1. TELE3113 Analogue and Digital
Communications – PCM
Wei Zhang
w.zhang@unsw.edu.au
School of Electrical Engineering and Telecommunications
The University of New South Wales
TELE3113 - PCM 16 Sept. 2009 p. -1

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)
Analog sampler Quantizer Encoder Digital
signal signal
1111111
1111110
1111101
1111100 1111101 1111001 1111001 1111011 1111101 1111110 1111101 1111001
1111011
1111010 c8 c7 c6 c5 c4 c3 c2 c1
x(t) 1111001
1111000
time
time time
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 set
of levels
Encoding: designates each quantized level by a digital code
TELE3113 - PCM 16 Sept. 2009 p. -2

3. Sampling
Consider an analog signal x(t) which is bandlimited to B (Hz), that is:
X ( f ) = 0 for | f |≥ B
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.
x(t) 1
f s ≥ 2B or Ts ≤
2B
Sampling rate fs=1/Ts
time
Sampling period Ts
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.
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4. 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.
Vmax xq(t) x(t)
Output
m7 ∆7
m7= 7∆/2
m6 ∆6
uniform
m6= 5∆/2
m5 ∆5 m5= 3∆/2
m4 ∆4 m4= ∆/2
m3 ∆3 time −4∆ −3∆ −2∆ −∆ ∆ 2∆ 3∆ 4∆ Input
m2 ∆2
−3∆/2
uniform
m1 ∆1 −5∆/2
m0 ∆0 −7∆/2
-Vmax
Sampling time
Uniform quantizer
2Vmax (midrise type)
For uniform quantization, ∆ i ( i =0 ,L,( L −1)) = ∆ =
L p. -4
TELE3113 - PCM 16 Sept. 2009

5. PCM - Encoding
Each 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)
Code Code Quantized
number level
4 x(t)
111 7 3.5
3
110 6 2.5
2
101 5 1.5
1
100 4 0.5
0
011 3 -0.5 time
-1
010 2 -1.5
-2
001 1 -2.5
-3
000 0 -3.5
-4
Sampler Sampled value -0.25 3.3 1.2 -2.8 -3.8 -2.1
Quantizer Quantized value -0.5 3.5 1.5 -2.5 -3.5 -2.5
Code number 3 7 5 1 0 1
Encoder
Binary code 011 111 101 001 000 001
Parallel to-serial converter
TELE3113 - PCM 16 Sept. 2009 p. -5
011 111 101 001 000 001

6. PCM Signal - Demodulation
At the receiver, the received pulses are distorted, need regeneration to
restore the ideal pulse (rectangular).
Regenerated
Decision signal
threshold
Distorted and time time Re-shaped, time
noise corrupted Sampling time Re-amplified,
(Re-timed)
Regeneration
The regenerated pulse stream will be separated into codes using serial-to-
parallel converter. 000
011
000011001011111001 001
3-bit 011
codes 111
001
The recovered codes are translated back into respective quantized levels.
The quantized levels are interpolated using low-pass filter to form the
recovered signal.
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7. 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 nfs
binary 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
1 Ts
(at least one sampling point for each bit or pulse, bit period= = )
nf s n
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.
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8. PCM Signal – SNR Analysis
 2
Vmax 
 
With L =2n, average SNRx (dB) = 4.77 + 20 log L − 10 log 
 x 2 (t )

 Vmax 
2
= 4.77 + 20 log 2 − 10 log
n 
 x (t ) 
2
 
 V2 
average SNRx (dB) = 4.77 + 6.02n − 10 log 2max 
 x (t ) 
  2
Vmax
If x(t) is a full-scale sinusoidal signal, i.e. x(t)=Vmaxcosωt , then x (t ) = x (t ) =
2 2
2
Thus,
average SNRx (dB) = 1.76 + 20 log L = (1.76 + 6.02n ) dB
If x(t) is uniformly distributed in the range [-Vmax,+Vmax], then pdf f(x)=1/(2Vmax),
Thus, average SNRx (dB) = 20 log L = 6.02n dB
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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9. 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
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