analog communication unit2 vtu

ANALOG COMMUNICATION (VTU) - 10EC53

UNIT - 2

AMPLITUDE MODULATION: Introduction AM: Time-Domain description, Frequency

Domain description. Generation of AM wave: square law modulator, switching modulator.

Detection of AM waves: square law detector, envelop detector. Double side band suppressed

carrier modulation (DSBSC): Time-Domain description, Frequency-Domain representation,

Generation of DSBSC waves: balanced modulator, ring modulator. Coherent detection of

DSBSC modulated waves, Costas loop.

TEXT BOOKS:

1. Communication Systems, Simon Haykins, 5thEdition, John Willey, India Pvt. Ltd, 2009.

2. An Introduction to Analog and Digital Communication, Simon Haykins, John Wiley India

Pvt. Ltd., 2008.

Special Thanks To:

Faculty(Chronological): Arunkumar G (STJIT), Prof Praveen Chitti (JGI), Raviteja B

(GMIT), Somesh HB (REVA ITM)

BY:

RAGHUDATHESH G P

Asst Prof

ECE Dept, GMIT

Davangere 577004

Cell: +917411459249

Mail: [email protected]

Quotes:

1. Live as if you were to die tomorrow learn as if you were live forever.

2. Commitment leads to action. Action brings your dream closer.

3. Success is moving from failure to failure without losing enthusiasm.

4. Attachment is the root cause of all misery. Possessiveness is nourishment for the ego.

5. I have failed over and over again that is why I succeed.

6. Train your mind to develop a taste for the good and the godly, not for money or material

gains.

Amplitude Modulation Raghudathesh G P Asst Professor

ECE Dept, GMIT [email protected] Page No - 1

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AMPLITUDE MODULATION

Communication:

Communication is exchange of information between 2 points i.e., means of conveying

the information from one point to other. The point from where the information is transmitted is

called transmitter and the point where the information is received is called receiver.

Basic Block Diagram of a Communication System:

The information source produces the information which will be in the form of voice, video and text. This information stored is electrical in nature (transmitter converts non-

electrical signal into electrical one).

The O/P of the information source is fed to the transmitter where a process called modulation is carried out in which the information signal is superimposed on the carrier

signal, the modulation is then sent through the channel to the destination.

The channel is medium through which the information signal travels. The communication through the channel may be wired or wireless communication.

If the information signal flows through wire or transmission lines, the communication system is said to be line or wired communication system. In line communication the

transmitter and receiver are connected through cables.

Ex: Telephony, Telegraphy. If the input signal flows through the open space in the form of electromagnetic or radio

waves it is called wireless or radio communication.

Ex: TV, Radio, Mobile. Maximum amount of noise interfere with the information signal in the channel; Noise is

some unwanted electromagnetic energy that interfere with information signal and tries to

corrupt it, due to noise the quality of information transmission will degrade. Once the

noise is added it cannot be separated from the information.



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The noise can be either natural (lighting & radiating from sun & stars) or man made (ignitions, welding, electric motion).Even though noise cannot be completely eliminated,

its effect can be reduced by using various techniques.

The output of channel is then given to receiver where a process called demodulation or detection is carried to extract the information signal from modulated carrier. The

information signal is then fed to the output devices, Such as loud speaker, monitor &

printer etc.

Classification of Communication System:

Baseband Communication and Modulation:

Based on the method used for signal transmission we can categories the communication

systems as:

1. Baseband transmission systems. 2. Communication systems using modulation.

Baseband Signals:

The information or the input signal to a communication system can be analog (sound,

picture) or it can be digital (computer data). The electrical equivalent of the original information

signal is known as the baseband signal.

Baseband Transmission System:

In systems like baseband transmission systems, the baseband signals (original

information signals) are directly transmitted.



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Ex.: 1. telephone networks where the sound signal converted into the electrical signal is placed

directly on the telephone lines for transmission (local calls).

2. Computer data transmission over the coaxial cables in the computer networks.

Thus the baseband transmission is the transmission of the original information signal as it is.

Limitations of Baseband Transmission:

1. The baseband transmission cannot be used with certain mediums. 2. Ex.: it cannot be used for the radio transmission where the medium is free space. This is

because the voice signal (in the electrical form) cannot travel long distance in air. It gets

suppressed after a short distance.

3. Thus, for the radio communication of baseband signals a technique called modulation is used.

Modulation:

Definition: Modulation is a process by which certain characteristics of a carrier (high frequency signal) are varied instantaneously in accordance with the

modulating/message/information signal (narrow band signal).

The 3 characteristics of a carrier are: 1. Amplitude. 2. Frequency. 3. Phase.

One of the above characteristics is varies in accordance with the modulating signal. Message/information signal is referred to as the modulating signal and the result of

modulation process is referred to as modulated signal.

Note: All the information is contained in the varying characteristics of the carrier but carrier itself does not contain any information.

The baseband signal will modify the amplitude or frequency or phase of the carrier in the process of modulation. Depending on which parameter of the carrier is changed the

modulation techniques are classified as follows :

1. Amplitude Modulation [AM]: Amplitude of the carrier is varied in accordance with the instantaneous amplitude of the baseband signal keeping its frequency and

phase constant.

2. Frequency Modulation [FM]: Frequency of the carrier is varied in accordance with the instantaneous amplitude of the baseband signal keeping the amplitude

and phase constant.



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3. Phase Modulation [PM]: Phase of the carrier is modified in accordance with the instantaneous amplitude of the baseband signal keeping the amplitude and

frequency constant.

Need/Necessity/Reason/Advantage of Modulation:

The low frequency signal such as voice, video and text cannot be directly transmitted using antenna because these low frequency signals are heavily attenuated in the space

and are corrupted by noise. So the message signals are always transmitted using a

technique called modulation.

In the process of modulation, the baseband signal is "translated" i.e. shifted from low frequency to high frequency.

The modulation process has the following advantages: 1. Reduction in the height of antenna 2. Avoids mixing of signals 3. Increases the range of communication 4. Multiplexing is possible 5. Improves quality of reception.

1. Reduction in height of antenna:

For the transmission of radio signals, the antenna height must be a multiple of ( ). Here is the wavelength. .

where : c is velocity of light and

f = the frequency of the signal to be transmitted.

The minimum antenna height required to transmit a baseband signal of f = 10 kHz is calculated as follows :

The antenna of this height is practically impossible to install.

Now consider a modulated signal at f = 1MHz. The minimum antenna height is given by,

This antenna can be easily installed practically. Thus modulation reduces the height of

the antenna.



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2. Avoids mixing of signals:

If the baseband sound signals are transmitted without using the modulation by more than one transmitter, then all the signals will be in the same frequency range i.e. 0 to 20 kHz.

Therefore all the signals get mixed together and a receiver cannot separate them from each other. So if each baseband sound signal is used to modulate a different carrier then

they will occupy different slots in the frequency domain (different channels). This is as

shown in Figure below. Thus modulation avoids mixing of signals.

