contents

  • the signal is prepared for transmission through the channel constraints
    • this includes modulating the signal
  • on the receiver end, the signal is demodulated
    • then is decoded to estimate the encoded information
    • estimate because there is error in
      • the encoding
      • the decoding
      • also noise is added in the channel during propagation

signal preparation

  • the signal flows through the following
    • signal in bitstream
    • scrambler makes it random and white
      • power spectral density is constant across full specrtrum
    • QAM mapper encodes bitstream to a complex valued symbol
      • resulting in \( a[n] \)
    • upsampler fits encoded signal to channel bandwidth
      • K times more samples
    • lowpass raised cosine remove the copies obtained after digital upsampling
      • cutoff frequency \( \frac{\pi}{K} \)

qam-transmitter

fig: QAM transmitter schematic

  • the signal thus obtained with a QAM decoder is \(b[n]\) \[ b[n] = b_r[n] + jb_i[n] \]
  • \(b[n]\) is a complex-valued baseband signal
  • having been subjected upsampling, its bandwidth is meets the carrier constraint
    • but does not sit in the bounds of the carrier’s bandwidth

complex baseband signal

fig: carrier bandwidth availability (green) and \(b[n]\) baseband (red)

modulation

  • modulation is the process of modifying the complex baseband that has the same size of the carrier bandwidth to sit exactly in the specified bandwidth
  • a complex baseband cannot be transmitted over a real physical channel
    • to fully reconstruct the information at the other end
  • so the complex baseband has to be made real before transmission

discrete-time domain

  • let \( \omega_c \) be the center frequency of the channel bandwidth
  • then, the modulated, real baseband is obtained as follows \[ \begin{align} s[n] & = Re\{ b[n] e^{j\omega n} \} \
    & = Re \{ (b_r[n] + jb_i[n])( \cos \omega_c n + j \sin \omega_c n ) \} \
    & = b_r[n] \cos \omega_c n - b_i [n] \sin \omega_c n \
    \end{align} \]
  • here,
    • the real part of the complex baseband is modulated with a cosine and
      • in-phase component
    • the imaginary part is modulated with a sine
      • quadrature component
    • both are at the carrier bandwidth central frequency
    • also, they are orthogonal to each other i.e. in quadrature
    • this is the source of the name QAM used to encode with the complex number symbols
  • \( s[n] \in \mathbb{N} \)
    • is used at the receiver end to recover the complex baseband signal

complex discrete-frequency domain

  • before modulation
    • real and imaginary component spectrums separated

complex-baseband-signal-spectrum

fig: carrier bandwidth availability (green); \(b[n]\) baseband real part (blue); \(b[n]\) baseband imaginary part (pink)

  • after modulation
    • real and imaginary modulated component spectrums separated
    • the real part is symmetric
    • the imaginary part is anti-symmetric

complex-baseband-signal-modulated

fig: carrier bandwidth availability (green); \(b[n]\) baseband modulated real part (blue); \(b[n]\) baseband modulated imaginary part (pink)

demodulation

  • demodulation is achieved by multiplying received signal by the carrier signal
  • in the QAM scenario, there are two carriers in the received signal
    • one sine and one cosine

in-phase part extraction

  • begin with multiplying by cosine to extract the real part of the received baseband

\[ \begin{align} s[n] \cos \omega_c n & = b_n[n] \cos^2 \omega_c n - b_i[n] \sin \omega_c n \cos \omega_c n \
& = b_r[n] \frac{1 + \cos 2 \omega_c n}{2} - b_i[n] \frac{2 \sin 2 \omega_c n}{2} \
& = \frac{1}{2} b_r[n] + \frac{1}{2} ( b_r[n] \cos 2 \omega_c n - b_i[n] \sin 2 \omega_c n ) \
\end{align} \]

  • the frequency component reveals one half of the real part of the transmitted and received signal
    • matched filter configuration: same raised cosine used at the transmitter is used at the receiver

received-baseband-signal-demodulated

fig: frequency domain of signal after multiplying with cosine wave (blue); raised cosine lowpass applied (green)

  • the raised cosine will eliminate everything but the real part of the transmitted baseband

real-baseband-signal-extracted

fig: recovered real part of transmitted baseband

quadrature part extraction

  • multiply by sine to extract the imaginary part of the transmitted baseband

\[ \begin{align} s[n] \sin \omega_c n & = b_r[n] \cos \omega_c n - b_i[n] \sin^2 \omega_c n \
& = - \frac{1}{2} b_i[n] + \frac{1}{2} ( b_r[n] \sin 2 \omega_c n - b_i[n] \cos 2 \omega_c n ) \
\end{align} \]

