ELG3175 Introduction to
Communication Systems
Quadrature and
Single Sideband AM
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:
sQAM (t )  Ac m1 (t ) cos 2f c t  Ac m2 (t ) sin 2f c t
• 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.
The spectrum of a QAM signal
• Le spectre d’un signal QAM est :
S QAM ( f ) 
Ac
A
A
A
M1 ( f  f c )  c M1 ( f  f c )  c M 2 ( f  f c )  c M 2 ( f  f c )
2
2
2j
2j
Demodulation of m1(t)
• Let us multiply sQAM(t) by Arcos2fct, which gives us:
Ar sQAM (t ) cos 2f ct 
Ac Ar m1 (t ) cos2 2f c t  Ac Ar m2 (t ) sin 2f ct cos 2f c t
Ac Ar
AA
AA

m1 (t )  c r m1 (t ) cos 4f c t  c r m2 (t ) sin 4f ct
2 22
basebandsignal
(cosAsinA = 0.5sin2A).
bandpasssignal centredat f  2 f c
Demodulation of m2(t)
• Similarly, if we multiply sQAM(t) by Arsin2fct, we get:
Ar sQAM (t ) sin 2f c t 
Ac Ar m2 (t ) sin 2 2f c t  Ac Ar m1 (t ) cos 2f c t sin 2f c t
Ac Ar
AA
AA

m2 (t )  c r m2 (t ) cos 4f c t  c r m1 (t ) sin 4f c t
2 22
baseband signal
bandpass signal centred at f  2 f c
QAM System
m1(t)
LPF
Arcos(2fct)
Accos(2fct)
HT
m2(t)
×
×
Km1(t)
×
+
Channel
HT
×
LPF
Km2(t)
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.
DSB-SC spectrum
M(f)
(1/2)M-(f)
-Bm
(Ac/4)M-(f+fc)
Bm
(Ac/4)M+(f+fc)
(Ac/2)mo
-fc-Bm
-fc
-fc+Bm
(1/2)M+(f)
mo
f
(Ac/4)M-(f-fc)
SDSB-SC(f)
fc-Bm
(Ac/4)M+(f-fc)
fc
fc+Bm
f
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):
S DSBSC ( f ),
SUSB ( f )  
0,

| f | f c
otherwise
• 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.
Spectrum of USB signal
M(f)
(1/2)M-(f)
(1/2)M+(f)
mo
-Bm
Bm
f
(Ac/4)M-(f+fc)
(Ac/2)mo
-fc-Bm
-fc
SUSB(f)
(Ac/4)M+(f-fc)
fc
fc+Bm
f
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.
1 | f | f c
H USB ( f )  
0 | f | f c
m(t)
×
Accos(2fct)
HUSB(f)
sUSB(t)
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:
sUSB (t ) 


m (t )e j 2f ct  A4c m (t )e  j 2f ct
(m(t )  jmh (t ))(cos2f c t  j sin 2f c t ) 
Ac
4 ( m(t )  jmh (t ))(cos2f c t  j sin 2f c t )
Ac
2 ( m(t ) cos 2f c t  mh (t ) sin 2f c t )
Ac
4
Ac
4
sUSB (t )  Am(t ) cos 2f c t  Am h (t ) sin 2f c t
USB Phase Discriminator Modulator
×
+
Acos2fct
m(t)
+
HT
Trans.
Hilbert
×
mh(t)
Asin2fct
-
sUSB(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:
S DSBSC ( f ),
S LSB ( f )  
0,

| f | f c
otherwise
LSB Spectrum
M(f)
(1/2)M-(f)
-Bm
Bm
(Ac/4)M+(f+fc)
(Ac/2)mo
-fc
-fc+Bm
(1/2)M+(f)
mo
f
(Ac/4)M-(f-fc)
SLSB(f)
fc-Bm
fc
f
LSB Modulation by frequency
discrimination
• We input the DSB-SC signal to a lowpass filter with
frequency response :
1 | f | f c
H LSB ( f )  
0 | f | f c
LSB modulation by phase
discrimination
• We can show that sLSB(t) is given by:
s LSB (t )  Am(t ) cos 2f c t  Am h (t ) sin 2f c t
Examples
• The messge is m(t) = cos(2fmt). Find the USB and LSB
signals for a carrier amplitude of A and carrier
frequency fc >> fm.
• Solution (phase discriminator)
• sUSB(t) = Acos(2fmt)cos(2fct)-Asin(2fmt)sin(2fct) =
(A/2)cos(2(fc-fm)t) + (A/2)cos(2(fc+fm)t) –
(A/2)cos(2(fc-fm)t) + (A/2)cos(2(fc+fm)t) =
Acos(2(fc+fm)t).
• Similarly we can show that sLSB(t) = Acos(2(fc-fm)t).
• Solution (discfrequency discrmination)
• sDSB-SC(t) = Accos(2fmt)cos(2fct).
• SDSB-SC(f) = (Ac/4)d(f-fc-fm)+(Ac/4)d(f+fc+fm)+(Ac/4)d(ffc+fm) +(Ac/4)d(f+fc-fm).
(Ac/4)
(Ac/4)
-fc-fm –fc+fm
(Ac/4)
fc-fm
(Ac/4)
fc+fm
therefore SUSB(f) = (Ac/4)d(f-fc-fm)+(Ac/4)d(f+fc+fm) and
sUSB(t) = (Ac/2)cos(2(fc+fm)t) = Acos(2(fc+fm)t)
similarly
SLSB(f) = (Ac/4)d(f-fc+fm)+(Ac/4)d(f+fc-fm) and
sLSB(t) = (Ac/2)cos(2(fc+fm)t) = Acos(2(fc+fm)t)
Demodulation of SSB
• Same as DSB-SC