UNIT-iii ct

7/30/2019 UNIT-iii ct

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The noise equivalent bandwidth of the filter is defined as the bandwidth of an ideal filter

at which the noise power passed by real filter and ideal filter is same.

5. Define Thermal noise. Give the expression for the thermal noise voltage across aregister.

Ans:

Thermal noise is the name given to the electrical noise arising from the random motion

of electrons in a conductor.

6. What is meant by figure of merit of a receiver?

Ans:

The ratio of output signal to noise ratio to channel signal to noise ratio is called figure of

merit. Figure of Merit = (SNR)o/(SNR)c = (So/No)/(Si/nNi)

7. Define Noise Figure.

Ans:

It is defined as the ratio of input signal to noise ratio to the output signal to noise ratio.

F = Input (s/n) / output (s/n).

8. Define Noise Temperature.

Ans:

The available noise power is directly proportional to temperature and it is independent

of value of resistance. this power specified in terms of temperatures is called as noise

temperature.


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9. What is partition Noise?

Ans:

In an electron tube having one or more positive grids, this noise is caused by irratic

partition of the cathode current among the positive electrodes. In a transistor, thepartition noise is created from the random fluctuation in the division of current between

the collector and base.

10. Define a random variable?

Ans:

Random variable is defined as a rule or mapping from the original sample space to anumerical sample space subjected to certain constraints. Random variable is also

defined as a function where domain is the set of outcomes s and whose range is R, is

the real line.

11. What is a random process?

Ans:

A Random process X (s,t) is a function that maps each element of a samples space into atime function called sample function. Random process is a collection of time functions.

12. What is a stationary random process?

Ans:

When the statistical properties of a process do not change with time, it is called

Stationary process.

13. A receiver connected to an antenna of resistance of 50 has an equivalent noise

resistance of 30. Find the receiver noise figure.

Ans:


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F = (1 + (Req / Ra)) = ( 1 + ( 30 / 50) ) = 1.6.

UNIT I

AMPLITUDE MODULATION SYSTEMS

PART A

1. State any two important spectral properties of periodic power signals.

Ans:

1. Linearity

2. Time shift.

3. Frequency Shift.

2. Define amplitude modulation. Give the expression for AM wave.

Ans:

It is the process by which the amplitude of the carrier wave is changed in Accordancewith the instantaneous value of the message signal.

3. A transmitter radiates 9 kW without modulation and 10.125 kW after modulation.

Determine depth of modulation.

Ans:

Modulation index m = 0.5
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4. A 500 W carrier is modulated to a depth of 60 percent. Calculate the Total power in

modulated wave.

Ans:

5. Define VSB transmission. Mention its application.

Ans:

Definition:

One of the sideband is partially suppressed and portion of the other sideband is

transmitted. This portion compensates the suppression of the sideband. It is called

vestigial sideband transmission.

Application:

VSB is mainly used in TV transmission

6. Give the applications of SSB-SC AM.

Ans:

1. Long distance point-to-point communications.

2. Mobile communications at frequencies below 30 MHz

3. It is also used in two way communications.

4. Multichannel communications in microwave.

7. Compare low level modulation and high level modulation.
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Ans:

8. An A M transmitter radiates 10 KW of power without modulation. With 0.65 depth of

modulation what is the radiated power?

Ans:

9. Draw the frequency spectrum of VSB.

10. A carrier is amplitude modulated to a depth of 75 percent. Calculate the total Power

in the modulated wave, if the carrier power is 40 watt.

Ans:

11. Calculate the percentage of power saving when the carrier and one of the Sidebandsare suppressed in an AM wave modulated to a depth of 100 percent.

Ans:
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12. SSB is suitable for speech signals and not for video signals. Why?

Ans:

To generate SSB,the message spectrum must have an energy gap centered at the origin.

This criteria is satisfied by speech signals, for which the energy gap is from -300 Hz to

+300 Hz(600 Hz wide). Hence any one sideband can be easily isolated with the help of

practical band pass filters. But in case of video signals, there is no such energy gap near

origin. In other words, the video signal extends from DC origin. Hence lower and upperside bands are practically joined at origin. Hence with the help of practical band pass

filters, it is not possible to isolate one sideband from other. Hence for video signals

vestigial sideband transmission is suitable, rather than SSB.

13. A 2MHZ carrier having an amplitude of 5 V is modulated by a 4 kHZ audio signal

having an amplitude of 2 V. Determine the modulation index.

Ans:

Modulation index m = Em / Ec = 2 / 5 = 0.4.

14. Define over modulation, under modulation and 100% modulation.

Under modulation. m1

15. Define modulation index of an AM signal.

Ans:
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It is defined as the ratio of the maximum modulating voltage to the maximum carrier

voltage. It is also called as Depth of modulation.

m = Vm/ Vc

16. Draw the spectrum of DSB.

17. Distinguish between low level and high level modulator.

18. Define FDM &frequency translation.

Ans:
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Frequency translation:

The single sideband modulation basically performs frequency translation.

it is also called frequency changing , mixing or heterodyning. the single sideband

modulation the message spectrum is shifted by an amount equal to the carrier frequencyfc

FDM:

For voice transmission, the frequency range of 300 to 3400 Hz is used. when more

number of such voice channels are to be transmitted, then each channel is given a fixed

frequency slot and then transmitted. This is called Frequency Division Multiplexing.

