analog communication unit3 vtu

8/12/2019 Analog Communication Unit3 Vtu

1/44

ANALOG COMMUNICATION (VTU) - 10EC53

UNIT - 3

SINGLE SIDE-BAND MODULATION (SSB): Quadrature carrier multiplexing, Hilbert

transform, properties of Hilbert transform, Pre-envelope, Canonical representation of band pass

signals, Single side-band modulation, Frequency-Domain description of SSB wave, Time-

Domain description. Phase discrimination method for generating an SSB modulated wave,

Frequency discrimination method for generating an SSB modulated wave, Demodulation of SSB

waves.

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), Ravitej B (GMIT), Somesh HB (REVA

ITM).

BY:

RAGHUDATHESH G P

Asst Prof

ECE Dept, GMITDavangere 577004

Cell: +917411459249

Mail:[email protected]

Quotes:

1. No one can make you feel inferior without your consent.

2. Everything we do affects other people.

3. Family isnt about whose blood you have. Its about who you care about.

4. I dont know the key to success, but the key to failure is trying to please everybody.

5. People will Hate you, Rate you, Shake you and Betray you. But How Strong you Stand iswhat makes You.

6. The saddest aspect of life right now is that science gathers knowledge faster than society

gathers wisdom.

SSB Modulation Raghudathesh G P Asst Professor

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


2/44

SINGLE SIDE-BAND MODULATION (SSB)

The transmission bandwidth of a DSB-SC system is 2 fmor 2W Hz. Thus, the number of

DSB-SC channels that can fit into the available channel bandwidth will be small.

If we wish to accommodate more number of channels within the same channel

bandwidth, then we can use a special technique called quadrature carrier multiplexing.

It is also called as quadrature amplitude modulation(QAM).

Quadrature Carrier Multiplexing (QCM) or Quadrature Amplitude Multiplexing (QAM):

The QAM technique enables two DSB-SC modulated waves to occupy the same

transmission bandwidth without mixing them with each other.

The two DSBSC waves have resulted from two different message signals and at the

receiver it is possible to separate them and obtain both the message signals.

For multiplexing the above block diagram is used. It consists of two product modulators.

To the 1stproduct modulator we give the first message signal m1(t) and a carrier wave which is locally generated using a local oscillator. This path is

known as I-Path. The output of the first product modulator is given by,

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

The second message is given to the 2nd

product modulator. The carrier wave of the 2nd

product modulator has same amplitude and frequency of the carrier wave that is used in

1stproduct modulator, but has a phase shift of 90 0as compared to the 1stcarrier wave.

This path is known as Q-Path. The output of the second product modulator is given by,


ECE Dept, GMIT [email protected] No - 2


3/44

-------------- (2) Outputs of 2 product modulators are added together to obtain the multiplexed signal as,

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

The equation (3) indicates that the transmission bandwidth of the multiplexing signal s(t)

is 2W Hz centered at the carrier frequent fc. Here W is the bandwidth of the message

signal or whichever is larger.Receiver of Quadrature carrier Multiplexing System (Demultiplexing):

The block schematic of a receiver of the quadrature carrier multiplexing system is shown

in Figure above.

There are two product modulators fed with same input signal s(t) and carrier signal

which are frequency and phase synchronized with the transmitter.

The output of the 1

st

product modulator is given by,

---------------- (1) Using trigonometric identities,




4/44

And

Thus,

The output of the 1st product modulator consists of pure m1(t) term along with higher

frequency terms.

By passing VI(t) through a low pass filter we get,

-------------- (2) The output of the 2stproduct modulator is given by,

Using trigonometric identities, And

Thus,

By passing VQ(t) through a low pass filter we get,

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

Thus, we have successfully multiplexed and demultiplexed m1(t) and m2(t).




5/44

NOTE:

For the satisfactory operation of the quadrature carrier multiplexing system, it is

necessary to maintain the correct phase and frequency relationships between the carrier

oscillators used in the transmitter and receiver parts of the system.

We can satisfy above requirements by using the Costas loop.

Why Fourier transforms are used?

The Fourier transform is useful for evaluating the frequency content of an energy signal,

or in a limiting case that of a power signal.

