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14 COMPARISONS OF VARIOUS ANALOG COMMUNICATION SYSTEMS

14 COMPARISONS OF VARIOUS ANALOG COMMUNICATION SYSTEMS

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9



8



7



6



Vc



Vm

mf =

fm



kf. Vm



mains same if modulating frequency is changed



no limitation, it is inversely

proportional to modulating



Fidelity



narrow bandwidth



bandwidth



frequency

AM has poor fidelity due to Fidelity is better due to wide



increased above 1



Modulation limitation. It cannot be



Depth of modulation re-



Depth of modulation have



Depth of modulation have



wide bandwidth



Fidelity is better due to



FM



than AM, but more than



Noise interference is less



Depth of



Noise interference is mini-



well as modulating frequency



mp = kp. Vm (radians)



mum



Noise interference is more



ing and carrier voltage



portional to both modulat- tional to modulating voltage as



Modulation index is pro- Modulation index is propor-



ma =



terference



Noise in-



index



Modulation



ANALOG AND DIGITAL COMMUNICATION



15



14



13



12



11



10



form



Output wave-



spectrum



Frequency



Vc



Amplitude



(fc-fm) fc



M aV c

2

(fc+fm)



MaVc

2



t



Radio, TV broadcasting,



complex

Radio and TV



broadcasting



complex



ments are less



Amplitude



0



f



I2(mf)



fc-2/m fc-fm fc fc+fm fc-2/m



I2(mf)

1



j (m )

j1(mf) 0 f j (m )



point communication



t



police wireless, point to



1



p



j (m )

j1(mp) 0 pj (m )



I2(mp)

fc-2/m fc-fm fc fc+fm fc-2/m



I2(mp)



0



Amplitude



systems



t



PM is used in some mobile



tion equipments are more



reception equip-



used



Applications



rior to that in FM

More complex



ter than that of PM

Transmission and recep-



is less

Transmission and



ratio

Equipment



ence is avoided



ference is avoided due to



interference is preSignal to noise ratio is infe-



Adjacent channel interfer-



Adjacent channel inter-



Adjacent channel



wide frequency spectrum

sent

interference

Signal to noise Signal to noise ratio Signal noise ratio is bet-



channel



Adjacent



ANALOG COMMUNICATION



ANALOG AND DIGITAL COMMUNICATION



SOLVED TWO MARKS

1. Define AM and draw its spectrum.



Amplitude of the carrier signal varies according to amplitude

variations



in



modulating



signal



is



known



as



amplitude



modulation. Spectrum: Figure shows the spectrum of AM signal. It

consists of carrier (ƒc) and two sidebands at ƒc ± ƒm .



maEc



Ec



2



2



fc-fm



maEc



fc



fc+fm



Figure Spectrum of AM wave

2. Why carrier frequencies are generally selected in HF range than

low frequency range?





The antenna size is very large at low frequencies. Such



antenna is practically not possible to fabricate. High carrier

frequencies require reasonable antenna size for transmission and

reception.





High frequencies can be transmitted using tropospher-



ic scatter propagation, which is used to travel long distances.





ANALOG COMMUNICATION



3. The equation of an AM wave is, eAM = 100[1 + 0.7 cos (3000t/2π)

+ 0.3cos(6000t/2π) sin(106t/2π)]

Find the amplitude and frequency of various sideband terms.

Solution: The given equation can also written as

eAM = [100 + 70cos(3000t/2π) + 30cos(6000t/2π)] sin(106t/2π)

Here

Em1 = 70 and ω1 = 3000/2π rad/sec

Em2 = 30 and ω2 = 6000/2π rad/sec

Ec = 100 and ωc = 106/2π rad/sec

Hence



m1 = Em1/Ec = 70/100 = 0.7

m2 = Em2/Ec = 30/100 = 0.3

Ec= 100

m1Ec

m1Ec

= 35

2

2 = 35

m2Ec

2



= 15



m2Ec

2



= 15



wc -6000/ 2p wc -3000/ 2p wc =106/ 2p w +3000/ 2p w +6000/ 2p

c

c



4.



5.



Calculate percentage modulation in AM if carrier amplitude

is 20 V and modulating signal is of 15V.

Solution:

HereEm= 15V

Ec = 20V

Modulation index, m = Em / Ec

= 15/20 = 0.75

Percentage modulation = m * 100



= 75%

Define detection (or) demodulation.