3. Increases the range of communication:

The frequency of baseband signals is low, and the low frequency signals can't travel a long distance when they are transmitted. They get heavily attenuated (suppressed).

The attenuation reduces with increase in frequency of the transmitted signals, and they travel longer distance.

The modulation process increases the frequency of the signal to be transmitted. Hence it increases the range of communication.

4. Multiplexing is possible:

Multiplexing means two or more signals can be transmitted over the same communication channel simultaneously. This is possible only with modulation.

The multiplexing allows the same channel to be used by many signals. Ex.:

1. Many TV channels can use the same frequency range, without getting mixed with each other.

2. Many radio stations broadcast the signals in same band simultaneously.



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5. Improves quality of reception:

With frequency modulation (FM), and the digital communication techniques like PCM, the effect of noise is reduced to a great extent. This improves quality of reception.

6. Allows adjustments in the bandwidth:

Bandwidth of a modulated signal may be made smaller or larger than the original signal.

7. Reduction in power requirement can be achieved.

Classification of Modulation System:

Various types of practically used modulation systems are as follows:

Amplitude Modulation:

Definition: Amplitude of the carrier is varied in accordance with the instantaneous amplitude of the baseband/modulating signal keeping its frequency and phase constant.

Ex.: A.M. is used in radio and picture transmission in T.V etc. For sinusoidal message signal and carrier signal the amplitude modulated signal is as

shown in the figure below



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Some of the important observation in the above figure: 1. The frequency of the sinusoidal carrier is much higher than that of the modulating

signal.

2. In AM the instantaneous amplitude of the sinusoidal high frequency carrier is changed in proportion to the instantaneous amplitude of the modulating signal. This is

the principle of AM.

3. The time domain display of AM signal is as shown in Figure above. This AM signal is transmitted by a transmitter. The information in the AM signal is contained in

the amplitude variations of the carrier of the envelope shown by dotted lines in

Figure above.

4. In the above figure the frequency and phase of the carrier remain constant. 5. Now let the AM waveform for m(t) be a square wave as shown below. Here, the

principal of AM remains same with the non sinusoidal modulating signal.



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Time Domain Description of AM:

Let the instantaneous modulating signal be represented by m(t). This is the modulating signal of arbitrary shape which may or may not be a sine wave.

Consider a sinusoidal carrier signal/wave be represented as,

--------------------- (1)

Here Ac = Peak amplitude of the carrier

fc = Carrier frequency in Hz, and

The standard form of AM wave is represented as,

-------------- (2)

In the above expression "ka" is a constant which is called as the modulation index or amplitude sensitivity of the modulator.

In the Equation (2), the term is called as the instantaneous amplitude of the modulated wave and it is denoted by a(t). It is also called as the envelope of the

AM wave.

------------------------------ (3)

The second term in Equation (2) is which shows the frequency of AM wave s(t) is same as the carrier frequency fc and the shape of AM wave is sinusoidal.

Percentage Modulation:

Definition: The maximum absolute value of multiplied by 100 is referred to as the percentage modulation.

Depending on the value of , there are two possible cases: Case I: Linear modulation: Here for all the values of t, thus will always be

non-negative.

Removing the mod sign, expression for the envelope is given as,



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for all (t) ------------------------ (4)

For this case the percentage modulation will be less than or equal to 100%. This type of modulation is known as linear amplitude modulation. The Amplitude modulated waveforms for this case are as shown below

Case II: Over modulation: Here for some values of t. Expression for the envelope for this case is given as,

As envelop will reverse the phase sometimes, as shown in the figure below.

For this case the percentage modulation will be greater than 100%. Overmodulation introduces a distortion. Thus it should be avoided.



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Important Observations:

The envelope of the AM wave has exactly the same shape and has one-to-one correspondence with the message signal if the percentage modulation is less than or equal

to 100 %.

No such correspondence is observed when the percentage modulation is greater than 100 %. Thus envelope distortion takes place and the AM wave is said to be overmodulated.

Due to above reason the shape of the AM wave envelope should be same as that of the modulating signal because this reduces the complexity of the demodulator circuit to a

great extent.

The above condition can achieved if the following 2 conditions are satisfied: 1. The percentage modulation should be less than 100 % to avoid the envelope

distortion.

2. The message bandwidth "W" of the modulating signal should be small as compared to carrier frequency fc which will help to visualize the envelope a(t)

satisfactorily.

Frequency Domain Description (Frequency Spectrum of AM Wave):

Below equation defines the equation for standard AM wave s(t) as a function of time.

---------------------- (1)

To develop the frequency description of the above AM wave, we will take the Fourier transform of both side of the equation (1).

Let S(f) denotes the Fourier transform of s(t). Let M(f) denotes the Fourier transform of the message signal m(t); here M(f) is the

message spectrum.

Thus,

---------------------- (2)

As and

Thus we get,



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-------- (3)

Here,

= Left shifted weighted delta function Shifted delta functions

= Right shifted weighted delta function

= M(f) shifted left by fc Shifted version of spectrum M(f)

= M(f) shifted right by fc

We can plot the spectrum of AM wave from Equation (3) as shown in Figure (A) below.

Here we select the triangular shape of the spectrum M(f) which is arbitrary. We can choose any other shape. Above figure (A) shows the double sided spectrum of AM wave,

but note that the negative frequencies are imaginary, and used for mathematical

convenience. The single sided spectrum of an AM wave is shown in Figure (B) below



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Spectral Description:

Figure (A) shows the spectrum of modulating signal M(f) and the double sided spectrum of the AM wave. Negative frequency spectrum does not actually exist.

Figure (B) shows the single sided spectrum of AM wave. Upper side band: The portion of the AM wave spectrum lying above the carrier

frequency is called as upper sideband or USB. The frequency components

corresponding to USB range from to , as shown in Figure (B)

Lower side band: The portion of AM wave spectrum which extends below the carrier frequency fc is called as the lower sideband LSB. The frequency

components corresponding to LSB;.range from to , as shown in

Figure (B)

Highest and Lowest Frequencies: From Figure (B) it is evident that the highest frequency in the spectrum of an AM wave is equal to .

----------------------- (4)

And the lowest frequency in the spectrum is equal to .

----------------------- (5)

Transmission Bandwidth (B):

The transmission bandwidth of an AM wave is defined as the difference between the highest and lowest frequency components present in the spectrum.

-

Single-Tone Modulation:

Consider a modulating wave/signal m(t) consists of a single tone or only one frequency component fm. Thus,

-------------------------- (1)

Here, Am = Amplitude of the modulating wave.

fm = Modulating frequency.



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In this case we need to find time and frequency domain characteristics of the resulting AM wave.

Time Domain Description: Equation for the generalized AM signal is

----------------------- (2)

Putting in equation (2) we get,

---------------- (3)

Here,

---------------- (4)

Equation (4) represents the time domain description of the single tone modulation. Here, = modulation index/depth/factor. Figure below shows the time domain description of the single tone modulation.

