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3F4 Pulse Amplitude Modulation (PAM)
Dr. I. J. Wassell
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
Modulator Channel Demodulator
Serial data symbols
ak
‘analogue’ channel pulses
Recovered data symbols
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
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
PAM Scheme
HC()
hC(t)
Symbol clock
HT() hT(t)
Noise N()
Channel
+
Pulse generator
ak Transmit filter
k
ks kTtatx )()(
k
Tk kTthatx )()(
Receive filter
HR(), hR(t)
Data slicer
Recovered symbols
Recovered clock
)()()( tvkTthatyk
k
Modulator
Demodulator
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)( kTtha Tk
Where hT(t) is the time domain pulse shape
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
k
ks kTtatx )()(
• Consequently, the filter output x(t) is a train of pulses, each with the required shape hT(t)
k
Tk kTthatx )()(
PAM
• Filtering of impulse train in transmit filter
Transmit
Filter
k
Tk kTthatx )()(
k
ks kTtatx )()(
)(thT
)(txs )(tx
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() and the receive filter HR().
• The overall frequency response is
H() = HT() HC() HR()
PAM• Hence the signal at the receiver filter output is
)()()( tvkTthatyk
k
Where h(t) is the inverse Fourier transform of H() and v(t) is the noise signal at the receive filter output
• Data detection is now performed by the Data Slicer
PAM- Data Detection• Sampling y(t), usually at the optimum instant
t=nT+td when the pulse magnitude is the greatest yields
nk
dkdn vtTknhatnTyy
))(()(
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
PAM- Data Detection
Data Slicer decision threshold = 0V
Signal at data slicer input, y(t)
Sample clock
Sampled signal, yn= y(nT+td)
Ideal sample instants at t = nT+td
TX dataTX symbol, ak
‘1’ ‘0’ ‘0’ ‘1’ ‘0’+A -A -A +A -A
Detected data ‘1’ ‘0’ ‘0’ ‘1’ ‘0’
td
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.
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.
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 non-linearly processed eg, squaring, to yield a component of the correct frequency.
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
nnk
dkdnn vtTknhathay
))(()(
– The first term depends only on the current symbol an
– The summation is an interference term which depends upon the surrounding symbols
Intersymbol Interference• Example
– Response h(t) is Resistor-Capacitor (R-C) first order arrangement- Bit duration is T
• For this example we will assume that a binary ‘0’ is sent as 0V.
Time (bit periods)
ampl
itud
e
Time (bit periods)
ampl
itud
e
Modulator input Slicer inputBinary ‘1’ Binary ‘1’
Intersymbol Interference• The received pulse at the slicer now extends
over 4 bit periods giving rise to ISI.
• The actual received signal is the superposition of the individual pulses
time (bit periods)
ampl
itud
e
‘1’ ‘1’ ‘0’ ‘0’ ‘1’ ‘0’ ‘0’ ‘1’
Intersymbol Interference• For the assumed data the signal at the slicer
input is,
• Clearly the ease in making decisions is data dependant
time (bit periods)
ampl
itud
e
Note non-zero values at ideal sample instants corresponding with the transmission of binary ‘0’s
‘1’ ‘1’ ‘0’ ‘0’ ‘1’ ‘0’ ‘0’ ‘1’
Decision threshold
Intersymbol Interference• Matlab generated plot showing pulse
superposition (accurately)
0 1 2 3 4 5 6 7 80
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
Decision threshold
time (bit periods)Received signal
Individual pulses
ampl
itud
e
Intersymbol Interference• Sending a longer data sequence yields the
following received waveform at the slicer input
Decision threshold
0 10 20 30 40 50 60 700
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0 10 20 30 40 50 60 700
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Decision threshold
(Also showing individual pulses)
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
Example R-C responseEye Diagram
Decision threshold
Optimum sample instant
h = eye height
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.90
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
h
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
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
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
nnk
knn vTknhahay
))(()0(
• If the received pulse is such that
0for 0
0for 1)(
n
nnTh
Nyquist Pulse Shaping• Then
nnn vay and so ISI is avoided
• This condition is only achieved if
TT
kfH
k
• That is the pulse spectrum, repeated at intervals of the symbol rate sums to a constant value T for all frequencies
Nyquist Pulse Shaping
T
f
H(f)
T
f
Why?• Sample h(t) with a train of pulses at times kT
k
s kTtthth )()()(
• Consequently the spectrum of hs(t) is
k
s TkHT
H )2(1
)(
• Remember for zero ISI
0for 0
0for 1)(
n
nnTh
Why?
