analog communication

1 Analog Communication

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Biyani's Think Tank

Concept based notes

Analog Communication

(B.Tech Vth Sem, EC)

Mukul Sharma

Asst. Professor (EC)

Deptt. of Engineering

Biyani International Institute of Engineering and Technology

Analog Communication 2


Published by :

Think Tanks

Biyani Group of Colleges

Concept & Copyright :

Biyani Shikshan Samiti

Sector-3, Vidhyadhar Nagar,

Jaipur-302 023 (Rajasthan)

Ph : 0141-2338371, 2338591-95 Fax : 0141-2338007

E-mail : [email protected]

Website :www.gurukpo.com; www.biyanicolleges.org

Edition : 2011

Leaser Type Setted by:

Biyani College Printing Department

While every effort is taken to avoid errors or omissions in this Publication, any mistake or

omission that may have crept in is not intentional. It may be taken note of that neither the

publisher nor the author will be responsible for any damage or loss of any kind arising to

anyone in any manner on account of such errors and omissions.



Preface

I am glad to present this book, especially designed to serve the needs of the students. The book has been written keeping in mind the general weakness in understanding the fundamental

concepts of the topics. The book is self-explanatory and adopts the Teach Yourself style. It is

based on question-answer pattern. The language of book is quite easy and understandable

based on scientific approach.

Any further improvement in the contents of the book by making corrections, omission and

inclusion is keen to be achieved based on suggestions from the readers for which the author

shall be obliged.

I acknowledge special thanks to Mr. Rajeev Biyani, Chairman & Dr. Sanjay Biyani, Director

(Acad.) Biyani Group of Colleges, who are the backbones and main concept provider and also

have been constant source of motivation throughout this Endeavour. They played an active role

in coordinating the various stages of this Endeavour and spearheaded the publishing work.

I look forward to receiving valuable suggestions from professors of various educational

institutions, other faculty members and students for improvement of the quality of the book. The

reader may feel free to send in their comments and suggestions to the under mentioned

address.

Mukul Sharma



Content

S.No Name of Topic

1 Amplitude Modulation

2 Frequency Modulation

3 Noise

4 Noise in AM, FM



Amplitude Modulation

Q.1 Definite of Communication:-

Ans. It is the Basic process of exchange information the communication process

essentially consists of three basic building blocks.

Transmitter:- A transmitter is physical system that transmit information.

Receiver:- It is physical system that receiver the information.

Channel:- I is the medium through which information takes place depending

upon the type of channel used for information exchange, there are two types of

communication.

a> Line or Wire Communication:- In this type of communication a physical

channel is created between transmitter and receiver through copper wires,

coaxial cables, optical fiber cable etc. before the information exchange can

take place.

Ex.:- basic telephone system, telegraphy etc.

b> Radio or Wireless Communication:- In this type of communication, there

is no physical connection between transmitter & receiver, the space is used

as a channel for information exchange.

Ex.:- Mobile Communication

Satellite Communication

Wireless phone

Broad Casting

Transmitter Channel Receiver



Modulation:-

Modulation is process of super imposing the information contents of a baseband

modulating signed on a high freq carried signed by altering its chartn. Ex.:-

amplitude, phase, freq.

C (t) = Phase = WCt

Amplitude Freq

Modulation process translates a low freq. Baseband signal into a high freq.

Band pass signed.

Q.2 Explain Need of modulation:-

Ans. (1) Avoid mixing of signal

-3 0 3 KHz f

0 3 kHZ

997 1003 1004 1110

Grass Band

AC WCt



0 3kHZ

The modulation process translates different baseband signals at different carries

frequency so that spectrum overlap does not take place & mixing of signal can be

avoided.

Allow multiplexing of signal:-

Multiplexing means transmission of two or more signal simultaneously over

same channel once the different baseband signal are translate at different

frequency now they can be simultaneously transmitted over the same channel

without any loss of information.

Reduce height of Antenna:-

(i) = c/

Practice antenna height = /4

1 = 15 KHz

= = 20,000 m

= = 5 km

(ii) = 15 MHz

= = 20 m

= 5 m

the high of antenna required for transmission & reception of Radio wave in

Radio transmission is a fusi of frequency used the min. height of antenna is given

a /4 from the above two examples it is clear that the signal should be



transmitted at highest frequency to achieve practical height of antenna

Increase Range of Communication:-

At low frequency, the radiation is poor & signal gets highly attenuated directly

and distance modulation effectively increase the frequency of signal to be

radiated and thus increase the distance over which signal can be transmitted

faithfully

Improves Quality of Reception:-

The signal communication using modulation techniques such as FM & PCM

reduces the effect of noise to great extent. Reduction in noise improves quality of

Reception.

