contents


  • from source to destination, a signal goes through many incarnations
  • it travels through a variety of analog communication channels
    • wireless, copper wire, optical fibre, et
    • each channel has a set of constraints the signal must adheres to
  • signal can undergo various transformations, such as conversion from analog to digital and later back to analog or signal regeneration

digital data throughput through time

  • Transatlantic Cable
    • 1866: 8 words per minute
      • \( \approx 5 \text{ } bps \)
    • 1956: AT&T, co-ax cable with 48 voice channels
      • \( \approx 3 \text{ } Mbps \)
    • 2005: Alcatel Tera10, fiber
      • 8.4 Tbps ( \( 8.4 \text{ } X \text{ } 10^{12} \) )
    • 2012: fiber
      • 60 Tbps
  • Voiceband modems
    • 1950s: Bell 202
      • 1200 bps
    • 1990s: V90
      • 56 Kbps
    • 2008; ADSL2+
      • 24 Mbps

success factors

distinctions of the DSP paradigm

  • integers are easy to regenerate
  • good phase control with digital filters
  • adaptive algorithms can be seamlessly integrated into DSP systems
    • procedures that adapt behavior in response to received signal

algorithmic nature of DSP synergizes with information theory

  • JPEG’s entropy coding
    • discrete cosine transform matched seamlessly to information theory techniques
    • techniques from different domains combined produces powerful algorithms
  • CD and DVD error corrections
    • encoding acoustic info matched to error corrector codes
    • even scratched cds and dvds play
  • trellis-coded modulation and viterbi decoding
    • analog channels are taken advantage of in communication systems with these techniques

hardware advancements

  • miniaturization of signal processing hardware
    • moore’s law
  • general-purpose platforms available
    • no need to make customized hardware for specific application
    • modular assembly of off-the-shelf hardware provides custom solutions
  • power efficiency
    • many communications channels can be processed by relatively small data centers

analog channel constraints

communication-system-signal-flow

fig: signal flow in a real world communication system

  • channels used:
    • air
    • copper
    • fiber
    • coax
    • copper
  • the signal goes through changes to be able to travel through the channel chain

channel capacity

  • unescapable limits of physical channels that affects the capacity of the channel
    • bandwidth constraints
      • signal will have to be limited to certain frequency band
    • power constraint
      • power over given bandwidth cannot be arbitrary
      • there will be limits
  • channel capacity is the the maximum amount of information that can be reliably delivered over a channel
    • usually in bits per second
  • the specifications of the channel is given to communication system engineers
    • to reliably transmit information over the channel

channel-capacity

fig: channel capacity representation

bandwidth and capacity
  • consider a situation where information encoded as a sequence of digital samples
    • is to be transmitted over a continuous-time channel
  • the digital sequence is interpolated with an interval \(T_s\)
  • if \(T_s\) is small:
    • more information per unit time - information is dense in time
    • trade-off is the bandwidth of signal will increase
      • bandwidth is proportional to \(T_s\)
      • \( \vert \Omega \vert > \Omega_N = \frac{\pi}{T_s} \)
power and capacity
  • all channels introduce noise
    • receiver has to guess what was transmitted
  • consider integers between 1 and 10 are being transmitted
    • and noise variance is 1
    • this causes a lot of guessing errors
    • the variance is in the range of the resolution of the numbers transmitted
  • transmit only odd numbers
    • less information i.e. lesser information
    • but lesser errors
    • snr is higher

channel capacity examples

AM radio channel
  • input is filtered in a lowpass operation
  • then a simple sinusoidal modulation is applied
  • the modulated signal is sent to an antenna where it is pushed into the radio spectrum

am-radio-channel

fig: AM radio channel

  • the AM spectrum:
    • 530 kHz - 1.7 MHz: scarce resource
    • each AM channel in the radio spectrum is 8kHz
    • each radio station gets allocated a specific channel
  • everybody has to share the radio bandwidth so the bands are regulated by law
    • specifically, power in the bandwidth is regulated
    • daytime/nighttime AM propagation behavior is very different
      • AM travels much further nights
    • too much AM power creates interference with other signals
    • too much power poses health hazards
telephone channel
  • called switched telephone network
  • phone connected to telephone exchange (CO - central office)
    • last mile cable
    • usually a couple km long
  • routing is done based on the number dialled by the incoming calls
  • exchange in the past was mechanical with rotary switches
    • now they are digital switches
  • the network maybe wires, fibres, satellites links and such

am-radio-channel

fig: AM radio channel

  • the channel itself is conventionally limited to 300Hz to 3000Hz
  • power limited to 0.2-0.7 ms forced by law
    • exact limit based on the hardware used
    • to protect hardware
  • SNR is high: around 30dB
    • analog cables in the ground
    • not a lot of interference