It provides mathematical basis for analyzing and designing the frequency selective filters

for the separation of signals on the basis of their frequency content.

Hilbert Transform:

A method of separating the signals based on phase selectivity, which uses phase shiftsbetween the appropriate signals (components) to achieve the desired separation.

The Hilbert transform is named afterDavid Hilbert.

It is a basic tool in Fourier analysis, and provides a concrete means for realizing the

harmonic conjugate of a given function or Fourier series.

Definition1:If the phase angles of all the components of a given signal are shifted by

900then the resulting function of time is called as "Hilbert transform" of the signal.

Here amplitude remains unchanged.

Definition 2:Hilbert transform is a method that introduces a phase shift of -900(

)

to all +ve frequencies and +900(+ ) to allve frequencies.

Consider a signal x(t) with Fourier transform X (f). The Hilbert transform of a signal x(t)

is denoted by and is given by,Hilbert transform

x(t)


http://en.wikipedia.org/wiki/David_Hilberthttp://en.wikipedia.org/wiki/David_Hilberthttp://en.wikipedia.org/wiki/David_Hilberthttp://en.wikipedia.org/wiki/David_Hilbert


6/44

Inverse Hilbert Transform:

We can recover back the original signal x (t) back from by taking the inverse Hilberttransform as follows:

We can say that x (t) and constitute a pair, Hilbert transform pair.

Frequency Description of Hilbert Transforms:

Applying Fourier transform on both sides,

Here, indicates signum function and is defined as,




7/44

Thus,

The conclusion from the above equation is that we may obtain the Hilbert transform

of

signal x ( t ) by passing x ( t) through a linear two port device whose transfer function is equal to

j sgn ( f) as shown in figure below.

Applications of Hilbert transform:

1. It is used to realize phase selectivity in the generation of special kind of modulation

called Single Side Band modulation.

2. It provides mathematical basis for the representation of band pass signals.




8/44

Note:Hilbert transform applies to any signal that is Fourier transformable.

Properties of Hilbert Transform:

Basics: We can obtain the Hilbert transform of any signal which is Fourier transformable.

We can obtain the Hilbert transform of energy signals as well as power signals.

The difference between Hilbert transform and the Fourier transform is that the H.T.

operates exclusively in the time domain.

The signal x (t) has been assumed to be a real valued signal.

Property 1: It states that the signal x (t) and its Hilbert transform have the sameamplitude spectrum.

Proof:

Fourier transform of .The magnitude of is equal to 1 for all the values of f. Therefore the amplitude spectrum of i.e. is equal to that of x (t) i.e. .

Property 2: This property states that if is the Hilbert transform of x (t) then theHilbert transform of is -x (t).Proof:

This statement suggests that the Hilbert transform is being taken twice as shown in Figure

below

We know that to take the H.T. of a signal is equivalent to passing it through a two port-

device which has a transfer function equal to - j sgn (f). The double Hilbert transform is therefore

equivalent to cascading of two such devices as shown.

The overall transfer function of such a cascade is given by, But and . Thus,




9/44

Hence the FT of output is, Thus the Hilbert transform of is equal to - x (t).

Property 3:x (t) and its Hilbert transform are orthogonal to each other.Proof:

We have to prove that, -------- (1)If a signal x (t) is multiplied with its Hilbert transform then we can write that,

-------- (2)

But, thus,

As,

Also,

The term inside the integration on RHS of this expression is equal to the product of an

odd function sgn (f) and an even function . Hence it is an odd function. The integration ofan odd function over the range will yield a zero value.

This expression shows that an energy signal x (t) and its H.T. are orthogonal over theentire interval ( ).Similarly it is possible to prove that a power signal x (t) and its Hilbert transform areorthogonal over one period.




10/44

Additional Properties:

1. The magnitude spectra of a signal x (t) and its Hilbert transform are identical.2. The Hilbert transform of an even function is odd and vice-versa.