Q. 3 Specify three Basic Properties:-

Ans. Three basic operation are performed on a signal

x (t) 1

1 0 3 E X (E) 1 : - 1



1

2 6 t

b> Time advance:-

if > 0

y (t) = x ( t + 4) 1

-5 -1 0

(ii) Time Reversal:-

y (t) = x (t)

y (t)

-3 1

(iii) Time Sealing:-

y (t) = x ( t)

a> Time compression:- > 1

y (t) = x (2 t)



- 3/2 t

b> Time expansion:-

0 < < 1

y (t) = x (t/2)

-1 3

-2 6

Unit impulse Signal:- [S (t)]

S (t) = 1 ; t = 0

0 : t 0

1

t =0 t



Q.4 Explain three Properties of impulse:- Ans.

1> Sealing Property:-

S ( = S (t)

2> Product Property:-

x (t) (t) = x (o). (t) x (t). (t-to) = x (to). (t-to)

x (t) x (o)

0 to

t = 0 t = t

3> Shifting Property:-

= x (t)

cos ot [ ( + ]

Cos t = +



j t

e

2 ( - )

2 ( - )

(

Q. 5 Explain Amplitude Modulation and derive equation for A.M Ans. In A.M. the amplit6ude of high frequency carries signal is varied in accordance

with instananes value of Base band modulating signal keeping frequency & phase carst. Edn. for A.M.:- m (t) = any modulating signal with max. Frequency n

C (t) = Ac os t

XAM (t) = A os t

A = (Ac + M (E)) XAM(E) = [Ac + M (E)] os t modulated signal

XAM(E) = Ac os t + m(t) os t

Let m (t) = Am os t

C (t) = Ac os t

A = Ac + Am os nt

XAM (t) = [Ac + Am os t

= Ac os c [ 1 + os nt]

XAM(t) = Ac os ct + MAc os ct os nt

[ 2 os A os B = os (A+B) + os (A-B)]

= m = modulation index



XAM (t) = Ac os ct + os ( c+ n) + os ( c n)

Carries USB LSB

Q.6 Explain time domain and frequency domain representation of A.M

Ans. Time domain Representation for AM:-

Am

m (t) t

- Am

Ac

C (t)

-Ac



Frequency Domain Representation:-

XAM (E) = AC os t + m (t) os t

XAM ( ) = Ac [ ( - c) + ( + )]

+ [M ( ) + ( + c)]

[ os t = [ ( - ) + ( + )]

Convolution in time domain equal multiplication in frequency domain

x (t) y (t) x ( ) Y ( )

x (t) y (t) [X, ( ) * Y( )] (t)

Unit impulse signal:- 1

(t) = 1 : t = 0

0 : otherwise

t = 0

Properties of impulse funn.:-

1> Scaling Property:-



(t) =

ii> Product Property:-

x (t) , (t) = x (0) . (t)

x (t) . (t-t) = x (t) (t t)

x (t) x (t)

x (0)

x (0)

iii) Shifting Property:-

x (t). (t) dt = x (t

x (t). (t-t) dt = x (t

M ( )



- om m

- ( c+ n) - c -( c- n)

the transmitted AM signal contains three components that is correct, USB, LSB.

the intimation simultaneously by upper and lower side band, the caries does

not contain any information. the transmission between of Am signal is m in red/see

Q. 7 Derive equation for Power Relation and current in Am Ans.

Pt = + +

= +

Pc =

PSB = = - = PC

Pt = PC +

Current Relation in AM:-

Pt = PC



P = IR

Pn = I

Pt = It ; PC = It

It = IC (

Transmission Efficiency (modulation efficiency):-

It gives the % of useful power in the total transmitted power & given by ratio of

total S.B. power to the total TA power.

= x 100%

=

Concept of modulation index:-

- It gives the depth to which modulation has

occurred.

- For m=0 of represent no modulation.

- For m=1 maximum modulation has occurred.

