Beamwidth - Communications Systems - Exam, Exams of Data Communication Systems and Computer Networks

Main points of this past exam are: Beamwidth, Noise Temperature, Receiver System, Antenna, Bandwidth, Required, Signal Power

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2012/2013

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Cork Institute of Technology
Bachelor of Engineering in Communications Systems, Award
NFQ Level 7
December 2006
Communications Systems
(Time: 2 Hours)
Answer four questions.
[All questions are worth equal marks]
Values for constants;
Speed of light in a vacuum c = 2.998 x 108 m s-1
Boltzmann’s Constant kB = 1.38 x 10-23 J K-1
Formulae attached.
Examiners: Mr. D. Denieffe
Dr. R. O’Dubhghaill
Dr. O. Gough
Q1 (a) Explain the terms (i) Noise Temperature and (ii) White Noise [10 marks]
(b) If an Antenna and Receiver system have the following characteristics:
Antenna TAntenna = 200 K
Bandwidth = 8 MHz
Stage 1: G = 12 dB, F = 1.4 dB
Stage 2: G = 20 dB, F3 = 8 dB
Determine the required Signal Power at the antenna input to achieve an output SNR of 20
dB. [15 marks]
Q2 (a) Explain the operation of a super-heterodyne receiver using suitable diagrams. In particular
explain how adjacent channel interference and image channel interference is rejected.
[11 marks]
(b) A standard AM transmitter has a total power output of 15 kW under 100% modulation. If
the modulation depth is reduced to 40%, determine the power content of each frequency
component. [6 marks]
(c) An FM modulator uses a VCO with sensitivity of 40 kHz/V and has a free running
frequency of 4 MHz.
(i) Determine the input amplitude of a 15 kHz signal that results in a modulation index of
2 at the output.
(ii) What bandwidth will this signal require? [8 marks]
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Cork Institute of Technology

Bachelor of Engineering in Communications Systems, Award

NFQ Level 7

December 2006

Communications Systems

(Time: 2 Hours)

Answer four questions. [All questions are worth equal marks] Values for constants; Speed of light in a vacuum c = 2.998 x 10 8 m s- Boltzmann’s Constant k (^) B = 1.38 x 10 -23^ J K - Formulae attached.

Examiners: Mr. D. Denieffe Dr. R. O’Dubhghaill Dr. O. Gough

Q1 (a) Explain the terms (i) Noise Temperature and (ii) White Noise [10 marks]

(b) If an Antenna and Receiver system have the following characteristics: Antenna TAntenna = 200 K Bandwidth = 8 MHz Stage 1: G = 12 dB, F = 1.4 dB Stage 2: G = 20 dB, F 3 = 8 dB Determine the required Signal Power at the antenna input to achieve an output SNR of 20 dB. [15 marks]

Q2 (a) Explain the operation of a super-heterodyne receiver using suitable diagrams. In particular explain how adjacent channel interference and image channel interference is rejected. [11 marks] (b) A standard AM transmitter has a total power output of 15 kW under 100% modulation. If the modulation depth is reduced to 40%, determine the power content of each frequency component. (^) [6 marks] (c) An FM modulator uses a VCO with sensitivity of 40 kHz/V and has a free running frequency of 4 MHz. (i) Determine the input amplitude of a 15 kHz signal that results in a modulation index of 2 at the output. (ii) What bandwidth will this signal require? [8 marks]

Q3 .(a) Explain the terms (i) Beamwidth (ii) Radiation Resistance as they can be applied to antennas. [10 marks] (b) A designer has some Bluetooth radio modules which operate at 2.4 GHz with transmitter power of +4dBm and have a maximum range of 10m. A customer has a requirement to extend the range to 150 m between fixed sites. The original antennas were simple quarter wave stubs with a nominal gain of 2.14 dB and it is proposed to achieve the new performance using small parabolic dish antennas. Determine the size of the dish required. [15 marks]

Q4 (a) Draw block diagrams for a QPSK transmitter and receiver and explain their operation. [12 marks] (b) The input data rate to a 16-QAM modulator is 48 Mbps. Determine the output baud rate and minimum transmission bandwidth. Compare these results with those for BPSK and 8- PSK systems.. [7 marks] (c) The curves shown in figure Q4 plot probability of error versus Eb /No for FSK and BPSK systems. If the available Eb /No is 10.5 dB determine the BER for both cases. What increase in signal power is required to ensure that the FSK system achieves the same level of performance as the PSK system? [6 marks]

0 5 10 15

10 -

10 -

10 -

10 -

10 -

10 -

10 -

100

E (^) b/No (dB)

P e

orr r

Figure Q

FSK
BPSK

Useful Formulae for Communications Systems

AM Wave v am ( t )= [ V c + Vm sin( ω mt )] sin( ω ct ).

Power in AM wave: (^)  

m^2 PT PC

Total modulation index: mt = m 12 + m 22 + m 22 + Λ

Diagonal Clipping:

m ω

RC ≤^1 − m^2

Negative Peak Clipping: DC

load AC R

R
R

mRR =

FM Wave: (^)  

( )= sin^ −∆ cos( t ) f

v t V t f m m

fm c ω c c^ ω

Carson’s Rule: B = 2 ( mf + 1 ) fm

Image Frequency Rejection Ratio:

si

s s

si f

f f

f

IRR Q

Nyquist Sampling: fs =2B

SQNR for sinusoid:

( ) 6. 02 1. 76

SQNRdB N b

SQNR M

SQNR for uniformly distributed signal:

SQNRdB N b

SQNR M M

Slope overload onset: f s

∆ ≥^2 π fA

Delta Modulator SQNR f B

f N

S (^) s q^2

3 8 2

π

Fourier coefficients (spectral amplitudes) of rectangular pulse train.:

2 sin 2 0

0

n

n

T

a V n

Antenna Gain:

2

λ

π Aeff G = Friis Formula:

4 r^2

P GA

PR T eff

Noise Figure of Cascade: = + − + − + Λ 2 1

3 1

G G
F
G
F F F