Digital Telephony
Digital Telephony
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Analog/digital systems
Analog signal
Digital signal
Discrete signal
Analog
SP
A/D converter
QuantizSampler
er
FsFmax Error is
Analog signal
introduced
-voltage
-speech
-pressure
or
-tape
data from -simulations
-digital devices
DSP
D/A
Digital Signal Processor
-digital computer
-dedicated dig. hw
-programmable hw
Digital signal
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Issues
 Reconstruction
accuracy
Conditions for perfect reconstruction
 Digital
signal is not just an approx.
representation of an analog signal
Could be generated digitally
The processing being performed may not be
realizable in analog
 The
theory of discrete time signal
processing is independent of continuous
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Digital vs. analog processing
 DSP implementations
are flexible,
programmable and modular
 More precise and repeatable
 Performance and cost effectiveness (riding
the microelectronics wave)
 Direct mapping of mathematical
expressions with less approximation
possible (enables sophisticated algorithms)
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Digital vs. analog ...
 Digital
hardware can be multiplexed better
than analog. Allows integration of multiple
operations and services on a h/w platform
 Digital storage is more reliable, cheaper and
more compact
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On the other hand
 Analog
SP still offers higher bandwidth
 Higher dynamic range
 Can be very low power
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Analog to Digital Conversion
 To
convert “real-world” analog signals to
digital signals for processing
 Sampling
 Quantizing and coding
Xa(t)
Analog signal
X [n]
Sampler
Quantizer
and Coder
Discrete signal
Digital Telephony
Xq[n]
Digital signal
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Sampling
 Uniform
One sample every T seconds (ideally)
x[n] = xa(nT),
-n
Sampling period: T
Sampling frequency: Fs=1/T
 Assume: xa(t) = Acos( 2pFt+q) = Acos(Wt+q)
 Then: x[n] = Acos[2pFnT+q] = Acos[WTn+q]
= Acos[w n+q], where w=WT is called the
normalized or discrete domain frequency
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f
= F/Fs must be rational in order for x[n] to be
periodic
If f = k/N, then x[n] is periodic with period N
 Now, xa(nT) = Acos(WTn+q)
= Acos((W+2pk/T)Tn+q)
is periodic in Wwith period 2p/T


Also, x[n] = Acos[w n+q] = Acos[(w+2pk) n +q]
is periodic in wwith period 2p
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
xs(t) = xa(nT) d(t- nT)
convert to
n=-

discrete
sequence
xa(t)
x[n] = xa(nT)

S(t) = d(t- nT)
n=-
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 Let
us look at the continuous time Fourier
transform of xs(t)
1
Xs(jW) = 2p Xa(jW) * S(jW)
S(jW) =
2p
T
Xs(jW) =
1
T

 d(W-kWs)
k=-

Xa(jW-kjWs)
k=-
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Xa(jW) must be bandwidth limited
 If the max frequency in Xa(jW) is WN, then
the sampling rate Ws2WN ensures no
information is lost due to aliasing
 This sampling rate is known as Nyquist rate
 A lower sampling rate causes a distortion of
the signal due to Aliasing
 If no Aliasing occurs, the signal can be
perfectly reconstructed by passing through
an ideal low pass filter with
 Thus,
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Reconstruction
Hr(jW)
Xs(jW)
-Wc
Wc
Ws>2WN
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Xr(jW) = Hr(jW) Xs(jW)
if WNWc(Ws-WN)
then Xr(jW) = Xc(jW)
Reconstruction
Frequency response of ideal reconstruction filter
Hr(jW) =
{
T, WcWWc
0, otherwise
T
-Wc
Wc
Impulse response of ideal reconstruction filter
sin pt/T
hr(t)=
pt/T
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Reconstruction
 Xr(jW)
= Hr(jW) Xs(jW)
 xr(t) = xs(t) * hr(t)

