Pulse amplitude modulation
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Pulse amplitude modulation

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Pulse amplitude modulation Pulse amplitude modulation Presentation Transcript

  • 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 ak Modulator Serial data symbols Channel ‘analogue’ channel pulses Demodulator 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 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)
  • 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
  • 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
  • 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
  • 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)
  • 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
  • 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
  • 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’
  • 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 nonlinearly 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 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
  • 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.
  • 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
  • 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
  • Intersymbol Interference • Matlab generated plot showing pulse superposition (accurately) 0.9 0.8 amplitude 0.7 Decision threshold 0.6 0.5 0.4 0.3 0.2 0.1 0 0 1 2 3 4 Received signal 5 6 7 8 time (bit periods) Individual pulses
  • Intersymbol Interference • Sending a longer data sequence yields the following received waveform at the slicer input 1 09 . 08 . 07 . Decision threshold 06 . 05 . 04 . 03 . 02 . 01 . 0 0 1 0 2 0 3 0 4 0 5 0 6 0 7 0 (Also showing individual pulses) 1 09 . 08 . 07 . Decision threshold 06 . 05 . 04 . 03 . 02 . 01 . 0 0 1 0 2 0 3 0 4 0 5 0 6 0 7 0
  • 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 response Eye Diagram 1 0.9 0.8 h = eye height 0.7 0.6 h 0.5 Decision threshold 0.4 0.3 0.2 0.1 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Optimum sample instant
  • 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 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
  • 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
  • Nyquist Pulse Shaping H(f) T f 0 T −2/Τ −1/Τ 0 1/Τ 2/Τ f
  • 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
  • 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
  • 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
  • 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
  • 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.
  • Causal Response Non-causal response T=1s 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 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
  • 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
  • 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 ( )
  • 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).
  • 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.
  • RC Pulse Shaping Normalised Spectrum H(f)/T Pulse Shape h(t) x=0 x = 0.5 x=1 f *T t/T
  • RC Pulse Shaping- Example 1 • Pulse shape and received signal, x = 0 (β = 0) 1 .4 1 .2 1 0 .8 0 .6 0 .4 0 .2 0 - . 02 - . 04 - . 06 0 1 2 3 4 5 6 7 8 • Eye diagram 2 1 .5 1 0 .5 0 - .5 0 1 0 0 .1 0 .2 0 .3 0 .4 0 .5 0 .6 0 .7 0 .8 0 .9
  • RC Pulse Shaping- Example 2 • Pulse shape and received signal, x = 1 (β = 1/2T) 12 . 1 08 . 06 . 04 . 02 . 0 - . 02 0 1 2 3 4 5 6 7 8 • Eye diagram 12 . 1 08 . 06 . 04 . 02 . 0 - .2 0 0 01 . 02 . 0 .3 04 . 05 . 06 . 07 . 08 . 09 .
  • 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 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
  • 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
  • BER f(vn) 0 dx a b vn
  • 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 f(yn|‘0’ sent) 0 V0 P(error|‘1’) f(yn|‘1’ sent) VT V1 P(error|‘0’) yn
  • 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 )
  • BER f(yn|‘0’ sent) 0 V0 VT f(vn) 0 VT - V0 P(error|‘0’) P(error|‘0’) yn vn
  • 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.
  • 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     
  • 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
  • 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
  • • 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
  • 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
  • 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 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
  • 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
  • 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