COMMUNICATION SYSTEMS

EE-351
 DEMODULATION OF AM SIGNALS
(CASE STUDY)
The AM signal can be demodulated coherently by a locally generated carrier. E.g.

[[A + m(t )]cos wct ]cos wct No benefit of sending carrier on the channel

There are two well known methods of demodulation of AM signals:

1) Rectifier detection 2) Envelope detection

Rectifier detector:

AM signal is applied to a diode and resistor circuit, the negative part of the the
AM wave will be suppressed.
The output across the resistor is the half wave rectified version of the AM signal
means multiplying AM with w(t).

2
 RECTIFIER DETECTOR (CASE
STUDY)

The rectified output VR

{[
vR = A + m(t ) cos w t w(t )
c
]}
1 2  1 1 
= [A + m(t )]cos w t  +  cos w t − cos 3w t + cos 5w t − ... 
c 2 π  c 3 c 5 c 

1
= [A + m(t )] + otherTerms
π 3
RECTIFIER DETECTOR (CASE
STUDY)

4
 ENVELOPE DETECTOR (CASE
STUDY – HOME TASK)
In an envelope detector, the output follows the envelope of the modulated signal.
The following circuit act as an envelope detector:

• During the positive cycle of the input signal, the diode conducts and the
capacitor C charges up to the peak voltage of the input signal.
•When input signal falls below this peak value, the diode is cut off. (because the
diode voltage which is nearly the peak voltage is greater than the input signal
voltage causing the diode to open ).
•At this stage the capacitor discharge at the slew rate (with a time constant RC)
• during the next positive cycle the process repeats. 5
 ENVELOPE DETECTOR (CASE
STUDY)

During each positive cycle the capacitor charges up to the peak voltage of the
input signal and then decays slowly until the next positive cycle.
This behavior of the capacitor makes output voltage Vc(t) follow the envelope of
the input signal.
Capacitor discharges during each positive peaks causes a ripple signal of 6
frequency wc at the output
 ENVELOPE DETECTOR (CASE
STUDY)
The ripple can be reduced by increasing the time constant RC so the capacitor
discharges very little between positive peaks of the input signals

Making RC too large, makes capacitor voltage impossible to follow the envelope.

Conditions:

RC should be large compared to 1/wc, but should be small compared to 1 2πB

Where B is the highest frequency in m(t)

Also requires a condition which is necessary for well defined envelope.

7
 ENVELOPE DETECTOR (CASE
STUDY)
The envelope detector output is with a ripple of frequency wc

The DC term A can be blocked by a capacitor or a simple RC high pass filter, and
the ripple may be reduced further by another low-pass RC filter.

8
 QUADRATURE AMPLITUDE
MODULATION
The DSB signals of AM require twice the bandwidth required for the baseband
signal!

Idea: Try to send two signals m1(t) and m2(t) simultaneously by modulating them
with two carrier signals of same frequency but shifted in phase by –π/2

The combined signal is m1 (t ) + m2 (t ) = m1 (t ) cos wc t + m2 (t ) sin wc t

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 QUADRATURE AMPLITUDE MODULATION
(CONT…)

Both modulated signals occupy the same band

• At the receiver the two baseband signals can be separated by using a second
carrier that is shifted in phase by –π/2

• The first signal m1(t) can be detected by a multiplication with 2cos(ωct) followed
by a low-pass filter

The second signal x2(t) can be detected accordingly by a multiplication with

sin(ωct) followed by a low-pass filter

10
 QUADRATURE AMPLITUDE
MODULATION (CONT…)

• Thus, two baseband signals, each of bandwidth B, can be simultaneously

transmitted over a channel with bandwidth 2B

• This principle is called quadrature amplitude modulation (QAM), because

the carrier frequencies are in phase quadrature.

