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TEST 1

EKT 231 COMMUNICATION SYSTEM

SEM 2 2012/2013

ANSWER ALL QUESTIONS

1. (a) Briefly discuss on the basic components of communication system.

[3 marks]

Transmitter

The transmitter is a collection of electronic components and circuits that

converts the electrical signal into a signal suitable for transmission over a given

medium.

The communication channel is the medium by which the electronic signal is sent

from one place to another.Types of media include

-Electrical conductors

-Optical media

-Free space

-system-specific media (e.g., water is the medium for sonar).

A receiver is a collection of electronic components and circuits that accepts the

transmitted message from the channel and converts it back into a form

understandable by humans.

Noise is random, undesirable electronic energy that enters the communication

system via the communicating medium and interferes with the transmitted

message

/40

(b) Ir. Abdullah, manager of XYZ Berhad is satisfied with the junior engineer’s

performance. Then, he gives another good project for him to accomplish. In this

project, there are three-stages system comprised of two amplifiers and one filter with

an input power of Pin = 0.05 mW and absolute power gains of Ap1 = 100, Ap2 = 0.1, and

Ap3 = 1000. Thus, he asks the junior engineer to calculate parameters as listed below in

order to check the feasibility of this project:

(i) the input power in dBm [1 marks]

Pin=10 log0 . 05mW

1mW=−13 . 01dBm

(ii) output power (Pout) in watts and dBm [3 marks]

AT = 100 x 0.1 x 1000 = 10000

Pout (watt) = 10000 x 0.05mW = 0.5W

Pout (dBm) = 10 log

0. 5W1mW

=26 . 99dBm

(iii) The dB gain of each of the three stages and the overall gain in dB.

[3 marks]

Ap1 = 10log 100 = 20dB

Ap2 = 10log 0.1 = -10dB

Ap3 = 10log 1000 = 30dB

AT = 20dB + (-10dB) + 30dB = 40dB

(c) Discuss what is modulation and why modulation is needed [3 marks]

A process of changing one or more properties of the analog carrier in

proportion to the information signal.

An electronic technique in which a baseband information signal

modifies a carrier signal (usually a sine wave) for the purpose of

frequency translation and carrying the information signal via radio.

Why modulation is needed?

To generate a modulated signal suited and compatible to the

characteristics of the transmission channel.

For ease radiation and reduction of antenna size

Reduction of noise and interference

Channel assignment

Increase transmission speed

(d) Describe what is multiplexing and de-multiplexing [2 marks]

Multiplexing (MUX or MPX) - the process of simultaneously

transmitting two or more baseband information signals over a single

communications channel.

Demultiplexing (DEMUX or DMPX) - the process of recovering the

individual baseband signals from the multiplexed signal.

2. (a) AM double-sideband full carrier (AM DSBFC) is the most commonly used and

the oldest and simplest form of AM modulation. Sometimes is also called

conventional AM or simply AM. Explain Amplitude Modulation (AM)

[2 marks]

The process of changing the amplitude of a relatively high frequency carrier signal in proportion with the instantaneous value of modulating signal (information)

(b) A telecommunication engineer is given a task to assess the output power levels of a

transmitter with respect to different kind of modulation modes with the same

intelligibility received. Given that the antenna transmits a 12.5 kW of total power at 90%

modulation of conventional AM, Compute the following:

i. The amount of carrier power, Pc delivers through the antenna.

[3 marks]

Pt=Pc+Pusb+Plsb

12.5 kW = Pc + (m2/4)Pc + (m2/4)Pc

= Pc [1 + m2/2]

Pc = (12.5 kW)/(1 + 0.902/2) = 8.896 kW

ii. the amount of power using Double-Side Band Suppressed Carrier (DSBSC).

[2 marks]

Pt = Pusb + Plsb = (m2/4)Pc + (m2/4)Pc = 3.603 kW

iii. the amount of power using Single-Side Band Suppressed Carrier (SSBSC).[1 mark]

Plsb = Pusb = (m2/4)Pc =1.80 kW

iv. the percentage power saving of SSBSC compares to Double-Side Band Full

Carrier (DSBFC) [2 marks]

Power saving = [(12.5 – 1.8)/12.5] x 100%

= 85.6%

(c) Study the given the modulated wave below and determine the following parameters;

vam( t )=12 sin(1000 πt )−3 .25cos (1020πt )+3 .25cos( 980πt )(i) Carrier signal frequency and modulating signal frequency [2 marks]

ωt=2πf c t=1000πt ,

f c=1000 πt2 πt

=500Hz

ωt=2π ( f c+ f m) t=1020πt ,

f m=1020 πt2πt

−f c=510−500=10Hz

(ii) Upper and lower side band frequency [2 marks]

f usf= f c+ f m=500+10=510Hzf lsf= f c− f m=500−10=490Hz

(iii) Modulation index and percentage modulation [1 marks]

mEc2

=3 .25V

m=3 . 25(2)12

=0 .542

M=m∗100 %=54 . 2%

3. (a) Input to an FM modulator with a modulation index m=1.5, is a modulating signal Vm(t)=Vm sin (2π1000t), and an unmodulated carrier Vc(t)=20 sin(2π500kt).Using the Bessel graph, determine

i. number of sets of significant side frequencies [1 mark]

n =4

ii. Their amplitudes [2 marks]

Vo = 0.51*20 = 10.2V1 = 0.56*20 = 11.2V2 = 0.23*20 = 4.6V3 = 0.06*20 = 1.2V4 = 0.01*20 = 0.2

iii. Illustrate the frequency spectrum showing their relative amplitudes [3 marks]

iv. Assume load resistance RL =50Ω; determine the unmodulated carrier power [1 mark]

Pc = Vc2/2R = 4 watts

v. the total power in the angle modulated wave. [3 marks]

50 50 50 50 5049494949

0.1.

4. 4.

1111.

10

1.0.

Pt = Po+P1+P2+P3+P4

= Vo2/2R + 2V12/2R +2V22/2R + 2V32/2R +2V42/2R

= 1.0404 + 2.5088 + 0.4232 + 0.0288 + 0.0008= 4.002Watts