James A. Ritcey 2012, Communications I
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Pulse Amplitude Modulation
Thus, p(t) is normalized (unit vector) and is orthogonal to its translated pulses, p(t + kT ). This
Pulse Amplitude Modulation or PAM is a lin- pulse does not exhibit inter-symbol interference
ear digital modulation technique that produces when the output of the matched filter is sampled
a baseband signal
at the optimal time.
X
As an example, the ideal minimum bandwidth
an p(t − nT )
(1)
s(t) =
min-BW pulse is
n
where an are the transmitted symbols, p(t) is the
p(t) = sinc(t/T ), P (f ) = T rect(f T )
(6)
pulseshape, and Rs = 1/T is the symbol or baud
∗
rate. The symbols are drawn from a discrete set, The Matched Filter response is H(f ) = P (f ),
which, except for amplitude scaling, yield a MF
or constellation, an ∈ A.
output, H(f )P (f ) = |P (f )|2 , which has same
Pulse Shapes Pulse Amplitude Modulation zero-crossings as p(t). This is unrealizable be(PAM) conveys information using a pulse shape cause the frequency response has no transition
band, and the pulse shape is infinite in both time
p(t) → P (f )
(2) directions. It can be approximated, or a better
pulse can be designed.
by sending symbol once every T seconds. The
The passband signal can be created from the
bandwidth usage depends on P (f ), and is always
baseband p be up-converting to carrier fo . The
defined as the one-sided bandwidth, or extent
transmitted signal appears
over the positive frequency axis. Various definiX
tions of extent may be appropriate - null-to-null,
s(t) =
an p(t − nT )
(7)
3dB, strict, etc. The pulse shape is usually real
n
valued, and exhibits a baseband spectra. The
sp (t) =Re s(t) exp(2πfo t)
(8)
pulse occupies bandwidth determined by the energy spectra
where the ak are (usually complex-valued) sym|P (f )|2
bols to be transmitted, but the p(t) remains realvalued. Note that the passband signal sp will
use twice the bandwidth of the baseband s. Also
(3) note that if the energy of p(t) is Ep , the energy
of p(t) cos(2πfo t) is Ep /2. We can take the realpart to obtain an explicit epxresison for passband
The normalized unit-energy pulse
PAM,
√
p̂ = p(t)/ E p
(4)
"
#
X
can be used to define a basis vector in signal
ssp (t) =
In p(t − nT ) cos(2πfo t)
(9)
space.
n
"
#
X
−
Qn p(t − nT ) sin(2πfo t) (10)
Matched Filter and Intersymbol InterferThe pulse will have energy
Z
Z
2
Ep = |p(t)| dt = |P (f )|2 df
n
ence. The pulse shape will be received using a correlator or matched filter structure. If where, the complex constellation points are exa matched correlator is used, we define a T- pressed in terms of the In-Phase and Quadrature
Nyquist pulse, with zero ISI, by
components.
Z
p(t) ? p(−t)|kT = p(t)p(kT + t)dt = δk (5)
an = In + jQn
James A. Ritcey 2012, Communications I
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Spectral Density of Baseband where b is the bit vector that addresses the symThe power spectral density of the sig- bol. This assume that log2 M is an integer.
When employed with the pulse shaping, we see
that
X
Ss (f ) = |P (f )|2 ×
Ra [k] exp(−2πjkf T )
1
k
Rs =
symbols/sec
(16)
T
and the symbol autocorrelation for random symlog2 M
Rb =
bits/sec, bps
(17)
bols ak is Ra [k] = E{an a∗n−k }. When a determinT
istic sequence ak is sent, the spectrum is
The constellation has various metrics. AssumX
2
2
ing
equally likely symbols
Ss (f ) = |P (f )| × |
ak exp(−2πjkf T )|
Power
PAM.
nal is
k
For example, if the pulse shape is the min-BW
M
1 X
|am |2
Ea =
M
(18)
d2ij = |ai − aj |2
(19)
m=1
p(t) = rect(t/T ), P (f ) = T sinc(f T )
dmin = min dij
(20)
i>j
which is a baseband spectra with a maximum at
f = 0.
The set of distances dij , i > j is called the disAssume the two pulse sequence a0 = +1, a1 =
tance spectra.