11
 AMPLITUDE MODULATION (SINGLE SIDEB AND
SSB)

• The DSB spectrum has two sidebands: USB and LSB

• Both USB and LSB contain complete information of the baseband
signal.
• A scheme in which only one sideband is transmitted is known as
single-sideband ( SSB) transmission.
• In SSB transmission the required bandwidth is half compared to DSB
signal.
• An SSB signal can be coherently (synchronously) demodulated. E.g.

For example multiplying USB signal by cos wct shifts its spectrum to
the left and right by wc
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 SINGLE SIDEBAND (SSB)
• Purpose : to reduce the bandwidth requirement of AM by one-half. This is
achieved by transmitting only the upper sideband or the lower sideband of
the DSB AM signal.
 SINGLE SIDEBAND SSB (CONT..)

Low pass filtering will give the required baseband signal at the
receiver.
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 SINGLE SIDEBAND SSB (CONT..)

Time domain representation of SSB signals:

ϕ SSB (t ) = m(t ) cos wc t ± mh (t ) sin wc t

mh (t ) Hilbert Transform of m(t) π
and delays the phase of each component by 2

Where minus sign applies to USB and the plus sign applies to LSB

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 EXAMPLE 4.7

Tone Modulation:

Find ϕ SSB (t ) for a simple case of tone modulation, that is, when the
modulating signal is a sinusoid m(t ) = cos wmt
Solution:
ϕ SSB (t ) = m(t ) cos wc t ± mh (t ) sin wc t

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 EXAMPLE 4.7

Hence

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EXAMPLE 4.7 P-174

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 GENERATION OF SSB SIGNALS

Two methods are generally used to generate SSB signals.

1) Sharp cutoff filters
2) Phase shifting networks

Selective Filtering Method:

• In this method the DSB-SC signal is passed through a sharp cutoff

filter to eliminate the undesired sideband.
• To obtain USB , the filter should pass all components above wc,
attenuated and completely suppress all components below wc
• Such an operation requires an ideal filter that is practically not
possible.
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 GENERATION OF SSB SIGNALS

• This method of generating SSB signal can be used when there is

some separation between the passband and stopband.
• In some application this can be achieved e.g. voice signals

Voice signals spectrum shows little power content at the

origin. Thus filtering the unwanted sideband is relatively
easy.
Tests have shown that frequency components
below 300Hz are not important.

600Hz transition region around the cutoff

frequency wc , makes filtering easy and
minimize the channel interference

20
 GENERATION OF SSB
SIGNALS(CONT…)
Phase-Shift Method:

The basis of this method is the following equation

ϕ SSB (t ) = m(t ) cos wc t ± mh (t ) sin wc t

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GENERATION OF SSB
SIGNALS(CONT…)

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 SSB DEMODULATION
Synchronous, SSB-SC demodulation
ϕ SSB ( t ) cos( ω c t ) = [m( t ) cos( ω c t )  jm h ( t ) sin( ω c t )]cos(n(ω c t ) = 1
2
[m(t )(1 + cos(ωc t ))  jm h (t ) sin(2ωc t )]

A lowpass filter can be used to get 1 m( t ).

2

SSB+C, envelop detection

ϕ SSB +C ( t ) = A cos( ω c t ) + [m( t ) cos( ω c t )  m h ( t ) sin( ω c t )]
An envelope detector can be used to demodulate such SSB signals .
What is the envelope of ϕ SSB +C ( t ) = ( A + m( t )) cos( ω c t )) + m h ( t ) sin( ω c t ) = E( t ) cos( ω c t + θ) ?

