1.
3F4 Pulse Amplitude Modulation
(PAM)
Dr. I. J. Wassell
2.
Introduction
• The purpose of the modulator is to convert
discrete amplitude serial symbols (bits in a
binary system) ak to analogue output pulses
which are sent over the channel.
• The demodulator reverses this process
ak
Modulator
Serial data
symbols
Channel
‘analogue’
channel pulses
Demodulator
Recovered
data symbols
3.
Introduction
• Possible approaches include
– Pulse width modulation (PWM)
– Pulse position modulation (PPM)
– Pulse amplitude modulation (PAM)
• We will only be considering PAM in these
lectures
4.
PAM
• PAM is a general signalling technique
whereby pulse amplitude is used to convey
the message
• For example, the PAM pulses could be the
sampled amplitude values of an analogue
signal
• We are interested in digital PAM, where the
pulse amplitudes are constrained to chosen
from a specific alphabet at the transmitter
5.
PAM Scheme
Modulator
xs (t ) =
ak
∞
k =−∞
k
Pulse
generator
Symbol
clock
Recovered
symbols
x(t ) =
∑ a δ (t − kT )
∞
∑ a h (t − kT )
k = −∞
k T
Transmit
filter
HT(ω) hT(t)
Demodulator
y (t ) =
∑ a h(t − kT ) + v(t )
k = −∞
Data
slicer
Recovered
clock
Channel
∞
hC(t)
k
Receive
filter
HR(ω), hR(t)
HC(ω)
+
Noise N(ω)
6.
PAM
• In binary PAM, each symbol ak takes only
two values, say {A1 and A2}
• In a multilevel, i.e., M-ary system, symbols
may take M values {A1, A2 ,... AM}
• Signalling period, T
• Each transmitted pulse is given by
ak hT (t − kT )
Where hT(t) is the time domain pulse shape
7.
PAM
• To generate the PAM output signal, we may
choose to represent the input to the transmit
filter hT(t) as a train of weighted impulse
∞
functions
xs (t ) =
∑ a δ (t − kT )
k = −∞
k
• Consequently, the filter output x(t) is a train of
pulses, each with the required shape hT(t)
x(t ) =
∞
∑ a h (t − kT )
k = −∞
k T
8.
xs (t ) =
PAM
∞
∑ a δ (t − kT )
k = −∞
x(t ) =
k
∞
∑ a h (t − kT )
k = −∞
k T
x(t )
xs (t )
Transmit
Filter
hT (t )
• Filtering of impulse train in transmit filter
9.
PAM
• Clearly not a practical technique so
– Use a practical input pulse shape, then filter to
realise the desired output pulse shape
– Store a sampled pulse shape in a ROM and read out
through a D/A converter
• The transmitted signal x(t) passes through the
channel HC(ω) and the receive filter HR(ω).
• The overall frequency response is
H(ω) = HT(ω) HC(ω) HR(ω)
10.
PAM
• Hence the signal at the receiver filter output is
y (t ) =
∞
∑ a h(t − kT ) + v(t )
k = −∞
k
Where h(t) is the inverse Fourier transform of H(ω)
and v(t) is the noise signal at the receive filter
output
• Data detection is now performed by the Data
Slicer
11.
PAM- Data Detection
• Sampling y(t), usually at the optimum instant
t=nT+td when the pulse magnitude is the
greatest yields
yn = y (nT + td ) =
∞
∑ a h((n − k )T + t ) + v
k = −∞
k
d
n
Where vn=v(nT+td) is the sampled noise and td is the
time delay required for optimum sampling
• yn is then compared with threshold(s) to determine
the recovered data symbols
12.
PAM- Data Detection
TX data
TX symbol, ak
Signal at data
slicer input, y(t)
‘1’ ‘0’ ‘0’ ‘1’ ‘0’
+A -A -A +A -A
Τ
0
td
Sample clock
Ideal sample instants
at t = nT+td
Sampled signal, 0
yn= y(nT+td)
Data Slicer decision
threshold = 0V
Detected data
‘1’ ‘0’ ‘0’
‘1’ ‘0’
13.
Synchronisation
• We need to derive an accurate clock signal at
the receiver in order that y(t) may be sampled at
the correct instant
• Such a signal may be available directly (usually
not because of the waste involved in sending a
signal with no information content)
• Usually, the sample clock has to be derived
directly from the received signal.
14.
Synchronisation
• The ability to extract a symbol timing clock
usually depends upon the presence of transitions
or zero crossings in the received signal.
• Line coding aims to raise the number of such
occurrences to help the extraction process.
• Unfortunately, simple line coding schemes often
do not give rise to transitions when long runs of
constant symbols are received.
15.
