Introduction to

Amplitude Modulation: DSB-SC

ELG3175 Introduction to Communication Systems

Lectures 7 & 8

Introduction to modulation

•  A message signal, m(t), is to be transmitted. •  Let us assume that this is a baseband signal with

bandwidth Bm. •  In other words, M(f) = 0 for |f| > Bm. •  To transmit the message, we modulate some parameter

of a carrier wave as a function of m(t). •  The carrier wave is a sinusoidal signal:

•  where Ac is the carrier amplitude, fc is the carrier frequency and !c is the carrier phase.

•  To simplify expressions, we will assume that !c = 0.

)2cos()( ccc tfAtc !" +=

Lectures 7 & 8

Why modulate?

•  We use modulation to transmit m(t) for the following reasons: –  The spectrum of m(t) may fall in a range of

frequencies that are not suitable for the channel. By modulating we can move the spectrum of the message signal to a range of frequencies that are appropriate.

–  Antenna lengths are proportional to the wavelength of the signal. Baseband signals require antenna lengths that are not practical. Modulating to a higher frequency reduces the length of the antenna needed to transmit or receive.

Lectures 7 & 8

Why modulate? (2)

•  To separate different signals that are to be transmitted. For example, by using different carrier frequencies, we can multiplex signals so that they can be separated at the receiver. This is called frequency division multiplexing (FDM)

Lectures 7 & 8

Double Sideband Suppressed Carrier (DSB-SC)

•  In amplitude modulation, (AM), the message, m(t) is used to modulate the instantaneous amplitude of the carrier.

•  In double sideband suppressed carrier (DSB-SC), only the information bearing modulated carrier is transmitted by multiplying the carrier with the message signal

•  Where we assume that fc >> Bm.

)2cos()()()()(

tftmAtctmts

cc

SCDSB

!=

="

Lectures 7 & 8

Example of a DSB-SC signal

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5-2

-1

0

1

2

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5-4

-2

0

2

4

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5-5

0

5

m(t)

c(t)

sDSB-SC(t)

Lectures 7 & 8

Fourier transform of a DSB-SC signal

•  The spectrum of a DSB-SC signal is given by:

)(2

)(2

)( cc

cc

SCDSB ffMAffMAfS ++!=!

Lectures 7 & 8

Spectrum of a DSB-SC signal

f

f

M(f)

Bm-Bm

SDSB-SC(f)

-fc-Bm -fc -fc+Bm fc-Bm fc fc+Bm

mo

(Ac/2)mo

(1/2)M+(f)(1/2)M-(f)

(Ac/4)M+(f-fc)(Ac/4)M-(f-fc)(Ac/4)M+(f+fc)(Ac/4)M-(f+fc)

Lectures 7 & 8

The two bands

•  M(f) = !M+(f)+ !M-(f), therefore m(t) = !m+(t) + !m-(t).

•  In DSB-SC, the upper sideband is the one where|f| > fc. Therefore the spectrum of the upper sideband is SUSB(f) = (Ac/4)M+(f-fc) + (Ac/4)M-(f+fc), which means that sUSB(t) = (Ac/4)m+(t)ej2"fct + (Ac/4)m-(t)e-j2"fct.

•  The lower sideband is the one |f| < fc. •  Therefore its spectrum is SLSB(f) = (Ac/4)M-(f-fc) + (Ac/

4)M+(f+fc). •  Therefore sLSB(t) = (Ac/4)m-(t)ej2"fct + (Ac/4)m+(t)e-j2"fct.

Lectures 7 & 8

Energy or power of a DSB-SC signal

•  We saw that if m(t) is an energy signal with energy Em, then Acm(t)cos2"fct is also an energy signal with energy (Ac

2/2)Em. •  Also, if m(t) is a power signal with power Pm, then

Acm(t)cos2"fct is also a power signal with power (Ac

2/2)Pm.

Lectures 7 & 8

Example 1

•  The modulating signal is: m(t) = a cos(2" fmt) •  Determine the DSB-SC signal, and its upper and lower

sidebands.

Lectures 7 & 8 Lectures 7 & 8

Example 2

•  The signal m(t) = sinc(t) is to be transmitted using DSB-SC. The carrier amplitude is 5V and its frequency is 25 Hz. –  What is the spectrum of the DSB-SC signal? –  What is the energy of sDSB-SC(t)?