Frequency Spectrum of the Single Tone AM Wave:

The frequency spectrum of AM wave tells us about which frequency components are present in the AM wave and what their amplitudes are. Consider the equation,

---------- (1)

--------- (2)



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But

thus equation (2) becomes,

------- (3)

Applying Fourier transform on both sides we get,

---------------- (4)

Figure below shows the double sided spectrum of single tone AM wave

Modulation Index:

In AM wave the modulation index () is defined as the ratio of amplitudes of the modulating and carrier waves and is given as below,

-------------- (1)

When the modulation index has values between 0 and 1 and no distortion is introduced in the AM wave.

If then is greater than 1 it will distort the shape of AM signal. The distortion is called as "over modulation."

The modulation index is also called as modulation factor, modulation depth, modulation coefficient or degree of modulation.

Modulation index is expressed as percentage it is called as "percentage modulation".



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------------------- (2)

Modulation Index Calculation Using the AM Wave:

To calculate modulation index "", we must represent Am and Ac in terms of Amax and Amin.

From the figure we can write,

----------------------- (1)

------------------------ (2)

Putting (1) in (2) we get,

Putting the value of Am and Ac we get,

--------------------- (3)



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Total Transmitted Power in single Tone AM:

In reality, the AM wave s( t ) is a voltage or current wave. An AM wave consists of carrier and two sidebands thus an AM wave will contain more

power than the power contained by an unmodulated carrier.

The amplitudes of the two sidebands are dependent on the modulation index . Hence the power contained in the sidebands depends on the value of . Hence the total power in

an AM wave is a function of the value of modulation index .

Total power in an AM wave is given by,

------------------------------- (1)

Here, Acarrier = rms value of the carrier.

AUSB = rms value of the USB sideband.

ALSB = rms value of the LSB sideband.

R = characteristic resistance of antenna in which the total power is dissipated.

Carrier Power (Pc):

------------------------ (2)

Here Ac = Peak carrier Amplitude.

Power in Two Sidebands:

Peak amplitude of each sideband is thus,

------------------- (3)

From equation (2) we have,

---------------------- (4)



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Total Power:

----------------------------- (5)

---------------------------- (6)

Equation above tells us about the relation between the total power of AM wave and the power contents of an unmodulated carrier.

With increase in the value of "", total power also increases. Pt will be maximum for = 1 and it will be 1.5 Pe.

Modulation Index in terms of Pt and Pc:

Consider the Equation

--------------------------- (1)

Modulation Index in terms of It and Ic:

The total power Pt of an AM wave can be expressed in terms of currents It which is rms current for AM wave and R be the characteristic impedance of an antenna through which

these currents flow. Then,

------------------------ (1)

The carrier power Pc of an AM wave can be expressed in terms of currents Ic which is rms current for unmodulated carrier and R be the characteristic impedance of an antenna

through which these currents flow. Then,



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---------------------- (2)

Equation (1) Equation (2) we get,

----------------------- (3)

We know that,

, Thus

----------------------- (4)

---------------------- (5)

Modulation index in terms of currents:

-------------------- (6)

Transmission Efficiency:

Transmission efficiency of an AM wave is the ratio of the transmitted power which contains information (i.e. the total sideband power) to the total transmitted power.

------------------------ (1)



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The percentage transmission efficiency is,

------------------------ (2)

Multitone Modulation or Modulation by Several Sinewaves:

Let us consider two modulating signal,

The total modulating signal will be the sum of these two in the time domain.

------------------------- (1)

Equation for AM modulated wave is,

---------------------- (2)

Substituting equation (1) in (2) we get,

[1+ ] cos2

cos2

cos2

But thus,

cos2

----------------- (3)

The Equation (3) shows that in the AM wave along with carrier there are four sideband components.



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There are two USB components at frequencies ( fc+f1 ) and (fc + f2) and two LSB components at frequencies (fc-f1) and (fc-f2). The frequency spectrum of AM wave is as

shown in Figure below

Thus for every modulating signal two sidebands are produced. The amplitude of the sidebands is proportional to the corresponding modulation index.

Total Power in Multitone AM Wave:

------------------ (1)

We know that

Thus, from above equations we can write as,

--------------------- (2)

Extending the concept to the AM wave with n number of modulating signals with modulating indices 1,2..,n the total power is given by,

------------------ (3)

Effective Modulation Index (t):



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Total Power transmitted is,

------------------- (A)

For the AM wave with two modulating signals let us assume that the effective modulation index is t. Therefore substitute t for in Equation (A) we get,

------------------------- (1)

Comparing equation (1) and equation (3) of previous derivation we get,

------------------ (2)

Bandwidth with several modulating signals:

Looking into the spectrum of AM wave shown in Figure below. As there are 2 modulating signals, the bandwidth of Multitone AM wave is given by,

-------------------- (1)

Transmission Efficiency ():

---------------------- (1)



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--------------------- (2)

Let , Then

= 0.333

Thus, only 33.333% of power is used and 66.66% is present in carriers.

Problems:

1. A modulating signal 10 sin (2 x 103t) is used to modulate a carrier signal 20 sin (2 x 104t).

Find the modulation index, percentage modulation, frequencies of the sideband components and

their amplitudes. What is the bandwidth of the modulated signal?

Solution:

The modulating signal ---------------- (1)

Here Am=10 v, fm=103 Hz = 1kHz.

The carrier signal, ---------------------- (2)

Here Ac=20 v, fc=104 Hz = 10 kHz.

Modulation Index:

Percentage Modulation:

Frequencies of the Sideband Components:



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Upper Sideband:

Lower Sideband:

Amplitudes of the Sideband Components:

Amplitude of sidebands (USB & LSB)

Bandwidth:

Bandwidth

2. A 10kW carrier wave is amplitude modulated at 80% depth of modulation by a sinusoidal

modulating signal. Calculate the sideband power, total power and the transmission efficiency of

the AM wave.

Solution:

Given: Pc = 10 kW, = 0.8

Sideband power:

Total Side Band power

Total Power:

Transmission Efficiency:

3. A transmitter transmits 10 kW of power without modulation and 12 kW after amplitude

modulation. What is the modulation index?

Solution:

Given: Pt = 12 kW, Pc = 10 kW



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4. The antenna current of an AM transmitter is 8 Amp. When only carrier is transmitted, but

increases to 9 Amp, when the carrier is modulated by a single sine wave. Find the percent

modulation. Find the total antenna current if the modulation index is changed to 0.9

Solution:

Given: Ic = 8 Amp, It = 9 Amp

When modulation index is changed to 0.9 then,

5. A transmitter radiates 10 kW power with the carrier unmodulated and 10.5 kW, when the

carrier is modulated by one sinusoidal signal. Calculate the modulation index. If another

modulating signal corresponding to 30% modulation is transmitted simultaneously determine the

total radiated power.