3. The Hilbert transform of a real signal is also real.

Problem1:Find the Hilbert transform of Solution:

Obtain the Fourier transform of x (t):

Calculate the Fourier transform of

But and

Obtain by taking IFT: Thus the Hilbert transform of a cosine function is equal to .Problem2:Prove that Hilbert transform of is .Solution:Obtain the Fourier transform of x (t):




11/44

Calculate the Fourier transform of

But and

Obtain

by taking IFT:

Thus the Hilbert transform of a cosine function is equal to .Problem3:Find the Hilbert transform of Solution:

Hilbert transform of x(t) is obtained by shifting the signal by -90othus,

Problem3:Find the Hilbert transform of Solution:

Given Hilbert transform of x(t) is obtained by shifting the signal by -90othus,




12/44

Pre-envelope:

Consider a real valued signal x (t), the pre-envelope x+ (t) for positive frequencies the

signal x (t) is defined as the complex valued function given by,

-------- (1) The pre-envelope is useful in handling the bandpass signals and systems.

Apply Fourier transform on both the sides,

But, thus,

--------- (2) As, ------------ (3)

Putting (3) in (2), thus we get,

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

Where X (0) is the value of X (f) at f = 0.

The above expression shows that the pre-envelope of a signal does not have any

frequency content for all the negative frequencies. [X+(f) = 0 for f < 0]. This is shown in

Figure below




13/44

Pre-envelope:

Consider a real valued signal x (t), the pre-envelope x- (t) for negative frequencies the

signal x (t) is defined as the complex valued function given by,

-------- (1) But, , thus, --------- (2) As,

------------ (3) Putting (3) in (2), thus we get,

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




14/44

Canonical Representations of Bandpass Signals:

The spectrum of a bandpass signal x (t) shown below is called as a bandpass signal if its

Fourier transform X (f) exists only in the band of frequencies of 2W centered about some

frequency fc. The frequency fcis called as carrier frequency. If 2W is small as compared

to fc then the signal is called as narrow band signal.

Pre-envelope of a narrow bandpass signal x (t) can be represented in canonical form as,

---------- (1) Here,

X+(t) = pre-envelop of x(t)




15/44

= complex envelope of the signal x(t) = Eulers form But x (t) represents the real part of the pre-envelope x+ (t). Hence the given bandpass

signal x (t) can be expressed in terms of the complex envelope as,

---------- (2) Since is a complex quantity we can express it as,

----------- (3) Where xI (t) and xQ(t) are both real valued low pass functions.

Putting equation (3) in (2) we get,

But Thus we get

--------- (4) In this expression xI(t) is the in phase component of the bandpass signal x (t) and xQ(t) is

the quadrature component of the signal.




16/44

Generation of In-phase and Quadrature components of x (t):

Both xI(t) and xQ(t) are low pass signals limited to the .band W f W. So we can

derive them from the band-pass signal x (t) using the method shown in the Figure below.

In Figure both the low pass filters are identical. The bandwidth of each filter is W. The

inphase component xI(t) is produced by multiplying x (t) with cos (2fct) and passing the

product through a low pass filter.

Quadrature component xQ(t) is obtained by multiplying x (t) with sin (2fct) and passing

the product through an identical low pass filter.

Reconstruction of x (t) from xI(t) and xQ(t):

The scheme to reconstruct the bandpass signal x (t) from xI (t) and xQ (t) is shown in

Figure below,

The inphase low pass signals xI(t) and xQ(t) are multiplied with the cos (2fct) and sin(2fct) respectively.

The resultant product terms are then subtracted to get the bandpass signal x (t).

The multiplication process of xI (t) and xQ (t) with the carriers is a linear modulation

process.




17/44

If fc is large enough, then the output bandpass function x (t) is referred to a passband

signalling waveform. This type of mapping of the inphase and quadrature components

into x (t) is known as passband modulation.

Single Sideband Modulation: The transmission bandwidth of standard AM as well as DSB-SC modulated wave is

2W Hzi.e. twice the message bandwidth W.

Thus, both these systems are bandwidth inefficient systems.

In both these systems, one half of the transmission bandwidth is occupied by the upper

sideband (USB) and the other half is occupied by the lower sideband (LSB).

But the most important thing is that the information contained in the USB is exactly

identical to that carried by the LSB. So by transmitting both the sidebands we are

transmitting the same information twice.

Thus, we can transmit only one sideband (USB or LSB) without any loss of information.

So it is possible to suppress the carrier and one sideband completely.