• Consequently hs(t)=(t)
• The spectrum of (t)=1, therefore
1)2(1
)( k
s TkHT
H
• Substituting f=/2 gives the Nyquist pulse shaping criterion
k
TTkfH )(
Nyquist Pulse Shaping
T
f
• No pulse bandwidth less than 1/2T can satisfy the criterion, eg,
Clearly, the repeated spectra do not sum to a constant value
Nyquist Pulse Shaping
• The minimum bandwidth pulse spectrum H(f), ie, a rectangular spectral shape, has a sinc pulse response in the time domain,
elsewhere 0
212T1-for )(
TfTfH
• The sinc pulse shape is very sensitive to errors in the sample timing, owing to its low rate of sidelobe decay
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 non-causal). No problem in a practical system because delays can be introduced to enable approximate realisation.
Causal Response
Non-causal response
T = 1 s
Causal response
T = 1s
Delay, td = 10s
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
2121
21 0
)21(4
cos
21
)( 2
TfT
Tf
TfT
TfT
fH
With, 0<<1/2T
RC Pulse Shaping• The general RC function is as follows,
H(f)
f (Hz)
T
0
T2
1 T2
1T2
1
T
1
2121
21 0
)21(4
cos
21
)( 2
TfT
Tf
TfT
TfT
fH
RC Pulse Shaping• The corresponding time domain pulse shape
is given by,
241
2cossin
)(t
t
tT
tT
th
• Now allows a trade-off between bandwidth and the pulse decay rate
• Sometimes is normalised as follows,
T
x
21
RC Pulse Shaping• With =0 (i.e., x = 0) the spectrum of the filter
is rectangular and the time domain response is a sinc pulse, that is,
TfTfH 21 )(
t
T
tT
th
sin
)(
• The time domain pulse has zero crossings at intervals of nT as desired (See plots for x = 0).
RC Pulse Shaping• With =(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,
TfTf
TfH 1 2
cos)( 2
tT
tT
th2
sinc4
1
1)(
22
• 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.
RC Pulse ShapingNormalised Spectrum H(f)/T Pulse Shape h(t)
x
x
x
f *T t/T
RC Pulse Shaping- Example 1
• Eye diagram
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9-1
-0.5
0
0.5
1
1.5
2
0 1 2 3 4 5 6 7 8-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
1.2
1.4
• Pulse shape and received signal, x = 0 ( = 0)
RC Pulse Shaping- Example 2
• Eye diagram
• Pulse shape and received signal, x = 1 ( = 1/2T)
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9-0.2
0
0.2
0.4
0.6
0.8
1
1.2
0 1 2 3 4 5 6 7 8-0.2
0
0.2
0.4
0.6
0.8
1
1.2
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
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)
BER• The received signal at the slicer is
nin vVy 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 2
v
BER2
2
2
22
1)( v
nv
v
n evf
Where f(vn) denotes the probability density function (pdf), that is,
dxxfdxxvx n )(}Pr{
and
b
an dxxfbva )(}Pr{
BER
dxvn
f(vn)
ba0
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
BER
yn
f(yn|‘0’ sent)
VT
0
f(yn|‘1’ sent)
P(error|‘0’)P(error|‘1’)
V1V0
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)
)()0|( oTn VVvPerrorP
1)1|()0|( PerrorPPerrorPP oe
• 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
BER
yn
f(yn|‘0’ sent)
VT
0
P(error|‘0’)V0
vn
f(vn)
VT - V0
0
P(error|‘0’)
BER
• 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.
oT VV v
oTnn
VVQdvvferrorP
)()0|(
Where,
z
xdxezQ 2
2
2
1)(
BER• Similarly for ‘1’ sent: an error occurs when yn<VT
– let vn=yn-V1, so when yn=V1, vn=0 and when yn=VT, vn=VT-V1.
– So equivalently, we get an error when vn < VT-V1
)()()1|( 11 TnTn VVvPVVvPerrorP
TVV v
Tnn
VVQdvvferrorP
1
1)()1|(
BER
ynVT
0
f(yn|‘1’ sent)
P(error|‘1’) V1
VT -V1vn
0
f(vn)
P(error|‘1’) -(VT -V1)
V1-VT
BER• Hence the total error probability is
Pe=Pr(‘0’sent and error occurs)+Pr(‘1’sent and error occurs)
1)1|()0|( PerrorPPerrorPP oe
11 P
VVQP
VVQP
v
To
v
oTe
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,
21VV
V oT
BER• Consequently,
ovv
oe VVh
hQ
VVQP
1
1min here, w22
• 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
BER Example• The received pulse h(t) in response to a
single transmitted binary ‘1’ is as shown,
Where,
h(0) = 0, h(T) = 0.3, h(2T) = 1, h(3T) = 0, h(4T) = -0.2, h(5T) =0
Bit period = T
t
h(t)
(V)
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).
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.
BER Example
• 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
t
‘1’ ‘1’ ‘0’‘X’ ‘X’‘X’
Optimum sample point for circled bit, amplitude = 1-0.3-0.2 = 0.5
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,
vve Q
hQP
2
1
2min Where v is the rms noise
at the slicer input
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