Synchronisation
• Some line coding schemes give rise to a
spectral component at the symbol rate
• A BPF or PLL can be used to extract this
component directly
• Sometimes the received data has to be nonlinearly processed eg, squaring, to yield a
component of the correct frequency.
16.
Intersymbol Interference
• If the system impulse response h(t) extends over
more than 1 symbol period, symbols become
smeared into adjacent symbol periods
• Known as intersymbol interference (ISI)
• The signal at the slicer input may be rewritten as
yn = an h(td ) + ∑ ak h((n − k )T + td ) + vn
k≠n
– The first term depends only on the current symbol an
– The summation is an interference term which
depends upon the surrounding symbols
17.
Intersymbol Interference
• Example
Modulator input
Binary ‘1’
1.0
0.5
0
2
4
6
Time (bit periods)
Slicer input
amplitude
amplitude
– Response h(t) is Resistor-Capacitor (R-C) first
order arrangement- Bit duration is T
Binary ‘1’
1.0
0.5
0
2
4
6
Time (bit periods)
• For this example we will assume that a
binary ‘0’ is sent as 0V.
18.
Intersymbol Interference
amplitude
• The received pulse at the slicer now extends
over 4 bit periods giving rise to ISI.
1.0
‘1’ ‘1’ ‘0’ ‘0’ ‘1’ ‘0’ ‘0’ ‘1’
0.5
0
2
4
6
time (bit periods)
• The actual received signal is the
superposition of the individual pulses
19.
Intersymbol Interference
amplitude
• For the assumed data the signal at the slicer
input is,
1.0
‘1’ ‘1’ ‘0’ ‘0’ ‘1’ ‘0’ ‘0’ ‘1’
0.5
Decision threshold
0
2
4
6
time (bit periods)
Note non-zero values at ideal sample instants
corresponding with the transmission of binary ‘0’s
• Clearly the ease in making decisions is data
dependant
22.
Eye Diagrams
• Worst case error performance in noise can be
obtained by calculating the worst case ISI over all
possible combinations of input symbols.
• A convenient way of measuring ISI is the eye
diagram
• Practically, this is done by displaying y(t) on a
scope, which is triggered using the symbol clock
• The overlaid pulses from all the different symbol
periods will lead to a criss-crossed display, with
an eye in the middle
24.
Eye Diagrams
• The size of the eye opening, h (eye height)
determines the probability of making incorrect
decisions
• The instant at which the max eye opening occurs
gives the sampling time td
• The width of the eye indicates the resilience to
symbol timing errors
• For M-ary transmission, there will be M-1 eyes
25.
Eye Diagrams
• The generation of a representative eye
assumes the use of random data symbols
• For simple channel pulse shapes with binary
symbols, the eye diagram may be
constructed manually by finding the worst
case ‘1’ and worst case ‘0’ and
superimposing the two
26.
Nyquist Pulse Shaping
• It is possible to eliminate ISI at the sampling
instants by ensuring that the received pulses
satisfy the Nyquist pulse shaping criterion
• We will assume that td=0, so the slicer input is
yn = an h(0) + ∑ ak h((n − k )T ) + vn
k≠n
• If the received pulse is such that
1
h(nT ) =
0
for n = 0
for n ≠ 0
27.
Nyquist Pulse Shaping
• Then
y n = a n + vn
and so ISI is avoided
• This condition is only achieved if
∞
k
∑ H f + T = T
k = −∞
• That is the pulse spectrum, repeated at
intervals of the symbol rate sums to a
constant value T for all frequencies
28.
Nyquist Pulse Shaping
H(f)
T
f
0
T
−2/Τ −1/Τ
0
1/Τ
2/Τ
f
29.
Why?
• Sample h(t) with a train of δ pulses at times kT
∞
hs (t ) = h(t ) ∑ δ (t − kT )
k = −∞
• Consequently the spectrum of hs(t) is
1
H s (ω ) = ∑ H (ω − k 2π T )
T k
• Remember for zero ISI
1
h(nT ) =
0
for n = 0
for n ≠ 0
30.
Why?
• Consequently hs(t)=δ(t)
• The spectrum of δ(t)=1, therefore
1
H s (ω ) = ∑ H (ω − k 2π T ) = 1
T k
• Substituting f=ω/2π gives the Nyquist pulse
shaping criterion
∑ H( f − k T) = T
k
31.
Nyquist Pulse Shaping
• No pulse bandwidth less than 1/2T can
satisfy the criterion, eg,
T
−2/Τ −1/Τ
0
1/Τ
2/Τ
f
Clearly, the repeated spectra do not sum to a constant value
32.