When only one sideband is transmitted, the modulation is referred to as single

sideband modulation. It is also called as SSB or SSB-SC modulation.

Frequency Domain Description:

Figure (a) below represents the spectrum M (f) of the message signal m (t). This spectrum

is limited to the band -W f W as shown in Figure (a) below.

The spectrum of DSB-SC wave which is obtained by taking the product of m (t) and c (t)

is shown in Figure (b) below. It contains the USB as well as LSB.

When only USB is transmitted by the SSB system, then the corresponding spectrum is as

shown in Figure (c) below.

When only LSB is transmitted, the frequency spectrum is as shown in Figure (d) below.




18/44

Transmission bandwidth of SSB-SC:

Since we are transmitting the frequencies only in the range (fc + W) or (fc - W), the

transmission bandwidth for the SSB-SC will be,

Or This is exactly half the bandwidth of the DSB-FC or DSB-SC modulated waves.

Time Domain Description of SSB Wave Having USB :

Consider an SSB modulated wave su (t) which contains only the upper sideband. The

spectrum of USB is shown in Figure (c) below and the spectrum of LSB is shown in

Figure (d) below,

The SSB wave su (t) is generated by passing the DSB-SC modulated wave through a

bandpass filter having a transfer function Hu(f) as shown in Figure below.




19/44

The DSBSC modulated wave is defined as,

--------- (1) Here,

= message signal = carrier signal The DSBSC signal of Equation (1) is a bandpass signal which contains only the in-phase

component. Hence the low pass complex envelope of the DSBSC modulated wave is

given by,

---------- (2) The SSB modulated wave is also a bandpass signal. But it contains the in-phase as well

as quadrature components.

Let the complex envelope of su (t) (SSB signal) be represented by su (t). Then the SSB

signal is represented mathematically in the time domain as follows,

--------- (3)Steps to find the expression of :

Replace the bandpass transfer function Hu(f) shown in Figure (a) above by an equivalent

low pass filter with transfer function

as shown in Figure (b) above. Thus, the

expression for as follows, --------- (4) DSBSC modulated wave be replaced by its complex envelope. The spectrum of such an

envelope is shown in Figure (c) above. This spectrum can be expressed mathematically as

follows,

---------- (5) Desired complex envelope is obtained by taking the inverse fourier transform of the

product and ,




20/44

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

Taking inverse Fourier transform of this expression we get,

---------- (7) Substitute the expression of in the expression of su (t) we get the mathematical

expression for the SSB signal in time domain as follows:

Simplifying the above expression we get,

--------- (8)

In phase component Quadrature component

Equation above shows that the SSB modulated contains only USB with an in phase

component and a quadrature component.




21/44

Time Domain Description of SSB Wave Having LSB :

Consider an SSB modulated wave sL (t) which contains only the lower sideband. The

spectrum of USB is shown in Figure (c) below and the spectrum of LSB is shown in

Figure (d) below,

The SSB wave sL (t) is generated by passing the DSB-SC modulated wave through a

bandpass filter having a transfer function HL(f) as shown in Figure below.

The DSBSC modulated wave is defined as,

--------- (1) Here, = message signal

= carrier signal

The DSBSC signal of Equation (1) is a bandpass signal which contains only the in-phasecomponent. Hence the low pass complex envelope of the DSBSC modulated wave is

given by,

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




22/44

The SSB modulated wave is also a bandpass signal. But it contains the in-phase as well

as quadrature components.

Let the complex envelope of sL(t) (SSB signal) be represented by sL (t). Then the SSB

signal is represented mathematically in the time domain as follows,

--------- (3)Steps to find the expression of :

Replace the bandpass transfer function HL(f) shown in Figure (a) above by an equivalent

low pass filter with transfer function as shown in Figure (b) above. Thus, theexpression for as follows,

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

DSBSC modulated wave be replaced by its complex envelope. The spectrum of such an

envelope is shown in Figure (c) above. This spectrum can be expressed mathematically as

follows,

---------- (5) Desired complex envelope

is obtained by taking the inverse fourier transform of the

product and , ------------ (6) Taking inverse Fourier transform of this expression we get,

---------- (7) Substitute the expression of in the expression of sL (t) we get the mathematical

expression for the SSB signal in time domain as follows:




23/44

Simplifying the above expression we get,

--------- (8)

In phase component Quadrature component

Equation above shows that the SSB modulated contains only LSB with an in phase

component and a quadrature component.

