[DSP] W08 - Receiver Design

contents

• receiver design
• delay compensation
• adaptive equalization
• adsl
  • channel propagation challenges
  • subchannel structure
  • discrete multitone modulation
  • ADSL schematic

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• signal picks up noise while propagating in the channel
• it also get distorted as the channel acts as some sort of filter,
  • that is not necessary lowpass or linear-phase
• interference occurs as well

• there might be parts of the channel that might assumed to be usable and actually not

• the receiver has to deal with a copy of the transmitted signal
  • very far from the idealized version used in the math for designing the transmitter
• adaptive filtering techniques enable digital receivers to cope with the distortions and the noise introduced by the channel
  • topics: advanced signal processing classes
• this is an overview
  • your ADSL receiver for instance
  • allows high data rates

receiver design

• following is the sound made by a dial-up internet modem
  • when connecting to the internet

• for graphical analysis of this sound, refer to receiver schematic below

fig: QAM receiver signal flow schematic

• baseband complex samples ˆb[n]$\hat{b}\left[n\right]$ are plotted on the complex plane
• if the input at the receiver is a signal ˆs[n]=cos((ωc+ω0)n)$\hat{s}\left[n\right]=\cos \left(\right({\omega }_c+{\omega }_0\left)n\right)$
  • the obtained baseband for this is ˆb[n]=ejω0n$\hat{b}\left[n\right]=e^{j{\omega }_{0}n}$
  • so they point on the unit circle on the argand plane
  • the angle between successive points will be ω0${\omega }_0$

pilot tones

• the receiver sends pilot tones
• pilot tones are simple sinusoids used to probe the channel
  • channel probing

• used to gauge the response at particular frequencies

• some components
  1. many sinusoids
     • which have abrupt phase changes
     • phase reversals are used as time markers
     • to estimate propagation delay of channel
  2. training sequence
     • known sequence is sent my transmitter
     • the receiver uses channel response to this known sequence to train an equalizer to offset channel effects
  3. handshake procedure between transmitter and receiver
     • just before core information transmission begins
     • low bitrate QAM transmission using only four points
       • 2 bits per symbol
       • parameters exchange of speed, constellation size etc
       • since only 4 points constellation,
         • so even in noisy conditions ensure vital information exchange
  4. data transmission proper

receiver function

• challenges faced at the receiver and measures taken to offset each challenge
  • interference
    • handshake and line probing
  • propagation delay
    • delay estimation
  • linear distortion
    • adaptive equalization
  • clock drifts between the receiver and the transmitter
    • timing recovery
    • advanced topic

main challenges

• challenge distortion

• time-varying discrepancies in clocks T′s=Ts$T_s^{\prime }=T_s$

• the channel is approximated as a linear filter in the continuous-time domain to begin analysis
  • D(jω)$D\left(j\omega \right)$: filter response
  • the filter is assumed to introduce all distortion and delays
• signal at receiver end: ˆs(t)$\hat{s}\left(t\right)$
  • delayed and distorted version of transmitted signal
• clock of transmitter: Ts$T_s$
• clock of receiver: T′s$T_s^{\prime }$
  • no guarantee these two are synchronized

fig: DAC at transmitter and ADC at the receiver

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delay compensation

• assuming the following are in sync
  • clock of transmitter: Ts$T_s$
  • clock of receiver: T′s$T_s^{\prime }$
• channel introduces a delay of d$d$ seconds
  • channel is a simple delay block
  • ˆs(t)=s(t−d)→D(jΩ)=e−jΩd$\hat{s}\left(t\right)=s(t-d)\to D\left(j\Omega \right)=e^{-j\Omega d}$
• we can write d=(b+τ)Ts$d=(b+\tau )T_s$ with b∈N$b\in N$ and |τ|<12$\left|\tau \right|<\frac{1}{2}$
• b$b$ is the bulk delay

• τ$\tau$ is the fractional delay

• bulk delay is simple to tackle
  • also called the integer delay
  • they are simply the delay that the channel adds to the signal
  • this does not sift the peaks of the data with respect to the sampling interval
• discontinuities in pilot tones help figure out bulk delay
  • impulses cannot be used as they are full band and get filtered out
• the fractional delay is more involved
  • it shifts the peaks with respect to the sampling intervals
  • interpolation is used to get the fractional delay compensation
• transmit b[n]=ejω0n$b\left[n\right]=e^{j{\omega }_{0}n}$
  • s[n]=cos((ωc+ω))n)$s\left[n\right]=\cos \left(\right({\omega }_c+{\omega }_{)}\left)n\right)$
• receive ˆs[n]=cos((ωc+ω0)(n−b−τ))$\hat{s}\left[n\right]=\cos \left(\right({\omega }_c+{\omega }_0\left)\right(n-b-\tau \left)\right)$
• after demodulation and bulk delay offset
  • ˆb[n]=ejω0(n−τ)$\hat{b}\left[n\right]=e^{j{\omega }_0(n-\tau )}$
• multiply by known frequency
  • ˆb[n]e−jω0n=e−jω0τ$\hat{b}\left[n\right]e^{-j{\omega }_{0}n}=e^{-j{\omega }_0\tau }$
• after offsetting bulk delay
  • ˆs[n]=s(n−τ)Ts$\hat{s}\left[n\right]=s(n-\tau )T_s$
• subsample values need to be computed
• in theory, compensate with a sinc fractional delay
  • h[n]=sinc(n=τ)$h\left[n\right]=sinc(n=\tau )$
• in practice use lagrange approximation
  • practical application of lagrange polynomials
• lagrange approximation is around n$n$
  • to compute x(n+τ)$x(n+\tau )$ with |τ|<12$\left|\tau \right|<\frac{1}{2}$

xL(n;t)=N∑k=−Nx[n−k]L(N)k(t) L(N)k(t)=n∏i=−N;i≠kt−ik−i  where =k=−N,…,N 

• x(n+τ)≈xL(n;τ)
• so, in summary
  • estimate the delay τ
  • compute the 2N+1 lagrangian coefficients
  • filter with the resulting FIR