Solution:

Given Pc = 10 kW, Pt = 10.5 kW and 2 = 0.3

Modulation index due to 1st signal:



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Total modulation index:

Total radiated Power:

6. The antenna current of an AM transmitter is 10 Amp when it is modulated to a depth of 30%

by an audio signal. It is increased to 11 Amp when another signal modulates the carrier. What

will be the modulation index due to second wave?

Solution:

Given: It1 = 10 Amp, m1 = 0.3, It2 = 11 Amp

Modulation index due to second wave:

Also,

7. The carrier wave is represented by the equation, . Draw the wave form of an

AM wave for = 0.5.



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Solution:

Given: , = 0.5

Thus, to draw the AM wave form 1st we have to find the amplitudes of carrier, modulating

signal, maximum and minimum amplitudes of AM wave.

Amplitude of carrier (Ac):

From given equation Ac = 10 v

Amplitude of Modulating Signal (Am):

Equation for modulation index is,

Maximum and Minimum Amplitudes of AM Wave:

Wave form of an AM wave for = 0.5 as below:

8. The carrier amplitude after AM varies between 4 v and 1 v. Calculate depth of modulation.

Solution:

Given: Amax = 4 v, Amin = 1 v



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9. The antenna current of AM broadcast transmitter modulates to the depth of 40% by an audio

sine wave of 11 Amp. It is increased to 12 Amp as a result of simultaneous modulation by

another audio sinewave. What is the modulation index due to the second wave?

Solution:

Given: 1=0.4, It1=11 Amp, It2=12 Amp.

Calculating the value of Carrier current (Ic):

Calculating Total Modulation index (t):

Calculating Modulation index (2):

10. A certain transmitter radiates 10 kW with carrier unmodulated and 12 kW when the carrier is

sinusoidaliy modulated. Calculate the modulation index. If another sinewave corresponding to

50% modulation is transmitted simultaneously determine the total radiated power.

Solution:

Given: Pc = 10 kW, Pt = 12kW, 2=0.5

Modulation index m1 is given by:



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Calculate the total modulation index t:

New transmitted power is:

11. A multitone modulating signal has the following time-domain form:

a. Given the time-domain expression for the conventional AM wave. b. Draw the amplitude spectrum for the AM wave obtained in part (a). Also find the

minimum transmission bandwidth.

Solution:

Time-domain expression for the conventional AM wave:

Substituting the expression of m (t) in the above equation,

We know that



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Applying Fourier transform on both sides,

Amplitude Spectrum of AM Wave is as shown below

Minimum transmission bandwidth:

12. Consider the message signal volts and the carrier wave

Volts.



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a. Give the time-domain expression for the resulting conventional AM wave for 75% modulation.

b. Find the power developed across a load of 100 due to this AM wave. c. Sketch to scale resulting AM wave for 75% modulation. d. Sketch the Spectrum of the Wave.

Solution:

Given: message signal volts

Carrier wave volts

Modulation index = 0.75

a. time-domain expression for the conventional AM wave is:

But

---------------- (1)

b. the power developed across a load of 100 :

Expanding the equation (1), we get

Expression for total power:

Here

c. Sketch of AM Wave:



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d. Spectrum of the Wave:

Time-domain expression for the AM wave is

--------- (2)

Taking Fourier transform on both sides for equation (2)

Spectrum:

13. An audio frequency signal 10sin2 (500)t is used to amplitude modulate a carrier of

50sin2(105)t. Consider modulation index as 0.2. Determine

a. Sideband frequencies b. Amplitude of each sideband c. Bandwidth required

Solution:

Sideband frequencies:



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Amplitude of each sideband:

Bandwidth required:

14. Am amplitude modulated signal represented by

v.

Determine the following:

a. Modulation index corresponding to each frequency b. Various frequency components present c. Sketch the line frequency spectrum d. Bandwidth

Solution:

Given:

-------- (1)

Rearranging the equation (1) we get

----------- (2)

Equation for the modulation due to two sine is given by

----------- (3)

Comparing equation (1) and (2), we get

Modulation index corresponding to each frequency:

1=0.5, 2=0.2

Various frequency components present:

fc=1 MHz, f1=1 KHz, f2= 2 kHz

Line frequency spectrum:

Consider equation (1)



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Applying Fourier transform on both sides to the above equation we get

Bandwidth:

Generation of AM Waves:

The circuit that generates the AM waves is called as amplitude modulator. There are two modulator circuits namely:

1. Square law modulator 2. Switching modulator

Both of these circuits use a nonlinear element like diode for their implementation. The above two circuits are used for low power modulation purpose.



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Square law modulator:

A square law modulator circuit is as shown in Figure above. It consists of the following 3 essential components:

1. A summer circuit for summing carrier source and modulating signal 2. A nonlinear device 3. A bandpass filter

The summer circuit is implemented by connecting modulating signal m(t) and carrier signal c(t) in series with each other

---------------- (1)

Their sum V1(t) is applied at the input of the nonlinear device, such as diode, transistor etc. The I/O relation for nonlinear device follows square law if the input to the non-linear

device is a low level signal. Thus, square law is given as follows,

-------------------- (2)

Here, a1, a2 = device constant

V2(t) = output of nonlinear device

V1(t) = input to the system

We use a single tuned or double tuned transformer as a band pass filter. Mathematical expression:

Putting equation (2) in equation (1)

----------- (3)



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Rearranging the terms we get,

In the above equation there are 3 frequency components: Two terms centered around carrier frequency fc. They are

.

Component frequency greater than fc. They are

.

Component frequency lower than fc. They are

.

The require term is centered around fc and components having frequency greater than and less than fc are unwanted terms.

By passing the signal through a bandpass filter which is tuned to fc we can select the unwanted term and can be eliminated.

Thus, output of the bandpass filter is,

Rearranging the above equation we get,

------------ (4)

Expression for standard AM wave is, ------------ (5)

Comparing the equations (4) and (5) we get,

Thus square law modulator produces an AM Wave.



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Switching Modulator:

The switching modulator is as shown in the figure above. Diode is used in the above circuit. It is assumed to be ideal i.e, when the input voltage is

greater than zero the device is closed and offers zero impedance. When input voltage is

less than or equal to zero volts the device is open and offers infinite impedance.

The circuit exploits the switching characteristics of the diode i.e, modulation is produces by switching on and off the diode.

Let m(t) be the message signal and c(t)=Ac cos2fct be the carrier wave. It is assumed that the amplitude of the carrier is much larger than that of the message

signal i.e., m(t)

Thus the output of the diode varies between zero and V1(t) at the rate of the carrier frequency,

The turning on and off the diode can be modeled as a rectangular train of pulses gp(t) where amplitude being one and frequency is fc as shown below

Mathematically the diode output can be represented as

-------- (2)

gp(t) can be represented using Fourier series as,

------- (3)

Substituting equation (1) and (3) in (2)

The odd harmonics components are unwanted and are removed by passing them through band pass filter. Thus, expanding the summation for n=1 we get



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The above equation contains higher order frequencies, lower order frequencies fc frequency terms. By passing the signal through a band pass filter we get the fc term.

