Quadrature and Single Sideband AM

ELG3175 Introduction to Communication Systems

Lecture 9

Quadrature Amplitude Modulation (QAM)

•  Spectral efficiency refers to the amount of information that we can transmit per unit bandwidth

•  DSB-SC transmita a signal with bandwidth Bm on a bandpass bandwidth of 2Bm.

•  QAM transmits two signals of bandwidth Bm on a bandpass bandwidth of 2Bm.

•  Therefore it uses bandwidth twice as efficiently.

QAM 2

•  A QAM signal is in the form:

•  Where –  Ac is the carrier amplitude –  m1(t) and m2(t) are independent information signals,

both with bandwidth Bm. –  fc is the carrier frequency and fc >> Bm.

tftmAtftmAts ccccQAM ππ 2sin)(2cos)()( 21 +=

The spectrum of a QAM signal

The Fourier Transform (i.e. spectrum) of a QAM signal is:

)(2

)(2

)(2

)(2

)( 2211 cc

cc

cc

cc

QAM ffMjA

ffMjA

ffMA

ffMA

fS +−−+++−=

Demodulation of m1(t)

•  Let us multiply sQAM(t) by Arcos2πfct, which gives us:

cff

crc

crcrc

ccrccrccQAMr

tftmAAtftmAAtmAAtftftmAAtftmAAtftsA

2at centred signal bandpass

21

signal baseband

1

22

1

4sin)(2

4cos)(2

)(2

2cos2sin)(2cos)(2cos)(

=

++=

+=

ππ

ππππ

Recall that cos2(A)= 0.5(1+cos(2A)) cosAsinA = 0.5sin2A

Demodulation of m2(t)

•  Similarly, if we multiply sQAM(t) by Arsin2πfct, we get:

cff

crc

crcrc

ccrccrccQAMr

tftmAAtftmAAtmAAtftftmAAtftmAAtftsA

2at centred signal bandpass

12

signal baseband

2

12

2

4sin)(2

4cos)(2

)(2

2sin2cos)(2sin)(2sin)(

=

+−=

+=

ππ

ππππ

QAM System

m1(t)

m2(t)

×

×

Accos(2πfct) HT + Channel

Km1(t)

Km2(t)

×

×

Arcos(2πfct) HT

LPF

LPF

Advantages and disadvantages

•  Can multiplex twice as much information on the same bandwidth as DSB-SC

•  Even more sensitive to carrier and phase errors compared to DSB-SC. –  Crosstalk.

Can we increase the bandwidth efficiency of DSB-SC?

•  We saw that DSB-SC uses both upper and lower sidebands.

•  QAM managed to double the bandwidth efficiency by using carriers in quadrature.

•  Can we remove one of the sidebands and save half the bandwidth used by DSB-SC?

Lecture 6

DSB-SC spectrum

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)

Motivation

•  From the previous slide, we see that SDSB-SC(f) = (Ac/4)M+(f-fc) + (Ac/4)M-(f-fc) + (Ac/4)M+( f+fc) + (Ac/4)M-(f+fc).

•  The spectrum of a DSB-SC signal has two “copies” of the positive pre-envelope of m(t) and two “copies” of its negative pre-envelope.

•  Actually, we would only need one of each to perfectly reconstruct m(t).

•  By eliminating one of the sidebands, we obtain single sideband (SSB) AM.

Upper Sideband (USB)

•  The upper sideband of a DSB-SC signal is one that has spectrum SUSB(f):

•  Compared to a DSB-SC signal that occupies a bandwidth of 2Bm, the spectrum of a USB signal occupies the frequency range fc < |f| < fc + Bm, therefore it has half the bandwidth of a DSB-SC signal.