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adaptive equalization

• measure to compensate for distortion
• let the channel distortion be D(z)
  • E(z) is the equalizer compensation to offset channel distortion
• in theory E(z)=1/D(z)
• but D(z) is not known

• D(z) may change over time during transmission

• hence the equalization E(z needs to adapt continuously
• following is the schematic of an adaptive equalizer

fig: core adaptive equalizer schematic

• the filter coefficient changes in time based on the error
  • obtained from the output with the transmitted signal
• the exact signal is sent by the transmitter
  • the receiver has the same copy to get the adaptive equalizer started
  • this is a bootstrapping technique
• there are some symbols that are common to both the transmitter and receiver together
  • this is called a training sequence
  • handshake 4 point QAM
  • the equalizer is initialized with this shared symbol set

fig: adaptive equalizer schematic in the big picture

• this process of bootstrapping is not error free
  • but a generally good place to get started
• details of adaptive signal processing is an advanced topic
  • needs more research, reading and understanding

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adsl

• ADSL: asymmetric digital subscriber line
• adsl receives signals on a copper wire channel
• DSLAM: digital subscriber line access multiplier

fig: telephone network overview

• last mile: copper wire connecting the home modem to the exchange (CO - central office)
• copper wire has a large bandwidth
• POTS: plain old telephone system
• the (A)symmetry in the bandwidth is the A of the ADSL

fig: adsl channel - copper wire bandwidth

channel propagation challenges

1. attenuation: the uneven curve across the bandwidth
   • physical wire imperfections
   • parasitic capacitance
2. electrical interference: large grey blog in a specific frequency region
   • running the vacuum for instance raises the noise floor of the copper channel
3. localized radio interference: the impulse at a specific frequency
   • ship-to-shore communications: 0 - 100 kHz
   • airplane communications: 100 - 500 kHz
   • AM radio band: 500 kHz +

fig: adsl channel propagation challenges

• the channel is divided into independent sub-channels
• different channels are treated separately
  • localized treatment in the receiver across all bands
  • the cleanest channels are used to send maximum data

subchannel structure

• allocate N sub-channels over the total positive bandwidth
• equal sub-channel bandwidth FmaxN
• equally spaced sub-channels with center frequency kFmaxN
  • k=0,…,N−1

digital design

• pick Fs=2Fmax
  • Fmax is high now
• center frequency for each subchannel
  • ωk−2πkFmax/NFs=2π2Nk
• bandwidth of each sub-channel 2π2N
• to send symbols over a subchannel
  • upsampling factor K≥2N

fig: subchannels of the adsl channel (N = 3)

• QAM modem is added on each channel
• decide on constellation size independently for each channel
  • clean channel gets high numbered constellation
  • noisy channel gets low numbered constellation
• noisy or forbidden sub-channels send zeros
• the structure of the communication scheme is sent to the receiver from the transmitter
  • part of handshake procedure
• classic modulation scheme is applied in each channel as per below schematic

fig: modem on each sub-channel

• the receiver modem bank has several modems in parallel
• each channel has two unique attributes
  • frequency of modulation
  • mappers symbols series

fig: modem bank

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discrete multitone modulation

• the modem banks maybe seen as oscillators whose output is summed to obtain a signal
  • each oscillator is scaled with an amplitude
  • and phase offset
  • this bank is run for N samples to get the signal
  • these are constant through the generation

fig: oscillator bank paradigm for modem bank

• in the modem scenario, the amplitude and phase change at every sample
  • they embed the complex symbol sequence
• with the discrete multitone modulation, adsl may be implemented with a simple inverse FFT
  • provided that the symbols can be help constant during the whole upsampling event
  • the modem structure can be mapped to the inverse DFT structure if this is done
• the ADSL trick:
  • instead of using a god lowpass filter use a the 2N-tap interval indicator h[n]=(1 for 0≤n≤2N0 otherwise ) 

fig: modem on each sub-channel

• oscillator in the modulator runs freely
• with simplification, in each chunk of 2N samples, the symbol is kept constant

fig: simplification of subchannel modem

• aggregate bandpass signal is calculated by c[n]=N−1∑k=0ak[⌊n2N⌋]ej2π2Nnk =2N∗IDFT2N{[a0[m] a1[m] …aN−1[m]0 0 …]}[n] m=⌊n2N⌋ 

fig: simplified subchannels’ modem bank

• final goal: calculate s[n] s[n]=Re{c[n]}=(c[n]+c∗[n])2

IDFT{[x0 x1 … xN−2 xN−2]}∗=IDFT{[x0 x1 … xN−2 xN−2]∗}

c[n]=2N∗IDFT{[a0[m] a1[m] … aN−1[m] 0 0…0]}[n]

• hence, since baseband always has read valued symbols

s[n]=N∗IDFT{[2a0[m] a1[m] … aN−1[m] a∗N−1[m] a∗N−2[m]…a∗1[m]]}[n]

ADSL schematic

fig: simplified subchannels’ modem bank

ADSL specs

• Fmax=1104kHz
• N = 256
• each QAM can send 0 - 15 bit per symbol
• forbidden channels 0 to 7
  • dedicated to voice
• channels 7 - 31: upstream data
• max theoretical throughput: 14.9 Mbps (downstream)

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references

• ecole
• itu adsl specs
• itu v.34 modem specs