Nyquist Pulse Shaping
• The minimum bandwidth pulse spectrum
H(f), ie, a rectangular spectral shape, has a
sinc pulse response in the time domain,
T
H( f ) =
0
for - 1 2T < f < 1 2T
elsewhere
• The sinc pulse shape is very sensitive to
errors in the sample timing, owing to its low
rate of sidelobe decay
33.
Nyquist Pulse Shaping
• Hard to design practical ‘brick-wall’ filters,
consequently filters with smooth spectral
roll-off are preferred
• Pulses may take values for t<0 (ie noncausal). No problem in a practical system
because delays can be introduced to enable
approximate realisation.
35.
Raised Cosine (RC) Fall-Off
Pulse Shaping
• Practically important pulse shapes which
satisfy the criterion are those with Raised
Cosine (RC) roll-off
• The pulse spectrum is given by
T
f ≤ 1 2T − β
π
H ( f ) = T cos 2
( f − 1 2T + β )
4β
f > 1 2T + β
0
With, 0<β<1/2T
1 2T − β < f ≤ 1 2T + β
36.
RC Pulse Shaping
• The general RC function is as follows,
H(f)
T
0
T
f ≤ 1 2T − β
π
H ( f ) = T cos 2
( f − 1 2T + β )
4β
f > 1 2T + β
0
1
−β
2T
1
2T
1
+β
2T
1
T
f (Hz)
1 2T − β < f ≤ 1 2T + β
37.
RC Pulse Shaping
• The corresponding time domain pulse shape
is given by,
π
sin t
t
T cos 2πβ
h(t ) =
π 1 −( 4 β ) 2
t
t
T
• Now β allows a trade-off between bandwidth
and the pulse decay rate
• Sometimes β is normalised as follows,
β
x=
1
2T
(
)
38.
RC Pulse Shaping
• With β=0 (i.e., x = 0) the spectrum of the filter is
rectangular and the time domain response is a sinc
pulse, that is,
H ( f ) =T
f ≤1 2T
π
t
sin
T
h(t ) =
π
t
T
• The time domain pulse has zero crossings at
intervals of nT as desired (See plots for x = 0).
39.
RC Pulse Shaping
• With β=(1/2T), (i.e., x = 1) the spectrum of the
filter is full RC and the time domain response is a
pulse with low sidelobe levels, that is,
πTf
H ( f ) = T cos
2
2
f ≤1 T
1
2π
h(t ) =
sinc
t
4 2
T
1 − 2 t
T
• The time domain pulse has zero crossings at
intervals of nT/2, with the exception at T/2
where there is no zero crossing. See plots for x
= 1.
40.
RC Pulse Shaping
Normalised Spectrum H(f)/T
Pulse Shape h(t)
x=0
x = 0.5
x=1
f *T
t/T
43.
RC Pulse Shaping- Example
• The much wider eye opening for x = 1 gives
a much greater tolerance to inaccurate
sample clock timing
• The penalty is the much wider transmitted
bandwidth
44.
Probability of Error
• In the presence of noise, there will be a finite
chance of decision errors at the slicer output
• The smaller the eye, the higher the chance that the
noise will cause an error. For a binary system a
transmitted ‘1’ could be detected as a ‘0’ and
vice-versa
• In a PAM system, the probability of error is,
Pe=Pr{A received symbol is incorrectly detected}
• For a binary system, Pe is known as the bit error
probability, or the bit error rate (BER)
45.
BER
• The received signal at the slicer is
yn = Vi + vn
Where Vi is the received signal voltage and
Vi=Vo for a transmitted ‘0’ or
Vi=V1 for a transmitted ‘1’
• With zero ISI and an overall unity gain, Vi=an,
the current transmitted binary symbol
• Suppose the noise is Gaussian, with zero mean
and variance σ v2
46.
BER
f ( vn ) =
1
2πσ v2
2
− vn
e
2
2σ v
Where f(vn) denotes the probability density
function (pdf), that is,
Pr{x < vn ≤ x + dx} = f ( x)dx
and
b
Pr{a < vn ≤ b} = ∫ f ( x)dx
a
48.
BER
• The slicer detects the received signal using
a threshold voltage VT
• For a binary system the decision is
Decide ‘1’ if yn≥ VT
Decide ‘0’ if yn<VT
For equiprobable symbols, the optimum threshold
is in the centre of V0 and V1, ie VT=(V0+V1)/2
49.
BER
f(yn|‘0’ sent)
0
V0
P(error|‘1’)
f(yn|‘1’ sent)
VT V1
P(error|‘0’)
yn
50.
BER
• The probability of error for a binary system can be
written as:
Pe=Pr(‘0’sent and error occurs)+Pr(‘1’sent and error occurs)
Pe = P (error | 0) Po + P (error | 1) P
1
• For ‘0’ sent: an error occurs when yn VT ≥
– let vn=yn-Vo, so when yn=Vo, vn=0 and when yn=VT, vn=VT-Vo.
