POCE Assignment1 Questions n Answers Set1 V1

8/10/2019 POCE Assignment1 Questions n Answers Set1 V1

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EKT 343

PRINCIPLES OF COMMUNICATION

ENGINEERING

Assignment 1-(answer set 1)

Due date : 27th

October 2014 (MONDAY)

------------------------------------------------------------------------------------------------You are required to answer all questions.

Penalty with deduction 1 mark per day will be given to those who submit the days later.

__________________________________________________________________________________________

NAME :_____________________________________________

MATRIC NO. :_____________________________________________

PROGRAM :_____________________________________________

Q1(30)

Q2(30)

TOTAL(60)

TOTAL(100%)


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Question 1. [CO1]

(a) What is the output of the ideal combiner for the following Figure 1? (4 marks)

Solution:

6dBm = 100.6

mW = 4 mW

3dBm = 100.3

mW = 2 mW

Combiner adds the signal power = 4 m + 2 m = 6 mW

In dBm = 10 log

= 7.76 dBm

(b) Given a four-stages system with an input power of Pin = 0.1mW. If the gain of Ap1= 500, while Ap2=2x10

-4Ap1, Ap4= 2Ap2and Ap3= 2Ap1. Calculate:

(i) the gain values of Ap2, Ap3 and Ap4. (3 marks)

Ap2= 2 x 10-4

Ap1= 2 x 10-4

(500) = 0.1

Ap3= 2 Ap1= 2 (500) = 1000

Ap4= 2 Ap2= 2 (0.1) = 0.2

(ii) the dB gain of each stage. (4 marks)

(iii) the overall gain in dB. (1 mark)

(iv) the output power in watt and dBm. (3 marks)

1000010)( 1040

unitlessAT

Combiner+6 dBm

+3 dBm

Output ? dBm


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(v) the possible components that you think constitutes in the system and justify your answer.

(1 mark)

Since values gain of Ap1and Ap3are greater than 1 then they could be considered as

amplifiers, while value gain of Ap2and Ap4are less than 1 then they could be considered

as filters

(c) An amplifier circuit having a noise figure of 9 dB and power gain of 25 dB is followed by a mixerhaving a noise figure of 16 dB and power gain of 30 dB. Calculate overall noise figure and equivalent

noise temperature at the input of the combination. (6 marks)

F1= 100.9

= 7.94328

F2= 101.6

= 39.8107

G1= 102.5

= 315.2277

F = F1+ = 7.94328 +

= 8.0660

Noise figure, NF = 10 log F

= 10 log (8.0660)

= 9.066 dB

Te= (F-1) T = (8.0660 -1 ) (290)

= 2049.14 K


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(d) A receiver with a 50 input resistance operates at a temperature of 29oC. The received signal is at 90MHz with a bandwidth of 5 MHz . The received signal voltage of 8V is applied to an amplifier with a

noise figure (NF) of 2.6 dB. Evaluate:

(i) The input noise power, Pn(2 marks)

T=29+273=302K

= 2.04x10-6

@ WattsxMxkTBPN1423 1008.2)5)(302)(1038.1(

(ii) The input signal power, PS(1 mark)

(iii) Input Signal to Noise Ratio in decibels , SNR (2 marks)

(iv) The noise factor F and SNR of the amplifier (2 marks)

NF= 10 log F(NOISE RATIO)

Noise ratio= F= 100.26

= 1.82

But NR = (SNR input)/(SNR output)

SNRoutput=(SNR input)/( NR)=61.54/( 1.82)=33.8

SNR output =SNR ampli fi er

(v) The noise temperature of the amplifier (1 mark)


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Question 2. [CO2]

(a) Under what condition that the overmodulation occur? Explain it. (3 marks)

If the amplitude of the modulating voltage is higher than the carrier voltage, mwill be greater

than 1, causing distortion. If the distortion is great enough, the intelligence signal becomes unintelligible.

(b) For a particular broadcast AM radio station with a carrier frequency of 80 sin (2 x 105t) and a

modulating signal of 20 sin (500 t), calculate the following:

(i) Modulation Index ( 2 marks)

Given, Vm(t)= 20 sin (500 t)

Vc(t)= 80 sin (2 x 105t)

Hence, modulation index, m =

(ii) Maximum and minimum amplitudes of the envelope (2 marks)

Vmax = Vc+ Vm

= 80 + 20

= 100 V

(iii) Sideband frequencies (4 marks)

kHzHzfm 25.02502

500

, kHzx

fc 1002

102 5

Hence upper sideband frequency = fc+ fm= 100.25 kHz

Hence lower sideband frequency = fc - fm = 99.75 kHz

(iv) Amplitude of each sideband frequency (1 marks)

Eusb/Elsb =

=

(v) Construct the mathematical expression of modulated signal Vam(t); (3 marks)

Vam (t) = 80 sin (2 x 10

5

t)- 10 cos [2 (100.25 k t) t] + 10 cos [ 2

(99.75k t)]

Vmin = Vc- Vm

= 80 - 20

= 60 V


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(vi) Illustrate and label the output envelope of the modulated signal, Vam(t) (4 marks)

(vii) Bandwidth (1 marks)

Bandwidth, B = 2 x fm = 2 x 250= 500 Hz

(viii) Total power delivered into a load of 600 (2 marks)

WR

VP cc 33.5

)600(2

80

2

22

Wm

PP cT 5.52

25.0133.5

21

22

(ix) Power saved if one of the sideband is suppressed and show the percentage power saving.

(3 marks)

Thus, if one sideband is suppressed,

Wm

PPm

PPPP cccLSBCNEWT 4167.54

25.0133.5

41

4

222

)(

WWPPPowersaved NEWTOLDT 0833.04167.55.5)()(

%5.1%1005.5

0833.0%100%

)(

xxP

PowersavedPower

OLDT

save d

Vm= 20V

Vc= 80 V

Vmin = 60 V

Vmax = 100 V


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(x) What are the drawbacks of Double sideband Full Carrier (DSBFC)? (3 marks)

Large power consumption, where carrier power constitutes >2/3 transmitted power.{remember :carrier does not contain any information}

Utilize twice as much bandwidthboth the upper and lower sideband actually contains sameinformation (redundant).

Noise is higher because the noise is proportional to the bandwidth

(c) The antenna current of AM broadcast transmitter modulated to a depth of 85% by an unmodulatedcarrier current that feed into a 50 antenna load impedance is 10 amperes.

(i) Calculate the modulated output current. (1 marks)

IT = 10

(ii) Calculate the total output power. (1 marks)

PT = (IT)2R = 11.67

2(50) = 6809 W

POCE Assignment1 Questions n Answers Set1 V1

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