Lectures 7 & 8

Solution

•  The DSB-SC signal is sDSB-SC(t) = 5sinc(t)cos(2"25t). •  Its spectrum is SDSB-SC(f) = (5/2)!(f-25) + (5/2)!(f+25). •  The signal sinc(t) is an energy signal with energy E=1,

therefore sDSB-SC(t) is also an energy signal with energy E = 25/2.

Lectures 7 & 8

The DSB-SC signal in our example

-2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5-5

-4

-3

-2

-1

0

1

2

3

4

5

Lectures 7 & 8

The spectrum of the DSB-SC signal in our example

-25 25 f

5/2

1 Hz

Lectures 7 & 8

Coherent demodulation of DSB-SC signals

•  The receiver must recover m(t) from the received signal.

•  If we ignore the effects of transmission (fading, propagation loss, interference, noise etc) then the received signal is sDSB-SC(t).

•  We wish to perform some operation on sDSB-SC(t) so as to obtain km(t), where k is a constant.

•  In coherent detection, we make use of the identity cos2(2"fct) = ! + ! cos(4"fct).

Lectures 7 & 8

Block diagram of demodulator

sDSB-SC(t)

Arcos(2!fct)

!x(t) Lowpass

Filter/FiltrePasse Bas

1

-Bm Bm f

(AcAr/2)m(t)sDSB-SC(t)

Arcos(2!fct)

!x(t) Lowpass

Filter/FiltrePasse Bas

1

-Bm Bm f

(AcAr/2)m(t)

Lectures 7 & 8

Coherent Detection

!!! "!!! #$!"!#$LPFby rejected

signal bandpass

2

LPF by the passedsignal baseband

2

2

)4cos()()()2(cos)(

)()()(

tftmtmtftmAA

tctstx

cAAAA

crc

rSCDSB

rcrc !

!

+=

=

= "

Lectures 7 & 8

Disadvantages of DSB-SC

•  The receiver must generate a replica of the carrier in order to demodulate a DSB-SC signal.

•  Any phase and/or frequency error will result in a distorted estimate of the message signal.

•  It is difficult to generate a perfect replica of the transmitted carrier.

•  A simple modification to the technique results in a less efficient transmission but simplifies the detection process greatly.

•  Conventional AM uses noncoherent demodulation. –  Detection is possible even with frequency and phase

errors.

Lectures 7 & 8

Conventional AM

•  Consider a message signal m(t) where M(f)=0 for f=0,

•  where Ac is the carrier amplitude, ka is the amplitude sensitivity, and fc is the carrier frequency.

•  Also, fc >> Bm where Bm is the bandwidth of m(t).

tftmkAts cacAM !2cos)](1[)( +=

Lectures 7 & 8

Modulation index

•  Let us assume that –mp " m(t) " mp. •  For conventional AM, |kam(t)| < 1, or -1 < kam(t) < 1. •  Therefore, 0< kamp < 1, or 0 < ka < 1/mp. •  The modulation index is µa = kamp. •  For conventional AM, 0 < µa < 1. •  Therefore for conventional AM, [1 + kam(t)] > 0.

Lectures 7 & 8

Example 1

•  We wish to transmit m(t) = cos2"t + 2cos2"(1.4)t using conventional AM. The carrier is c(t) = 5cos2"10t. –  Find the value of ka so that the modulation index is

0.5. •  Solution •  We can show that -3 " m(t) " 3, therefore mp = 3.

therefore for µa = 0.5, ka = 1/6. •  The resulting AM signal is

–  sAM(t) = 5[1+(1/6)m(t)]cos2"10t

Lectures 7 & 8

-2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5-3

-2

-1

0

1

2

3

-2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5-10

-5

0

5

10

m(t)

sAM(t)

Lectures 7 & 8

The envelope of an AM signal

•  The instantaneous amplitude of sAM(t) is Ac[1+kam(t)]. •  This instantaneous amplitude is called the signal’s

envelope. •  The message signal m(t) can be extracted directly from

the envelope of sAM(t).