Single Tone Modulation transmitting only USB:

Let the modulating sinusoidal signal m (t) be represented as,

------------- (1) The Hilbert transform of this signal can be obtained by passing it through a - 90 phase

shifter. So the Hilbert transform is given by,

--------- (2) The SSB wave with only USB is given by,

------------ (3) Putting Equation (1) and (2) in (3) we get,




24/44

----------- (4) This expression shows that the SSB wave consists of only the upper sideband of

frequency (fc+fm).

Single Tone Modulation transmitting only LSB:

Let the modulating sinusoidal signal m (t) be represented as,

------------- (1) The Hilbert transform of this signal can be obtained by passing it through a - 90 phase

shifter. So the Hilbert transform is given by,

--------- (2) The SSB wave with only USB is given by,

------------ (3) Putting Equation (1) and (2) in (3) we get,




25/44

----------- (4) This expression shows that the SSB wave consists of only the lower sideband of

frequency (fc-fm).

Methods of Generation of SSB Modulated Wave:

Two methods are discussed:1. Frequency Discrimination Method

2. Phase Discrimination Method

Frequency Discrimination Method (Filtration Method):

This method can be used for generating the SSB modulated wave if the message signal

satisfies the following conditions:

1. The message signal should not have any low frequency content. The audio signals

posses this property, e.g. the telephone signal will have a frequency range extending

from 300 Hz to 3.4 kHz. The frequencies in the range 0-300 Hz are absent.2. The highest frequency in the spectrum of the message signal i.e. W Hz should be

much smaller than carrier frequency fc.

System Block Schematic:

Figure shows the block diagram of an SSB modulator which operates on the principle of

frequency discrimination.




26/44

This modulator consists of a product modulator, carrier oscillator and bandpass filter

designed to pass the desired sideband.

At the output of the product modulator, we get the DSB-SC modulated wave which

contains the two sidebands only.

The bandpass filter will pass only one of these sidebands and produce the SSB modulated

wave at its output.

Design of bandpass filter:

Therefore the design of bandpass filter must be based on satisfying the following two

conditions:

1. Passband of the BPF should occupy the same frequency range as that occupied by the

spectrum of the desired SSB modulated wave.

2. The width of the guard band which separates the passband from stop band be twice

the lowest frequency component of the message signal.

Guard band = 2f1Hz

Types of filters:

The conditions mentioned above are satisfied only by the highly selective filters having

high Q factor (typically in the range 1000 to 2000).

Two Stage SSE Modulator:

When the carrier frequency is very high as compared to the message frequency, the SSB

modulated wave occupies the frequency band which is much higher than that of themessage signal.

Under such operating conditions it becomes extremely difficult to design a bandpass filter

that passes the desired sideband and attenuates the unwanted sideband.

Then we have to use the two stage SSB modulator of Figure below which uses multiple

modulation process.




27/44

The message signal m (t) modulates the carrier fl, to produce a DSBSC signal. This signal

is passed through the band pass filter 1 to produce an SSB modulated signal.

Let the sideband frequencies be (f1+ fm1) to (f1+ fm2) assuming that USB is selected.

The output of BPF 1 is then used to modulate another carrier f2which is higher than f1.

Then at the output of the second product modulator, we get another DSB-SC signal the

spectrum of which is as shown in Figure below.

Notice that the guard band between the highest LSB frequency and the lowest USB

frequency is increased.




28/44

This will make the filter design easy. Thus the two stage SSB modulator will simplify the

filter design.

Why to modulate at low frequency?

The filter used to suppress the unwanted sideband must have a flat pass band and veryhigh attenuation outside the passband. And the filter response must change from zero

attenuation to full attenuation over a range of about 600 Hz.

To fulfill these requirements at very high operating frequency the Q of the tuned circuit

must be very high.

This is practically not achievable. Hence the modulation is carried out at low frequency

and then the frequency is raised by means of up conversion

Types of filters used:

The LC filters, crystal filters, ceramic or mechanical filters can be used for removing the

unwanted sideband. The crystal or ceramic filters are cheap but technically better only

above 1 MHz operating frequency. The mechanical filter has the best properties i.e. small

size, good passband and good attenuation characteristics.