= [k=-xa(kT) d(t-kT)]*hr(t)

k=-xa(kT) hr(t-kT)

sin p(t-nT)/T
k=-xa(kT) p(t-nT)/T
=
=
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xa(t)
xs(t)
hr(t)
xr(t)
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Sampling theorem
 If
the highest frequency contained in a
signal xa(t) is W0 and the signal is uniformly
sampled at a rate WsW0, then xa(t) can be
exactly recovered from its sample values
using the interpolation function
sin pt/T
hr(t)=
pt/T
and then xa(t) = k=-xa(kT) hr(t-kT), where
{xa(kT) } are the samples
of xa(t), and

T=2p/Ws
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Quantization and coding
 Quantization:
Converting discrete time signal to digital
xq(n) =Q [x(n)]
D
Quantization
step
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Q(x)
3D
D
D
-D/
-7D/ -5D/ -3D/
-D
D/ 3D/ 5D/
7D/
x
-D
-3D
-4D
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Quantization
 Rounding: Assign x[n] to the closest
quantization level
 Quantization
error
eq[n] = xq[n] - x[n]
-D/eq[n] D/
Uniformly distributed
 mean
=0
 variance
= D/1
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Quantization
of quantizer: xmax-xmin
 Quantization levels: m
 Assuming uniform quantization
 Range
xmax-xmin
D=
m-1
= Xm/ (m-1)
where Xm = (xmax-xmin)/2 is called the
full-scale level of the A/D converter
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Coding
 Coding
is the process of assigning a unique
binary number to each quantization level
 Number of bits required  log2m
 Alternatively, given b+1 bits
D=(xmax-xmin)/2b+1 =Xm /2b
 For A/D
devices, the higher Fs and m, the
less the error (and the more the cost of the
device)
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x(n)
xq(n)
Quantizer
x(n)
xq(n)
+
eq(n)
 Assuming
dynamic range of A/D converter
is larger than signal amplitude
2
2 2
SNR = 10 log10(sx/se) = 10 log10(sx/(D/1))
2
2
2b
= 10 log10(12.2 sx/(Xm))
=6.02b +10.8 + 20 log10(Xm/sx)
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Signal to Quantiiation Noise Ratio (dB)
Uniformly Encoded PCM
13
12
11
10
9
8
80
60
Number of bits
per sample
40
20
-40
-30
-20
-10
0
dB
X/Xm
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Example
 What
is the minimum bit rate that a uniform
PCM encoder must provide to encode a
high fidelity audio signal with a dynamic
range of 40 dB? Assume the fidelity
requirements dictate passage of a 20-kHz
bandwidth with a minimum signal-to-noise
ratio of 50 dB. For simplicity, assume
sinusoidal input signals.
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Companding

Companded PCM with analog compression and expansion
Compressed
Digital
Codewords
Input
Signal
A/D
Compression
D/A
Linear PCM
Encoder
F ( x) = sgn( x)
Linear PCM
Decoder
ln( 1   x )
ln( 1   )
1
-1
F ( y ) = sgn( y) [(1   )| y|-1 ]

Output
Signal
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Expansion
-1  x  1
-1  y  1
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Segment Approximation
Uniform quantization
111
110
101
100
011
010
001
000
Input Sample Values
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T1 Channel Bank
1
2
A/D
T1 transmission
Line
Analog
Inputs
D/A
24
•Eigth bits per PCM code word
•companding functions with mu=255
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Performance of a 55Encoder
Signal-to-quantization noise ratio (dB)
40
30
20
1 7
noise power =
pi qi2