- For m>1 overlapping of envelop take place

which results in envelop distortion this condition is known as over

modulation & should always be avoided.

m = 1 ; Am = Ac

It = IC

= z 100%



Amin ; Am Ac = 0

m = 1

t



M > 1

t

(i) Pt = Pc

For m = 1

(ii) = x 100%

= = 33.33%

Simultaneous modulation by several line work:-

M1 (t) = Am1 s m1t

M2 (t) = Am2 m2t

m (t) = m1 (t) + m2(t) M (t) = Am1 s m1t + Am2 s m1t

C (t) = Ac s m1t

A = Ac + Am1 s m1t + Am2 m2t

XAM(t) = [Ac + Am1 s m1t + Am2 s m1t] s ct

XAM (t) = Ac

M1 = , m2 =

XAM (t) = Ac s ct + M1 Ac s ct. s mt + m2 Ac s ct + s mt

Pt = 1.5 Pc



XAM (t) = Ac s ct + s ( c + m1)t

causes USB + s ( c - m1)t s ( c + m)t

LSB USB s ( c - m)t

LSB

Generalized Result-

When a high frequency signal is simultaneously modulated by several line

waves of maximum frequency m1, m, m3

then

a> for each addition of modulating signal a pair of sideband getting added in

the resultant A.M. signal that is why amplitude modulation in also known

as linear modulation.

b> The transmission between for resu.. A.M signed given by

B.W. = 2 max [ m1, m, m3 ..]

Power Relation:-

Pt = + + + +

Pt = + +

Pc =



Ps.B1 = = Pc.

Ps.B2 = = Pc.

Pt = Pc

DSB-SC (Double side band suppress carries)

Transmission of full A.M signal is not advisable because

a> Since carries { } also transmitted that does not contain any information.

b> For m=1, 2/3 power appears in the carries which is complete wastage

So instead of transmitting full A.M signal the carries is suppressed before

transmission such type of modulation known as DSB-SC modulation.

m (t) y (t)

c (t)

y (t) = m (t). C (t)

= m (t). Ac os ct

y ( ) = [M ( - c) + M ( + c)]

Pt = Pc Mr = +



1 M ( )

- m 0 m

- ( c + m) - c - ( c- m) c- m c c+ m

Ring Modulated:-

Diagram:-



(i) Positive half cycle

During positive half cycle of carries signed D1 & D2 are on and D3 D4 are ON and

D1 & D2 are off but in any case the Net O/P is zero because currents are in opp

direction

(ii) Negative half cycle



During positive half cycle of carries signed D3& D4are on and D3 D4 are ON and

D1 & D2 are off but in any case the Net O/P is zero because currents are in opp

direction

m (t) c (t) O/P

+ ve +Ve +Ve D1 & D2 ON

+ ve -Ve - Ve D3 & D4 ON

- ve +Ve -Ve D1 & D2 ON

- ve -Ve +Ve D3 & D4 ON



Power saving in DSB-SC modulation :=>

Total power saved = Pc

%age power saving =

= x 100%

X 100% = 66.67%

X 100% = 88.88%

SSB-SC (Single Side Band Suppress Carries):-

In case of DSB-SC modulation, both the side bands are transmitted which

contains same information so DSB-SC transmission is further redundant. Instead

of transmitting both the SB we can suppress one SB as well to achieve maximum

efficiency. Such type of modulation in which are

Q.8 Describe DSB-SC (Double Side Band Suppress Carries) and method to

modulate and demodulate it

Ans. In case of DSB-SC modulation both the sidebands are transmitted which contains

same information so DSB-SC transmission is further redundant.

Instead of transmitting both the SB we can suppress one SB as well to achieve

maximum efficiency. Such type of modulation in which either upper or lower

sideband is transmitted is known as SSB-SC modulation.

There are two methods for generation of SSB-SC signal



(i) Frequency Discrimination method:-

It is also known as fitted method

m (t) DSB-SC SSB-SC

Signal Signal

a(t) = Ac os ct

- ( c+ m - c - ( c- m) c- m c c + n

Disadvantage of frequency discrimination methods:

In case of frequency discrimination method the band pass filled should be as

ideal as possible but ideal filters are not practically possible because they are

unstable system so this method can only be used when upper & lower S.B do not

meet at caries frequency such as voice signal.

(ii) Phase Discrimination method:-

m (t) y1 (t)

s ct

M (t) O/P

M (t)

BPF

Hilbert

Transformations



M (t)

Sin ct

XSSB (t) = m (t) s ct (t) sin ct

m (t) = mt

(t) = sin mt

XSSB (t) = os ct mt sin ct. mt

= os ( c- m)t

= os ( c- m)t

Disadvantage of phase Discrimination method:- - The phase discrimination method is based upon 90 phase shift of the

modulating signal but for higher modulation frequency. It is very difficult to generate a phase shift of exact 90 this method can be used only for low modulation frequency up to flow kHz.