Thus, output of switching modulator is

------------ (4)

Rearranging the terms we get,

--------------- (5)

Expression for standard AM wave is, ------------ (6)

Comparing the equations (5) and (6) we get,

In this wave AM wave is generated using switching modulator.

Detection of AM Waves:

The process of detection or demodulation is a method of recovering the message signal from a received modulated wave/signal.

The process of detection is exactly opposite to that of modulation and is as shown in the figure.

Standard AM can be demodulated using following methods: 1. Square law detector 2. Envelope detector

Square law detector:

Square law demodulator utilizes a square law modulator for the purpose of the demodulation.



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It contains two parts: 1. Non Linear Device (NLD). 2. Low pass filter.

The output of the non linear device can be represented as

----------------- (1)

Where V1(t) = input of NLD (amplitude modulated wave).

V1(t) = Ac(1+kam(t))cos2fct

Substituting V1(t) in equation (1) we get

Out these terms the only desired term is which is due to the term due which it is termed as square law detector.

The only desired term is obtained by passing the above signal through a low pass filter we get,

------------------- (2)

Distortion in the detector output:

Another term which passes through the low pass filter to the load is

.



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This term is an unwanted signal and will give rise to the signal distortion. The ratio of desired signal to the undesired one is given as follows,

--------------- (3)

We should maximize the ratio in order to minimize the distortion. We can achieve this by choosing | | small as compared to unity for all value of t. but if is small then the

AM wave is weak.

Envelope Detector:

It is a simple and very efficient device which is suitable for the detection of a narrowband AM signal. A narrowband AM signal is one in which the carrier frequency fc is much

higher as compared to the bandwidth of the modulating signal.

An envelope detector produces an output signal that follows the envelope of the input AM signal exactly.

The envelope detector is used in all the commercial AM radio receivers. The circuit diagram of the envelope detector is shown in Figure below. It consists of a

diode and an RC filter.

Operation Logic: The standard AM wave is applied at the input of the detector.

During the +ve half cycle of the input: The detector diode is forward biased. The filter capacitor C connected across the load resistance R charges to almost

the peak value of the input voltage.



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As soon as the capacitor charges to the peak value, the diode stops conducting.

The capacitor will discharge through R between the positive peaks as shown in the Figure below.

During the -ve half cycle of the input: The diode is reverse biased The capacitor will discharge through R.

Waveform Description: The input-output waveforms for the envelope detector are shown in Figure below. It shows the charging/discharging of the filter capacitor and the approximate output

voltage.

It can be seen from these waveforms, that the envelope of the AM wave is being recovered successfully.

Design Criteria: 1. Fast charging 2. Slow Discharging

The capacitor charges through D and Rs when the diode is on and it discharges through R when the diode is off.

Fast charging: The charging time constant Rs C should be short as compared to the carrier period .

--------------- (1)

Slow Discharging: The discharging time constant RC should be long enough so that the capacitor discharges

slowly through the load resistance R.

This time constant should not be too long which will not allow the capacitor voltage to

discharge at the maximum rate of change of the envelope.



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--------------- (2)

Here W = Maximum modulating frequency

Distortions in the Detector Output: There are two types of distortions which can occur in the detector output. They are:

1. Diagonal clipping 2. Negative peak clipping

Diagonal clipping:

This type of distortion occurs when discharge time constant

is too large.

If the discharge time is too long then the detection circuit cannot follow the fast charge in the envelop, resulting in a portion in the ve half cycle lossed or clipped as shown in the

above figure.

Negative peak clipping:

This type of distortion occurs due to a fact that the modulation index on the output side of the detector is greater than that on its input side.



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Thus, at higher depths of modulation of the transmitted signal,the overmodulation (more than 100% modulation) may take place at the output of the detector.

The negative peak clipping will take place as a result of this overmodulation as shown in Figure above.

Solution: The distortion in envelop detector can be avoided by selecting the right values of

charging time constant

and discharging time constant

.

Advantages of AM:

1. AM transmitters are less complex. 2. AM receivers are simple, detection is easy. 3. AM receivers are cost effective. 4. AM waves can travel a long distance. 5. Low bandwidth.

Disadvantages of AM (DSBFC):

The AM signal is also called as "Double Sideband Full Carrier (DSBFC)" signal. The two main

disadvantages of this technique are:

1. Power wastage takes place. 2. AM needs larger bandwidth. 3. AM wave gets affected due to noise.

These are explained as follows:

The carrier signal in the DSBFC system does not convey any information. The information is contained in the two sidebands only. But the sidebands are images of each

other and hence both of them contain the same information. Thus all the information can

be conveyed by only one sideband.

Power wastage due to DSBFC transmission:

The total power transmitted by an AM wave is given by ------------------ (1)

-------------- (2)

Out of the three terms in Equation (1), carrier component does not contain any

information and one sideband is redundant. So out of the total power

the

wasted power is given by:



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Bandwidth requirement of DSBFC:

The BW of DSBFC system is 2fm. This is due to the simultaneous transmission of both the sidebands, out of which only one is sufficient to convey all the information. Thus the

BW of DSBFC is "double" than actually required. Therefore DSBFC is a "bandwidth

inefficient" system.

Effect of Noise:

When the AM wave travels from the transmitter to receiver over a communication channel, noise gets added to it.

The noise will change the amplitude of the envelope of AM in a random manner. As the information is contained in the amplitude variations of the AM wave, the noise will

contaminate the information contents in the AM. Hence the performance of AM is very

poor in presence of noise.

Applications of AM:

1. Radio broadcasting. 2. Picture transmission in a TV system.

Problems:

15. using the message signal

, obtain the expression for AM wave when the

percentage modulations are

a. 50% b. 100% c. 125%

Also sketch the modulated AM wave for the above percentage modulation.

Solution:

Time domain expression for the AM wave is

------------- (1)

Modulation index for the above equation is

------------- (2)



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Given

----------------- (3)

To determine maximum value of m(t), we should differentiate the equation (3) and equate it to

zero thus,

Putting the value t in equation (3) we get,

Putting the value in equation (2) we get,

Putting the value in equation (1) we get,

---------------- (4)

Case 1: 50% =0.5 thus,

Case 2: 100% =1 thus,

Case 3: 125% =1.25 thus,

Sketching the AM wave:

We will prepare a table for m(t) which contains the value of m(t) for different values of t

Thus,

t 0 0.2 0.4 0.6 0.8 1.0 2.0 3.0 4.0 5.0

m(t) 0 0.192 0.345 0.44 0.49 0.5 0.4 0.3 0.235 0.192



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Thus, maximum value of m(t)=0.5

Now let us calculate Ac for different values of :

We know that

but here Am = 0.5

For =0.5,

For =1,

For =1.25,



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16. A 250 W carrier of 1000 kHz is simultaneously modulated by sinusoidal signals of 2 kHz, 6

kHz and 8 kHz with modulation indices of 35 %, 55 % and 75 respectively. What are the

frequencies present in the modulated wave, Effective modulation index and what is the radiated

power?

