Video Codec for Multimedia Communications

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COSC 3213 Computer Networks I Instructor Dr.
Amir Asif Department of Computer Science York
University Section M Topics 1. Digital
Information 2. Digital Communications versus
Analogue Communications 3. Channel Capacity
Nyquist and Shannon Bounds. 4. Time and Frequency
Domain Representations of Signals Garcia
Sections 3.1 3.5
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Digital Information

• Recall that the lowest layer in OSI is the
  physical layer.
• The physical layer deals with the transfer of raw
  bits and is the focus of our attention in the
  next few lectures.
• Communication systems can be classified in two
  categories
• Analogue Communication System dealing with
  analogue signals
• Digital Communication System dealing with digital
  signals
• Analogue signals are defined for the entire
  duration and can have any value.
• Digital signals are defined at fixed instants and
  can only have one of the pre-selected set of
  values.
• Information can be classified in two categories
• Block-oriented information usually arranged
  nicely in contiguous blocks. Includes data files,
  BW documents, and images
• Stream information usually arising from a natural
  process such as audio and video.

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Block Oriented Information (1)
InformationType Format Raw Data Size Data Compression Compressed Data Size(CR) Applications
Text Files ASCII Kbytes to Mbytes Lossless Compression such as Dictionary or Arithmetic codes. Compress, Zip, and variations (2 6) Disk Storage, Modem transmission
Scanned BW documents A4 (8.5 x 11 in) page _at_ 200 x 100 pixels / inch overhead 256 Kbytes CCITT Group 3 fax standard 1554 KB (1D)5 35 KB (2D)(5 50) Storage,Fax Transmission
Color images 8 x 10 inch photo scanned _at_ 400 pixels / in 38.4 Mbytes JPEG standard 1.2 8 Mbytes(5 30) Image storage or transmission
Activity 1 Show that the raw data size of the
color image (last row entry) is indeed
38.4Mbytes/s?
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Block Oriented Information (2)

• Data compression seeks to exploit the redundancy
  present in the data to encode information in a
  fewer number of bytes.
• Compression ratio (CR) is the ratio between the
  size of raw data to the size of the compressed
  data.
• Why Data Compression?
• Transmission time of Color image (raw) on V.90
  modem 12 minutes
• Transmission time of JPEG encoded Color image on
  V.90 modem 24 seconds
• Data Compression schemes can be classified in two
  categories
• Lossless Compression No loss of information but
  compression ratio limited.
• Lossy Compression Controlled loss of information
  for higher compression.
• Bound of Lossless Compression is provided by
  Shannon's Source encoding theorem which states
  that the minimum size of compressed data without
  loss of information equals the entropy of the
  source generating the data.

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Streamed Information
InformationType Format Raw Data Size Data Compression Compressed Data Size Applications
Voice 4 kHz voice 64 kbps PCM 64 kbps Digital telephony
ADPCM 16 - 32 kbps Voice mail, Telephony
Linear Prediction 8 - 16 kbps Cellular Telephony
Audio 16 24 kHz audio 512-748 kbps MPEG MP3 32 384 kbps MPEG audio
Video QCIF (176 x 144) or CIF (352 x 288)_at_ 10 - 30 fps 2 36.5 Mbps ITU H.261 coding 64 kbps 1.544 Mbps Video conferencing
720480 pels/frame _at_ 30 fps 249 Mbps ISO MPEG 2 2 6 Mbps NTSC TV, DVD
19201020 pels/frame _at_ 30 fps 1.6 Gbps ISO MPEG 2 19 38 Mbps HDTV
Activity 2 Show that the raw data size of HDTV
(last row entry) is indeed 1.6Gbps?
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Why Digital Communications (1)?
Digital Communications results in an improved
Signal to Noise ratio (SNR) as compared to
analogue communications.
Example of an Analogue Communication System
Note that the tx signal can not be recovered from
the rx signal due to attenuation and
distortions. Attenuation must be compacted so
that the received signal (1) is strong enough
for the receiver to detect the signal and (2)
maintains a level sufficiently higher than noise
for correct bit detection.
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Why Digital Communications (2)?
Example of a Digital Communication System
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Analog Signals

• Analog signals are classified in two categories
• Periodic Signals that repeat themselves
• Aperiodic or nonperiodic signals that do not
  repeat themselves.