Detection is the process of extracting modulating signal

(message signal or original information or base band) from the



ANALOG AND DIGITAL COMMUNICATION



modulated carrier. Different types of detectors are used for

different types of modulations.

In other words, demodulation or detection is the process by

which the message is recovered from the modulated signal at receiver.The devices used for demodulation or detection are called

demodulation or detectors.

6.



Define the term modulation index for AM.

Modulation index is the ratio of amplitude of modulating signal

(Em) to amplitude of carrier (Ec).

i.e.

m = Em / Ec



7.



State Carson’s rule of FM bandwidth.

Carson's rule' approximates the bandwidth necessary to

transmit an angle modulated wave as twice the sum of the

peak frequency deviation and the highest modulating signal

frequency.



Carson’s rule of FM bandwidth is given as,



BW = 2(δ + ƒm (max))

Here δ is the maximum frequency deviation and ƒm (max) is the

maximum signal frequency.



8.



Differentiate between narrow band FM and wideband FM.

In narrowband FM, the frequency deviation is very small. Hence

the frequency spectrum consists of two major sidebands like

AM. Other sidebands are negligible and hence they can be neglected. Therefore the bandwidth of narrowband FM is limited

only to twice of highest modulating frequency.

If the deviation in carrier frequency is large enough so that other

sidebands cannot be neglected, then it is called wideband FM.

The bandwidth of wideband FM is calculated as per Carson’s

rule.



ANALOG COMMUNICATION



9.



Define frequency modulation.

Frequency modulation is defined as the process by which the

frequency of the carrier wave is changed in accordance with the

instantaneous value of the message signals.



10. Define modulation index for FM.

Modulation index is defined as the ratio of maximum frequency

deviation to the modulating frequency.

Modulation index mf = δ/ƒm

11. Define frequency deviation.

Frequency deviation is the change in frequency that occurs in

the carrier when it is acted on by a modulating signal frequency.

The frequency deviation is typically given as the peak frequency

shift in Hertz (Δf).

12.



What is the effect of increasing modulation index in FM?

In FM, the total transmitted power always remains constant.

But with increased depth of modulation, the required bandwidth

is increased.



13.



Why is FM superior to AM in performance?

i). In AM system the bandwidth is finite. But FM system has infinite number of sidebands in addition to a single carrier.

ii). In FM system all the transmitted power is useful whereas in

AM most of the transmitted power is used by the carrier.

iii). Noise is very less in FM; hence there is an increase in the

signal to noise ratio.



14.



Define instantaneous phase deviation

The instantaneous phase deviation is the instantaneous change

in phase of the carrier at a given instant of time and it indicates



ANALOG AND DIGITAL COMMUNICATION



how much the phase of the carrier is changing with respect to

the reference phase.

16.



Define angle modulation

Angle modulation is defined as the process by which the frequency or phase of the carrier wave is changed in accordance

with the instantaneous value of the message signals.



17.



Define PM.

In phase modulation, the phase of the carrier varies according

to amplitude variations of the modulating signal. The PM signal

can be expressed mathematically as,

ePM = Ec sin(ωct+ mpsinωmt)

Here mp is the modulation index for phase modulation. It is given as, mp = Φm



Here Φm is the maximum value of phase change.



18.



What is the need for pre-emphasis in FM transmission?

The noise has greater effect on higher modulating

frequencies than on lower ones. The effect of noise on higher

frequencies can be artificially boosting them at the transmitter

and correspondingly attenuating them at the receiver. Thus is

pre-emphasis.



19.



What are the advantages of FM over AM?

a. The amplitude of FM is constant. It is independent of depth of

modulation. Hence transmitter power remains constant in FM

whereas it varies in AM.

b. Since amplitude of FM is constant, the noise interference is

minimum in FM. Any noise superimposing amplitude can be

removed with the help of amplitude limits. Whereas it is difficult

to remove amplitude variations due to noise in AM.



ANALOG COMMUNICATION



c. The depth of modulation has limitation in AM. But in FM the

depth of modulation can be increased to any value by increasing

the deviation. This does not cause any distortion in FM signal.

d. Since guard bands are provided in FM, there is less possibility of

adjacent channel interference.

e. Since space waves are used for FM, the radius of propagation

is limited to line of sight. Hence it is possible to operate several

independent transmitters on same frequency with minimum interference.

f. Since FM uses UHF and VHF ranges, the noise interference is

minimum compared to AM which uses MF and HF ranges.