Solution:

Given: Pc = 250 W, fc = 1000 kHz, fml = 2 kHz, fm2 = 6 kHz, fm3 = 8 kHz, m1 = 0.35, m2 = 0.55,

m3 = 0.75.

Frequencies present in the modulated wave:

The modulated wave consists of the following frequency components:

Carrier fc = 1000 kHz.

Sidebands of fm1 i.e. fc+ fm1 = 1002 kHz and fc - fm1 = 998 kHz.

Sidebands of fm2 i.e. fc + fm2 = 1006 kHz and fc - fm2 = 994 kHz.

Sidebands of fm3 i.e. fc + fm3 = 1008 kHz and fc - fm3 = 992 kHz.

Effective modulation index:

Radiated power:

17. An AF signal 20 sin (2 x500 t) is used to amplitude modulate a carrier of 50 sin ( 2 x 105

t ). Calculate :

a. Modulation index b. Sideband frequencies c. Amplitude of each sideband frequency d. Bandwidth required e. Total power delivered into a load of 600 .

Solution:

Given: Modulating signal: x ( t ) = 20 sin ( 2 x 500 t ) Am=20v, fm=500 Hz.

Carrier signal : c ( t) = 50 sin (2 x 105 t ) Ac=50, fc = 105 Hz= 100 kHz.

Modulation index ():



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Sideband frequencies:

Upper sideband fUSB = fc+ fm = 100.5 kHz,

Lower sideband fLSB = fc-fm = 99.5 kHz.

Amplitude of each sideband:

Bandwidth required:

Total power delivered into a load of 600 :

18. A carrier wave c(t) = 4 sin (2 x500x103t) is amplitude modulated by an audio wave x(t) =

0.2 sin 3 [ ( 2 x 500 t )] + 0.1 sin 5 [ ( 2 x 500 t) ]. Determine the upper and lower sidebands

and sketch the complete spectrum of the modulated wave. Estimate the total power in the

sidebands.

Solution:

Given:

(i) The carrier is, c ( t ) = 4 sin (2 x 500 x 103 t )

Ac= 4, fc = 500 x 103 Hz.

(ii) The modulating signal is, x ( t) = 0.2 sin 3 [ 2 x 500t ] + 0.1 sin 5 [ 2 x 500t ]

Thus, it consists of two sinewaves. The peak amplitude of the first sinewave is "0.2" and its

frequency is fml = 1500 Hz. The peak amplitude of the second sinewave is 0.1" and its

frequency fm2 = 2500 Hz.

(a) USB and LSB of the AM signal:

The USB and LSB corresponding to first modulating signal are at,



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USB1 = (fc + fml) = 500 kHz + 1.5 kHz = 501.5 kHz.

LSB1 = (fc - fml) = 500 kHz - 1.5 kHz = 498.5 kHz.

The USB and LSB corresponding to the second modulating signal are at,

USB2 = (fc + fm2) = 500 kHz + 2.5 kHz = 502.5 kHz

LSB2 = (fc - fm2) = 500 kHz- 2.5 kHz = 497.5 kHz

(b) Modulation index of individual modulating signals:

Modulation index for the first signal,

Modulation index for the second signal,

(c) Sideband amplitudes:

Amplitudes of USB1 and LSB1 will be:

Amplitudes of USB2 and LSB2 will be:

(d) Complete spectrum of AM signal:

(e) Total power in the sidebands:

Here, t = total modulation index thus,

Also carrier power,



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Thus, total power in the sidebands

19. The rms antenna current of AM transmitter increases by 15% over its unmodulated value,

when sinusoidal modulation by 1 kHz is applied. Determine the modulation index.

Solution:

Let It = rms antenna current of modulated signal.

Ic = rms antenna current of unmodulated signal.

Thus,

We have,

20. In AM signal, power in upper sideband is 500 W for 100 % modulation. Determine the

power in carrier.

Solution:

Given: PUSB = 500 W, = 1

Thus,

21. An AM transmitter radiates 9 kW of carrier power and delivers at its output 10.124 kW of

power. What is the depth of modulation? If the same carrier is modulated with a sinewave of 40

% modulation then find the total transmitted power. Now if both the signals simultaneously

modulate this carrier then what is the resultant transmitted power?

Solution:

Given: Pc= 9 kW, Pt = 10.124 kW



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Calculate depth of modulation ():

Total transmitted power for =0.4:

Pt when both signals modulate simultaneously:

Total modulation index is,

The total transmitted power,

22. A complex modulating waveform consisting of a sine wave of amplitude 3V and frequency

1000 Hz plus a cosine wave of amplitude 5V and frequency 3000 Hz, amplitude modulates a

carrier of 500 kHz and 10V peak. Plot the spectrum of modulated wave and determine the

average power when the modulated wave is fed into a 50 load.

Solution:

Given: Am1 = 3V, Am2 = 5V, Ac = 10V

fm1 = 1 kHz, fm2 = 3kHz, fc = 500kHz and RL = 50

Plotting the spectrum of modulated waveform:

The spectrum of AM signal consists of three frequency components namely the carrier, upper

sideband and lower sideband.

The frequencies of these components are:

Carrier

Upper sidebands and



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Lower sidebands and

The modulation indices of the two signals are,

The amplitudes of the upper and lower sidebands are given by,

Amplitudes of USB1 and LSB1 is:

Amplitudes of USB2 and LSB2 is:

The complete spectrum is as shown below,

To calculate the average power dissipated in the load:

Also,



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23. An AM signal with a carrier of 1 kW has 200 Watts in each sideband. What is the percentage

of modulation?

Solution:

Given: Pc = 1000 W, PUSB = PLSB = 200 W

24. A sinusoidal carrier Vc=100cos(2x105t) is amplitude modulated by a sinusoidal voltage

Vm=50cos(2x103t) upto a modulation depth of 50 %. Calculate the amplitude and frequency of

each sideband and the rms voltage of the modulated carrier.

Solution:

Given: Ac = 100 V, Am = 50 V, = 0.5, fc = 100 kHz, fm = 1 kHz

Sideband frequencies fUSB = 101 kHz

fLSB = 99 kHz

Amplitude of each sideband =

rms voltage of the modulated carrier:

------ (1)

Let

----- (2)

Here AAM= RMS value of modulated carrier

----- (3)

Putting (2) and (3) in (1)



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25. A sinusoidal carrier has amplitude of 10 V and frequency 30 kHz. It is amplitude modulated

by a sinusoidal voltage of amplitude 3V and frequency 1 kHz. Modulated voltage is developed

across a 50 resistance.

a. Determine the modulation index. b. Write the equation for modulated wave. c. Plot the modulated wave showing maxima and minima of waveform. d. Draw the spectrum of modulated wave. e. Calculate the total average power. f. Calculate the power carried by the sidebands.