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Periodic Signals

• Periodic signals repeat over time, i.e.,
• Aperiodic signals do not repeat themselves
  regularly.

s(t) Asin(2pf0t f0)
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Periodic Signals Sine wave
Activity 3 Identify the amplitude, fundamental
frequency, and phase of the sinusoidal signals?
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Frequency Representation of Periodic Signals (1)
Periodic signals with fundamental frequency f0
1/T Hz may be represented by the Fourier Series,
defined as
Time Domain Frequency Domain


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Frequency Representation of Periodic Signals (1)

• Fourier Series says that any periodic signal with
  the fundamental frequency of f0 can be
  represented as a linear combination of a
  sinusoidal wave with the fundamental frequency of
  f0 and the higher order harmonics of the
  sinusoidal wave with fundamental frequencies
  (kf0) for (2 k 8).
• As an example, consider the sum of the first two
  harmonics of
• Including more terms will make the approximation
  closer to a square wave.

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Frequency Representation of of Aperiodic Signals
(1)
Non-periodic signals may be represented by the
Fourier Transform, defined as
Time Domain Frequency Domain


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Streamed Data Analogue to Digital Conversion

• There are two steps involved in converting an
  analogue signal to a digital signal
• Sampling obtain the value of signal every T
  seconds.
• Choice of T is determined by how fast a signal
  changes, i.e., the frequency content of the
  signal
• Nyquist Sampling theorem says

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Analogue to Digital Conversion

• There are two steps involved in converting an
  analogue signal to a digital signal
• Quantization approximate signal to certain
  levels. Number of levels used determine the
  resolution.

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Example
First row of streamed information detailing voice
(from previous table) Voice maximum frequency
4 kHz voice Sampling rate (1 / T) gt 2
4000 or 8000 samples/second Sampling period (T)
1 / 8000 125 microseconds For digital
telephony, no. of levels (L) used in the uniform
quantizer are 256 Number of bits required to
represent a level log2(L) log2 (256) 8
bits Data rate 8000 8 or 64
kbps Activity 4Repeat for stereo music system
that contains a maximum frequency of 22 kHz. The
number of levels used by the uniform quantizer
are 64K. Remember there are 2 channels (L R) in
a stereo system. How much data will be generated
in one hour?
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Pulse Code Modulation (PCM)

• PCM (sampling followed by quantization) is used
  to digitize voice signals in telephony.
• Voice signal is band limited to 4 kHz (Sampling
  rate 8 ksamples/s)
• 8-bit nonuniform quantizer is used to quantize
  each sample (Data rate 64 kbits/s)
• It can be shown that the SNR for PCM (6m 10)
  dB

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Fundamental Problem

• Fundamental Question How fast (maximize data
  rate) and reliably (minimize errors) digital
  transmission can occur through a channel?
• Depends upon a number of factor
• Amount of energy present in the signal
• Noise properties of the channel
• Distance for signal to propagate
• Bandwidth (BW) of the transmission medium
• Bandwidth determines the range of frequencies
  that can be transmitted through a channel.
• Consider a sinusoidal wave
• Frequency present in the wave f0 Hz or 2pf0
  radians/s
• Apply s(t) at the input of the channel and
  measure the amplitude of the output
• Calculate the ratio of the amplitude of the
  output to that of the input (referred to as
  Amplitude response function) that provides a
  measure of the Bandwidth

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Communication Channels Frequency Domain
Characterization (1)

• Communication channels can be characterized
  either in the frequency domain or time domain
• To obtain the frequency domain characterization,
  apply a sinusoidal signal at the input
• and measure the output
• Amplitude Response A(f) is the ratio of the
  output amplitude to input amplitude (Aout / Ain)
  as a function of frequency.
• Phase Shift is the variation in f(f ) as a
  function of frequency.

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Communication Channels Frequency Domain
Characterization (2)

• Examples of Amplitude and Frequency-response
  functions
• Bandwidth (BW) is the range of frequencies
  passed by the channel.
• Attenuation is the reduction in signal power as
  in propagates through the channel.
• Attenuation in dB 10 log10(Pin / Pout)
• Activity 5 Show that the attenuation of the
  above channel is -201og10(A ( f ))?

