DIGITAL COMMUNICATIONS

1
DIGITAL COMMUNICATIONS

• Part I Source Encoding(Chapter 6)

2
Why digital?

• Ease of signal generation
• Regenerative repeating capability
• Increased noise immunity
• Lower hardware cost
• Ease of computer/communication integration

3
Basic block diagram
Info source
Source encoder
Channel encoder
Digital modulation
CH
channel
Output transducer
Source decoder
Channel decoder
Digital demod
4
Some definitions

• Information source
• Raw datavoice, audio
• Source encoderconverts analog info to a binary
  bitstream
• Channel encodermap bitstream to a pulse pattern
• Digital modulator RF carrier modulation of bits
  or bauds

5
A bit of history

• Foundation of digital communication is the work
  of Nyquist(1924)
• Problemhow to telegraph fastest on a channel of
  bandwidth W?
• Ironically, the original model for communications
  was digital! (Morse code)
• First telegraph link was established between
  Baltimore and Washington in 1844

6
Nyquist theorem

• Nyquist theorem, still standing today, says that
  over a channel of bandwidth W, we can signal
  fastest with no interference at a rate no more
  than 2W
• Any faster and we will get intersymbol
  interference
• He further proved that the pulse shape that
  achieves this rate is a sinc

7
Signaling too fast

• Here is what might happen when signaling exceeds
  Nyquists rate

Transmittted bitstream
Received bitstream
Pulse smearing could have been avoided if
pulses had more separation, I.e. bitrate reduced
8
Shannon channel capacity

• Claude Shannon, a Bell Labs Mathematician proved
  in 1948 that a communication channel is
  fundamentally speed-limited. This limit is given
  by
• CWlog2(1P/NoW) bits/sec
• Where W is channels bandwidth, P signal power
  and No is noise spectral density

9
Implications of channel capacity

• If data rate is kept below channel capacity, RltC,
  then t is theoretically possible to achieve
  error-free transmission
• If data rate exceeds channel capacity, error-free
  transmission is no longer possible

10
First step toward digital comm sampling theorem

• Main question can a finite number of samples of
  a continuous wave be enough to represent the
  information? OR
• Can you tell what the original signal was below?

11
How to fill in the blanks?

• Could you have guessed this? Is there a unique
  signal connecting the sample

12
Sampling schemes

• There are at least 3 sampling schemes
• Ideal
• Flat-top
• Sample and hold

13
Ideal sampling

• Ideal sampling refers to the type of samples
  taken. Here, we are talking about impulse
  like(zero width) samples.

Ts
14
Ideal sampler

• Multiply the continuous signal g(t) with a train
  of impulses

g(t)
g?(t)?g(nTs) ?(t-nTs)
??(t-nTs)
Ts
15
Key question

• What is the proper sampling rate to allow for a
  perfect reconstruction of the signal from its
  samples?
• To answer this question, we need to know how g(t)
  and g?(t) are related?

16
Spectrum of g?(t)

• g?(t) is given by the following product
• g?(t)g(t)?g(t-nTs)
• Taking Fourier transform
• G?(f) G(f)fs??(f-nfs)
• Graphical rendition of this convolution follows
  next

17
Expanding the convolution

• We can exchange convolution and summation
• G?(f)G(f)fs??(f-nfs) fs ? G(f) ? (f-nfs)
• Each convolution shifts G(f) to f nfs

G(f)
G(f) ? (f-nfs)
nfs
18
G?(f)final result

• Spectrum of the sampled signal is then given by
• This is simply the replication of the original
  continuous signal at multiples of sampling rate

G?(f)fs ? G(f-nfs)
19
Showing the spectrum of g?(t)

• Each term of the convolution is the original
  spectrum shifted to a multiple of sampling
  frequency

G(f)
G?(f)
fs
fs 2fs
20
Recovering original signal

• It is possible to recover the original spectrum
  by lowpass filtering the sampled signal

G?(f)
fs
W
fs 2fs
LPF
W
-W
21
Nyquist sampling rate

• In order to cleanly extract baseband (original)
  spectrum, we need sufficient separation with the
  adjacent sidebands
• Min. separation can be found as follows

G?(f)
fs-wgtW
fs
fsgt2W
W
fs
22
Sampling below Nyquistaliasing

• If signal is sampled below its Nyquist rate,
  spectral folding, or aliasing, occurs.