Power Saving in SSB-SC:-

Pt =

Pt =

Total power served =

% power saving = X 100%

= X 100%

For m =1

= 5/3 x 100% = 83.33%

Demodulation of A.M. Signal-



The recovery of the baseband signal, a process which is referred to as

demodulation or detection there are two method for detection

i. Synchronous detection:-

In this method a local carries is generated at the receiving and where

phase is exactly synchronize with the transmitted carries phase the

received signal is multiplied by locally generated carries signal & the

product is passes through a low pass filter to detect

Original baseband

y1 (t) y (t)

os ct

i. Detection of A.M. signal:-

XAM (t) = [AC + m (t) ] os ct

y1 (t) = [AC + m (t)] os ct

y1 (t) = [AC + m (t)]

y1 (t) = + + + os ct

y (t) = (AC + m (t))

1 M ( )

- m m

LPF



(2 c+ n) -2 c -(2 c- m) - m m 2 c- m 2 c 2 c+ c

HLPF( )

1

- m m

When the phases of transmitted locally generated carries are not synchronies we

always obtain a distorted signal at O/P of detector & it = 90 y (t) = 0 this

condition is known as Quardature null effect.

(ii) A synchronous detection:-

It is also known as envelop detector or diode detector.

RD

D

A.M C R V0(t) => To LPF



AC

Lets initially assume that the O/P of fixed amplitude. The capacitor

charges to the peak positive voltage of carries.

- Detection of DSB-SC Signal-

XDSB(t) = Ac m(t) os ct

y1(t) = Ac m(t) os ct

=

y1(t) = +

y (t) =

Detection of SSB-SC signal:-

XSSB(t) = m (t) os ct (t) sin ct

Y1 (t) = m (t) s (t) sin s ct

y1(t) = [1 + s2 ct] (t) sin2 ct

y (t) =

y (t) y (t)

LPF c = m



os ( ct+ )

XDSB(t) = Ac m (t) s ct

y1(t) = Ac m (t) s ct + s( ct + )

y1(t) = [ s ct + ) + os ]

y (t) = os

if = /2 y(t) = 0

- It is necessary to include the resistor R so that the capacitor may

discharge

- If RC is very-very high

SSB-SC (Single Side Band Suppress Carries):-

In case of DSB-SC modulation both the side bands are transmitted

which contains same information so DSB-SC transmission is further

redundant

Instead of transmitting both the SB we can suppress one SB as well to

achieve maximum efficiency such type of modulation in which either

upper or lower sideband is transmitted is known as SSB-SC signal.

i. Frequency Discrimination Method:-

It is also known as fitted method.

m (t) DSB-SC SSB-SC

BPF



Signal Signal

c (t) = Ac s ct

- ( c+ m - c - ( c- m) c- m c c+ m

HBPF ( )

- ( c+ m) - c c c+ m

- c - ( c+ m) c- m c

XSSB(

- ( c+ m) - c c c+ m



XSSB(

- ( c+ m) - c- m

c- m c c+ m

USB

Q.9 Explain VSB (Vestigial Side Band) Modulation and its advantage over SSB

Ans. SSB modulation is well suited for transmission of voice because of frequency gap

that exist in the spectrum of voice signals between O & few hundred Hz when

the baseband signal contains significant components at extremely low frequency

as in case of T.V. picture signal, the upper & lower side band meet at carries

frequency this means that use of SSB modulation is in appropriate for



transmission of such base band signals. Due to difficulty of isolating one

sideband

In VSB modulation one sideband is passed almost completely whereas just a

trace or vestige of unwanted sideband is retained specifically the vestige part of

unwanted sideband components for the part removed from the desired side

band.

DSB-SC

M (t) s (t)

Signal

C (t) = s ct

Y (t)

S (t) y (t)

s ct

XDSB(t) = m (t) s ct

S (t) = [m(t) s ct ] * h (t)

S ( ) = [S ( - c) + ( + c)] h ( )

Y1 (t) = S (t) s ct

Y1 ( ) = [S ( - c) + ( + c)]

Y1 ( ) = [M ( - c c) + M ( - c + c) H ( - c)

DSB Bitta

LPF = n



+ [M ( + c c) + M ( + c + c) H ( + c)

Y ( ) = [M ( ) [H ( - c) + H ( - c)]

H ( - c) + H ( + c) = const.

In T.V. transmission the picture signal is VSB modulated & the sound signal is frequency modulated & the total transmission B.W. 7 MHz *+(5.75 MHz for picture signal and 0.25 MHz for sound signal)

A.M. Receiver Q.10 Explain different Function of Receiver

To collect the electromagnetic waves transmitted by the transmitters. To select desired signal & reject all other, this is known as selectivity of the Receiver. To amplify the selected modulated signal this is known as sensitivity of the receiver. To defect baseband modulating signal from the modulated R.F. signal. To amplify the modulated signal as to operate the loud speaker.

