Source coding

DIGITAL COMMUNICATIONS

Part I: Source Encoding

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

©2000 Bijan Mobasseri 2

Basic block diagram

Info Source Channel Digital
source encoder encoder modulation

channel CH

Output Source Channel Digital
transducer decoder decoder demod


Some definitions
• Information source
– Raw data:voice, audio
– Source encoder:converts analog info to a binary
bitstream
– Channel encoder:map bitstream to a pulse
pattern
– Digital modulator: RF carrier modulation of
bits or bauds


A bit of history
• Foundation of digital communication is the
work of Nyquist(1924)
• Problem:how 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

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

Signaling too fast
• Here is what might happen when signaling
exceeds Nyquist’s rate

Transmittted bitstream

Received bitstream

Pulse smearing could have been avoided if pulses
had more separation, I.e. bitrate reduced


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
C=Wlog2(1+P/NoW) bits/sec
• Where W is channel’s bandwidth, P signal
power and No is noise spectral density


Implications of channel capacity
• If data rate is kept below channel capacity,
R<C, then t is theoretically possible to
achieve error-free transmission
• If data rate exceeds channel capacity, error-
free transmission is no longer possible


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?


How to fill in the blanks?
• Could you have guessed this? Is there a
unique signal connecting the samples?


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


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

Ts


Ideal sampler
• Multiply the continuous signal g(t) with a
train of impulses
g(t)
gδ(t)=Σg(nTs) δ(t-nTs)

Σδ(t-nTs)

Ts

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 gδ(t) are related?


Spectrum of gδ(t)
• gδ(t) is given by the following product
gδ(t)=g(t)Σδ(t-nTs)
• Taking Fourier transform
Gδ(f)= G(f)*{fsΣδ(f-nfs)
• Graphical rendition of this convolution
follows next


Expanding the convolution
• We can exchange convolution and
summation
Gδ(f)=G(f)*{fsΣδ(f-nfs)= fs Σ {G(f)* δ (f-nfs)}
• Each convolution shifts G(f) to f= nfs
G(f)
G(f)* δ (f-nfs)}

nfs


Gδ(f):final result
• Spectrum of the sampled signal is then
given by
Gδ (f)=fs Σ {G(f-nfs)

• This is simply the replication of the original
continuous signal at multiples of sampling
rate


Showing the spectrum of gδ(t)
• Each term of the convolution is the original
spectrum shifted to a multiple of sampling
frequency G(f)

Gδ(f)
fs

fs 2fs

Recovering the original signal
• It is possible to recover the original
spectrum by lowpass filtering the sampled
signal
Gδ(f)
fs

W
fs 2fs
LPF

-W W


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
Gδ(f)
fs fs-w>W

fs>2W
W fs


Sampling below Nyquist:
aliasing
• 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

fs<2W

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


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?


Modeling sample-and-hold
• The result of sample-and-hold can be
simulated by writing the sampled signal as
s(t)=Σm(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


A system’s view
• It is possible to come up with a system that
does sample-and-hold.
h(t)

X h(t)
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
Ts


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


Spectrum of the sample-and-hold
signal
• Sample-and-hold signal is generated by
passing an ideally sampled signal, mδ(t),
through a filter h(t). Therefore, we can write
s(t)= mδ(t)*h(t)
or
S(f)= Mδ(f)H(f)
what we have available Contains message M(f) Known( it is a sinc)


Is message recoverable?
• Let’s look at the individual components of
S(f). From ideal sampling results
Mδ(f)=fsΣM(f-kfs)
Mδ(f)


Problems with message recovery
• The problem here is we don’t have access to
Mδ(f). If we did, it would be like ideal
sampling
• What we do have access to is S(f)
S(f)= Mδ(f)H(f)
• We therefore have a distorted version of an
ideally sampled signal

Example message
• Let’s show what is happening. Assume a
message spectrum that is flat as follows
M(f)

-W W

Mδ(f)

fs 2fs


Sample-and-hold spectrum
• We don’t see Mδ(f). We see Mδ(f)H(f). Since
h(t) was a square pulse of width Ts, H(f) is
sinc(fTs) . M (f).
δ

W
f
H(f)

1/Ts=fs f

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


Illustrating distortion
Mδ(f)

W
fs 2fs

want to recover this H(f)

1/Ts=fs f

Sample and hold signal.
If lowpass filtered, the original What is actually recovered
Message is not recovered


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


Does it make sense?