Lowpass filtering will not recover the baseband
spectrum intact as a result of spectral folding
fslt2W
23
Sample-and-hold

• A practical way of sampling a signal is
  sample-and-hold operation. Here is the
  ideasignal is sampled and its value held until
  the next sample

24
Issues

• Here are the questions we need to answer
• What is the sampling rate now?
• Can the message be recovered?
• What price do we pay for going with a practical
  approach

25
Modeling sample-and-hold

• The result of sample-and-hold can be simulated by
  writing the sampled signal as
• s(t)?m(nTs)h(t-nTs)
• Where h(t) is a basic square pulse and m(t) is
  the baseband message

This is a square pulse h(t) scaled by
signal Sample at that point, ie m(nTs)h(t-nTs)
Ts
26
A systems view

• It is possible to come up with a system that does
  sample-and-hold.

h(t)
h(t)
X
Ts
Ideal sampling
Each impulse generates a square pulse,h(t), at
the output. Outputs are also spaced by Ts this we
have a sample-and- hold signal
27
Message reconstruction

• Key question can we go back to the original
  signal after sample-and-hold ?
• This question can be answered in the frequency
  domain

28
Spectrum of the sample-and-hold signal

• Sample-and-hold signal is generated by passing an
  ideally sampled signal, m?(t), through a filter
  h(t). Therefore, we can write
• s(t) m?(t)h(t)
• or
• S(f) M?(f)H(f)

Known( it is a sinc)
Contains message M(f)
what we have available
29
Is message recoverable?

• Lets look at the individual components of S(f).
  From ideal sampling results
• M?(f)fs?M(f-kfs)

M?(f)
30
Problems with message recovery

• The problem here is we dont have access to
  M?(f). If we did, it would be like ideal sampling
• What we do have access to is S(f)
• S(f) M?(f)H(f)
• We therefore have a distorted version of an
  ideally sampled signal

31
Example message

• Lets show what is happening. Assume a message
  spectrum that is flat as follows

M(f)
W
-W
M?(f)
fs 2fs
32
Sample-and-hold spectrum

• We dont see M?(f). We see M?(f)H(f). Since h(t)
  was a square pulse of width Ts, H(f) is sinc(fTs)
  .

M?(f).
W
f
H(f)
f
1/Tsfs
33
Distortion potential

• The original analog message is in the lowpass
  term of M?(f)
• H(f) through the product M?(f)H(f) causes a
  distortion of this term.
• Lowpass filtering of the sample-and-hold signal
  will only recover a distorted message

34
Illustrating distortion
M?(f)
W
2fs
fs
H(f)
want to recover this
f
1/Tsfs
Sample and hold signal. If lowpass filtered, the
original Message is not recovered
What is actually recovered
35
How to control distortion?

• In order to minimize the effect of H(f) on
  reconstruction, we must make H(f) as flat as
  possible in the message bandwidth(-W,W)
• What does it mean? It means move the first zero
  crossing to the right by increasing the sampling
  rate, or decreasing pulse width

36
Does it make sense?

• The narrower the pulse, hence higher sampling
  rate, the more accurate you can capture signal
  variations

37
Variation on sample-and-hold

• Contrast the two following arrangements

Ts
?
sample period and pulse width are not the same
38
How does this affect reconstruction?

• The only thing that will change is h(t) and hence
  H(f)

M?(f)
W
f
H(f)
want to recover this
different zero crossing
1/?
f
Sample and hold signal. If lowpass filtered, the
original Message is not recovered
What is actually recovered
39
How to improve reconstruction?

• Again, we need to flatten out H(f) within (-W,W).
  and the way to do it is to use narrower pulses
  (smaller ?)

40
Sample-and-hold converges to ideal sampling

• If reducing the pulse width of h(t) is a good
  idea, why not take it to the limit and make them
  zero?
• We can do that in which case sample-and-hold
  collapses to ideal sampling(impulses are zero
  width pulses)

41
Pulse Code Modulation

• Filtering, Sampling, Quantization and Encoding

42
Elements of PCM Transmitter

• Encoder consists of 5 pieces
• Transmission path

Continuous message
Quantizer
Encoder
LPF
Sampler
Regenerative repeater
Regenerative repeater
43
Quantization

• Quantization is the process of taking continuous
  samples and converting them to a finite set of
  discrete levels

?
1.52
1.2
.86
-0.41
44
Defining a quantizer

• Quantizer is defined by its input/output
  characteristics continuous values in, discrete
  values out

out
out
in
in
Output remains constant Even as input varies over
a range
Midtread type
Midrise type
45
Quantization noise/error

• Quantizer clearly discards some information.
  Question is how much error is committed?

q(m)
Message(m)
Quantized message (v)
Errorqm-v
46
Illustrating quantization error
Sampled quantized
v3
?
Quantization error
v2
v1
? quantizer step size
47
More on ?

