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Pulse Amplitude Modulation dan Pulse Code Modulation. oleh Risanuri Hidayat. t. 0. T. t. 0. t. 0. T. 0. t. t. 0. Pulse Amplitude Modulation. Low Pass. Band Pass. 0. 0. *make sure that Sinc function is big and flat by reduce time. By using very narrow. Demodulation of PAM. - PowerPoint PPT Presentation
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Pulse Amplitude Modulationdan
Pulse Code Modulation
oleh Risanuri Hidayat
190/4 /2 3
-PAM PCM 2
Pulse Amplitude Modulation tf
t
t
t
t
t
n
imp nTttf
tftftf imps
n
nTtnTf
T
T
trecttr
2T 2T
trtft sPAM
n
nTtrnTf
2
sincT
TR
nimp T
n
TF
22
imps FFF 2
1
n T
nF
T
21
0
0
F
B2 B2
0
B2
0
B2B2
T
2
T
2
T
2
T
2
T
0
RFF sPAM
n
RT
nF
T 21
190/4 /2 3
-PAM PCM 3
Demodulation of PAM
0
Low Pass Band Pass
ns T
nF
TF
21
LowpassnFT
0;1
0
T
T
RFF sPAM
n
RT
nF
T 21
RFT
n 1
;0
By using very narrow 1.0
T
t
T
trect*make sure that Sinc function
is big and flat by reduce time
190/4 /2 3
-PAM PCM 4
-Multiplexing PAM TDM
LPF
LPF
LPFClock
Pulsegenerator
Pulsegenerator
Sampler
Sampler
switch
tf1
tf2
PAM1
PAM2
PAM-TDM before filtering
PAM-TDM to thetransmission line
Switch : determining the synchronization and sequence of the channelsClock : determine the timing of the overall systemPulse generator : produces narrow rectangular pulses to drive the sampler
190/4 /2 3
-PAM PCM 5
Nyquist Sampling
T
2
TTx Two signal are sampled equally
Tx=the time interval between adjacent channels or samples
For n channel
n
TTx
Nyquist interval for one signal
190/4 /2 3
-PAM PCM 6
Bandwidth Requirement
S1 S2 S3 S4
tTx
BF-BF0
Tx
With condition of equal BW and sampling equallytherefore the Total BW Requirement is n x BW
If BW of each channel is not equalTherefore the Total BW Requirement is n x largest BW
Fx B
T2
1
xF T
B2
1
190/4 /2 3
-PAM PCM 7
Transmitting Analog sign al in Digital format
• Advantages Immunity to noise : with some amount of noise digital can withstand
while analog failed to provide virtually error free transmission.
Reduce signal to noise ratio
Error control coding ; parity check, Hamming code make more reliable
Signal can be completely regenerated at intermediate regenerator for long haul system.
More compatible with computer system for signal processing and digital memories for data storage.
More elaborate code can be used.
Ideal for integrated services digital network (ISDN)
190/4 /2 3
-PAM PCM 8
Pulse Code Modulation• m(t) is sampled, each sample value is rounded off to
the nearest allowable value. This value is digitally encoded as a sequence of binary digits
• There are three process of Digitization 1. Sampling
2. ( )Quantization devide into level of voltage oo ooooooooo ooooo oo ooooo oooo() oooo ooo oo o oooooooo ooooooooo oooooo
3. Encoding Each quantization level is encoded into N binary digits
NM 2 MN 2logWhere N is the number of binary digit per code word
M is the number of quantization level
190/4 /2 3
-PAM PCM 9
Quantization
0
-V
V
t
v M Steps
Sampling Signal
v
VM
2 Where M = no. of steps = quantization stepv
190/4 /2 3
-PAM PCM 10
Encoding
0
-V
V
t
000
001
010
011
100
101
110
111
NM 2
1 1 1 1 1 0 1 0 0 0 1 1 0 1 0 1 0 0 1 0 1 1 1
190/4 /2 3
-PAM PCM 11
Quantization Error
0
2
v
2
v 2
v
Quantization Noise
2
v2
v
vv1
Uniform distribution
tftfte Q 22
vte
v
190/4 /2 3
-PAM PCM 12
Signal to Noise Ratio
lPowerErrorSigna
rSignalPoweSNR
Q
dttfT
PT
21lim
2
2
22 1T
T
dtteT
te
dvvpv2
2
v2
v
vv1
The average power
Time average
Continuous RV
time
190/4 /2 3
-PAM PCM 13
Signal to Noise Ratio[1]
2
2
2
2
322
3
11v
v
v
v
v
vdv
vvte
883
1 33 vv
v
12
2v
M
Vvtf
vte
tfSNR
Q
2;
12 222
2
tfMV
2224
12
2
2
22 33 M
V
tfM AvgPower
PeakPower
tf
V
2
2
where
190/4 /2 3
-PAM PCM 14
Signal to Noise Ratio[2]
23MSNR
Q
In dB )(log10log203log10 101010 dBMSNRQ
)(log10log2077.4 1010 dBM
Encoding : each quantization level is encoded into N binary digit
NM 2
MN 2log No.of level
No.of binary digit per code word
b
aab
10
10
log
loglog
190/4 /2 3
-PAM PCM 15
- TDM PCM (E1 standard)
300 3.4k
3.1kvoice time
0 500 800300 3.4k
MAN WOMAN
kHzFs 8 (2x3.4k=6.8k)Nyquist
Each sample is quantized and encoded into 8 bits
sec1258
18
kHzTkHzFs
sFBit rate = x 8 = 64kbps ; with 32 channels sec25.4886432
1n
kT
1channel = 8 bits therefore 8 x 488.25nsec = 3.9 sec
Total bit length = 8 bit x 32 channels = 256 bits
MbpsDataRate 048.2sec125
256
nTimePropagatio
LengthBit