Advantages of frequency discrimination method:

1. This method gives the adequate sideband suppression. The sideband filter also helps to

attenuate the carrier.

2. The bandwidth is sufficiently flat and wide.

Disadvantages:

1. Due to the inability of the system to generate SSB at high radio frequencies, the

frequency up conversion is necessary.

2. Low audio frequencies cannot be used as the filter becomes bulky.

3. Two expensive filters are to be used one for each sideband.




29/44

Phase Discrimination Method for the SSB Generation (Hartley Method):

The block diagram of Phase Discrimination Method or Hartley method of SSB generation

is as shown in the figure below,

This system uses two balanced product modulators M1and M2and two 900phase shifting

networks.

The message signal m (t) and a carrier signal Accos2fct is directly applied to the product

modulators M1, , producing a DSB-SC wave,

--------- (1) The Hilbert transform of m (t) and 900 phase shifted carrier wave are applied to the

product modulators M2, , producing a DSB-SC wave,

---------- (2) The output of the product modulators M1and M2 are then applied to the adder and the

output is give by,

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




30/44

Advantages of phase discrimination method:

1. It can generate the SSB at any frequency so the frequency up converter stage is not

required.

2. It can use the low audio frequencies as modulating signal. (In filter method this is not

possible)3. It is easy to switch from one sideband to the other.

Disadvantage:

The design of the 90 phase shifting network for the modulating signal is extremely

critical. This network has to provide a correct phase shift of 90 at all the modulating

frequency which is practically difficult to achieve.

Note: SSB not used for broadcasting because of the following reason:

1. As the SSB transmitter and receiver require excellent frequency stability, a smallfrequency shift in the system can result in degradation in the quality of the transmitted

signal. Thus it is not possible to transmit a good quality music using the SSB system.

2. It is not possible to design a tunable receiver oscillator with very high frequency stability.

Now with the advent of the frequency synthesizers, this has become possible. But such

receivers are too expensive. These are the reasons why SSB is not generally used in the

broadcasting applications.

Comparison between different Sideband Suppression Methods:

SlNo

Parameter Frequencydiscrimination

method

Phase discrimination method

1 Method to cancel theunwanted sideband

Using a filter By shifting AF and RF signals to BMby 90

2 Design of 90 shifter atmodulating frequency

Not Applicable Design is critical

3 Possibility of SSB

generation at anyfrequency

Not possible to

generate at anyfrequency.

Possible.

4 Need of up conversion. Needed Not Needed

5 Use of low modulating

frequencies

Not possible possible

6 Need of linear amplifiers Needed Needed

7 Critical points in systemdesign

Filter characteristics,its size and weight,

cutoff frequency

Design of 90 phase shifter formodulating frequency. Symmetry of

balanced modulators




31/44

Demodulation of SSB waves:

The SSB receivers are normally used for professional or commercial communications.

special requirements of SSB receivers are as follows:

1. High reliability2. Excellent suppression of adjacent signals

3. High signal to noise ratio

4. Ability to demodulate SSB

Coherent SSB Demodulation:

To recover the modulating signal from the SSB-SC signal, we require a phase coherent or

synchronous demodulator. The product modulator is a type of coherent SSB demodulator.

The block diagram of the coherent SSB-SC demodulator is as shown in the Figure above. The product modulator is having two inputs. One input is SSB modulated wave S(t) and

another input is a locally generated carrier then the output of product modulator is passed

through low-pass filter having a bandwidth of fmfinally at the output of filter we get the

modulating signal back.

The output of product modulator is given by ------------ (1)But

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

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




32/44

-------- (3)

When v (t) is passed through the filter, it will allow only the first term to pass through

and will reject all other unwanted terms.

Thus at the output of the filter we get the scaled message signal and the coherent SSB

demodulation is achieved.

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

Phase error in coherent detection:

In the coherent detection explained, it is assumed that the ideal operating conditions in

which the locally generated carrier is in perfect synchronization.

inpractice a phase error may arise in the locally generated carrier wave. The detector

output will get modified due to phase error as follows:

-------- (5) In the above expression, the plus sign corresponds to the SSB input signal with only USB

whereas the negative sign corresponds to SSB input with only LSB.

