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Pulse Code Modulation
The advantages of digital communication systems(cf. analogue communication) Easier to store as a pattern of 1's and 0's
Increased Immunity
non-linearities
Easier to process in computers and digital signalprocessors
Can be coded for security and error correction purposes Several digital signals can easily be interleaved
(multiplexed) and transmitted on one channel
Noisy digital signals can be regenerated more effectivelythan analogue signals can be amplified.
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0000
1111
1110
1101
1100
1011
1010
1001
0001
0010
0011
0100
0101
0110
0111
A brief aside about ADCs
0000 0110 0111 0011 1100 1001 1011
Numbers passed from ADC to computer to represent analogue voltage
ADCs are used to convert an analogue input voltage into a number that can
be interpreted as a physical parameter by a computer.
Resolution=1 part in 2n
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Sampling The input signal is sampled prior to digitisation and an
approximation to the input is reconstructed by the digital-to-
analogue converter:
Sampling Digitisation code, modulate
Transmission
Wire/optical fibre
Aerial/free-space
input
FilteringDigital-to-analogue
conversionDemodulate, Decode
output
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Sampling an analogue signal Prior to digitisation, signals must be sampled
With a frequency fs=2B=1/T
ADC converts the height of each pulse into binary representation
Sampling involves the multiplication of the signal by a train of samplingpulses
vi(t) vo(t)
T TT
time time
1
0
vi(t)
Input Signal
WaveformSampled Signal
Waveform
(i)(ii) Sampling (iii)
...011101010011010011...
Samples are coded for transmission
(iv)
Analogue-to-digital
Converter
Digital-to-analogue
Converter
Sampled signal is recovered
(v)
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Sampling as multiplication by a sampling
waveform:
vs(t)
time
T
X1
0
Vi(f) V
o(f)
vs(t)
T TT
timetime
vi(t)
time
" ample-
a d-Hold" Sampling pulse is
short enough so thatcan normallyconsidered have zero
duration DAC, however
produces pulseslength T
Multiplication = Amplitude modulation
Amplitude modulation produces sidebands
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Sidebands produced by multiplication with acarrier That is, amplitude modulation
[c
[c+ [
s[
c-[
s[
s0
s
1/2cA
s
LSB USBBaseba d
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Sidebands at each harmonic of the sampling pulse
Digital-to-analogue conversion involves recovery of the baseband How?
What is the minimum value of fs for which there is no overlap of the Harmonics withthe baseband?
vs(t)
Vi(f) V
o(f)
freq
Vi(f)
B
fs= 1/T 2f
sfreq
Vo(f)
fs/2
B
armonics of vs(t)
(iv)
(v)
(vi)
time
T
X1
0
vs(t)
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fs= 1/T 2f
sfreq
Vo(f)
fs/2
B
GuardBand
B fs-B
H(f)1
0freq
Reconstructionfilter
Originalsignal
armonics of vs(t)
(vi)
freq
Vi(f)
If the sidebands do not overlap the signalcan be recovered
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Practical sampling the "Sample-and-hold" system:
This is Nyquists theorem For a signal of bandwidth B Hz, the minimum sampling
rate is 2B samples/s
T TTtime
Sample-
and- old
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Effect of sampling rate
sampling at more than the Nyquist Rate
Sample
sidebands
(i) Oversample - F s 4B
FsB
Fs- B Fs+ B
0freq
spectr m
Original signal
can easily be recovered
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Sampling at the Nyquist Rate
cannot build an ideal filter -
(ii) Nyquistsam ling- s = 2B
FsB
Fs-B Fs+B
0freq
Original signal can just be recoveredwith ideal filter
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Undersampling
produces aliasing distortion!
(iii) Undersam ling-F s = 1.5B
FsB
Fs-BFs+B
0freq
Original signal
cannot be recovered
Alias
signal
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Aliasing-time domain
Oversampled signal
Reconstructed signal
Undersampled signal
Reconstructed signal
Sampling:aliasing & Nyquist:time domain
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The Anti-alias
(Pre-sampling)
filter
ensures that
sampling
obeys the
Nyquisttheorem
fs= 1/T freq
Vo(f)
B
Aliasing orFoldback
distortion
Originals ectrum
freq
fs/2
fs/2
fs= 1/T freq
Vo(f)
B
Filtereds ectrum
fs/2
(ii) Anti-aliasfilter
bandwidth less than fs/2
high frequency components removed
(iii) Guardband
created
1st pair of
sampling sidebands
fs-B
H(f)
1
0
(i) Signalbandwidth > fs/2
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Examples
For the compact disc (Audio CD) themaximum signal frequency is 20 kHz and thesampling rate is 44.1 kHz.