• ? Controls how fine samples are quantized.
  Equivalently, ? controls quantization error.
• To determine ? we need to know two parameters
• Number of quantization levels
• Dynamic range of the signal

48
? for a uniform quantizer

• Let sample values lie in the range ( -mmax,
  mmax). We also want to have exactly L levels at
  the output of the quantizer. Simple math tells us

max
L levels
?2mmax/L
min
49
Quantization error bounds

• Quantization error is bounded by half the step
  size

Level 2
Error q
?
Error q
Level 1
qlt?/2
50
Statistics of q

• Quantization error is random. It can be positive
  or negative with equal probability.
• This is an example of a uniformly distributed
  random variable.

Density function f(q)
1/?
q
?/2
-?/2
51
Quantization noise power

• Any uniformly distributed random variable in the
  range (-a/2 to a/2) has an average
  power(variance) given by a2/12.
• Here, quantization noise range is ?, therefore

?2q ?2/12
52
Signal-to-quantization noise

• Leaving aside random noise, there is always a
  finite quantization noise.
• Let the original continuous signal have power
  Pltm2(t)gt and quantization noise variance(power)
  ?2q
• (SNR)qP/ ?2q12P/ ?2

53
Substituting for ?

• We have related step size to signal dynamic range
  and number of quantization levels
• Therefore, signal to quantization noise(sqnr)
• sqnr(SNR)q3P/m2maxL2

?2mmax/L
54
Example

• Let m(t)cos(2?fmt). What is the signal to
  quantization noise ratio(sqnr) for a 256-level
  quantizer
• Average message power P is 0.5, therefore
• sqnr(3x0.5/1)25629830450dB

55
Nonuniform quantizer

• Uniform quantization is a fantasy. Reason is that
  signal amplitude is not equally spread out. It
  occupies mostly low amplitude levels

56
Solutionnonuniform intervals

• Quantize fine where amplitudes spend most of
  their time

57
Implementing nonuniform quantizationcompanding

• Signal is first processed through a nonlinear
  device that stretches low amplitudes and
  compresses large amplitudes

Large amplitudes pressed
output
Low amplitudes stretched
input
58
A-law and ?-law

• There are two companding curves, A-law and ?-law.
  Both are very similar
• Each has an adjustment parameter that controls
  the degree of companding (slope of the curve)
• Following companding, a uniform quantization is
  used

59
Encoder

• Quantizer outputs are merely levels. We need to
  convert them to a bitstream to finish the A/D
  operation
• There are many ways of doing this
• Natural coding
• Gray coding

60
Natural coding

• How many bits does it take to represent L-levels?
  The answer is
• nlog2L bits/sample
• Natural coding is a simple decimal to binary
  conversion
• 0000
• 1001
• 2010
• 3011
• .
• 7111

Encoder output(3 bits per sample
Quantizer levels(8)
61
Gray coding

• Here is the problem with natural coding if
  levels 2(010) and 1(001) are mistaken, then we
  suffer two bit errors
• We want an encoding scheme that assigns code
  words to adjacent levels that differ in at most
  one bit location

62
Gray coding example

• Take a 4-bit quantizer (16 levels). Adjacent
  levels differ by juts one bit

0 0 0 0 1 1 0 0 0 0 2 0 1 0 0 3 0 1 0 1 4
1 1 0 1 .
63
Quantizer word size

• Knowing n, we can refer to n-bit quantizers
• For example, if L256 with n8bits/sample
• We are then looking at an 8-bit quantizer

64
Interaction between sqnr and bit/sample

• Converting sqnr to dB provides a different
  insight. Take 10log10(sqnr)
• sqnrkL2 where k3P/m2max
• In dB
• (sqnr)dB?20logL ?20log2n
• (sqnr)dB ?6n dB

65
sqnr varies linearly with bits/sample

• What we just saw says higher sqnr is achieved by
  increasing n(bits/sample).
• Question then is, what keeps us from doing that
  for ever thus getting arbitrarily large sqnrs?