Lectures 7 & 8

Overmodulation

•  If µa > 1, we say that sAM(t) is overmodulated. •  An overmodulated signal cannot be detected using the

noncoherent method that is used for conventional AM. •  Take our previous example with ka = 0.8. •  In this case, µa = kamp = 0.8#3 = 2.4. •  Therefore, sometimes kam(t) < -1 leading to Ac[1+kam

(t)] < 0.

Lectures 7 & 8

-2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5-2

-1

0

1

2

3

4

-2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5-20

-10

0

10

20

1+0.8m(t)

5[1+0.8m(t)]cos2!20t

Lectures 7 & 8

Spectrum of Conventional AM signals

•  We can express the AM signal as :

•  Its Fourier transform is SAM(f) = F{sAM(t)} which is given by:

tftmkAtfAts cacccAM !! 2cos)(2cos)( +=

)(2

)(2

)(2

)(2

)(

cac

cac

cc

cc

AM

ffMkAffMkA

ffAffAfS

++!

+++!= ""

Lectures 7 & 8

Spectrum of AM signal

•  Assuming that m(t) has no DC component, then M(f-fc) has no spectral component at f = fc and M(f+fc) has no spectral component at f = -fc.

f

f

M(f)

Bm-Bm

SAM(f)

-fc-Bm -fc -fc+Bm fc-Bm fc fc+Bm

mo

(Acka/2)mo

Ac/2 Ac/2

Lectures 7 & 8

Example 2

•  Find the spectrum of sAM(t) = 5[1+(1/6)m(t)]cos2"10t where m(t) = cos2"t + 2cos2"(1.4)t

•  SOLUTION •  The signal m(t) has spectrum M(f) = !#(f-1) + !#(f

+1) + #(f-1.4) + #(f+1.4). Therefore the spectrum of sAM(t) is:

)6.8(125)4.11(

125)9(

245)11(

245)6.8(

125

)4.11(125)9(

245)11(

245)10(

25)10(

25

)10(125)10(

125)10(

25)10(

25)(

++++++++!+

!+!+!+++!=

++!+++!=

fffff

fffff

fMfMfffS AM

"""""

"""""

""

Lectures 7 & 8

SAM(f)

f

Lectures 7 & 8

Power of a conventional AM signal

•  Let’s assume that m(t) is a power signal. •  If m(t) has no DC component, then we can find that:

•  where Ac2/2 is the power of the carrier and Ac

2ka2Pm/2 is

the power of the component that carries the message.

22

222macc

sPkAAP +=

Lectures 7 & 8

Efficiency of a conventional AM signal

•  The efficiency of a modulation scheme is the ratio of the power dedicated to the transmission of message to the total power of the transmission.

•  Therefore the efficiency of conventional AM is:

( )mama

maccmac

PkPk

PkAAPkA

22

22222

1222

+=

!!"

#$$%

&+='

Lectures 7 & 8

Noncoherent detection

•  We know that the envelope of sAM(t) is Ac[1+kam(t)]. •  An envelope detector is a circuit that outputs the

envelope of a bandpass signal. •  A half-wave rectifier with a capacitor filter is a simple

envelope detector. •  If fc >> Bm, the output of a halfwave rectifier

ressembles a sampled version of the envelope.

sAM(t)! Halfwave !rectifier! x(t)!

Lectures 7 & 8

-2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5-3

-2

-1

0

1

2

3

-2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5-10

-5

0

5

10

m(t)

sAM(t)

Lectures 7 & 8

-2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5-1

0

1

2

3

4

5

6

7

8

x(t)

Lectures 7 & 8

Simple envelope detector

•  Applying a low pass filter at the output of the halfwave rectifier should give us a signal that ressembles Ac[1+kam(t)].

•  Placing a capacitor at the output of the rectifier provides a simple solution.

•  The power of vr(t) is inversely proportional to fc.

sAM(t)+

-

+Ac[1+kam(t)] + vr(t)

-

Lectures 7 & 8

Output of envelope detector compared to actual envelope.

-2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5-1

0

1

2

3

4

5

6

7

8

Lectures 7 & 8

Obtaining m(t) from the envelope

•  If we neglect vr(t), the output of the envelope detector is Ac + Ackam(t).

•  Passing this through a device that blocks DC components, such as a transformer, we have Ackam(t) at the output as long as m(t) has no DC component.

sAM(t)détecteurd’enveloppe Ackam(t)

Envelope !detector!

Lectures 7 & 8