Solution:

Given: Ac = 10 V, Am = 3 V, fc = 30 kHz, fm = 1 kHz, RL = 50

a. equation for modulated wave:

b. Thus Equation is given as,

c. Plot the modulated wave showing maxima and minima of waveform:

d. spectrum of modulated wave:

Sideband frequencies



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Amplitude of each sideband =

e. total average power:

f. power carried by the sidebands:

26. For an AM DSBFC envelope with +Vmax = 20 V and +Vmin = 4 V, determine the following:

a. Peak amplitude of the carrier b. Modulation coefficient and percentage modulation c. Peak amplitude of the upper and lower side frequencies

Solution:

Given: Type of modulation: AM (DSBFC), Envelope: + Vmax = 20 V, + Vmin = 4 V

a. Peak amplitude of the carrier:

The peak amplitude of the modulating signal is,

The peak amplitude of the carrier is,



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b. Modulation coefficient and percentage modulation:

c. Peak amplitude of the upper and lower side frequencies:

Peak amplitude of USB or LSB =

27. Suppose nonlinear devices are available for which the output current io, input voltage vi, are

related by where al and a3 are constants. Explain how these devices could be

used to produce a DSBSC wave.

Solution:

Given:

Let , thus i0 becomes,

------------- (1)

As,

and

Thus,

------------ (2)

Taking Fourier transforms on both sides of the above equations we get,



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Spectrum Plotting

Double Side Band Suppressed Carrier (DSB-SC):

Definition: Type of amplitude modulation in which the carrier is suppressed/removed before

transmission.

OR

Type of transmission where only two sidebands are transmitted without the carrier.

Time-Domain description:

Let m(t) be the message signal band limited to W Hz. C(t)=Accos2fct be the carrier wave. Then DSB-SC modulated wave can represent by,

---------------- (1)

The DSB-SC waveform undergoes a phase reversal at the zero crossing point of the message signal and thus the envelop of DSB-SC is completely different from that of

DSB-FC as shown below.



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Frequency Domain Description:

DSB-SC modulated wave in time domain is given as,

Applying Fourier transform on both-sides of the equation,

----------- (1)

The above expression represent the frequency spectrum of the DSB-SC signal as shown below.

From the above spectrum we can clearly see that the DSB-SC comprises only two side bands and there is no carrier wave.

The upperside band is located from fc to fc+W. The lowerside band is located from fc to fc-W. The highest frequency component fH in DSC-SC modulated wave is given by,

The lowest frequency component fL in DSC-SC modulated wave is given by,

Transmission Bandwidth of the DSB-SC modulation is given as follows,

Thus, the bandwidth required for DSB-SC wave is same as the standard wave.



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Note:

DSB-SC modulated wave can be generated simply multiplying m(t) and c(t) using a multiplier circuit as shown below

The product of the modulator circuit is the DSB-SC modulated wave. This multiplier circuit is also known as Product modulator.

Single Tone DSB-SC Modulation:

In single tone DSB-SC modulation, the message signal m ( t) will be a sinusoidal signal and it is represented as,

----------- (1)

Here Am = amplitude and fm = frequency of the modulating signal

Time Domain Description:

The DSB-SC modulated wave is obtained by,

This is the time domain representation/description of the single tone DSB-SC wave. The figure below shows the DSB-SC modulated wave.



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Frequency Domain Description:

The time domain representation/description of the single tone DSB-SC wave is

Applying Fourier transform on both-sides of the equation,

This is the spectrum of .DSB-SC modulated wave, which contains the delta functions at frequencies ( ) and ( ) as shown in Figure below.

The spectrum shows that the DSB-SC wave consists only of the two sidebands (USB and LSB) at frequencies ( ) and ( ) respectively.

Transmitted Power (Pt):

Transmission Efficiency ():



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Power Saving:

Due to the suppression of carrier, a lot of power saving takes place in DSB-SC. At 100% modulation, = 1 the percent power saving is given by (Pc/1.5 Pc) i.e. 66.66% shown as

below

The total power in AM wave,

At 100 % depth of modulation = 1,

At 50 % depth of modulation = 0.5,

Transmission bandwidth:

The transmission bandwidth of DSB-FC with a single tone modulation is 2fm where fm is the frequency of the modulating signal. Thus transmission bandwidth of DSB-SC is same

as that of the standard AM wave.

Differentiate between DSB-FC and DSB-SC signals:

The time domain displays of DSB-FC signal with = 100 % and DSB-SC signal look exactly the same. The only difference between them is that in the time domain display of

DSB-SC the carrier undergoes 1800 phase shift. This is how we can identify the DSB-SC.

Generation of DSB-SC Waves:

There are two forms of product modulators to Generation of DSB-SC namely: 1. Balanced modulators 2. Ring modulator



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Balanced Modulator (BM):

The block diagram of a balanced modulator (BM) is shown in Figure above. It consists of two standard amplitude modulators arranged in the balanced configuration, so as to

suppress the carrier completely.

Both the standard amplitude modulators are fed from a common carrier wave Both the standard amplitude modulators are fed with a same message signal but with a

phase reversal as shown in Figure.

The outputs of the standard AM modulators are fed to a summer circuit to get the difference of two standard AM waves and this is the DSB-SC modulated wave.

Let s1(t) represent the output of the modulator with a no-phase reversal message signal as input.

------------ (1)

Let s2(t) represent the output of the modulator with a phase reversal message signal as input.

------------ (2)

The difference of summer circuit is

------------ (3)

The above equation is the DSB-SC modulated wave. Taking Fourier transform on both sides for equation (3) we get,



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-------------- (4)

Spectrum of above equation is as shown below,

Ring Modulator:

Figure above shows the circuit diagram of a diode ring modulator. It consists of

1. Four diodes or Diode Bridge. 2. An audio frequency transformer T1 3. An RF transformer T2

T1 and T2 are selected in such a way that they are balanced. The diode bridge consists of 4 diodes D1 to D4 connected in series back to back forming a

ring structure thus, the name Ring modulator.

The carrier signal is assumed to be a square wave with frequency fc and it is connected between the center taps of the two transformers.

Ac is greater than Am so that the switching of the diodes is dependent upon only on the carrier wave c (t).

The DSB-SC output is obtained at the secondary of the RF transformer T2. Operation of the circuit:



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During +ve half cycle of c(t): Diodes D1 and D2 are turned on while the diodes D3 and D4 are turned off.

Thus, m(t) is multiplies by +1 (amplitude of c(t)). Thus,

During -ve half cycle of c(t): Diodes D1 and D2 are turned off while the diodes D3 and D4 are turned on.