  • the frequency band looks similar to the demodulation of the in-phase part
  • the core signal is extracted with a raised cosine lowpass

design example

  • to be explored is a system that enables encoding and transmission of complex-valued sequence over a real-valued channel

QAM transmitter

  • the signal in a QAM transmitter is processed as follows:
    • signal in bitstream
    • scrambler makes it random and white
      • power spectral density is constant across full specrtrum
    • QAM mapper encodes bitstream to a complex valued symbol
      • resulting in \( a[n] \)
    • upsampler fits encoded signal to channel bandwidth
      • K times more samples
    • lowpass raised cosine remove the copies obtained after digital upsampling
      • cutoff frequency \( \frac{\pi}{K} \)
    • the filtered signal is multiplied with complex exponential whose frequency is the central frequency of carrier bandwidth
      • this results in a complex passband signal
    • the real part of the complex baseband is extracted along with modulation
    • this is sent to the DAC which propagates it into the channel

qam-transmitter

fig: QAM transmitter signal flow schematic

QAM receiver

  • goal of the receiver to obtain the original bitstream which is the core information that was transmitted
  • an ideal QAM receiver processes the received signal to retrieve that as follows:
    • analog signal is received from the channel
    • this analog signal is sampled with appropriate sampling rate
    • signal is split into two parts to modulate with cosine and sine separately
      • cosine demod results in the real part
      • sine demod results in the imaginary part
    • both demodulated signals go through their own lowpass filter
      • matched filter configuration: same lowpass used at the transmitter
    • the imaginary component is multiplied with complex root to and summed with the real part to construct an estimate of the transmitted baseband
    • this is then subjected to downsampling
    • thus obtained complex symbol sequence is passed through a slicer
      • the bit chuck associated with the symbol is obtain so
    • these chunks are assembled into a sequence and passed into a descrambler
      • this recovers the original bitstream

qam-receiver

fig: QAM receiver signal flow schematic

voiceband modem application

channel specifications
  • analog telephone channel
    • \(F_{min} = 450 Hz; F_{max} = 2850 Hz \)
    • usable bandwidth: \( W = 2400 Hz \)
    • center frequency: \( F_c = 1650 Hz \)
    • pick \( F_s = 3 * 2400 = 7200 Hz \)
      • so K = 3
    • \(\omega_c = 0.458 \pi \)
bandwidth constraint
  • sampling theorem states that the sampling frequency is to be higher than twice the maximum frequency
    • so atleast \( F_{max} * 2 = 2850 Hz = 5700 Hz \)
  • upsampling also has to be considered, so sampling frequency must also be an integer multiple of the channel bandwidth
    • channel bandwidth \(W = 2850 - 450 = 2400 Hz \)
    • with center frequency \(F_c = 1650 Hz \)
    • if upsamling factor is chosen to be three, then \( K = 3\)
    • so sampling frequency \(F_s = 3 * W = 23 * 2400 = 7200 Hz \)
      • this satisfies the sampling theorem frequency criteria as well
    • in the digital domain, \( \omega_c = 0.458 \pi \)
      • this is the modulating frequency
power constraint
  • maximum SNR: \( 22 dB \)
  • pick \( P_{err} = 10^{-6} \)
  • using QAM, find M (number of bits per signal) \[ M = \log_2 \Bigg( 1 - \frac{3}{2} \frac{10^{\frac{22}{10}}}{ln(10^{-6})} \Bigg) \approx 4.1865 \]
  • so pick M = 4 and use 16-point constellation
    • 4 points in each quadrant
  • final data-rate is \( WM = 4 * 2400 Hz = 9600 \) bits per second
    • W: baud rate (bandwidth of the channel)
theoretical channel capacity

  • capacity formula based on signal bandwidth and SNR \[ C = W \log_2 (1 + SNR) \]

  • only gives upper bound on the amount of information that can be sent over the channel

  • doesn’t actually state how to build a communication system to meet this specification

  • for the previously designed scheme
    • \( C \approx 17500 \) bps
    • this hits half the channel capacity
  • the gap can be narrowed with encoding techniques
  • this topic needs a more thorough study of information theory

references