˆ d )

r ( t )= A 2 Px ( t − Td ) g ( t − Td )cos[ω 0 ( t − Td )+ φ ]

BPSK Data Demodulator

Fig. Q2_(c)

Filter

(c) Describe the demodulation process illustrated in Fig.Q2_(c) and discuss in detail the action of the correlator in the synchronization process. (9.33 %) [33.33 %]

Q3 (a) Draw the block diagram of a parallel-search acquisition system for Direct

Sequence Spread Spectrum receiver synchronization and discuss its operation. If λ chips of the pN code, with timing epoch Tc, are examined during a correlation period with a probability P (^) D of correct detection show that the average acquisition time of the DS receiver is given by

P D

Tc Tacq = λ (14%)

(b) Draw the block schematic of the RASE method for direct sequence acquisition and compare this with the parallel search method in part (a). What are the advantages and disadvantages of this type of acquisition? (8%)

Local pN Code Generator

g ^ t + T 2^ c +τ 

Early Correlator

Despreading Correlator Z(t) (^) Data Demodulator g(t+ τ ) BPF

BPF Late Correlator

g ^ t − T 2^ c +τ 

VCO (^) FilterLoop

Square Law Detector

Square Law Detector

Data O/P

Y( τ ) Σ

Received DS-BPSK Signal

r ( r )= A 2 px ( t ) g ( t )cos (ω 0 t +φ)

Data: x(t) pN Code: g(t)

Fig.Q3 (c) Direct Sequence Delay Locked Loop Tracker