Due to the presence of the Hilbert transform in the output, the detector output willsuffer from the phase distortion.

Such a phase distortion does not have serious effectswith the voice communication as

human ear is relatively insensitive to phase distortion (This error is called Donald

Duck effect)

But in the transmission of music and videoit will have intolerable effects.

The advantages of SSB over DSB-FC signal:

1. Less bandwidth requirement as SSB requires a BW of fm. This will allow more number of

signals to be transmitted in the same frequency range.

2. Lots of power saving. This is due to the transmission of only one sideband component. At

100 % modulation, the percent power saving is 83.33 %.




33/44

3. Reduced interference of noise. This is due to the reduced bandwidth. As the bandwidth

increases, the amount of noise added to the signal will increase.

4. Bulkier filters are replaced by smaller filter.

Disadvantages of SSB:1. The generation and reception of SSB signal is complicated.

2. The SSB transmitter and receiver need to have excellent frequency stability. A slight

change in frequency will hamper the quality of transmitted and received signal. SSB

therefore is not generally used for the transmission of good quality music. It is used for

speech transmission.

3. Selective filtering is to be done to get the original signal back.

Applications of SSB:

SSB transmission is used in the applications where the power saving and low bandwidthrequirements are important. The application areas are

1. Land and air mobile communication

2. Telemetry

3. Military communications

4. Navigation and amateur radio

Many of these applications are point to point communication applications.

Note:

Among the family of AM systems, SSB modulation is optimum with regard to noise

performance as well as bandwidth conservation.

Problems:

3. Calculate the percent power saving for the SSB signal if the AM wave is modulated to a depth

of (a) 100 % and (b) 50 %.

Solution:

In case of SSB, Carrier and one sideband are suppressed. Therefore only one sideband istransmitted. Thus,




34/44

For 100 % modulation (=1)

For 50 % modulation (=0.5)

4. Consider a two stage modulator shown in Figure below. The input signal consists of voice

signal occupying the frequency band 0.3 to 3.4 kHz. The two oscillator frequencies have the

values f1 = 100 kHz and f2= 10 MHz. Specify the following:

a. Sidebands of DSBSC modulated wave appearing at the two product modulator output.

b. Sidebands of SSB modulated wave appearing at the two BPF outputs.

c. The passband and guard bands of the two BPF.

d. Draw the spectrum of the signal at each stage assuming suitable M(f).

Solution:

Let the us assume spectrum m(t) be a rectangular form.

Output of PM1:

The PM1 output consists of two sidebands as follows:

LSB100 kHz - 300 Hz to 100 kHz3.4 kHz

99.7 kHz to 96.6 kHz

USB100 kHz + 300 Hz to 100 kHz + 3.4 kHz

100.3 kHz to 103.4 kHz




35/44

Output of BPF1:

Assume that this BPF passes only the USB.

S1(t) = 100.3 kHz to 103.4 kHz

Output of PM2:Output of PM2 consists of the following two sidebands.

USB10 MHz + 100.3 kHz to 10 MHz + 103.4 kHz

10.1003 MHz to 10.1034 MHz

LSB10 MHz - 100.3 kHz to 10 MHz103.4 kHz

9.8997 MHz to 9.8966 MHz




36/44

Output of BPF2:

Let the BPF2 select the upper sideband from the PM2 output.

BPF2 output10.1003 MHz to 10.1034 MHz.

Passbands of the two BPF:

1. Passband of BPF1: 100.3 kHz to 103.4 kHz.

2. Passband of BPF2: 10.1003 MHz to .10.1034 MHz

Guardband of BPF1:

The guard band of BPF 1 extends from the lowest frequency of the USB to the highest frequency

of LSB.

Guardband of BPF 1 = 97 kHz to 100.3 kHz.