66
Cost factor

• We can increase number of bits/sample hence
  quantization levels but at a cost
• The cost is in increased bandwidth but why?
• One clue is that as we go to finer quantization,
  levels become tightly packed and difficult to
  discern at the receiver hence higher error rates.
  There is also a bandwidth cost

67
Basis for finding PCM bandwidth

• Nyquist said in a channel with transmission
  bandwidth BT, we can transmit at most 2BT pulses
  per second
• R(pulses/second)lt2BT(Hz)
• Or
• BT(Hz)gtR/2(pulses/second)

68
Transmission over phone lines

• Analog phone lines are limited to 4KHz in
  bandwidth, what is the fastest pulse rate
  possible?
• Rlt2BT2x40008000 pulses/sec
• Thats it? Modems do a bit faster than this!
• One way to raise this rate is to stuff each pulse
  with multiple bits. More on that later

69
Accomodating a digital source

• A source is generating a million bits/sec. What
  is the minimum required transmission bandwidth.
• BTgtR/2106/2500 KHz

70
PCM bit rate

• The bit rate at the output of encoder is simply
  the following product
• R(bits/sec)n(bits/sample)xfs(samples/sec)
• Rnfs bits/sec

quantized
Encoded at 5 bits/sample
1 0 1 1 0 1
71
PCM bandwidth

• But we know sampling frequency is 2W.
  Substituting fs2W in Rn fs
• R2nW (bits/sec)
• We also had BTgtR/2. Replacing R we get
• BTgtnW

72
Comments on PCM bandwidth

• We have established a lower bound(min) on the
  required bandwidth.
• The cost of doing PCM is the large required
  bandwidth. The way we can measure it is
• Bandwidth expansion quantified by
• BT/Wgtn (bits/sample)

73
Bandwidth expansion factor

• Similar to FM, there is a bandwidth expansion
  factor relative to baseband, i.e.
• ?BT/Wgtn
• Lets say we have 8 bits/sample meaning it takes
  , at a minimum, 8 times more than baseband
  bandwidth to do PCM

74
PCM bandwidth example

• Want to transmit voice (4KHz ) using an 8-bit
  PCM. How much bandwidth is needed?
• We know W4KHz, fs8 KHz and n8.
• BTgtnW8x400032KHz
• This is the minimum PCM bandwidth under ideal
  conditions. Ideal has to do with pulse shape used

75
Bandwidth-power exchange

• We said using finer quantization (more
  bits/sample) enhances sqnr because
• (sqnr)dB ?6n dB
• At the same time we showed bandwidth increases
  linearly with n. So we have a trade-off

76
sqnr improvement

• Lets say we increase n by 1 from 8 to 9
  bits/sample. As result, sqnr increases by 6 dB
• sqnr ?6x8 ?48
• sqnr ?6x9 ?54

6dB
77
Bandwidth increase

• Going from n 8 bits/sample, to 9 bits/sample,
  min. bandwidth rises from 8W to 9W.
• If message bandwidth is 4 KHz, then
• BT64 KHz for n8
• BT72 KHz for n9

8 KHz or 12.5 increase
78
Is it worth it?

• Lets look at the trade-off
• Cost in increased bandwidth12.5
• Benefit in increased sqnr 6dB
• Every 3 dB means a doubling of the sqnr ratio. So
  we have quadrupled sqnr by paying 12.5 more in
  bandwidth

79
Another way to look at the exchange

• We provided 12.5 more bandwidth and ended up
  with 6 dB more sqnr.
• If we are satisfied with the sqnr we have, we can
  dial back transmitted power by 6 dB and suffer no
  loss in sqnr
• In other words, we have exchanged bandwidth for
  lower power

80
Similarity with FM

• PCM and FM are examples of wideband modulation.
  All such modulations provide bandwidth-power
  exchange but at different rates. Recall ?BT/W
• FM.SNR?2
• PCM..SNR22?

Much more sensitive to beta, Better exchnage
81
Complete PCM system design

• Want to transmit voice with average power of 1/2
  watt and peak amplitude 1 volt using 256 level
  quantizer. Find
• sqnr
• Bit rate
• PCM bandwidth

82
Signal to quantization noise

• We had
• sqnr3P/m2maxL2
• We have L256, P1/2 and mmax1.
• sqnr9830450 dB

83
PCM bitrate

• Bit rate is given by
• R2nW (bits/sec)2x8x400064 Kb/sec
• This rate is a standard PCM voice channel
• This is why we can have 56K transmission over the
  digital portion of telephone network which can
  accomodating 64 Kb/sec.