AM Band 540-1650 KHz fm : 5 KHz B.W. : 10 KHz I.F. : 455 KHz

ANGLE MODULATION

In angle modulation the phase angle of high frequency carries signal is varied in accordance with instantaneous value of modulating signal keeping amplitude const.

C (t) = Ac s ct

c (t) = ct

=

=



UNIT 2 Q.11 Explain Frequency Modulation and phase modulation and derive equation for

it. Ans. In F.M. the frequency of high frequency carries signal is varied in accordance

with instantaneous value of modulating signal keeping amplitude const. phase is varied indirectly. Phase Modulation :=> In P.M. the phase is varied directly in accordance with instateous value of modulating signal keeping amplitude const. frequency is varied indirectly.

F.M. n (t) = any arbitrary modulating signal

with maximum frequency m

c (t) = Ac s ct

c (t)= ct

c (t) = Ac s c (t0

i = c + Kf m (t)

i (t) =

i (t) =

i (t) = ct + Kf

AFM(t) = Ac s ( ct + Kf

P.M. m (t) = any arbitrary modulating

signal c (t) = Ac s ct

c (t) = ct

C (t) = Ac s c (t)

XPM (t) = Ac s i (t)

i (t) = c (t) KP m (t)

XPM(t) = Ac s [ ct) + KP m (t)

XPM(t) = Ac s [ ct + KP m (t)]

XFM (t) = Ac s i (t)



Q.12 Derive Relation between FM & PM signal:- Ans.

XFM(t) = Ac s [ ct + Kf

XPM(t) = Ac s [ ct + KP m (t)]

m (t) m (t) PM

Ac Ac s ( ct + Kf

Ac s [ ct + KP m (t)]

m (t) FM

Ac s ( ct + Kf

Sinusoidal FM m (t) = Am s nt

c (t) = Ac s ct

c (t) = ct

XFM(t) = Ac s i (t)

i (t) = c + s mt

i (t) =

i (t) = s mt]dt

i (t) = ct +

i (t) = ct + FM sin mt

XFM (t)= Ac s ( ct + FM sin mt

Sinusoidal PM m (t) = Am s nt

c (t) = Ac s ct

c (t) = ct

XPM(t) = Ac s i (t)

XPM (t) = Ac s ( ct + PM sin mt

FM

PM

Modulation Index = =



Q.13 Different between FM and PM

FM = i (t) = ct + sin mt

= ct +

i (t) = = c + Kf AM s mt

PM = i (t)PM = ct + KP AM s mt

i (t) = = c KP AM m sin mt

/PM = KP AM m

PM = = KP AM

Types of Frequency Modulation:-

Depending upon value of , there are two types of frequency modulation.

i. Narrowband F.M.:-

When is very small

os = 1

0

Sin =

0

/FM = Kf Am

m



XFM (t) = Ac os ( ct + sin mt)

= Ac [ os ct ( sin mt) sin ct sin ( sin mt)

os ( sin mt) = 1

0

Sin ( sin mt) = sin mt

0

XNBFM (t) = Ac [ os ct - sin ct sin mt]

= Ac os ct - [ os ( c- m)t os ( c- m)t]

= Ac os ct - os ( c- m)t + os ( c- m)t

Ac

c- m

c c + m

Q.14 Explain Wide Band FM derive equation for wide band FM



When the value of modulation index FM is large than a large number of

sidebands are produced hence the B.W of FM is large.

XFM (t) = A s ( ct + FM sin mt)

This equation may be considered as Real part of exponential phases.

FM (T) = A.e j ( ct + FM sin mt)

= A.e j ( ct e j FM sin mt

The coefficient Xn is given by

Xn = FM SinX-nx)dx

nth order Bessel function

Xn = Jn (mf)

Since the changing in frequency causes the time required to complete one half

of a cycle to differ from time required to complete next half cycle, so the actual

wave is a disturbed sinusoidal oscillation.