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


Variation on sample-and-hold
• Contrast the two following arrangements

Ts

τ
sample period
and pulse width
are not the same


How does this affect
reconstruction?
• The only thing that will change is h(t) and
hence H(f) Mδ(f)

W
f
want to recover this H(f) different zero crossing

1/τ f
Sample and hold signal.
What is actually recovered
If lowpass filtered, the original
Message is not recovered


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 τ)


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)


Pulse Code Modulation

Filtering, Sampling, Quantization and
Encoding

Elements of PCM Transmitter
• Encoder consists of 5 pieces

Continuous
LPF Sampler Quantizer Encoder
message

• Transmission path

Regenerative Regenerative
repeater repeater


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


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


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

Message(m) Quantized message (v)
q(m)

Error=q=m-v

Illustrating quantization error
Sampled
quantized

v3

∆
v2 Quantization error

v1

∆ quantizer step size


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


∆ 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


Quantization error bounds
• Quantization error is bounded by half the
step size
Level 2

∆
Error q

Error q
Level 1

|q|<∆/2


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


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

σ2q= ∆2/12


Signal-to-quantization noise
• Leaving aside random noise, there is always
a finite quantization noise.
• Let the original continuous signal have
power P=<m2(t)> and quantization noise
variance(power) σ2q
(SNR)q=P/ σ2q=12P/ ∆2


Substituting for ∆
• We have related step size to signal dynamic
range and number of quantization levels

∆=2mmax/L
• Therefore, signal to quantization
noise(sqnr)
sqnr=(SNR)q=[3P/m2max]L2


Example
• Let m(t)=cos(2πfmt). 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)2562=98304~50dB


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


Solution:nonuniform intervals
• Quantize fine where amplitudes spend most
of their time


Implementing nonuniform
quantization:companding
• Signal is first processed through a nonlinear
device that stretches low amplitudes and
compresses large amplitudes
Large amplitudes pressed

output

Low amplitudes stretched

input


A-law and µ-law
• There are two companding curves, A-law
and µ-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


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


Natural coding
• How many bits does it take to represent L-
levels? The answer is
n=log2L bits/sample
• Natural coding is a simple decimal to binary
conversion
0…000
1…001
Quantizer levels(8) 2…010 Encoder output(3 bits per sample
3…011
……….
7…111


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


Gray coding example
• Take a 4-bit quantizer (16 levels). Adjacent
levels differ by juts one bit
0…0001
1…0000
2…0100
3…0101
4…1101
………….

Quantizer word size
• Knowing n, we can refer to n-bit quantizers
• For example, if L=256 with n=8bits/sample
• We are then looking at an 8-bit quantizer


Interaction between sqnr and
bit/sample
• Converting sqnr to dB provides a different
insight. Take 10log10(sqnr)
• sqnr=kL2 where k=[3P/m2max]
• In dB
(sqnr)dB=α+20logL= α+20log2n
(sqnr)dB= α+6n dB


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
sqnr’s?


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 ©2000 Bijan Mobasseri 66

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)<2BT(Hz)
Or
BT(Hz)>R/2(pulses/second)


Transmission over phone lines
• Analog phone lines are limited to 4KHz in
bandwidth, what is the fastest pulse rate
possible?
R<2BT=2x4000=8000 pulses/sec
• That’s 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


Accomodating a digital source
• A source is generating a million bits/sec.
What is the minimum required transmission
bandwidth.
BT>R/2=106/2=500 KHz


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)
R=nfs bits/sec
quantized

101101 Encoded at 5 bits/sample


PCM bandwidth
• But we know sampling frequency is 2W.
Substituting fs=2W in R=n fs

R=2nW (bits/sec)
• We also had BT>R/2. Replacing R we get
BT>nW


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/W>n (bits/sample)


Bandwidth expansion factor
• Similar to FM, there is a bandwidth
expansion factor relative to baseband, i.e.
β=BT/W>n
• Let’s say we have 8 bits/sample meaning it
takes , at a minimum, 8 times more than
baseband bandwidth to do PCM


PCM bandwidth example
• Want to transmit voice (~4KHz ) using an
8-bit PCM. How much bandwidth is
needed?
• We know W=4KHz, fs=8 KHz and n=8.
BT>nW=8x4000=32KHz
• This is the minimum PCM bandwidth under
“ideal” conditions. Ideal has to do with
pulse shape used

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


sqnr improvement
• Let’s say we increase n by 1 from 8 to 9
bits/sample. As result, sqnr increases by 6
dB
sqnr= α+6x8= α+48
+6dB
sqnr= α+6x9= α+54


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
BT=32 KHz for n=8 +4 KHz or 12.5% increase
BT=36 KHz for n=9


Is it worth it?
• Let’s look at the trade-off:
– Cost in increased bandwidth:12.5%
– 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


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


Similarity with FM
• PCM and FM are examples of wideband
modulation. All such modulations provide
bandwidth-power exchange but at different
rates. Recall β=BT/W
2
• FM…….SNR~β
• PCM…..SNR~22β Much more sensitive to beta,
Better exchnage


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


Signal to quantization noise
• We had
sqnr=[3P/m2max]L2
We have L=256, P=1/2 and mmax=1.

sqnr=98304~50 dB


PCM bitrate
• Bit rate is given by
R=2nW (bits/sec)=2x8x4000=64 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.