84
PCM bandwidth

• We can really talk about minimum bandwidth given
  by
• BTminnW8x400032 KHz
• In other words, we need a minimum of 32 KHz
  bandwidth to transmit 64 KB/sec of data.

85
Realistic PCM bandwidth

• Rule of thumb to find the required bandwidth for
  digital data is that bandwidthbit rate
• BTR
• So for 64 KB/sec we need 64 KHz of bandwidth
• One hertz per bit

86
Differential PCM

• Concept of differential encoding is of great
  importance in communications
• The underlying idea is not to look at samples
  individually but to look at past values as well.
• Often, samples change very little thus a
  substantial compression can be achieved

87
Why differential?

• Lets say we have a DC signal and blindly go
  about PCM-encoding it. Is it smart?
• Clearly not. What we have failed to realize is
  that samples dont change. We can send the first
  sample and tell the receiver that the rest are
  the same

88
Definition of differential encoding

• We can therefore say that in differential
  encoding, what is recorded and ultimately
  transmitted is the change in sample amplitudes
  not their absolute values
• We should send only what is NEW.

89
Where is the saving?

• Consider the following two situations
• The right samples are adjacent sample differences
  with much smaller dynamic range requiring fewer
  quantization levels

2
0
-0.4
2
2
2
2
-0.4
0.4
-0.8
0.4
1.6
0.8
1.6
1.6
0.8
90
Implementation of DPCMprediction

• At the heart of DPCM is the idea of prediction
• Based on n-1 previous samples, encoder generates
  an estimate of the nth sample. Since the nth
  sample is known, prediction error can be found.
  This error is then transmitted

91
Illustrating prediction

• Here is what is happening at the transmitter

To be trasmited
Prediction error
Past samples(already sent)
Prediction of the Current sample
Only Prediction error is sent
92
What does the receiver do?

• Receiver has the identical prediction algorithm
  available to it. It has also received all
  previous samples so it can make a prediction of
  its own
• Transmitter helps out by supplying the prediction
  error which is then used by the receiver to
  update the predicted value

93
Interesting speculation

• What if our power of prediction was perfect? In
  other words, what if we could predict the next
  sample with no error?. What kind of communication
  system would be looking at?

94
Prediction error

• Let m(t) be the message and Ts sample interval,
  then prediction error is given

Prediction error
95
Prediction filter

• Prediction is normally done using a weighted sum
  of N previous samples
• The quality of prediction depends on the good
  choice of weights wi

96
Finding the optimum filter

• How do you find the best weights?
• Obviously, we need to minimize the prediction
  error. This is done statistically
• Choose a set of weights that gives the lowest (on
  average) prediction error

97
Prediction gain

• Prediction provides an SNR improvement by a
  factor called prediction gain

98
How much gain?

• On average, this gain is about 4-11 dB.
• Recall that 6 dB of SNR gain can be exchanged for
  1 bit per sample
• At 8000 samples/sec(for speech) we can save 1 to
  2 bits per sample thus saving 8-16 Kb/sec.

99
DPCM encoder

• Prediction error is used to correct the estimate
  in time for the next round of prediction

quantizer
encoder
Prediction error

Input sample

Prediction error
-

Prediction
Updated prediction
N-tap prediction
100
Delta modulation (DM)

• DM is actually a very simplified form of DPCM
• In DM, prediction of the next sample is simply
  the previous sample

Prediction error
Estimate of
101
DM encoder-diagram
out
?
in
-?
1-bit quantizer
Prediction error(??)

Input sample

Prediction error
-

Prediction
Updated prediction
Delay Ts
102
DM encoder operation

• Prediction error generates ? at the output of
  quantizer
• If error is positive, it means prediction is
  below sample value in which case the estimate is
  updated by ? for the next step

103
Slope overload effect

• Signal rises faster than prediction ? too small

samples
Ts
?
predictions
initial estimate
104
Steady state granular noise

• Prediction can track the signal prediction error
  small

Two drops to reach the signal
?
105
Shortcomings of DM

• It is clearly the prediction stage that is
  lacking
• Samples must be closely taken to insure that
  previous-sample prediction algorithm is
  reasonably accurate
• This means higher sample rates