The higher mathematical analysis using Bessel function shows that the putting

the value of Xn in equation (1)

e j t sin mt = Jn ( t)e jn mt

XFM (t) = Ae j ct Jn ( t)e jn mt

= A Jn ( t)e j ( mt )t

Real part provide FM signal



XFM (t) = A Jn ( t) os ( mt )

Bessel functions amp. Properties

i. Jn ( t) = Jn ( t) for even n

Jn ( t) = - Jn ( t) for odd n

Making use of property

XFM (t) = A J0 ( f (t) os ct + AJ1 ( f) [ os ( c + m)t

- os ( c - m)t] + AJ2 ( f) [ os ( c + m)t

+ os ( c - m)t] + AJ3 ( f) [ os ( c + m)t

- os ( c - m)t] + AJ3 ( f) [ os ( c + m)t

Since the spectrum of wideband F.M

The higher sideband are neglected, it does not affect the quality of transmission

so the practical transmission B.W for wideband FM & wideband PM signals is

given by Carsons Rule.

Rad/sec

Hz

Observations:-

- In F.M. the total transmitted power is the carries power & is always

remain const.

(B.W)FM = 2 ( + m)

= 2 ( + m)

= 2 (2 ( + 1) m



C (t) = Ac os ct

XFM (t) = Ac os ct + sin mt)

- The modulation index determine how many sideband component

hence signification amplitude.

- The co-efficient occasionally have negative value signaling a 180 phase

for that pair of side band

Q.15 Explain different Generation method of F.M. wave

Ans. There are two methods for generation of F.M. waves.

i. Direct method:- The basic concept of F.M. is c in accordance with

modulating signal m () the carries is generated by an LC Oscr. In an LC

Oscr. The frequency of Oscr is

CV = Voltage Variable capacities (Varicap)

Co = Fixed capacities

Semiconductor diodes when operated in R.B. have chat suitable to

permit their use as voltage variable capacitor the modulating signal.

Voltage across CV the capacitance of CV changes and causes

corresponding change in frequency.

Any oscr. Whose frequency is controlled by the modulating signal

voltage is called a voltage controlled oscr or VCO

FET and PIN can also used in place of VCO.

Disadvantage of Direct Method:-

Most LC oscr are not stable enough to provide a carries signal the carries frequency usually vary due to temp. Variation, humidity, ageing of component etc. so instead of using LC oscr, a crystal oscr must be used but since they provide highly stable carries frequency so only a



very small frequency. Deviation is possible that is why indirect method of FM generation is used.

Indirect method (Armstrong method):-

NBFM m (t) NBFM

A Balance modulator is employed to generate the DSB-SC signal using sin ct

as the carries as the carries of the modulator this carries is than shifted in phase by 90 and when added to the balance modulator output thereby forms an NBFM signal Ac s ct

NBFM

Ac sin ct

Integrator NBFM Frequency Multiplies

Crystal Oscr

90 Phase shift

Crystal Oscr

Addu.

Balance

Modular



Ac sin ct sin mt

y (t) = xn (t)

y (t) = xn (t)

x (t) = Ac os ( ct + sin mt)

y (t) = Ac os ( ct + sin mt)

= [1+ os ct + sin mt)]

Detection of F.M. Wave:-

FM demodulator is basically a frequency to amplitude convertor. It is expected

to convert the frequency variations in FM wave at its input into amplitude

variations at its output to recover the original modulating signal.

Q.16 Explain different FM Demodulation technique

Ans. The process of expecting signal from FM modulator wave is called FM

Demodulation

- In FM change in carries frequency shows change in amplitude of message

signal.

- Frequency Discriminator senses rising amplitude of message signal by

sensing the changes of carries frequency.

- O/P of frequency Discriminator is fed to AM detector that expects the

message signal from AM.

Frequency

c n c

n

n m

n



c +

c -

Voltage An -Am i = c + K m (t)

FM frequency Amp m (t) In variation modulation Disadvantage of Slope detector:-

i. The ckt is non-liner so harmonies are generated ii. The discriminator also respond for amplitude variation of i/p

signal Balanced Slope Detector:-

+

FM in m (t)

-

Frequency Selective N/W

Envelop detector

Slope Ckt 1

Envelop

detector

Slope Ckt 2

Envelop

detector



Voltage

c -

c +

Useful range

Working:-

This ckt is also known as triple tuned ckt as three tuned ckt are used. The

i/p tuned ckt is tuned at carries frequency c T1 is tuned at c + & T2 is

tuned at c -

When fin = c than gain provided by T1 & T2 will be same hence the O/P due

to T1 is same as due to T2 but in phase opposition so net O/P voltage is zero.



Phase Locked Loop:-

The Phase Lock Loop (PLL) is a f/b system which may be used to extract a

baseband signal from a frequency modulated carries. The basic component

building blocks of a Phase Locked Loop are-

i. Phase detector

ii. Low pass filter

iii. Voltage Control Oscr (VCO)

Vs Ve Ve

fc

fo Vo

if an input signal Vs of frequency fc is applied to the PLL the phase

detector compares the phase & frequency of incoming signal to that of the

O/P of VCO.