12 i =0
Piecewise linear
8 bit 255
33
27
22
8 bit 255
7 bit 100
10
-70
-60
-50
-40
-30
-20
-10
Signal Power of sinewave (dBm0)
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dB
29
Total Noise Power
40
30
Pe = 10 -8
20
1
Pe =010 -6
-70
-60
-50
40 dB range of possible signals
Signal-to-total noise noise ratio
Pe = 0
30 dB required for good
communication
Pe = 10 -5
Pe = 10 -3 15 dB at which persons fin
communication difficult
Pe = 10 -4
-40
-30
-20
-10
0 3
Signal Power relative to full-load signal (dBm0)
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dB
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Error Performance
 Fewer
than 10% of 1 min intervals to have
BER worse than 10E-6
 Fewer than 0.2% of 1 sec intervals to have
BER worse than 10E-3
 92% error free sec
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DS1 Signal Format
bits in 125 s 1 = 125 10 -6
8000
 193 x 8000 = 1.544 Mbs
 Bit “robbing” technique used on each sixth
frame to provide signaling information
 (8x24)+1=193
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Plesiochronous Transmission Rates
Japanese Standard
North America
Standard
97728 kbits/s
European Standard
564992 kbits/s
x4
x4
97728 kbits/s
274176 kbits/s
x3
x6
32064 kbits/s
139264 kbits/s
x4
x3
44736 kbits/s
x5
x4
x7
6312 kbits/s
x4
1544 kbits/s
x24
34368 kbits/s
8448 kbits/s
x3
2048 kbits/s
x4
x30
64 kbits/s
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Plesiochronous Digital Hierarchy
MULTIP
LEXING
LEVELS
(DS)
# OF VOICE NORTH
EUROPE JAPAN
CHANNELS AMERICA
0
1
0.064
1
24
1.544
30
2
(4xDS1)
0.064
0.064
1.544
2.048
48
3.152
3.152
96
6.312
6.312
120
8.448
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Multiplexing # OF VOICE
Levels
CHANNELS
480
3 (7xDS2)
NORTH
AMERICA
672
44.376
1344
91.053
EUROPE
JAPAN
34.368
32.064
1440
4 (6xDS3)
97.728
1920
4032
139.264
274.176
5760
397.200
7680
565.148
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Plesiochronous Digital Hierarchy
 The
output of the M12 multiplexer is
operating 136 kbs faster than the agragate
rate of four DS1 6.312 vs 4x1.544=6.176
 M12 frame has 1176 bits, i.e. 294-bit
subframes ; each subframe is made of up of
49-bits blocks; each block starts with a
control bit followed by a 4x12 info bits
from four DS1 channels
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Makeup of a DS2 Frame
M1 01 02 03 04
C1 01 02 03 04
F0 01 02 03 04
C2 01 02 03 04
C3 01 02 03 04
F1 01 02 03 04
Bit stuffing
M1 01 02 03 04
C1 01 02 03 04
F0 01 02 03 04
C2 01 02 03 04
C3 01 02 03 04
F1 01 02 03 04
4
M bits (O11X X=0 alarm)
 C=000,111 bit stuffing present/absent
 nominal stuffing rate 1796 bps, max 5367
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Regenerative Repeaters
Amplifier
Equalizer
Input
 Spacing
Pair
diameter
(mm)
Regenerator
Output
Timing
recovery
between adjacent repeaters
0.9
Loop
attenuation
at 1 MGHz
(dB/km)
12
Loop
Maximum Total
Max
resistance distance
Repeaters distance
(km)
system
(W/km)
(km)
60
3
18
54
0.8
16
100
2.25
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Digital Transmission Systems
Designati Administra Bit
on
tion
Rate
Line Code
Media
T1
AT&T
1.544
AMI/B8ZS
Twisted pair 6000 ft
CEPT1
CCITT
2.048
B4ZS
Twisted pair 2000 m
T1C
AT&T
3.152
Bipolar
Twisted pair 6000 ft
T148
ITT
4B3T
Twisted pair 6000 ft
9148A
GTE
2.37
ternary
3.152
Twisted pair 6000 ft
T1D
AT&T
3.152
T1G
AT&T
6.443
1-DD
duobinary
1+D
duobinary
4-level
LD-4
Canada
274.176 B3ZS
Coax
1900 m
T4M
AT&T
274.176 Polar
Coax
5700 ft
Digital Telephony
Repeater
Spacing
Twisted pair 6000 ft
Twisted pair 6000 ft
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PCM System Enhancements
 North America
Superframe of 12 DS0’s has a sync sequence
101010 for odd (001110 for even frames)

Extended superframe
24 frames - (4 S bits for frame allignment
signal); 6 S bits for CRC-6 check; the rest 12
constitute 4 kbs data link
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