Transcript Should Coded Modulation Use Nyquist Pulses?

Should Coded Modulation Use
Nyquist Pulses?
John B Anderson
High Speed Wireless Center and
Electrical & Information Technology Dept.
Lund University, Sweden
Outline
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Faster than Nyquist signaling
Capacity of signals with a PSD
Linear modulation with orthogonal pulses
Extend to FTN signaling
Dmin, capacities and performance of FTN
The outcome: Coded linear modulation based on nonorthogonal
pulses has superior information rates
Lund University / Department of Electrical and Information Technology
Shannon’s Square PSD Capacity
P/2W
-W

W log 1
2

0
P

N 0W 

bits/s
W
Consideration of Shannon’s argument shows:
• As W grows, P/W is fixed if Es and Eb are fixed (NB: Es=P/2W) .
• So log[1+P/WNo] data bits/s are carried in each +Hz of the square PSD .
• This is a flow of bits, per Hz-s.
• For fixed P/No, flow is maximized by the square PSD.
• But the square PSD stems from sinc pulses.
Lund University / Department of Electrical and Information Technology
Capacity at Different Energies & Bandwidths
per Data Bit
Physical channel has bandwidth W, power P and Es = P/2W
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What is a Faster than Nyquist Signal?
Lund University / Department of Electrical and Information Technology
Faster-than-Nyquist Signaling (Mazo, 1975)
• Baseband signals of form
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s(t )   an h(t  nT )
n
Reduce symbol time to τT ≤ T
Euclid. min. distance preserved for   Μazo limit
Asymptotic error performance thus unaffected
Allows higher data bits/Hz
More bandwidth efficiency at same energy
• The new form:
s(t )   an h(t  nT )
n
Lund University / Department of Electrical and Information Technology
FTN Example
Send 1 -1 1 – 1- 1
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Information Rates for Un-Square PSDs
Now suppose the signals have a PSD that is not
square.
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AWGN Information Rate for a PSD

 2P

C   log 2 1
| H ( f )|2 df
 N0

0
2
PSD = P | H ( f )| , power P


Generalization of W log 1 P
2
N 0W 


Low power example:
Fix W at 1 Hz
P
N
1
Low bit density coding
0
0.5
Square PSD, C=1 b/Hz-s
30% rtRC, C=1.03 b/Hz-s
-1
0
1
Lund University / Department of Electrical and Information Technology
Increase the Power
...but not the bandwidth. So, more bits/Hz-s.
P
500
N
High bit density coding
1000
0
Square PSD, C=9.97 b/Hz-s
30% rtRC, C=11.81 b/Hz-s
-1
0
1
In the limit P → ∞,
C
for 30% rtRC → 1.3
_____________
C for square PSD
(i.e., 1 + excess bandwidth factor)
Lund University / Department of Electrical and Information Technology
Why Does This Happen?
Reason 1:
C is linear in W, but log in P
Reason 2: Moving PSD parts by Nyquist’s reflection principle*
can only increase C
*
A
0
*
*A pulse is T-orthogonal if its PSD is antisymmetrical about (A/2, 1/2T)
The effect grows stronger with bit density (noticeable > 3 bits/Hz-s)
Lund University / Department of Electrical and Information Technology
More about Bit Densities ...
• Binary antipodal signaling: 1 data bit / (T s)(1/2T Hz) = 2 bits/Hz-s
• QPSK:
2 bits / (T s)(1/T Hz) = 2 bits/Hz-s
• Convolutional R=1/2 coding + above: 1 bit/Hz-s
• DTV, 64QAM + R=1/2 coding:
(1/2) 6 / (T s)(1/T Hz) = 3 bits/Hz-s
Lund University / Department of Electrical and Information Technology
Capacity and FTN with Linear Modulation
Lund University / Department of Electrical and Information Technology
Coded Orthogonal Pulse Linear Modulation
s(t ) 
Es /T
 an h(t  nT )
n
• Most modulations/coded modulations are this, with h T-orthogonal
• Average PSD of s(t) is
E s / T  | H ( f )|2, even if h(t) not T-orthogonal
• Form codes as subsets of these signals, having same PSD
• Capacity for s(t) signal set is same for any orthogonal h(t).
So it must be the square PSD C.
• But
C sq  C rc
!!
Lund University / Department of Electrical and Information Technology
What Does Sq PSD C < RC PSD C Mean for Linear
Modulation?
• Orthogonal h(t) cannot achieve C for PSD shape unless sinc h
• Can Faster than Nyquist linear modulation ? ?
s(t ) 
E s / T
 anh(t  nT ),  1
n
• Let’s try an FTN h(t) !
• Same h(t), coming faster
- h(t) not  T orthogonal
- Higher bits/Hz-s, by 1 /
- Same average PSD
Lund University / Department of Electrical and Information Technology
Dmin and Capacity in FTN Signaling
• Fact: Consider set of s(t) signals made from a n  1
For sinc pulse, d 2min  2,   .802 (25% more bits/Hz-s, same PSD, Pe)
For rtRC pulse, d 2min  2,   .703 (42% more)
• Theorem: FTN codewords can achieve the PSD capacity
(Ph.D. Thesis, F. Rusek, 2007; Rusek & Anderson, IT Trans., Feb. 2009)
For 30% rtRC, need
 1/1.3
• In fact, codes based on binary FTN can have higher rate than any
code based on orthogonal pulses.
Lund University / Department of Electrical and Information Technology
FTN Constrained Capacities
Under bounds to capacity for binary-data FTN based on 30% rtRC pulse (red).
Nyquist C (green) has no input constraint. Top C is capacity for PSD (blue).
( Rusek & Anderson,
Trans. IT, Feb. 09)
Lund University / Department of Electrical and Information Technology
Shannon BER Bound and 2 Systems at 3 Bits/Hz-s
FTN: A. Prlja, 2008
TCM: W. Zhang, 1995
Lund University / Department of Electrical and Information Technology
What about Sinc Pulses?
• No gain in C available if h(t) is a sinc !
(but sinc FTN linear modulation is better than sinc without FTN)
• But is coding based on sinc of practical interest?
• Can show: For finite frames, sinc is not an optimal use of time
and bandwidth
(does not give the best time - frequency occupancy)
Lund University / Department of Electrical and Information Technology
Conclusions
• Signal capacity must be computed from true PSD at high bit
density
• Extra information rate available with non-sinc pulses
• It exceeds that available with any use of orthogonal pulses
• FTN linear modulation can achieve this capacity
Lund University / Department of Electrical and Information Technology
Tack!
Lund University / Department of Electrical and Information Technology