If the two signal differ in frequency and/or phase, an error voltage Ve is

generated the phase detector is basically a multiplies & produces as control

voltage Vc to VCO.

Vs (t) = A os [ ct + (t)]

(t) = 0

Vo (t) = B os ( 0t + )

Ve (t) = AB os ( ct. os ( 0t + )

Ve (t) = os ( c + o) t + os [( 5 - o) t 0]

Vc = os

Phase

detector LPF

VCO



= /2

The signal Vc shifts the signal frequency in a direction to reduce the frequency difference between fs & f0. One this action start we say that the signal is in capture range the VCO continue to change frequency till its o/p frequency is exactly the same as the i/p signal frequency the ckt is then said to be locked exact locking is achieved when the two signals are having same frequency & phase difference of /2 between them.

- Let the input signal makes on abrupt change the abrupt frequency

change causes the phase (t) to begin to increase linearly with time.

- The phase different at the comparator i/p will generate a positive O/P V0 which increase the frequency of the VCO.

- Thus the O/P voltage is proportional to the frequency change as required in an FM demodulates.

Pre-emphasis & De-emphasis:=>

FM Band ; 88 108 MHz m ; 15 KHz

nex ; 75 KHz

B.W. ; 200 KHz Guard Band ; 20 KHz I.F. ; 10.7 MHz

Sn (f)

Y2

Vc = 0



F

SX (f)

F

At higher modulating frequency the effect of noise is more pronounced as

compared to low frequency the power contents becomes very small as a

result it is desirable to increase the amplitude at higher modulating

frequency before modulation. This boosting of higher modulating

frequency in accordance with a pre arranged curve is known as pre-

emphasis.

At the receiving and after demodulation, the same level of higher

modulating signal frequency is maintained by using do emphasis.



UNIT-3

NOISE

Q.17 What is noise and how noise can be classified

Noise may be defined as any unwanted form of energy tending to interfere with

propel & easy reception & reproduction of wanted signal.

- Depending upon whether noise is internal or external to the system

there are two types of noise.

i. External noise:- The noise which to the system is known as

extender noise

Ex.: atmosphere noise extraterrestrial noise

ii. Internal noise:- The noise which is generated within system is

known as internal noise. There are five types of internal noise

Thermal noise:- This type of noise arises due to constitutes and

electric current flow in the register, the direction current flow is

random & it has zero mean value.

- Noise power available from a register is directly proportional to its

absolute temp. In addition this power is also directly proportional to

the B.W. over which noise is measured.

P T

P B

P TB

K = Boltzmann const.

= 1.30 x 10- 23 J/K

T = Absolute temp in K

Pn = KTB



B = Bandwidth in Hz

Thermal noise is also known as Jhonsons noise

Thermal noise is a zero mean Gaussian random variable

- It is also known as white noise

- As white light contain equal

amount of all frequency within the visible band of electromagnetic

radiation. Similarly thermal noise has a uniform power spectral density

over useful range of common.

Sn (f)

/2 |=KT|

0 f

/2 x 2B = B

=Two sided PSD

- B 0 B F



The auto ccerelation function of thermal noise is an impulse at T = 0

Rn ( )

= 0

Eqt ckt of Resistor as noise generator:-

RL = R

= Pn = KTB

V



High frequency or transit time noise:-

In semiconductor devices, the is the transitive is the time taken by the

carriers to cross a junction, when the signal frequency is high periodic time

of signal becomes very small & hence may be comparable to transitive of

carriers in such situations some of the carriers may diffuse back to the

source the noise produce due to this condition is known as high frequency

or transit time noise.

Signal to noise ratio:-

It is defined as the ratio of signal power to the noise power & is given by -

Noise Figure (F):

PSi Psi

Pi P0

For better system performance the noise figure should be as low as

possible.

PSC PSO = GPsi

Pni = KTB PSO = GPni

System

Ampulation C PM

PNi = Pni + Pna



Pno = FKTBG

= PNi = FKTB

Pni + Pna =FKTB

KTB + Pna = FKTB

Pni = FKTB KTB

Q.18 Derive Equivalent Noise Temp. for any amplifier

Ans. It is defined as temp. of the Ampr (system) at which the noise contribution by

Ampr becomes thermal noise power

Pna = (F-1) KTB

(F-1) KTB = KTeB

Cascading of Ampr.:-

Te = T (T-

1)

Ampler

Ampler

Ampler

+ .