19. What do you meant by image frequency?

Ans:

The local oscillator frequency (f0), input signal frequency (f) and IF (f) are related as, f0

fs = f i.e., f0 = fs + fi

If some other frequency, f = fsi + 2fi appears at the input of mixer, then it produces fi at

the output of the mixer. fsi- image frequency.

20. Why we need modulation?

Ans:

Needs for modulation:

1.Ease of transmission

2.Multiplexing

3.Reduced noise

4.Narrow bandwidth

5.Frequency assignment


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6.Reduce the equipment's limitations.

21. When a signal m(t)=3 cos(2 103t) modulates a carrier c(t)=5 cos( 6t), find the

modulation index and translate bandwidth if the modulation is AM.

Ans:

Modulation index m = Em / Ec = 3 / 5 = 0.6.

Bandwidth = 2 fm = 2 kHz

UNIT II

ANGLE MODULATION SYSTEMS

PART A

1. Compare narrow band FM and Wideband FM.

Ans:

2. What is pre-emphasis? Why is it used?

Ans:
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The premodulation filtering in the transistor, to raise the power spectral density of the

base band signal in its upper-frequency range is called pre emphasis (or pre distortion)

Pre emphasis is particularly effective in FM systems which are used for transmission of

audio signals.

3. Define phase modulation.

Ans:

It is a type of modulation, used in communication systems, in which the phase of a

carrier wave is varied by an amount proportional to the instantaneous amplitude of the

modulating signal.

4. What is Narrowband FM?

Ans:

Narrowband FM has only a single pair of significant sidebands.

5. What are the applications of phase locked loop?

Ans:

Phase-locked loops are widely used in , radio, telecommunications,computer and other

electronic applications.

6. State the frequency in an FM system is 500 Hz and modulating voltage is 3

V,modulation index is 60. Calculate maximum deviation and bandwidth.

Ans:

Modulation index m = /fm

60 = / 500

= 60 x 500 = 30 kHz
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Bandwidth BW = 2 = 2 x 30 = 60 kHz.

7. Mention advantages of angle modulation over amplitude modulation.

Ans:

1. 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.

2. Since amplitude of FM is constant, the noise interference is minimum in FM.

3. FM uses UHF and VHF ranges, the noise interference is minimum compared to AM

which uses MF and HF ranges.

8. A 80MHz carrier is frequency modulated by sinusoidal signal of 1V amplitude and the

frequency sensitivity is 100Hz/V. Find the approximate bandwidth of the FM waveform

if the modulating signal has a frequency of 10KHz.

Ans:

2 ( + fm ) = 2 ( 100 + 10000 ) = 20.2 kHz.

9. What is frequency deviation in FM?

Ans:

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 (f).

10. Differentiate FM and AM.


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11. What is the bandwidth required for an FM wave in which the modulating frequency

signal is 2 KHz and the maximum frequency deviation is 12 Khz?

Ans:

Bandwidth = 2 ( + fm ) = 2 ( 12 + 2 ) = 28 kHz.

12. A carrier wave of frequency 100 MHz is frequency modulated by a signal 20sin(200

x 103t) What is bandwidth of FM signal if the frequency sensitivity of the modulation is

25kHz/v.

Ans:

Bandwidth = 2 ( + fm ) = 2 ( 500 + 100 ) = 1.2 MHz.

UNIT-V

INFORMATION THEORY

PART A

1. Define information rate.

Ans:
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If the time rate at which source X emits symbols is r symbols per second. The

information rate

R of the source is given by R=r H(X) bits/second H(X)- entropy of the source .

2. Define entrophy?

Ans:

Entropy is the measure of the average information content per second. It is given by the

expression H(X)=I P(xi)log2P(xi) bits/sample.

3. What is a prefix code?

Ans:

In prefix code, no codeword is the prefix of any other codeword. It is variable length

code. The binary digits are assigned to the messages as per their probabilities of

occurrence.

4. What is channel capacity of Binary Synchronous with error probability of 0.2?

5. Define mutual information.

Ans:

Mutual information I(X,Y) of a channel is defined by I(X,Y)=H(X)-H(X/Y) bits/symbol

H(X)- entropy of the source H(X/Y)- conditional entropy of Y.

6. State Shanon Hartley theorem.

Ans:

The capacity C of a additive Gaussian noise channel is


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C=B log2(1+S/N)

B= channel bandwidth ,S/N=signal to noise ratio.

7. State any four properties of entropy.

Ans:

1.I(X,Y)=I(Y,X)

2.I(X,Y)>=0

3.I(X,Y)=H(Y)-H(Y/X)

4.I(X,Y)=H(X)+H(Y)-H(X,Y).

