Wireless Communications

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Wireless Communications
2
Wireless Communications

• Wireless is more and more widely deployed
  cellphone, wireless LAN,
• The fundamental fact is that if the sender sends
  a sine wave, the receiver will receive a sine
  wave at the same frequency. But with
• A different phase
• A new amplitude
• How do you design communication schemes based on
  that?

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Wireless Communications

• AM stronger signal when 1, weaker signal when
  0.
• FM faster waveform when 1, slower signal when
  0.
• BPSK 0 degree when 0, 180 degree when 1.

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Basic Wireless System

• The very basic wireless communication system
• Sender given the bit stream, convert it to the
  baseband waveform by a Low Pass Filter, then
  multiply with the carrier waveform (2.4GHz if
  802.11g, 1.8GHz if some cellphones), and send.
• Receiver given the signal received from the
  antenna, multiply it with a locally generated
  carrier, send it to the low pass filter,
  regenerate the baseband waveform.

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Basic Wireless System

• The sender cannot send the square waveform but
  has to send I(t) which is bandwidth limited from
  0Hz to some cutoff frequency BHz. The signal in
  the air in this simple system occupies frequency
  range f-B,fB.
• Because cos(2pf1t)cos(2pft) cos2p(f1tf)t
  cos2p(f1-f)t (constant dropped). So basically
  one frequency in the baseband will become two
  frequencies.
• The f-B,fB must be within the frequency band
  allocated for this system (around 20MHz in
  802.11g, around 25MHz in GSM) because the medium
  is shared

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Basic Wireless System

• How can the receiver regenerate the baseband
  waveform? A simplified explanation
• The sender sends I(t)cos(2pft).
• Assume there is no phase difference, the receiver
  multiplies I(t)cos(2pft) with cos(2pft), and gets
  
• I(t)cos2(2pft) I(t)1/2 1/2cos(4pft).
• Then, after the low pass filter, what is left is
  the low frequency component I(t).
• http//math2.org/math/trig/identities.htm

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Two Orthogonal Channels

• The sender sends I(t)cos(2pft) Q(t)sin(2pft).
• The receiver multiplies the received signal with
  cos(2pft) , and will get
• I(t)cos(2pft) Q(t)sin(2pft) cos(2pft)
  I(t)1cos(4pft) Q(t) sin(4pft)
• (constant dropped), and after the LPF, will have
  I(t).
• At the same time, the receiver also multiplies
  the received signal with sin(2pft) , and will get
  
• I(t)cos(2pft) Q(t)sin(2pft) sin(2pft) I(t)
  sin(4pft) Q(t) 1-cos(4pft)
• (constant dropped), and after the LPF, will have
  Q(t).
• So, the sender can send TWO baseband waveforms at
  the same time. Each baseband waveform can carry
  one voltage value, so a symbol is a point on the
  two-dimensional plane. We often use a complex
  number to represent the symbol, where the value
  in I(t) is the real part and the value in Q(t) is
  the imaginary part.

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BPSK, QPSK, QAM

• BPSK is using only one channel. And in this
  channel, only two possible voltages.
• Quadrature phase-shift keying (QPSK) is using
  both channels. In each channel, only two
  voltages.
• Quadrature amplitude modulation (QAM) is using
  both channels. In each channel, multiple
  voltages. If 4 levels of voltage, it is 16QAM. If
  it 8 levels of voltage, 64QAM.

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After Getting the Baseband Signal

• The baseband signal is a continuous waveform.
• You have to run an algorithm on that to convert
  the continuous waveform into bits.
• Main constraint of the algorithm should run
  very fast, i.e., capable of making a decision
  every symbol time, must be implemented in
  hardware, cost is issue.

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Issues

• How to offset the phase difference from the
  sender to the receiver?
• How to take samples at the right time?
• How to get rid of the residual values from nearby
  symbols, i.e., dealing with multipath?
• In wireless channels, when the sender sends one
  waveform, the receiver will receive many copies
  of it, with different phase offset, because the
  signal travels multiple paths with reflecting
  surfaces like the metal door, wall, etc.
• When taking a sample, the sample could be the
  best for the strongest path, but may not be for
  other paths. So some residual voltage from the
  previous symbol will show up.

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Exercise

• Consider an 802.11g device A transmitting at
  channel 1 which is 2.412GHz. There is another
  802.11g device B tuned to channel 6 which is
  2.437GHz. Assume B is in transmission range of A.
  Why wont B respond to As transmission?
• (a).Because Bs antenna cannot respond to
  waveforms centered at 2.412GHz.
• (b).Because Bs local sine wave is at 2.437GHz,
  and after multiplying it with the received
  waveform, the result is 0.
• (c). Because Bs upper layer software will filter
  out packets from other channels.
• (d). None of the above.

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Exercise

• An 802.11g channel has bandwidth of 22MHz. Which
  of the following statements is true?
• (a).Due to the bandwidth constraint, the highest
  data rate of 802.11g cannot be more than 22Mbps.
• (b).The bandwidth constraint limits the highest
  achievable speed, but the actual speed also
  depends on the strength of the channel.
• (c). An 802.11g transmitter transmits at the
  center of the channel frequency so the bandwidth
  limit has no effect on the speed.
• (d). None of the above.

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Exercise

• Which of the following statements about the
  baseband signal is true?
• (a). The baseband signal refers to the signal
  after the low pass filer.
• (b). The baseband signal is bandwidth-limited.
• (c). Both of the above.
• (d). None of the above.

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Exercise

• In wireless transmissions, we often use the terms
  sample and symbol. Which of the following
  statements is true?
• (a). A symbol is a complex number. A BPSK symbol
  is either -1 or 1.
• (b). A sample is a complex number. A BPSK sample
  is either -1 or 1.
• (c). Both of the above.
• (d). None of the above.

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Exercise

• With GNU Software Defined Radio, the transmitter
  and the receiver can operate at 500K symbols per
  second. Which of the following statement is true?
• (a).It means that we can send one symbol per 2
  microseconds.
• (b).It means that our data rate is 1Mbps if we
  use QPSK.
• (c).Both of the above.
• (d). None of the above.

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Some Important Concepts

• Bandwidth How fast can signals change. Or, the
  width of the frequency range that can pass the
  system.
• Noise Will be added to the signal. Random, but
  some statistics can be known with which we make
  our detection rules.
• Data rate How fast can we send bits, can be
  measured in bps. Upper bounded by the Shannons
  theorem given the bandwidth and noise.
• Propagation delay How long does it take for the
  symbol to reach to the destination.
• Symbol A point on the plane to represent one
  bit or multiple bits.
• Sample A reading of the received signal at some
  time instant. Usually a corrupted version of the
  symbol.
• Filter Some device or software that can remove
  certain frequency components in the received
  waveform.
• Carrier A sine wave to be multiplied with the
  baseband waveform.
• Baseband waveform The waveform that is a
  smoothed version of the square waveform after the
  low pass filter. To be multiplied with the
  carrier.