UNIT-4

Narrow Band Noise:-

/2 Sn ( f )

HBPF ( f )

fm 0 fm

(g+ (t))

Pre envelop:- the pre envelop of a band pass signal g (t) is given by g+ (t) = g (t) + j (t)

g+ (t) = g (t) + j (t)

G+ ( ) = G ( ) + j (-jsgn ( ) G ( )

G+ ( ) = G ( ) + sgn ( ) G ( )

(t) = g+ (t) e jafct

G+ (f fc)



(f)

fm +fm

x (t) = os (2 fct + (-)) calculate

its pre envelop & complex envelop

Xf (t) = os (2 fct + (+ )) +j sin (2 fct + )

Xf (t) = ej (2 fct + )

(t) = ej (2 fct + ) j2 fct

S (t) = e at os [( c + ) t] u (t)

The Hilbert transform of a low part signal multiplies with s ct is equal

to same low pass signal multiplied by Hilbert transform of signal

S+ (t) = e-at u (t) [ os ( c + )t + j sin ( c + )

(t) = ej (t)

S+ = e-at u (t) ej ( c + )t



(t) = e-at u (t) ej ( c + )t e- j ct

= e-at u (t) e- j ct

X ( )

1/2

- ( ct m) - c - ( c- m) c- m - c c m



j

- j

c- m c c m

- ( ct m) - c - ( c- m)



Given the power spectral density of noise wave form N (t), the power

spectral density n (t) os 2 fCt is carries at as follows-

Divide Sn (t) by 4 shifts the divided plot to the L.H.S. as well as to the

R.H.S. by fc & add the shifted plots.

Noise performance of continuous wave modulation :->

Si (t) Sd (t) S0 (t)

n (t) N0 (t)

os 2 fCt

i. Noise performance of SSB-SC system

Sd (t) = m (t). os ct

sin2 ct

m (t)

RX ( ) os 2 r fc

[Sx (f-fc) + SX (f + fc)]

BPF LPF fc=fm



S0 (f )

/2

- (fC + f m) - fC fC fC + m f

Sn0 (f)

/8

- fm 0 fm

=

PSO =

PSi

=



Noise performance of DSB-SC system:-

Si (t) = Ac m (t) s ct

Sd (t) = AC m(t) os ct

s ct ]

Noise performance of full A.M. system:-

XAM (t) = [Ac + m (t)] s ct

] s ct

= Ac [ 1 + x (t) ] s ct

Sd (t) = AC [ 1 + x (t)] s ct

]



- fm 0 fm

2

- fm 0 fm

f OM for SSB-SC = 1

f OM for DSB-SC = 1

FOM for full AM < 1

XAM (t) = [Ac + m (t)] s ct

= +



= Ac [1 + x (t)] s ct

Let m (t) = AM s mt

XAM (t) = Ac [1 + m s mt] s ct

x (t) = m s ct

For 100% modulation

(FOM)AM

Figure of merit in F.M.:

*

Pre-emphasis & De-emphasis

FM Band : 88-108 MHz fm : 15 KHz fmex : 75 KHz

B.W : 200 KHz Guard Band : 20 KHz i.f. : 10.7 MHz

(FOM)FM =



Sn (f)

/2

0 f

SX (f )

0 f

At higher modulating frequency the effect of noise

Pre-emphasis:

At the receiving and after demodulation, the same level of higher modulating

signal frequency is maintained by using de-emphasis.

m (t)

m (t)

Pre-

emphasis

FM Detector De-

emphasis



Vi (S) R1 R2 V0 (S)

R1 C = 75 H sec

= 2.122 KHz

Gain (dB)

20 dB/dec

If an input signal is subjected system whose frequency response H ( ) a phase

spectrum which is a non-linear fun of than the differ frequency components

present in the signal will be shifted in a mann that results in a change in their

relative phases when all these frequency component are aided we objective a

signal that may look considerable different from the I/p signal.



H ( ) = e j to

|H ( )|= |

< H ( ) = - t0

< H ( )

- t0

So for a system with frequency response H ( ) = |H( )| < ( )

Phase delay:- (Te) the delay provide

Group delay (tg):- the delay prove at group of frequency is known as group

delay & given by tg.

if ( ) varies linearly with frequency than group delay & phase delay not only

const. but are also equal.



-

X =4 ; y = 0

4} = erfc (0) =

4

The noise is added to the system with a power spectral density Sn (f) the

sum of signal & noise forms the i/p of a filter with frequency response H (f)

calculate

a) SNR at the i/p of the signal

P = 2 f dt

= 2

300



b) The filter response is as shown below

H (f)

1

-2 0 2 f

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