Guardband of BPF2:

Similarly the guardband of BPF2 extends from 9.8997 MHz to 10.1003 MHz

5. Find the Hilbert transform of the rectangular pulse below.

Solution:

Hilbert transform of x(t) is

But the time domain description of x(t) i.e, rectangular pulse is,

Thus,




37/44

6. Let Su(t) represents the SSB signal obtained by transmitting only upper sideband and represents its Hilbert transform. Show that,

And

Solution:Equation for SSB modulation with Upper sideband Su(t) is given as below,

------- (1)Multiplying Equation (1) by we get,

----- (2)




38/44


39/44

------- (5)

Now Multiplying Equation (3) by

----- (6)Subtracting equation (6) from (5) we get,

7. Consider a message signal m(t) with the spectrum shown in Figure below. The message

bandwidth W = 1 kHz. This signal is applied to a product modulator, together with a carrier

wave Accos(2fct), producing the DSB-SC 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) the carrier frequency fc= 1.25 kHz and

(b) the carrier frequency fc= 0.75 kHz.

(c) What is the lowest carrier frequency for which each component of the modulated signal s(t) isuniquely determined by m(t)?

Solution:

a. For carrier frequency fc = 1.25 kHz the spectrum of the message signal m(t), the product

modulator s(t) and coherent detector v(t) is as given below




40/44

b. For carrier frequency fc = 0.75 kHz the spectrum of the message signal m(t), the product

modulator s(t) and coherent detector v(t) is as given below

c. To avoid sideband-overlap, the carrier frequency fcmust be greater than or equal to 1kHz. The

lowest carrier frequency is thus 1 kHz for each side band of the modulated wave s(t) to be

uniquely determined by m(t).

Question Bank

1. What is Hilbert transform? Obtain the Hilbert transform of the signal .December 2012 (04 M) , July 2014 (06 M)

2. With a neat block diagram, explain the generation of SSB wave using phase

discrimination method. December 2012 (08 M)

3. Consider a two stage modulator shown in Figure below. The input signal consists of

voice signal occupying the frequency band 0.3 to 3.4 kHz. The two oscillator frequencies

have the values f1 = 100 kHz and f2= 10 MHz. Specify the following:




41/44

a. Sidebands of DSBSC modulated wave appearing at the two product modulator

output.

b. Sidebands of SSB modulated wave appearing at the two BPF outputs.

c. The passband and guard bands of the two BPF.

4.

December 2012 (08 M)

5. Derive the expression for SSB modulated wave for which upper sideband is retained.


6. Derive the expression for SSB modulated wave for which lower sideband is retained in

time domain. July 2014 (08 M)

7. Figure below shows the block diagram of a two stage SSB modulator. The input signal

m(t) consists of a voice signal occupying the frequency band 0.3 to 3.0 kHz. The two

carrier frequencies are f1= 100 kHz and f2= 10 MHz.

8.

Evaluate the following:a. The sidebands of DSB-SC modulated waves at the output of the product

modulators.

b. The sidebands of the SSB modulators at the output of band pass filters.

c. The passbands and the guardbands of the two bandpass filters.

d. Sketch the spectrum of the signal at each stage. [Assume suitable m(f)]


9. With neat block diagram, write a note on quadrature carrier multiplexing. June 2012 (08

M)

10.

With frequency spectrum and equations, generate SSBSC wave by using (USB) phaseshift method. June 2012 (06 M)

11.Explain the operation of quadrature carrier multiplexing scheme with transmitter and

receiver diagrams. June 2013 (08 Marks)

12.With a block diagram approach, explain the phase discrimination method for generating

SSB modulated wave. June 2013 (08 Marks), July 2014 (06 Marks)




42/44

13.Explain the demodulation of SSB waves with a block diagram and mathematical

expressions. June 2013 (04 Marks)

14.What is the significance of single side band modulation? Give the frequency domain

description of the same. January 2014 (04 Marks)

15.Explain with block diagram a frequency discrimination method (two stage) for generatingSSB modulated wave. January 2014 (08 Marks)

16.Consider the message signal m(t) containing components at 100,200 and 400 Hz. This

signal is applied to an SSB modulator together with a carrier at 100 kHz, with only the

upper sideband retained. In the coherent detector used to recover m(t), the local oscillator

supplies a sinewave of frequency 100.02 kHz. Determine the frequency components of

the detector output. January 2014 (08 Marks)




43/44




44/44


analog communication unit3 vtu

Documents