communication system design

  • everything is kept digital until signal needs transmitting through a physical channel
  • all-digital paradigm
    • all signal operations are performed in the digital domain
  • signal transmittance is done in physical channels
    • signal has to be modified to adhere to channel capacity
    • this is done at the transmitter
  • the channel capacity challenge looks similar to filter design challenge

channel-capacity

fig: channel capacity specs - similar to filter design specs

  • the physical channel capacity is converted to digital specs
  • the nyquist frequency i.e. the minimum sampling frequency is chosen based on the bandwidth

nyquist-frequency

fig: selecting the sampling frequency, given channel capacity

  • in the digital domain, maximum frequency is set to \(\pi\) \[ \Omega = 2\pi \frac{F}{F_s} \]

digital-spectrum

fig: converting to digital domain

transmitter design

  • a set of hypothesis is adhered to when designing transmission systems
common transmitter hypothesis
  • map bitstream into a sequence of symbols \( a[n] \)
    • using a mapper
    • eg: mapping bits to decimal value
  • model \(a[n]\) as a white random sequence
    • assume bitstream is completely random
    • incase of orderly sequence i.e. silence stretch in music
      • scrambler operation on bitstream ensures randomness
      • scrambling is a superficial processing tool
      • is completely invertible at receiving end
  • then convert sequence \(a[n]\) to a continuous-time signal
    • output must be within channel constraints

transmitter-design-scheme

fig: transmitter design scheme


bandwidth control

challenge #1: bandwidth constraint
  • the scrambled bitsream is random and white
  • its power spectrum density is simply its variance
  • so has constant power over the entire frequency spectrum
    • however the power has to fit only the frequency bandwidth of the specs of the analog channel

power-bandwidth-relationship

fig: distribution of power of scrambled sequence and spec bandwidth

  • bandwidth constraint demands control over the support of the signal
  • the full-frequency-band of the scrambled signal needs to be shrinked to the channel bandwidth

multirate techniques

  • tools for bandwidth controls
  • key idea of multirate technique:
    • increase or decrease the number of samples in a discrete-time signal
  • equivalent to going to continuous-time and re-sampling
    • staying the digital world is cleaner
    • so the process of conversion and back-conversion is mimicked digitally
upsampling via continuous-time
  • in continuous-time, upsampling scheme looks like the following

fig: interpolation and higher rate sampling - upsampling through continuous-time domain

  • for \( K = 3 \)

fig: consider a sequence that needs upsampling

fig: interpolated sequence in continuous-time

fig: 3 times higher frequency sampling

fig: resulting upsampled frequency

  • choice of interpolation interval is arbitrary
  • for \(T_s = 1\), the interpolated signal is \[ x_c = \sum_{m = -\infty}^{\infty} x[m] sinc(t-m) \]
  • the upsampled signal is \[ \begin{align} x^{\prime} & = x_c \bigg (\frac{n}{K} \bigg) \
    & = \sum_{m = -\infty}^{\infty} x[m] sinc \bigg( \frac{n}{K} - m \bigg) \
    \end{align} \]

  • in the frequency domain, the process looks like the following

fig: spectrum of input signal (top); spectrum of interpolated signal (middle); spectrum of re-sampled i.e. upsampled signal (bottom)

  • note how the upsampled signal takes up a smaller bandwidth than the signal it was upsampled from
upsampling in digital-time
  • keeping in mind the all-digital paradigm, it is desired to upsample within the digital domain

  • here, the number of samples needs to be increased by K
  • such that \( x_U[m] = x[n] \) when \(m \) is multiple of \(K\)
    • to be able to recover original signal
  • for lack of better strategy, put zeros elsewhere
    • insert \( K - 1 \) zeros in between consecutive samples
  • example: for \( K = 3 \) \[ x_U[m] = \ldots x[0], 0 , 0 , x[1] , 0 , 0 , x[2] , 0 , 0 \ldots \]

fig: sequence to be upsampled

fig: upsampled signal (\(K = 3\))

  • the fourier transform of the upsampled sequence is \[ \begin{align} X_U(e^{j\omega}) & = \sum_{m = - \infty}^{\infty} x_U[m] e^{-j\omega m} \
    & = \sum_{n = - \infty}^{\infty} x[n] e^{-j\omega nK} \
    & = X_U(e^{j\omega K}) \
    \end{align} \]

fig: spectrum of signal to be upsampled (top); periodic spectrum of signal to be upsampled (middle); spectrum of upsampled signal (bottom)

recovery with downsampling
  • the resulting upsampled signal has K copies in the full \( [-\pi,\pi] \) bandwidth spectrum
    • \( K = 3 \) in this case
  • a lowpass filter has to be applied to extract only one copy in the center
    • with a cutoff frequency \( \frac{\pi}{K} \)
  • recovery in time domain
    • insert \( K - 1\) zeros after every sample
    • ideal lowpass filtering with \( \omega_c = \frac{\pi}{K} \)

\[ \begin{align} x^{\prime} & = x_U*sinc\bigg(\frac{n}{K}\bigg) \
& = \sum_{i = -\infty}^{\infty} x_U[i] sinc\bigg( \frac{n-1}{K} \bigg) \
\text{ substitute } & i = mK \
& = \sum_{m = -\infty}^{\infty} x[m] sinc\bigg( \frac{n}{K} - m \bigg) \
\end{align} \]