Thus, m(t) is multiplies by -1 (amplitude of c(t)). Thus,

Waveforms are as shown below,

Time domain Description: Using Fourier series, c(t) can be represented as,

Expanding the above equation we get,

Generic equation for DSB-SC is,

Thus,



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Frequency Domain Description: Time domain expression for DSB-SC wave using ring modulator is,

Taking Fourier transform on both sides for equation we get,

Its spectrum is as shown below,

From the spectrum it is clear that to obtain DSB-SC wave the output of the ring modulator has to passed to the bandpass filter with the centre frequency of fc Hz and

bandwidth of 2W Hz.

The coherent detector for the DSB-SC signal:

Vo(t)

DSB-SC signal s(t) m(t)

c(t)=cos(2fct)

Product Modulator

Local Oscillator

Low Pass Filter



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The coherent detector for the DSB-SC signal is shown in Figure above. The DSB-SC wave s (t) is applied to a product modulator in which it is multiplied with the locally

generated carrier cos (2fet).

Here the locally generated carrier is exactly coherent or synchronized in both frequency and phase, with the original carrier wave c (t) used, to generate the DSB-SC wave. This

method of detection is therefore called as coherent detection or synchronous detection.

The output of the product modulator is applied to the low pass filter which eliminates all the unwanted frequency components and produces the message signal as proved below.

Analysis of Coherent Detection: Let the output of the local oscillator be given by,

--------- (1)

Thus its amplitude is 1 (unity), frequency is fc and the phase difference is arbitrary equal to equation (1). This phase difference has been measured with respect to the

original carrier c (t) at the DSB-SC generator.

Hence the output of the product modulator is given by,

--------- (2)

Here,

c'(t) = Local carrier = cos (2 t +)

Thus, ---------- (3)

------- (4)

Equation (4) shows that the output of product modulator i.e. m (t) consists of two terms.

The first one represents the message signal m (t) with an amplitude of

. So this

is the wanted term. The second term is an unwanted one.

Signal m (t) is then passed through a low pass filter. Which allows only the first term to pass through and will reject the second term. Hence the filter output is given by,

--------- (5)



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Thus output voltage of the coherent demodulator is proportional only to the message signal m (t) if the phase error cos is constant.

Effect of Phase Error on the Demodulated Output:

Consider the expression for the output of coherent detector,

In this expression represents the phase error and the amplitude of demodulated output is

.

Hence the amplitude of demodulated output is maximum and equal to

when = 0 &

the amplitude is zero when = 900 or /2 radians. This effect is called as the

quadrature null effect of the coherent detector. Here "quadrature" term represents the

phase difference of 900 or /2 radians.

In other words the phase error attenuates the demodulator output. In practice the phase error varies randomly with time due to the random variations taking

place in the communication channel.

So cos will vary randomly and the detector output also will vary in a random manner. This is undesirable.

Hence circuitry must be provided in the detector to keep the locally generated carrier c'(t) in perfect synchronism, in both frequency and phase, with the original carrier c (t).

Costas Loop:



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Costas loop is a method of obtaining a practical synchronous receiver system suitable for demodulating DSB-SC waves as shown in the figure above.

The receiver consists of two coherent detectors supplied with the same DSB-SC modulated wave but with separate carrier waves which are 90

0 apart from each other.

The detector in the upper point is referred to as the in-phase channel or I-channel and lower part is referred to as quadrature channel or Q-channel.

These two detectors are coupled together to form a ve feedback system designed in such a manner as to maintain the phase of the local oscillator in synchronous with the carrier

wave of transmitter.

Operation of Circuit: Case I:

When the local oscillator signal is in phase with the carrier wave Accos2fct (used at the transmitter end) the I channel output will be

But When =0 then cos=1 Thus,

The Q channel output will be,

But When =0 then sin = 0 Thus,

Case II: When the local oscillator phase changes by a small angle then the Q-

channel modulates an output which is proportional to .

The output of Q and I channels are combined in the phase discriminator. The phase discriminator consists of a multiplier and low pass filter which

generates a DC signal, which is then applied to the voltage controlled

oscillator.

This DC voltage rectifies the phase difference to produce the DSB-SC wave.

Formulas:

DSB-FC:

AM Equation:



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Question Bank VTU

1. Define Demodulation. Explain the envelop detection method for AM wave with circuit

diagram and waveforms. December 2008 (6 M), July 2013 (08 M)

2. Explain the generation of DSBSC using ring modulator. December 2008 (6 M), July

2013 (08 M)

3. Consider a message signal m(t) with a spectrum shown below. The message bandwidth w=1 kHz. This signal is applied to a product modulator, together with a carrier wave Ac

cos(2fct), producing the DSBSC modulated signal S(t). The modulated signal is next

applied to a coherent detector. Assuming perfect synchronism between the carrier waves

in the modulator and detector. Determine the spectrum of the detector output when:

a. fc= 1.25 kHz b. fc= 0.75 kHz

What is the lowest carrier frequency for which each component of the modulated signal

S(t) is uniquely determined by m(t). December 2008 (6 M)

4. Define standard form of amplitude modulation and explain the time and frequency domain expression of AM wave. June 2009 (6 M)

5. Explain with the help of a neat sketch, how a square law modulator is used to generate

AM with relevant equations and spectrum. June 2009 (8 M), June 2014 (8 M)

December 2012

6. A carrier wave volts is amplitude modulated by an audio wave volts. Determine the upper and lower

sideband and sketch the complete spectrum of the modulated wave. Estimate the total

power in the sideband. June 2009 (6 M)

7. Show that a square law can be used for the detection of an AM wave. May 2010 (6 M)

8. Consider the message signal volts and the carrier wave Volts.

a. Find the power developed across a load of 100 due to this AM wave. b. Sketch to scale resulting AM wave for 75% modulation. May 2010 (6 M)

9. Explain the method of obtaining a practical synchronous receiving system with DSBSC modulated waves using costas loop. May 2010 (8 M), December 2012 (08 M)June 2014

(8 M)

10. An audio frequency signal is used to amplitude modulate a carrier of . Assume modulation index as 0.5 find:

a. Side band frequencies. b. Amplitude of each side band. c. Bandwidth required.



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d. Total power delivered to a load 100 June 2014 (06 M)

11. Explain the operation of coherent detection of DSBSC modulating wave and show that

the overall output

. December 2012 (06 M)

12. Consider a signal and the carrier wave 13. Write an expression for the resulting AM wave for 75% modulation in time domain.

a. Draw the spectrum of AM wave. b. Sketch the resulting wave for 75% modulation. December 2012 (06 M)

14. An AM wave is given by is applied to the system, show below.

Assume that the message signal m(t) is limited to the |W|fc and that the carrier frequency

fc> 2W show that m(t) can be obtained from the square-rooter output V3(t).

December 2012 (06 M)

15. What is the significance of DSBSC modulation? Explain with time domain description. June 2013 (04 M).

16. Using the message signal

.Determine and sketch the modulated wave for

AM whose % modulation are i) 50% ii) 100% iii) 125% December 2013 (05 M)

17. What do you mean by coherent detection? What is quadratude null effect? December 2013 (04 M)



Ragh

udath

esh G

P



Ragh

udath

esh G

P