( )
1

{Recall Acos( α ) + Bsin( α ) = A 2 + B 2 2

cos( α + θ), θ = −tan -1( B
A
))
E(t) = (( A + m( t )) 2 + m h2 ( t )) 2 = (( A 2 + m 2 ( t )) + m h2 ( t ) + 2Am( t )) 2
1 1

= A 1 + A 2 +
2m( t ) 
2
m2 ( t ) mh ( t )
A2
+ A 

≈ A + m( t ) for A >> m(t) , A >> m h (t) .
The efficiency of this scheme is very low since A has to be large.
 SSB VS. AM
• Since the carrier is not transmitted, there is a reduction by 67% of the transmitted power (-
4.7dBm). --In AM @100% modulation: 2/3 of the power is comprised of the carrier; with the remaining
(1/3) power in both sidebands.
• Because in SSB, only one sideband is transmitted, there is a further reduction by 50% in transmitted
power
• Finally, because only one sideband is received, the receiver's needed bandwidth is reduced by one half--
thus effectively reducing the required power by the transmitter another 50%
• (-4.7dBm (+) -3dBm (+) -3dBm = -10.7dBm).
• Relative expensive receiver
 VESTIGIAL SIDEBAND (VSB)
• VSB is a compromise between DSB and SSB. To produce SSB signal from DSB
signal ideal filters should be used to split the spectrum in the middle so that
the bandwidth of bandpass signal is reduced by one half. In VSB system one
sideband and a vestige of another sideband are transmitted together. The
resulting signal has a bandwidth > the bandwidth of the modulating
(baseband) signal but < the DSB signal bandwidth.
DSB

−ωc ωc ω→
0
Φ SSB (ω) SSB (Upper sideband)

−ωc ωc ω→
0
Φ VSB (ω) VSB Spectrum

−ωc ωc ω→
Filtering scheme for the generation of VSB modulated wave.
 VSB TRANSCEIVER

m(t) ϕ VSB (ω) e(t)

ϕ VSB (ω) M(ω)
H i ( ω) LPF
Ho(ω)

2cos( ω c t )
2cos( ω c t )
Transmitter Receiver

M( ω) is bandlimite d to 2πB rad/sec

ϕ VSB ( ω) = [M( ω − ω c ) + M( ω + ω c )]H i ( ω)
E( ω) = [ Φ VSB ( ω − ω c ) + Φ VSB ( ω + ω c )]
= [H i ( ω − ω c )M( ω − 2ω c ) + H i ( ω + ω c )M( ω) + H i ( ω − ω c )M( ω) + H i ( ω + ω c )M( ω + 2ω c )]
High freq. term High freq. term
∴ M( ω) = E( ω)H o ( ω) = [Hi ( ω + ω c ) + Hi ( ω − ω c )]M( ω)H o ( ω)
+ [Hi (ω − ω c )M(ω − 2ω c ) + Hi (ω + ω c )M(ω + 2ω c )]H o (ω)
Lowpass filter removes this.
Thus we should have [Hi (ω + ωc ) + Hi (ω − ωc )]Ho (ω) = 1 for ω ≤ 2πB
1
OR H o (ω) =
Hi ( ω + ω c ) + Hi ( ω − ω c )
 OTHER FACTS ABOUT VSB
• Envelope detection of VSB+C
• Analog TV:
• DSB, SSB and VSB
• DSB bandwidth too high
• SSB: baseband has low frequency
component, receiver cost
• Relax the filter and baseband
requirement with modest increase
in bandwidth
( A ) I D E A L I Z E D M AG N I T U D E S P E C T RU M
OF A TRANSMITTED TV SIGNAL.
( B ) M AG N I T U D E R E S P O N S E O F V S B
S H A P I N G F I LT E R I N T H E R E C E I V E R .
COMPARISON
 RADIO FREQUENCY BANDS

Classification Band Initials Frequency Range Characteristics

Extremely low ELF < 300 Hz

Ground wave
Infra low ILF 300 Hz - 3 kHz

Very low VLF 3 kHz - 30 kHz

Low LF 30 kHz - 300 kHz

Medium MF 300 kHz - 3 MHz Ground/Sky wave

High HF 3 MHz - 30 MHz Sky wave

Very high VHF 30 MHz - 300 MHz

Ultra high UHF 300 MHz - 3 GHz

Space wave
Super high SHF 3 GHz - 30 GHz

Extremely high EHF 30 GHz - 300 GHz

Tremendously high THF 300 GHz - 3000 GHz