  • downsampling is used to recover the upsampled sequence on the received side \[ x[n] = x_U[nK] \]
  • the filter used for lowpass has to either be ideal or fulfill the interpolation property
    • i.e. impulse response of filter should be the delta function
  • downsampling of generic signals more complicated than upsampling
    • aliasing has to be dealt with as some samples are begin discarded

fitting the transmitter spectrum

  • the channel imposes a frequency constraint
    • that signal frequency must only be between \(F_{min} \) and \(F_{max} \)
    • this has to be translated to the digital domain

nyquist-frequency

fig: selecting the sampling frequency, given channel capacity

  • let \(W = F_{max} - F_{min}\)
    • difference on the bandwidth limits in the positive half of the frequency spectrum
  • pick \(F_s\) such that \(F_s > 2 F_{max}\)
  • \(F_s = KW, \text{ } K \in \mathbb{N} \)

  • in the digital domain,
    • \( \omega_{max} - \omega_{min} = 2\pi \frac{W}{F_s} = \frac{2\pi}{K} \)
    • may simply be upsampled by K
    • the width obtained so will fit over the channel bandwidth
  • upsampling does not change the data rate
  • W symbols per seconds are produced and transmitted
  • this is then upsampled by K which is equal to the sampling frequency
  • W is the baud rate of the system
    • is equal to the available positive bandwidth
    • fundamental data rate of the system

transmitter-with-upsampling-to-fit-bandwidth

fig: transmitter schematic after upsampling to fit channel bandwidth

  • in the frequency spectrum this process looks as follows

transmitter-with-upsampling-to-fit-bandwidth

fig: scrambled white random signal

transmitter-with-upsampling-to-fit-bandwidth

fig: upsampled lowpass filtered signal

transmitter-with-upsampling-to-fit-bandwidth

fig: modulated to fit carrier bandwidth

raised cosines

  • FIR filters will be used in lowpass filtering after upsampling

  • sinc lowpass filter is difficult to implement in practice

    • the raised cosine filter is used instead

transmitter-with-upsampling-to-fit-bandwidth

fig: raised cosine lowpass filter frequency response

  • has a parameter to tune the gentleness of the transiton from passband to stopband

transmitter-with-upsampling-to-fit-bandwidth

fig: gentler raised cosine filters - tunable parameter

  • raised cosine filter is an ideal filter, but
    • is easier to approximate than the sinc filter
    • also satisfies the interpolation property of filters
  • even short approximations can provide a good response because of the nature of its impulse response

transmitter-with-upsampling-to-fit-bandwidth

fig: impulse response of raised cosine filter


power control

  • transmission reliability
    • transmitter sends a sequence of symbols \(a[n]\)
    • receiver obtains a sequence \( \hat{a}[n] \)
      • this is only an estimate of the transmitted signal
      • noise has leaked into the symbol
    • even if channel doesn’t distort signal, noise cannot be avoided
      • \( \hat{a}[n] = a[n] + \eta[n] \)
    • when noise is large
  • encoding schemes
    • PAM: pulse amplitude modulation
    • QAM: quadrature amplitude modulation

probability of error

  • depends on
    • power of the noise w.r.t power of signal
    • decoding strategy
    • the transmission symbols alphabet
  • increasing throughput does not necessarily increase the error
transmission symbols alphabet
  • a randomized bitstream is coming in
  • some upsampled and interpolated samples need to be sent over the channel
  • how are samples obtained from bitstream?
mapper
  • splits incoming bitstream into chunks
  • assign a symbol \( a[n] \) from a finite alphabet \( \mathcal{A} \) to each chunk
slicers
  • receives a value \( \hat{a}[n] \)
  • decide which symbol from \( \mathcal{A} \) is closest to \( \hat{a}[n] \)
  • piece back together the corresponding bitstream
two-level signalling
  • given:
    • \(G\): amplitude of signal
    • \( \sigma_0 \): standard deviation of noise
  • probability of error

\[ \begin{align} P_{err} & = erfc \bigg( \frac{G}{\sigma_0} \bigg) \
& = erfc \bigg( \frac{\sigma_s}{\sigma_0} \bigg) \
& = erfc(\sqrt{SNR}) \
\end{align} \]