8. Give the expressions for channel capacity of a Gaussian channel.

Ans:

Channel capacity of Gaussian channel is given as, C = B log2 (1 + (S/N))

9. Define the entropy for a discerte memory less source.

Ans:

The entropy of a binary memory-less source H(X)=-p0log2p0-(1-p0)log2(1-p0) p0-

probability of symbol 0,p1=(1- p0) =probability of transmitting symbol 1.

10. Define the significance of the entropy H(X/Y) of a communication system where X is

he transmitter and Y is the receiver.

Ans:


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H(X/Y) is called conditional entrophy. It represents uncertainty of X, on average, when

Y is known.In other words H(X/Y) is an average measure of uncertainty in X after Y is

received.

H(X/Y) represents the information lost in the noisy channel.

11. An event has a six possible outcomes with probabilities 1/2, 1/4, 1/8, 1/16, 1/32, 1/32.

Find the entropy of the system.

Ans:

H = Pk log2(1/Pk)

= () log2 2 + () log2 4 + (1/16) log2 16 + (1/32) log2 32 + (1/32) log2 32

= 1.5625.

12. When is the average information delivered by a source of alphabet size 2, maximum?

Ans:

Average information is maximum, when the two messages are equally likely i.e., p1 = p2

= 1/2. Then the maximum average information is given as,

Hmax = 1/2 log2 2 + 1/2 log2 2 = 1 bit / message.

13. Name the source coding techniques.

Ans:

1. Prefix coding

2. Shanon-fano coding

3. Huffman coding.


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14. Write down the formula for mutual information.

Ans:

The mutual information is defined as the amount of information transferred when Xi is

transmitted Yj is received. It is represented by I( Xi , Yj ) and it is given as,

I(Xi,Yj) = log ( P(Xi/Yj)/ P(Xi) ) bits.

15. Write the expression for code efficiency in terms of entropy.

Ans:

Redundancy = 1 - code efficiency. Redundancy should be as low as possible.

16. Is the information of a continuous system non negative? If so, why?

Ans:

Yes,the information of a continuous system is non- negative. The reason is that

I(X;Y)>= 0 is one of its property.

UNIT-IV

PERFORMANCE OF CW MODULATION SYSTEMS

PART A

1. What do you understand by Capture Effect in FM?

Ans:

When the interference signal and FM input are of equal strength, the receiver fluctuates

back and froth between them .This phenomenon is known as the capture effect.


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2. What is sensitivity and selectivity of receivers?

Ans:

Selectivity:

Selectivity of a receiver is defined as its ability to select the desired signals among the

various signals.

Sensitivity:

It is defined as a measure of its ability to receive weak signals

3. What is the purpose of pre emphasis and de emphasis in FM?

Ans:

The psd of noise at the output of FM receiver usually increases rapidly at high

frequencies but the psd of message signal falls off at higher frequencies. This means the

message signal does not utilise the frequency band in efficient manner. Such more

efficient use of frequency band and improved noise performance can be obtained with

the help of pre emphasis and de-emphasis.

4. Define SNR.

Ans:

It is defined as the ratio of signal power to the noise power.

5. How to achieve threshold reduction in FM receiver?

Ans:

When the carrier to noise ratio reduces tocertain value, tyhe message information is lost.

The performance of the envelope detector deteriorates rapidly and it has no proportion

with carrier to noise ratio.


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6. This is called thershold effect.

Ans:

As the input noise power is increased the carrier to noise ratio is decreased the receiver

breaks and as the carrier to noise ratio is reduced further crackling sound is heard and

the output SNR cannot be predicted by the equation. This phenomenon is known as

threshold effect.

7. What is meant by FOM of a receiver?

Ans:

The ratio of output signal to noise ratio to channel signal to noise ratio is called figure of

merit.

Figure of Merit = (SNR)o/(SNR)c = (So/No)/(Si/Ni)

8. What is the SNR for AM with small noise case?

Ans:

(So/No)/(Si/Ni) = (Ka2P)/ (1+ Ka2 P)

9. What is threshold effect with respect to noise?

Ans:

When the carrier to noise ratio reduces to certain value, type message information is

lost.

The performance of the envelope detector deteriorates rapidly and it has no proportion

with carrier to noise ratio. This is called thershold effect.


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10. What is FM threshold effect?

Ans:

As the input noise power is increased the carrier to noise ratio is decreased the receiver

breaks and as the carrier to noise ratio is reduced further crackling sound is heard andthe output SNR cannot be predicted by the equation. This phenomenon is known as

threshold effect.

11. Define Pre-emphasis and De-emphasis.

Ans:

Pre-emphasis

The premodulation filtering in the transistor, to raise the power spectral density of the

base band signal in its upper-frequency range is called pre-emphasis (or pre distortion)

Pre emphasis is particularly effective in FM systems which are used for transmission of

audio signals.

De-emphasis

The filtering at the receiver to undo the signal pre-emphasis and to suppress noise is

called de-emphasis.

UNIT-iii ct

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