The Nyquist Sampling Rate is 40 kHz
Hence the guard band is 4.1 kHz wide.
In the telephone system (see Section 5.8),
the speech signal has a bandwidth up to 3.4kHz and a sampling rate of 8 kHz,
The Nyquist Sampling Rate is 6.8 kHz
Hence the guard band is 1.2 kHz wide.
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Regeneration vamplification:
Gain of amplifiers equals loss in transmission lines
SNR analog: S/kN
SNR digital: S/N In practice finite S/N means there will be a low level of bit errors
Some accumulation of bit-error noise with repeaters, but much lower levelthan with analogue amplification
1 2 k
Atte atio A i eac sectio , Gai A i eac amplifier
xA xAxAo rce est
N
S/N
N
A(S+N)
S/2N
N N
S/kN
A(S+2N)
S/(k-1)N
O tp t SNR
(i) Analog e Channel
1 2 k
Atte atio A i eac sectio , Gai A i eac repeater
xA xAxASo rce est n
N
S/N
N
AS
S/N
N N
S/NAS
S/N
O tp t SNR
(ii) Digital Channel
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A Pulse-Code Modulation communicationsystem
"PCM"
Anti-alias
filter
Sampleand
HoldADC
The channel
codi g, mod latio , repeaters,
demod latio , decodi g
DACReconstr ction
Filter
Sig al
I
Sig al
O t
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A digital communication system - "PCM" Anti-alias Filter*
Digitiser/Sample-and-Hold circuit* Analogue-to-Digital Converter*
Coding- Source codingfor data compression,
Line codingfor signalling efficiency
Errorcoding to reduce the effect oferrors
Modulator
Physical Channel (with repeaters if necessary)* Copper cables
Fibre Optic cables
Radio
Sonar
Recording medium
Demodulator
Decoder (Source-, Line- and Error-)
Digital-to-Analogue Converter*
Reconstruction Filter*
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Time-division Multiplexing "TDM" Allocate interleaved time-slots to each signal
Assemble the binary coded samples into Frames: 2-channel time-division multiplexing scheme:
Frame n Frame n+1
Slot1 Slot2 Slot1 Slot2
Channel1Sample1
Channel2Sample1
Channel1Sample2
Channel2Sample2
Frame n Frame n+1
slot 1 slot 2 slot 1 slot 2
Sampling and
A/D conversionDigitised samples
Frame length 1/fs
Slot length =No. of channels
Frame length
Analogue
inputs
Channel1
Channel2
10111001
Sample 2
10111010
Sample 1
00110000
Sample 2
00101111
Sample 1
Sample rate f s 10111010001011111011100100110001
T
wo channels share a single physical channel Cost?
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The 32-channel PCM Transmission system
30speech signals plus two control channels
for signalling and synchronising:
Signal bandwidth 3.4 kHz
Sampling rate 8 kHz
Hence frame length?
Sample size 8 bits/sample
Hence bit rate from each signal 64 kbit/s
32 channels Hence each time slot3.906Qs
1/(8000*32)
Overall data rate 2.048 Mbit/s
8000*32*8
125Qs
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1
2
3
4 1
2
3
4
13
14
15
1661
62
63
64
1
2
3
4
2.048 Mbit/s
8.448 Mbit/s
34.368 Mbit/s
139.264 Mbit/s
x16 x4
(32 channels x 64
= 2048 channels)
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A number of frames can be time-division multiplexed
together in a TDM heirachy. 4 frames of 32 channels
= 128 basic PCM channels,
Has data rate of 4 x 2.048 Mbit/s = 8.192 Mbit/s
8.448Mbit/s including extra signalling bits
4 x 128 = 512 channels
Has data rate = 4 x8.192 Mbit/s (+ signalling bits) = 34.368 Mbit/s
etc
Up to a multiplex of32768 channels with an overall data
rate of2.48832G
bit/
s.
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1
2
3
4 1
2
3
4
13
14
15
1661
62
63
64
1
2
3
4
2.048 Mbit/s
8.448 Mbit/s
34.368 Mbit/s
139.264 Mbit/s
x16 x4
(32 channels x 64
= 2048 channels)
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Spectrum of a train of pulses:
Samplingp lsespectr mfrequency
Harmonic spacing 1/T
N llspacing 1/ X
Pulse
idt X
Sampling pulse
period T
time
Samplingp lse waveform
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