106
Multiplexing

• Concurrent communications calls for some form of
  multiplexing. There are 3 categories
• FDMA(frequency division multiple access)
• TDMA(time division multiple access)
• CDMA(code division multiple access)
• All 3 enjoy a healthy presence in the
  communications market

107
FDMA

• In FDM, multiple users can be on at the same time
  by placing them in orthogonal frequency bands

guardband
user 1 user 2 user N
TOTAL BANDWIDTH
108
FDMA exampleAMPS

• AMPS, wireless analog standard, is a good example
• Reverse link(mobile-to-base) 824-849MHz
• Forward link 869-894 MHz
• channel bandwidth30 KHz
• total channels 833
• Modulation FM, peak deviation 12.5 KHz

109
TDMA

• Where FDMA is primarily an analog standard, TDMA
  and CDMA are for digital communication
• In TDMA, each user is assigned a time slot, as
  opposed to a frequency slot in FDMA

110
Basic idea behind TDMA

• Take the following 3 digital lines

frame
111
TDM-PCM
TDM-PAM
TDM-PCM(bits)
quantizer and encoder
quantizer and encoder
channel
lpf
decoder
lpf
112
Parameters of TDM-PCM

• A TDM-PCM line multiplexing M users is
  characterized by the following parameters
• data rate(bit or pulse rate)
• bandwidth

113
TDM-PCM Data rate

• Here is what we have
• M users
• Each sampled at Nyquist rate
• Each sample PCMd into n bit words
• Total bit rate then is
• RM(users)xfs(samples /sec/user)xn(bits/sec)
• nMfs bits sec

114
TDM-PCM bandwidth

• Recall Nyquist bandwidth. Given R pulses per
  second, we need at least R/2 Hz.
• In reality we need more (depending on the pulse
  shape) so
• BTRnMfs Hz

115
T1 line

• Best known of all TDM schemes is ATTs T1 line
• T1 line multiplexes 24 voice channels(4KHz) into
  one single bitstream running at the rate of 1.544
  Mb/sec. Lets see how

116
T1 line facts

• Each of the 24 voice lines are sampled at 8 KHz
• Each sample is then encoded into 8 bits
• A frame consists of 24 samples, one from each
  line
• Some data bits are preempted for control and
  supervisory signaling

117
T1 line structureall frames except 1,7,13,19...
channel 24
channel 2
channel 1
1
2
3
4
5
6
7
8
1
2
3
4
5
6
7
8
1
2
3
4
5
6
7
8
information bits (8-bits per sample)
FRAME(repeats)
118
Inserting non-data bits

• In addition to data, we need slots for signaling
  bits (on-hook/off hook, charging)
• Every 6th frame (1,7,13,19..) is selected and the
  least significant bit per channel is replaced by
  a signaling bit

channel 24
channel 2
channel 1
1
2
3
4
5
6
7
1
2
3
4
5
6
7
1
2
3
4
5
6
7
119
Framing bit

• Timing is of utmost significance in T1. We MUST
  be able to know where the beginning of each frame
  is
• At the end of each frame a single bit is added to
  help with frame identification

channel 24
channel 2
channel 1
1
2
3
4
5
6
7
8
1
2
3
4
5
6
7
8
1
2
3
4
5
6
7
8
F
information bits (8-bits per sample)
120
T1 frame length

• How long is one frame?One revolution generates
  frame

sampled at 8KHz
rotates at 8000 revs/sec.
24
frame length1/8000 125 microseconds
121
T1 bit rate per frame

• Data rate
• 8x24192 bits per frame
• Framing bit rate
• 1 bit per frame
• Total per frame
• 193 bits/frame

122
Total T1 bit rate

• We know there are 8000 frames a sec. and there
  are 193 bits per frame. Therefore
• T1 rate193x80001.544 Mb/sec

123
Signaling rate component

• Not all 1.544 Mb/sec is data. Since every 8th bit
  per channel of every 6th frame was replaced by
  signaling bit, we have
• signaling rate(8000)(1/6)1333 bits/sec

124
TDM hierarchy

• It is possible to build upon T1 as follows

64 kb/sec
DS-2
DS-1
1st level multiplexer
2nd level multiplexer
3rd level multiplexer
DS-3
24
DS-0
DS-3 44.736 Mb/sec
DS-2 6.312 Mb/sec
7 lines
DS-1 1.544 MB/sec
125
Recommended problems

• 6.2
• 6.15
• 6.17