⎩⎨⎧ >

= −

otherwise,0|| ),(

)( cSCDSBUSB

fffSfS

Spectrum of USB signal

f

f

M(f)

Bm-Bm

SUSB(f)

mo

(Ac/2)mo

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

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

-fc-Bm -fc fc fc+Bm

USB Modulation using frequency discrimination •  We can produce USB modulation by two methods:

Frequency discrimination or phase discrimination. •  For frequency discrimination, we use a high pass filter on

a DSB-SC signal.

m(t) ×

Accos(2πfct)

HUSB(f) sUSB(t)m(t) ×

Accos(2πfct)

HUSB(f) sUSB(t)

⎩⎨⎧

≤

≥=

c

cUSB ff

fffH

||0||1

)(

USB Modulation by phase discrimination

•  Let us consider the USB signal’s spectrum. •  SUSB(f) = (Ac/4)M+(f-fc) + (Ac/4)M-(f+fc). •  Taking the inverse Fourier Transform we get:

)2sin)(2cos)(()2sin2))(cos()((

)2sin2))(cos()(()()()(

2

4

4

24

24

tftmtftmtfjtftjmtmtfjtftjmtm

etmetmts

chcA

cchA

cchA

tfjAtfjAUSB

c

c

c

cccc

ππ

ππ

ππ

ππ

−=

−−

+++=

+= −−+

tftAmtftAmts chcUSB ππ 2sin)(2cos)()( −=

USB Phase Discriminator Modulator

m(t)

×

Trans.Hilbert ×

Acos2πfct

HT

+

-

+ sUSB(t)

Asin2πfct

mh(t)

Lower Sideband Modulation (LSB)

•  The lower sideband of a DSB-SC signal is the part of the spectrum where |f|<fc.

•  Therefore the spectrum of an LSB signal is SLSB(f) which is given by:

⎩⎨⎧ <

= −

otherwise,0|| ),(

)( cSCDSBLSB

fffSfS

LSB Spectrum

f

f

M(f)

Bm-Bm

SLSB(f)

-fc -fc+Bm fc-Bm fc

mo

(Ac/2)mo

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

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

LSB Modulation by frequency discrimination

•  We input the DSB-SC signal to a lowpass filter with frequency response :

⎩⎨⎧

>

≤=

c

cLSB ff

fffH

||0||1

)(

LSB modulation by phase discrimination

•  We can show that sLSB(t) is given by:

sLSB (t) = A m(t) cos2! fct + A mh (t) sin2! fct

Examples

•  The message is m(t) = cos(2πfmt). Find the USB and LSB signals for a carrier amplitude of A and carrier frequency fc >> fm.

•  Solution (phase discriminator)

•  sUSB(t) = Acos(2πfmt)cos(2πfct)-Asin(2πfmt)sin(2πfct) = (A/2)cos(2π(fc-fm)t) + (A/2)cos(2π(fc+fm)t)

- (A/2)cos(2π(fc-fm)t) + (A/2)cos(2π(fc+fm)t) = Acos(2π(fc+fm)t).

•  Similarly we can show that sLSB(t) = Acos(2π(fc-fm)t).

Examples

•  The message is m(t) = cos(2πfmt).

•  Solution (frequency discrimination)

•  sDSB-SC(t) = Accos(2πfmt)cos(2πfct). •  SDSB-SC(f) = (Ac/4)δ(f-fc-fm)+(Ac/4)δ(f+fc+fm)+(Ac/4)δ(f-

fc+fm) +(Ac/4)δ(f+fc-fm). •  SUSB(f) = (Ac/4)δ(f-fc-fm)+(Ac/4)δ(f+fc+fm) and •  sUSB(t) = (Ac/2)cos(2π(fc+fm)t) = Acos(2π(fc+fm)t)

-fc-fm –fc+fm fc-fm fc+fm

(Ac/4) (Ac/4) (Ac/4) (Ac/4)

similarly SLSB(f) = (Ac/4)δ(f-fc+fm)+(Ac/4)δ(f+fc-fm) and sLSB(t) = (Ac/2)cos(2π(fc+fm)t) = Acos(2π(fc+fm)t)

Demodulation of SSB

•  Same as DSB-SC

Questions

•  1. Why don’t we hear much about SSB these days?

•  2. Our examples for QAM used 2 streams (m1(t) and m2(t)). How could you use QAM with single source stream m(t)?

Lecture 6