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Communication Channels Frequency Domain
Characterization (3)
Activity 6 What are the bandwidth of the
channels with the following amplitude-response
functions?
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Communication Channels Time Domain
Characterization

• Time Domain characterization of a channel is
  determined by applying an impulse at the input of
  the channel and measuring the output.
• where h(t) is called the impulse response.
• Impulse response and amplitude-response function
  are related by the Fourier transform. Knowing
  one, the other can be calculated.
• Given the impulse response (or the
  amplitude-response function), the output for any
  given input can be calculated.

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Baseband Transmission (1)

• Baseband Transmission is the transmission of
  digital information over a lowpass channel.
• Two parameters used to characterize the
  performance of a communication system
• Data rate in bps Number of bits transmitted per
  second.
• Error rate Fraction of bits received in error.
• Consider a binary lowpass channel using Polar NRZ
  representation for bits
• The communication system is designed in such a
  way that the response to a single pulse is p(t)

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Baseband Transmission (2)

• Overall response to the binary data 101101 is
• How does the receiver detects bits from r(t)?
• Sample r(t) at t 0, T, 2T, 3T
• If r(kT) gt 0, then bit 1 was transmitted at t
  kT
• If r(kT) lt 0, then bit 0 was transmitted at t
  kT
• For t 0, we get
• The second term involving the summation results
  in Intersymbol Interference (ISI).
• ISI causes overlapping between neighbouring
  pulses and therefore degrades the ability of the
• receiver in detecting the transmitted bits from
  the received signal. ISI is a nuisance.
• How do we reduce ISI? Select a pulse p(t) that
  contributes zero Intersymbol Interference.

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Channel Capacity (1)
Width of pulse (T) 1 / (2W) where W BW of the
channel Maximum transmission rate 2W pulses /
second
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Channel Capacity (2)

• Let us calculate the bit rate for a channel with
  BW W Hz.
• If bandwidth is W Hz, then minimum width of pulse
  1 / 2W seconds.
• Nyquist Signaling Rate 2W pulses/s (maximum
  data rate assuming a noiseless channel).

• Binary Transmission 1 bit per pulse gt
  Transmission rate 1 / duration of pulse 2W
  bps
• M-level Transmission
• No. of bits represented by one pulse log2(M)
  m
• Nyquist Signaling Rate in bps m (1 / duration
  of pulse) 2mW 2W log2(M) bps
• Increasing m, increases the transmission rate
  !!!! Is there an upper limit?

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Channel Capacity (3)

• Channel Capacity (C) is the maximum bit rate
  supported by a channel.
• Can the channel capacity C be made infinite by
  increasing m?
• No! There are other constraints introduced by
  noise and channel interference.

More Errors
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Channel Capacity (4)

• By increasing m, the difference between adjacent
  levels is reduced affecting SNR
• Reduction in SNR affects the Channel Capacity
  (C).
• Shannon Channel Capacity theorem provides an
  upper bound on the channel capacity in terms of
  bandwidth for a noisy channel
• C W log2 (1 SNR) bps
• Recall that the Nqyuist theorem provided the
  upper bound on the channel capacity for a
  noiseless channel. The Shannon theorem provides
  the upper bound for a noisy channel.
• Shannon theorem provides no indication of levels.
  For M 2m levels or symbols, the channel
  capacity based on the Shannon theorem is given by
• C W log2 (1 SNR) / m symbols/s
• Activity 7 Calculate the channel capacity of a
  dial-in modem that has a BW of 3400 Hz if the
  best SNR possible in the modem is 40dB. Recall
  that
• SNR in dB 10 log10(SNR on a linear scale).

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Channel distortion

• Probability of error in presence of additive
  White Gaussian noise (AWGN)
• where d is the distance between levels and s is
  the standard deviation of noise.
• Activity 8 Consider a dial-in modem that uses
  Pulse Shift Keying (PSK) of a sinusoidal wave
  having a maximum amplitude of 5V. The bandwidth
  of the twisted pair wire used is limited to
  3400Hz. Assuming that the noise introduced by the
  channel is AWGN with a variance (s2) of 2.25.
  Calculate
• the signal-to-noise ratio (SNR) for the channel
  in dB.
• the channel capacity C of the twisted pair as a
  channel.
• the probability of error Pe for binary
  transmission.
• Answers (a) 7.45 dB (b) 2.78 kbps (c) 2.43 x 10-6