−1 is used. Then the energy spectrum is
A normalized constellation has average energy
X
per symbol E√a = 1. For such a normalized conSs (f ) =|P (f )|2 × |
ak exp(−2πjkf T )|2
stellation A, E×A has average energy per symk
(11) bol, Es = E. Letting am = Im + jQm we get the
passband signal
2
2
=|P (f )| × |a0 + a1 exp(−2πjf T )| (12)
X
=|P (f )|2 × |2 sin(f T )|2
(13)
s(t) =
Ik p(t − kT ) cos(2πfo t)
(21)
k
X
This has a null at DC, Ss (f = 0) = 0, so
−
Qk p(t − kT ) sin(2πfo t)
(22)
the transmitted sequence can change the in-band
k
spectra. In-band refers to the fact that the second factor is (1/T )− periodic. For passband
Assuming an AWGN channel w(t) with specPAM, the spectra is translated to the carrier fre- tral level No /2 and a unit energy receive filter,
quency, fo .
the T-samples have variance σ 2 = No /2.
The performance can be estimated using the
NNUB
- nearest neighbor union bound, which
Constellations The constellation A is a set of
symbols am . They can be real-valed or complex gives the symbol error probability
valued, and listed as
dmin
PSE ≈ N N Q
(23)
2σ
A = {a1 , . . . , aM }.
(14)
The size of the constellation is M = |A|, which where N N is the average number of nearest
can be indexed by any of log2 M bits through a neighbors to the constellation points. Typically this term is approximated by the numbit-labeling, or constellation mapping
ber of neighbors for the dmin pair of constellaam = µ(b), b = [b1 . . . blog2 M ]
(15) tion points. We can use this as the exact error
James A. Ritcey 2012, Communications I
rate. This should be expressed, using the average energy of the constellation, in terms of Eb /No
where
Es = Eb log2 M
(24)
√
√
and 2σ = 2No assuming the unit energy receive filter. Here Q(x) can be evaluated from
Q(x) = Prob{t > x}, t ∼ N (0, 1)
Z ∞
2
dt
e−x /2
2
e−t /2 √ ∼ √
=
2π
x 2π
x
Q(x) = (1/2) ∗ erfc(x/sqrt(2))
2
3
p
√
The distance spectra is ( 3E/2) × [1, 1, 2], so
the minimum distance, when scaled to energy E
is
p
dmin = 3E/2
(32)
Thus,
dmin
=
2σ
p
3E/2
2No
(33)
(25) and the error rate is bounded/approximated by
!
r
(26)
3Es
PSE ≤ 2Q
(34)
4No
(27)
Examples
PBE
2
= Q
2
r
2Eb 3 log2 3
No
8
!
(35)
QPSK This M = 4 constellation is
which gives a loss of 10 log10 (3log2 3/8) =
√ jmπ/4
−2.26dB.
A = { Ee
, m = 0 : M −1, M = 4} (28)
The bit labeling could not be easily achieved
since log2 (3) is not an integer, which is why this
This is normalized√to Es = E, and basic geomeconstellation is non-standard.
try gives dmin = 2E Therefore, the error rate
is
Analyzing a Constellation Here are the
!
r
r !
steps.
2Es
Es
= 2Q
(29)
PSE = 2Q
2No
No
1. Determine your transmitted signals and
Converting to Eb /No and computing the PBE
assuming a Gray code gives
!
r
2Eb
2
(30)
PBE = Q
2
No
This is the same as BPSK, when Eb /No is used.
To convert from per symbol to per bit, simply
use dimensional analysis,
Es = Eb log2 M
their dimension. Plot as points in signal
space
2. Use a scaled matched filter receiver to generate the signal space. Use unit vectors to
assure that the noise gain through the MF
is unity.
3. Determine the variance of the noise at the
MF output, usually
σ 2 = No /2
with unit vector and orthogonal MFs.
Non-standard Constellation The constellation has only M = 3 points in the complex
plane (draw it!) at 0, +1, +j, with average energy E = (02 + 12 + |j|2 )/3 = 2/3, so a scaled
constellation is
p
A = ( 3E/2) × {0, 1, j}
(31)
4. Compute the minimum distance dmin
5. Assemble to get the nearest neighbor union
bound (NNUB) from equation 23.
6. Convert to Eb /No and compare to BPSK