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Prof. Brian L. Evans Dept. of Electrical and Computer Engineering The University of Texas at Austin. Quadrature Amplitude Modulation (QAM) Transmitter. Introduction. - PowerPoint PPT Presentation
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EE345S Real-Time Digital Signal Processing Lab Spring 2006
Lecture 15
Quadrature Amplitude Modulation (QAM) Transmitter
Prof. Brian L. Evans
Dept. of Electrical and Computer Engineering
The University of Texas at Austin
15 - 2
Introduction
• Digital Pulse Amplitude Modulation (PAM) modulates digital information onto amplitude of pulse and may be later modulated by sinusoid
• Digital Quadrature Amplitude Modulation (QAM) is two-dimensional extension of digital PAM that requires sinusoidal modulation
• Digital QAM modulates digital information onto pulses that are modulated ontoAmplitudes of a sine and a cosine, or equivalently
Amplitude and phase of single sinusoid
15 - 3
Amplitude Modulation by Cosine
• Example: y(t) = f(t) cos(c t)Assume f(t) is an ideal lowpass signal with bandwidth 1
Assume 1 < c
Y() is real-valued if F() is real-valued
• Demodulation: modulation then lowpass filtering• Similar derivation for modulation with sin(0 t)
0
1
-
F()
0
Y()½
-c - -c + c
c - c + c
½Fc½Fc
Review
15 - 4
Amplitude Modulation by Sine
• Example: y(t) = f(t) sin(c t)
Assume f(t) is an ideal lowpass signal with bandwidth 1
Assume 1 < c
Y() is imaginary-valued if F() is real-valued
• Demodulation: modulation then lowpass filtering
Y()j ½
-c - -c + c
c - c +
c
-j ½Fcj ½Fc
-j ½
0
1
-
F()
Review
15 - 5
Digital QAM Modulator
Serial/ParallelMap to 2-D constellation
Impulse modulator
Impulse modulator
Pulse shaping gT(t)
Local Oscillator
+
90o
Pulse shaping gT(t)
d[n]an
bn
a*(t)
b*(t)
s(t)
1 J
MatchedDelay
Matched delay matches delay through 90o phase shifter
bit stream
15 - 6
Phase Shift by 90 Degrees
• 90o phase shift performed by Hilbert transformer
cosine => sine
sine => – cosine
• Frequency response of idealHilbert transformer:
)(2
1)(
2
1) 2cos( 000 fffftf
)(2
)(2
) 2sin( 000 ffj
ffj
tf
)sgn()( fjfH
0 if1
0 if0
0 if1
)sgn(
x
x
x
x
15 - 7
Hilbert Transformer
• Magnitude responseAll pass except at origin
• For fc > 0
• Phase responsePiecewise constant
• For fc < 0)2sin()
22cos( tftf cc
))(2sin()2
)(2cos(
))2
2(cos()2
2cos(
tftf
tftf
cc
cc
f
)( fH
-90o
90o
f
|)(| fH
)sgn()( fjfH
15 - 8
Hilbert Transformer
• Continuous-time ideal Hilbert transformer
• Discrete-time ideal Hilbert transformer
h(t) =
1/( t) if t 0
0 if t = 0h[n] =
if n0
0 if n=0
nn )2/(sin2 2
Even-indexed samples are zero
t
h(t)
)sgn()( fjfH )sgn()( jH
n
h[n]
15 - 9
Discrete-Time Hilbert Transformer
• Approximate by odd-length linear phase FIR filterTruncate response to 2 L + 1 samples: L samples left of
origin, L samples right of origin, and originShift truncated impulse response by L samples to right to
make it causalL is odd because every other sample of impulse response is 0
• Linear phase FIR filter of length N has same phase response as a delay of length (N-1)/2(N-1)/2 is an integer when N is odd (here N = 2 L + 1)
• How would you make sure that delay from local oscillator to sine modulator is equal to delay from local oscillator to cosine modulator?
15 - 10
Performance Analysis of PAM
• If we sample matched filter output at correct time instances, nTsym, without any ISI, received signal
where the signal component is
v(t) output of matched filter Gr() for input ofchannel additive white Gaussian noise N(0; 2)
Gr() passes frequencies from -sym/2 to sym/2 ,where sym = 2 fsym = 2 / Tsym
• Matched filter has impulse response gr(t)
)()()( symsymsym nTvnTsnTx
dianTs nsym )12()( for i = -M/2+1, …, M/2
v(nT) ~ N(0; 2/Tsym)
4-PAM
d
-d
3 d
-3 d
15 - 11
Performance Analysis of PAM
dnTwgnTv r
)()()(
22 )()()( dnTwgEnTvE r
})()()()({ 212211
ddnTwgnTwgE TT
212121 )}()({)()( ddnTwnTwEgg TT
TdGdg
sym
sym
rr
22/
2/
2222 )(2
1)(
Noise power
T = TsymFiltered noise
2 (1–2)
15 - 12
Performance Analysis of PAM
• Decision errorfor inner points
• Decision errorfor outer points
• Symbol error probability
-7d -5d -3d -d d 3d 5d 7d
O- I I I I I I O+
T
dQdnTvPePI 2))(()(
T
dQdnTvPePO
))(()(
T
dQdnTvPdnTvPePO
))(())(()(
Td
QM
MeP
MeP
MeP
M
MeP OOI
)1(2)(
1)(
1)(
2)(
8-PAM Constellation
15 - 13
Performance Analysis of QAM
• Received QAM signal
• Information signal s(nT)
where i,k { -1, 0, 1, 2 } for 16-QAM
• Noise, vI(nT) and vQ(nT) are independent Gaussian random variables ~ N(0; 2/T)
)()()( nTvnTsnTx
dkjdibjanTs nn )12( )12( )(
)( )()( nTvjnTvnTv QI
15 - 14
Performance Analysis of QAM
• Type 1 correct detection
))(&)(()(1 dnTvdnTvPcP QI
))(())(( dnTvPdnTvP QI
)))((1))()((1( dnTvPdnTvP QI
2))(21( Td
Q
3
3
3
3
2 2
2
2
2 2
2
211
1 1
I
Q
16-QAM
)(2 Td
Q
)(2 Td
Q
15 - 15
Performance Analysis of QAM
• Type 2 correct detection
• Type 3 correct detection
))(&)(()(2 dnTvdnTvPcP QI
))(())(( dnTvPdnTvP QI
))(1))((21( Td
QTd
Q
))(&)(()(3 dnTvdnTvPcP QI
))(())(( dnTvPdnTvP QI
2))(1( Td
Q
3
3
3
3
2 2
2
2
2 2
2
211
1 1
I
Q
16-QAM
15 - 16
Performance Analysis of QAM
• Probability of correct detection
• Symbol error probability
22 ))(1(164
))(21(164
)( Td
QTd
QcP
))(1))((21(168
Td
QTd
Q
)(49
)(31 2 Td
QTd
Q
)(49
)(3)(1)( 2 Td
QTd
QcPeP
15 - 17
Average Power Analysis
• PAM and QAM signals are deterministic• For a deterministic signal p(t), instantaneous
power is |p(t)|2
• 4-PAM constellation points: { -3 d, -d, d, 3 d }– Total power 9 d2 + d2 + d2 + 9 d2 = 20 d2
– Average power per symbol 5 d2
• 4-QAM constellation points: { d + j d, -d + j d,d – j d, -d – j d }– Total power 2 d2 + 2 d2 + 2 d2 + 2 d2 = 8 d2
– Average power per symbol 2 d2
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