ELG3175 Introduction to Communication Systems Lecture 9yongacog/courses/elg3175/Lecture9-10.pdf ·...
Transcript of ELG3175 Introduction to Communication Systems Lecture 9yongacog/courses/elg3175/Lecture9-10.pdf ·...
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Quadrature and Single Sideband AM
ELG3175 Introduction to Communication Systems
Lecture 9
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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.
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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 +=
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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 +−−+++−=
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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
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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
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QAM System
m1(t)
m2(t)
×
×
Accos(2πfct) HT + Channel
Km1(t)
Km2(t)
×
×
Arcos(2πfct) HT
LPF
LPF
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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.
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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
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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)
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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.
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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
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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
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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
)(
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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)()( −=
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USB Phase Discriminator Modulator
m(t)
×
Trans.Hilbert ×
Acos2πfct
HT
+
-
+ sUSB(t)
Asin2πfct
mh(t)
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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
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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)
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LSB Modulation by frequency discrimination
• We input the DSB-SC signal to a lowpass filter with frequency response :
⎩⎨⎧
>
≤=
c
cLSB ff
fffH
||0||1
)(
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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
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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).
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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)
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-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)
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Demodulation of SSB
• Same as DSB-SC
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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