  • probability of error decays exponentially with the signal-to-noise-ratio
  • this exponential decay trend is the norm in communication systems
    • while the absolute error might change depending on the constants

probability-error-function

fig: two-level signalling probability error function

takeaways
  • increase signal gain G (amplitude) to reduce probability of error
  • increasing G increase the power
  • keep in mind power cannot exceed channel constraint

pulse amplitude modulation (PAM)

  • mapper
    • split incoming bitstream into chunks of \(M\) bits
    • chunks define a sequence of integers \( k[n] \in { 0,1,\ldots, 2^M-1 } \)
    • \( a[n] = G( (-2^M + 1) + 2 k[n] ) \)
      • odd integers around zero
      • G: gain factor
  • slicer
    • \( a^{\prime}[n] = arg \text{ } \min_{a\in\mathcal{A}} [ \vert \hat{a} [n] - a \vert ] \)
error analysis
  • result very similar to two-level signal error analysis
  • bi-level signalling is PAM with M = 1
example: M = 2, G = 1
  • distance between points is 2G
    • points are indicated by the diamond markers
  • using odd integers creates a zero-mean sequence

pam-example

fig: pulse amplitude modulator; M = 2; G = 1

quadrature amplitude modulation (QAM)

  • mapper:
    • split incoming bitstream into chunks of \(M\) bits
      • \(M\): even
    • use \(\frac{M}{2}\) bits to define a PAM sequence \(a_r[n] \)
    • use the remaining \(\frac{M}{2} \) bits to define an independent PAM sequence \(a_i[n] \)
    • \( a[n] = G(a_r[n] _ ja_i[n]) \)
    • so the resulting
  • slicer:
    • \( a^{\prime} = arg \text{ } min_{a\in\mathcal{A}}[ \vert \hat{a}[n] - a \vert ] \)
example: M = 2, G = 1

qam-example

fig: quadrature amplitude modulator; M = 2; G = 1

example: M = 4, G = 1

qam-example

fig: quadrature amplitude modulator; M = 4; G = 1

example: M = 8, G = 1

qam-example

fig: quadrature amplitude modulator; M = 8; G = 1

probability of error
  • the slicer adds the square area centered around the symbol

qam-example

fig: quadrature amplitude modulator - error analysis

  • each received symbol \(\hat{a} \) is the sum of the original sample and an error \( \hat{a}[n] = a[n] + \eta[n] \)
  • the error term \( \eta[n] \) is assumed to be
    • a complex value
    • with gaussian distribution of equal variance
    • in both real and complex components
  • the probability of error is given by \[ \begin{align} P_{err} & = P[ \vert Re(\eta[n]) \vert > G ] + P[ \vert Im(\eta[n]) \vert > G ] \
    & = 1 - P[ \vert Re(\eta[n] ) \vert < G \wedge \vert Im(\eta[n]) \vert < G ] \
    & = 1 - \int_D f_\eta(z) dz \
    \end{align} \]

  • so now the slicer area is converted to a circle from being a square

qam-error-probability

fig: quadrature amplitude modulator - error analysis

  • if noise variance is assumed to be \( \frac{\sigma_0^2}{2} \)
    • in both real and imaginary component
  • probability of error is approximated by \[ P_{err} \approx \exp^{\frac{-G^2}{\sigma_0^2}} \]

  • to obtain probability of error as a function of SNR:
    • compute power of signal
  • assume all symbols equiprobable and independent, variance of signal is \[ \begin{align} \sigma_s^2 & = G^2 \frac{1}{2^M} \sum_{a\in\mathcal{A}} \vert a \vert^2 \
    & = G^2 \frac{2}{3} (2^M - 1) \
    \end{align} \]

  • plugging this back into the equation of error probability \[ P_{err} \approx \exp^{\frac{-G^2}{\sigma_0^2}} \approx \exp^{ -3 * SNR * 2^{(M+1)} } \]

qam-error-probability

fig: plotting error probability on a log-log scale

  • probability of error increases with number of points, because the area associated with each point becomes smaller
    • the decision at the receiving end in classifying a received sample to a symbol become harder
    • because it could belong to more other symbols than when the number of symbols is less
QAM design procedure
  • pick a probability of error that is acceptable (i.e. \(10^{-6} \))
  • find out the SNR imposed by the channel’s power constraint
  • find \(M\) using \( M = log_2 \bigg( 1 - \frac{3}{2} \frac{SNR}{\ln(p_e)} \bigg) \)
    • size of symbol constellation
    • might need rounding
      • to even number
  • final throughput (data rate) will be \(MW\)
    • M: size of symbol constellation
    • W: baud rate
regroup-1
  • bandwidth constraint fit discussed
  • given power constraint, the bits/symbol can be calculated
  • we know the theoretical throughput of the transmitter
  • how to transmit complex symbols over a real channel?

references