tele3113 wk8wed
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16 Sept. 2009TELE3113 - PCM p. -1
TELE3113 Analogue and Digital Communications – PCM
School of Electrical Engineering and TelecommunicationsThe University of New South Wales
16 Sept. 2009TELE3113 - PCM p. -2
Analog-to-Digital Conversion (PCM)
Goal: To transmit the analog signals by digital means better performance
convert the analog signal into digital format (Pulse-Code Modulation)
Sampling: a continuous-time signal is sampled by measuring its amplitude at discrete time instants.
Quantizing: represents the sampled values of the amplitude by a finite setof levels
Encoding: designates each quantized level by a digital code
sampler EncoderQuantizerAnalog signal
Digital signal
x(t)
time
11111111111110111110111111001111011111101011110011111000
1111101 1111001111111011110111111001 11110011111101 1111101
c1c2c3c4c5c6c7c8
time
time
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Sampling
Consider an analog signal x(t) which is bandlimited to B (Hz), that is:
BffX ≥= ||for 0)(
The sampling theorem states that x(t) can be sampled at intervals as large as 1/(2B) such that the it is possible to reconstruct x(t) from its samples, or the sampling rate fs=1/Ts can be as low as 2B.
timeSampling period Ts
Sampling rate fs=1/Ts
x(t)
Minimum required sampling rate=2B (Nyquist rate) i.e. 2B samples per second
Sampling rate should be equal or greater than twice the highest frequency in the baseband signal.
BTBf ss 2
1or 2 ≤≥
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Quantization
After the sampling process, the sampled points will be transformed into a set of predefined levels (quantized level) Quantization
Assume the signal amplitude of x(t) lies within [-Vmax ,+Vmax], we divide the total peak-to-peak range (2Vmax) into L levels in which the quantized levels mi (i=0,…,(L-1)) are defined as their respective mid-ways.
LV
Liimax
))1(,,0(
2=∆=∆ −= LFor uniform quantization,
∆ 2∆ 3∆ 4∆ Input−4∆ −3∆ −2∆ −∆
m7= 7∆/2
m6= 5∆/2
m5= 3∆/2
m4= ∆/2
−3∆/2
−5∆/2
−7∆/2
Output
Uniform quantizer(midrise type)
uniform
unifo
rm
Vmax
m7
m6
m5
m4
m3
m2
m1
m0
-Vmax
∆0
∆1
∆2
∆3
∆4
∆5
∆7
∆6
Sampling time
x(t)xq(t)
time
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PCM - EncodingEach quantized level is translated into a binary code (digital).
For L (=2n) quantized levels, number of bits per code is n (=log2L).
The binary code is then converted into a sequential string of pulses for transmission Pulse code modulation (PCM)
4
3
2
1
0
-1
-2
-3
-4
7
6
5
4
3
2
1
0
Code number
Quantized level
time
3.5
2.5
1.5
0.5
-0.5
-1.5
-2.5
-3.5
x(t)
Sampled value -0.25 3.3 1.2 -2.8 -3.8 -2.1
Quantized value -0.5 3.5 1.5 -2.5 -3.5 -2.5
Code number 3 7 5 1 0 1
Binary code 011 111 101 001 000 001
111
110
101
100
011
010
001
000
Code
011 111 101 001 000 001
Sampler
Quantizer
Encoder
Parallel to-serial converter
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PCM Signal - Demodulation
At the receiver, the received pulses are distorted, need regeneration to restore the ideal pulse (rectangular).
timetimeSampling time
Decision threshold
timeDistorted and noise corrupted
Re-shaped, Re-amplified, (Re-timed)
Regenerated signal
Regeneration
The regenerated pulse stream will be separated into codes using serial-to-parallel converter.
The recovered codes are translated back into respective quantized levels.
The quantized levels are interpolated using low-pass filter to form the recovered signal.
000011001011111001
000 011 001 011 111 001
3-bit codes
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PCM Signal - Sampling Rate
For L (=2n) quantized levels, each sample point of the message signal is quantized into n bits. Then, n (=log2L) binary pulses must be transmitted for each sample point of the message signal.
If the message bandwidth is B and the sampling rate is fs (≥2B), then nfsbinary pulses must be transmitted per second as each sample point will be translated into a n-bit code.
minimum sampling rate for the PCM signal is R=nfs(at least one sampling point for each bit or pulse, bit period= )
minimum bandwidth for PCM signal = nfs /2 ≥ nB
Minimum required bandwidth for PCM is proportional to the message signal bandwidth, B, and the number of bits per code (symbol), n.
nT
nfs
s
=1
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PCM Signal – SNR Analysis
−+=
−+=
−+=
)(log1002.677.4)( average
)(log102log20774
)(
log10log20774)( average
2
2max
2
2max
2
2max
txVndBSNR
txV.
txVL.dBSNR
x
n
xWith L =2n,
If x(t) is a full-scale sinusoidal signal, i.e. x(t)=Vmaxcosωt , thenThus,
If x(t) is uniformly distributed in the range [-Vmax,+Vmax], then pdf f(x)=1/(2Vmax),
2)()(
2max22 Vtxtx ==
( ) dB 02.676.1log2076.1)( average nLdBSNRx +=+=
Thus, dB 02.6log20 )( average nLdBSNRx ==
The SNR can be improved by 6dB when one more bit is used in the code, but the bandwidth required for the PCM signal will be getting larger (as minimum bandwidth for PCM signal is nB where B is the message signal bandwidth).
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PCM - Applications
Telephone System:Voice bandwidth ~ 4kHz minimum sampling rate: 8kHz
8000 samples per second
8-bit PCM is used 8 bits per sample 8x8000 =64kbits per second
For modem application, only 7 bits are for data, the other is an overhead bit
56kbit/s
Compact Disk (CD) Audio:
Each of the two stereo signals is sampled at 44.1kHz
16-bit PCM is used 16x44.1k = 0.7056Mbits per second per stereo channel
100% overhead (error correction code) 1.411Mbit/s per stereo channel
One CD can record 1 hour music, total number of bits (for 2 stereo channels) = 1.411M x 2 x 3600 = 10.16Gbits