20.



A 107.6 MHZ carrier is frequency modulated by a 7 kHZ sine

wave. The resultant FM signal has a frequency of 50 kHZ

Determine the modulation index of the FM wave.



Here δ = 50 kHZ and ƒm = 7 kHZ .



Modulation index = δ/ƒm = 50/7 = 7.142

21.



If the rms value of the aerial current before modulation is

12.5 A and during modulation is 14 A, calculate the percentage of modulation employed, assuming no distortion.

Here Itotal = 14 A and Ic = 12.5 A.







m=







=



22.



2



2



(



(



I2total

I2C

142



12.52



-1



-1



)



)









= 0.71



An AM broadcast transmitter radiates 9.5 KW of power with

the carrier unmodulated and 10.925 KW when it is sinusoidally modulated. Calculate the modulation index.



Ptotal = 10.925 KW, Pc = 9.5 KW



ANALOG AND DIGITAL COMMUNICATION











m=







m=



2



(

(



Ptotal

PC



-1



)

)



10.925



2

-1

9.5



= 0.54



23. A broadcast radio transmitter radiates 5 KW power when the

modulation percentage is 60%. How much is the carrier power?

Ptotal = 5 KW, m = 0.6, Pc =?



Ptotal = Pc







(



5 KW = Pc



1+



(



1+



m2

2



)



0.62

2



)



Pc = 4.237 KW.



24.

What are two major limitations of the standard form of

amplitude modulation?

1) Most of the power is transmitted in the carrier. Hence AM is less

efficient.

2) Because of amplitude variations in AM signal, the effect of noise

is more.

25.



The antenna current of an AM transmitter is 8 A when only

carrier is sent, but it increases to 8.96 A when the carrier

is modulated by a single tone sinusoid. Find the percentage

modulation.





Here Itotal = 8.96 A and Ic = 8 A.



ANALOG COMMUNICATION





Itotal = Ic



1+



1+







8.96 = 8







m = 0.713



26.



m2







2

m2

2



If a modulated wave with an average voltage of 20 Vp changes in amplitude 5 V, determine the maximum and minimum

envelope amplitudes and the modulation coefficients.









Emax = 20 + 5 = 25 V

Emin = 20 – 5 = 15 V







Modulation index =





=

27.







Emax - Emin

Emax + Emin

25-15

25 +15



= 0.25



A carrier is frequency modulated with a sinusoidal signal of

2 kHZ resulting in a maximum frequency deviation of 5 kHZ

. Find 1) Modulation index 2) Bandwidth of the modulating

signal.

Given data: Modulating frequency ƒm = 2 kHz



Maximum frequency deviation δ = 5 kHz

1) Modulation index = mf = δ/ƒm





=



5 x 103

2 x 103



= 2.5



2) Bandwidth of the modulating signal



BW = 2(δ + ƒm (max))



Here ƒm (max) is the maximum modulating frequency, which

is given as 2 kHz .

Hence,



ANALOG AND DIGITAL COMMUNICATION



28.





BW = 2(5 x 103 + 2 x 103) = 14 KHz

Calculate the bandwidth of commercial FM transmission assuming Δƒ = 75 kHz and W = 15 kHz.





Here δ = Δƒ = 75 kHz

And ƒm (max)) = W = 15 kHz



BW = 2(δ + ƒm (max))



= 2[75+15] kHz = 180 kHz

29.



Define bandwidth efficiency.

It is the ratio of the transmission bit rate to the minimum

bandwidth required for a particular modulation scheme. It is

denoted as Bη and given by



Transmission bit rate (bps)

Bh=

Minimum bandwidth (Hz)

bits/s

bits

bits/s

=

=

=



cycles/s

cycle

Hz



30.



What is the required bandwidth for FM signal in terms of

frequency deviation?

BW = 2 (Df + fm) Hz

Where,



Df

fm



31.



= Peak frequency deviation (Hz)

= Modulating-signal frequency (Hz)



Distinguish between FM and PM.



Sl. No

FM

1

VFM(T) = Vc cos (wct + mf sin

2



PM

VPM(t) = Vc cos (wct +mp cos



wmt)

Associated with the change



wmt)

Associated with the changes



in fc, there is some phase



in phase there some change



change



in fc.



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