PCM bandwidth
• We can really talk about minimum
bandwidth given by
BT|min=nW=8x4000=32 KHz
• In other words, we need a minimum of 32
KHz bandwidth to transmit 64 KB/sec of
data.


Realistic PCM bandwidth
• Rule of thumb to find the required
bandwidth for digital data is that
bandwidth=bit rate
BT=R
• So for 64 KB/sec we need 64 KHz of
bandwidth
One hertz per bit


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


Why differential?
• Let’s 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 don’t change. We can send
the first sample and tell the receiver that the
rest are the same

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.


Where is the saving?
• Consider the following two situations
2 0 -0.4
2 2 2 2 0.4 -0.8 -0.4
1.6 0.8 0.4
1.6 1.6

0.8

• The right samples are adjacent sample
differences with much smaller dynamic
range requiring fewer quantization levels


Implementation of
DPCM:prediction
• 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


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


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


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?


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

ˆ
e(nTs ) = m(nTs ) − m (nTs )

Prediction error


Prediction filter
• Prediction is normally done using a
weighted sum of N previous samples
N
m (nTs ) = ∑ wi m(( n − i )Ts )
ˆ
i =1

• The quality of prediction depends on the
good choice of weights wi


Finding the optimum filter
• How do you find the “best” weights?
• Obviously, we need to minimize the
prediction error. This is done statistically
Min { e 2 ( nTs )}
over w

• Choose a set of weights that gives the
lowest (on average) prediction error


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

σ2
M message power
Gp = 2 =
σe prediction error power


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.


DPCM encoder

Input sample + Prediction error
+ quantizer encoder
Prediction
- error

Prediction +
Updated prediction
N-tap
prediction

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

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


DM encoder-diagram
out
∆
in

-∆
Input sample +
+ 1-bit Prediction error(±∆)
Prediction quantizer
-
error
Prediction +
Updated prediction
Delay Ts


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


Slope overload effect
• Signal rises faster than prediction: ∆ too
small samples
Ts

∆ predictions

initial estimate


Steady state: granular noise
• Prediction can track the signal; prediction
error small
Two drops to reach the signal

∆


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


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


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


FDMA example:AMPS
• AMPS, wireless analog standard, is a good
example
– Reverse link(mobile-to-base): 824-849MHz
– Forward link: 869-894 MHz
– channel bandwidth:30 KHz
– total # channels: 833
– Modulation: FM, peak deviation 12.5 KHz


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


Basic idea behind TDMA
• Take the following 3 digital lines

frame


TDM-PCM

TDM-PAM TDM-PCM(bits)

quantizer and quantizer and
encoder encoder

channel

lpf
decoder
lpf


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


TDM-PCM Data rate
• Here is what we have
– M users
– Each sampled at Nyquist rate
– Each sample PCM’d into n bit words
• Total bit rate then is
R=M(users)xfs(samples /sec/user)xn(bits/sec)
=nMfs bits sec


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
BT=R=nMfs Hz


T1 line
• Best known of all TDM schemes is AT&T’s
T1 line
• T1 line multiplexes 24 voice
channels(4KHz) into one single bitstream
running at the rate of 1.544 Mb/sec. Let’s
see how


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


T1 line structure:
all frames except 1,7,13,19...

channel 1 channel 2 channel 24

1234567812345678 12345678

information bits (8-bits per sample)

FRAME(repeats)


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

1234567 1234567 1234567


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

1234567812345678 12345678F

information bits (8-bits per sample)


T1 frame length
• How long is one frame?One revolution
generates frame
sampled at 8KHz

rotates at 8000 revs/sec.

24

frame length=1/8000=
125 microseconds


T1 bit rate per frame
• Data rate
– 8x24=192 bits per frame
• Framing bit rate
– 1 bit per frame
• Total per frame
– 193 bits/frame


Total T1 bit rate
• We know there are 8000 frames a sec. and
there are 193 bits per frame. Therefore

T1 rate=193x8000=1.544 Mb/sec


Signaling rate component
• Not all 1.544 Mb/sec is data. In every 6th
frame, we replace 24 data bits by signaling
bits. Therefore
signaling rate=
(8000 frames/sec)(1/6)(24 bits)=32 Kbits/sec


TDM hierarchy
• It is possible to build upon T1 as follows

64 kb/sec
DS-2
DS-1
DS-3
1st level 2nd level 3rd level
24 multiplexer multiplexer multiplexer

DS-0

DS-2: 7 lines DS-3:
DS-1: 44.736 Mb/sec
6.312 Mb/sec
1.544 MB/sec


Recommended problems
• 6.2
• 6.15
• 6.17


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