– So equivalently, we get an error when vn VT-V0
≥
P (error | 0) = P (vn ≥ VT −Vo )
51.
BER
f(yn|‘0’ sent)
0
V0
VT
f(vn)
0
VT - V0
P(error|‘0’)
P(error|‘0’)
yn
vn
52.
BER
P(error | 0) =
∞
∫
VT −Vo
Where,
∞
Q( z ) = ∫
z
VT − Vo
f (vn )dvn = Q
σ
v
−x 2
1
e 2 dx
2π
• The Q function is one of a number of
tabulated functions for the Gaussian
cumulative distribution function (cdf) ie the
integral of the Gaussian pdf.
53.
BER
• Similarly for ‘1’ sent: an error occurs when y n<VT
– let vn=yn-V1, so when yn=V1, vn=0 and when yn=VT, vn=VTV1.
– So equivalently, we get an error when vn < VT-V1
P (error | 1) = P (vn < VT −V1 ) = P (vn > V1 −VT )
P (error | 1) =
∞
∫
V1 −VT
V1 − VT
f (vn )dvn = Q
σ
v
54.
BER
0
P(error|‘1’)
P(error|‘1’)
f(yn|‘1’ sent)
VT V1
f(vn)
yn
0
vn
VT -V1
-(VT -V1)
V1-VT
55.
BER
• Hence the total error probability is
Pe=Pr(‘0’sent and error occurs)+Pr(‘1’sent and error occurs)
Pe = P (error | 0) Po + P (error | 1) P
1
VT − Vo
V −V
Po + Q 1 T
Pe = Q
σ
σv
v
P1
Where Po is the probability that a ‘0’ was sent and P1 is the
probability that a ‘1’ was sent
• For Po=P1=0.5, the min error rate is obtained when,
Vo +V1
VT =
2
56.
• Consequently,
min
e
P
BER
V1 − Vo
h
= Q
2σ = Q 2σ
v
v
where, h = V1 − Vo
• Notes:
– Q(.) is a monotonically decreasing function of its
argument, hence the BER falls as h increases
– For received pulses satisfying Nyquist criterion, ie
zero ISI, Vo=Ao and V1=A1. Assuming unity overall
gain.
– More complex with ISI. Worst case performance if h
is taken to be the eye opening
57.
BER Example
• The received pulse h(t) in response to a
single transmitted binary ‘1’ is as shown,
1.0
0.8
h(t) 0.6
(V) 0.4
0.2
0
−0.2
Where,
Bit period = T
0
Τ
2Τ
3Τ
4Τ
t
5Τ
h(0) = 0, h(T) = 0.3, h(2T) = 1, h(3T) = 0, h(4T) = -0.2, h(5T) =0
58.
BER Example
• What is the worst case BER if a ‘1’ is received as
h(t) and a ‘0’ as -h(t) (this is known as a polar
binary scheme)? Assume the data are equally likely
to be ‘0’ and ‘1’ and that the optimum threshold
(OV) is used at the slicer.
• By inspection, the pulse has only 2 non-zero
amplitude values (at T and 4T) away from the ideal
sample point (at 2T).
59.
BER Example
• Consequently the worst case ‘1’ occurs
when the data bits conspire to give negative
non-zero pulse amplitudes at the sampling
instant.
• The worst case ‘1’ eye opening is thus,
1 - 0.3 - 0.2 = 0.5
as indicated in the following diagram.
60.
BER Example
1.0
0.8
0.6
0.4
0.2
0
−0.2
−0.4
−0.6
−0.8
−1.0
‘X’
‘1’
‘X’
‘1’
‘0’
‘X’
−0.3
−0.2
Optimum sample point for circled bit, amplitude = 1-0.3-0.2 = 0.5
0
Τ
2Τ
3Τ
4Τ
5Τ
6Τ
7Τ
8Τ
t
• The indicated data gives rise to the worst case ‘1’ eye
opening. Don’t care about data marked ‘X’ as their pulses
are zero at the indicated sample instant
61.
BER Example
• Similarly the worst case ‘0’ eye opening is
-1 + 0.3 + 0.2 = -0.5
• So, worst case eye opening h = 0.5-(-0.5) = 1V
• Giving the BER as,
min
e
P
h
1
= Q
2σ = Q 2σ
v
v
Where σv is the rms noise
at the slicer input
62.
Summary
• For PAM systems we have
– Looked at ISI and its assessment using eye diagrams
– Nyquist pulse shaping to eliminate ISI at the
optimum sampling instants
– Seen how to calculate the worst case BER in the
presence of Gaussian noise and ISI
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