3.3 Pulse code modulation
3.3 Pulse code modulation
1
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Pulse Code Modulation
Analog waveform
Sampling
Quantizing
Coding & Streaming
t1: -4.88 v
-5 v
1000 0101=-5
t2: +21.43 v
21 v
0001 0101=21
…, 0001 0101, 1000 0101, …
2
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Pulse Code Modulation Three basic operations
x(t)
Sampling Quantizing Encoding PCM transmitter (A/D conversion) ~ xn xn
sampler
quantizer
~ x(t) 3
s(t)
encoder
channel
Low-pass filter
decoder Your site here
PCM Quantizing Approximating the analog sample values by using finite number of levels
Uniform quantizing Quantizing error quantizing noise 4
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Pulse Code Modulation Encoding
The quantized analog sample values are replaced by n-bit binary code
M =2
E.g. three-bit Gray code for M=8 levels Quantized Sample Voltage
Gray Code Word (PCM output)
+7 +5 +3 +1 -1 -3 -5 -7
110 111 101 100
5
000 001 011 010 Your site here
n
Bandwidth of PCM signals The bit rate of PCM signal is
R = nf s Example. Design of a PCM signal for telephone system Assume: An analog audio voice-frequency (VF) telephone signal band: 300Hz ~ 3400 Hz The minimum sampling frequency is 2×3.4 = 6.8 ksample/sec. actually, using sampling frequency of 8 ksamole/sec. Bit rate:
R= fs (samples/s)×n(bits/sample) = 8 k sample/s ×8 bits/sample = 64 kbps 6
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Bandwidth of PCM signals The bandwidth of (serial) binary PCM waveforms depends on:
The bit rate The waveform pulse shape used to represent the data For rectangular pulse, the first null bandwidth is
B PCM = R = nf s
Example. the result for the case of the minimum sampling: Number of quantizer levels, M
Length of the PCM, n(bit)
Bandwidth of PCM signal (the first null bandwidth)
2
1
2B
4
2
4B
8
3
6B
256
8
16B
……
7
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Effects of noise Two main effects produce noise or distortion:
Bit errors in the recovered PCM signal . (channel noise, improper channel filtering, ISI etc. ) Quantizing noise that is caused by the M-step quantized at PCM transmitter
8
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Effects of noise
x(t)
PCM transmitter (A/D conversion) ~ x x n
sampler
n
quantizer
encoder
0101110……
channel
0101010……
~ x(t)
9
Low-pass filter
decoder
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Effects of noise Under certain assumptions, the ratio of the recover analog peak signal power to the total average noise power is: 2 3M S = 1 + 4 ( M 2 − 1) Pe N pk out
The ratio of the average signal power to the average noise power is 2 S M = 1 + 4 ( M 2 − 1) Pe N out 10
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Effects of noise If Pe=0 (no ISI), the peak SNR resulting form only quantizing errors is S 2 = 3M N pk out
The average SNR due only to S 2 quantizing error is= M N out
6-dB rule S = 6.02n + α N dB 11
α=4.77 for the peak SNR, α=0 for the average SNR. Your site here
Effects of noise This equation points out the significant performance characteristic for PCM:
An additional 6-dB improvement in SNR is obtained for each bit added to the PCM word. Assumptions: ① No bit errors ② the input signal level is large enough to range over a significant number of quantizing levels 12
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Performance
13
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Example SA3-2 (page 219) PCM signal bandwidth and SNR In a communications-quality audio system, an analog voice-frequency (VF) signal with a bandwidth of 3200 Hz is converted into a PCM signal by sampling at 7000 samples/s and by using a uniform quantizer with 64 steps. The PCM binary data are transmitted over a noisy channel to a receiver that has a bit error rate (BER) of 10-4. What is the null bandwidth of the PCM signal if a polar line code is used? What is the average SNR of the recovered analog signal at the14 receiving end? Your site here
Nonuniform Quantizing Characteristic voice analog signal
Nonuniform amplitude distribution The granular quantizing noise will be a serious problem if uniform quantizing is used.
Solution: nonuiform quantizing Nouniform Quantizing: a variable step size is used
15
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Nonuniform Quantizing Method:
passing the analog signal through a compression (nonlinear) amplifier and then into a PCM circuit that uses uniform quantizer.
Analog Signal
A Compression (nonlinear) Amplifier
Nonuniform PCM (uniform quantized) Quantizing signal
µ-Law and A-Low 16
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Nonuniform Quantizing µ-Law
w2 (t ) =
ln (1 + µ w1 (t ) ) ln (1 + µ )
1.0
1.0
0.8
0.8 Compression quantizer characteristic
0.6 0.4
0 ≤ w1 (t ) ≤ 1 μ=225
μ=100
0.6
Uniform quantizer characteristic
0.4
0.2
μ=0 μ=1
0.2
μ=5
0 0
0.2
0.4
0.6
0.8
1.0
0
0.2
0.4
0.6
0.8
(a) M=8 Quantizer Characteristic 17
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1.0
Nonuniform Quantizing A-Law
A w1 (t ) , ln (1 + µ ) w2 (t ) = 1 + ln (A w1 (t ) ) , 1 + ln A 1.0
0
0.2
0.4
0.6
1 0 ≤ w1 (t ) ≤ A 1 ≤ w1 (t ) ≤ 1 A
0.8
1.0
0.8 0.6 A=100
0.4
A=87.6 A=5
0.2 A=1 0
0
0.2
A=2
0.418
0.6
0.8
1.0 Your site here
Compression characteristics In practice, the smooth nonlinear characteristics of μ-Law and A-Low are approximated by piecewise linear chords 19
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Output SNR Following 6-dB law: S = 6 .02 n + α N dB where
α = 4 .77 − 20 log (V / x rms )
Uniform quantizing
α = 4.77 − 20 log[ln(1 + µ )]
μ-Law companding
α = 4.77 − 20 log[1 + ln A]
A-Law companding
20
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Comparison of output SNR
21
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1
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Pulse Code Modulation
Analog waveform
Sampling
Quantizing
Coding & Streaming
t1: -4.88 v
-5 v
1000 0101=-5
t2: +21.43 v
21 v
0001 0101=21
…, 0001 0101, 1000 0101, …
2
Your site here
Pulse Code Modulation Three basic operations
x(t)
Sampling Quantizing Encoding PCM transmitter (A/D conversion) ~ xn xn
sampler
quantizer
~ x(t) 3
s(t)
encoder
channel
Low-pass filter
decoder Your site here
PCM Quantizing Approximating the analog sample values by using finite number of levels
Uniform quantizing Quantizing error quantizing noise 4
Your site here
Pulse Code Modulation Encoding
The quantized analog sample values are replaced by n-bit binary code
M =2
E.g. three-bit Gray code for M=8 levels Quantized Sample Voltage
Gray Code Word (PCM output)
+7 +5 +3 +1 -1 -3 -5 -7
110 111 101 100
5
000 001 011 010 Your site here
n
Bandwidth of PCM signals The bit rate of PCM signal is
R = nf s Example. Design of a PCM signal for telephone system Assume: An analog audio voice-frequency (VF) telephone signal band: 300Hz ~ 3400 Hz The minimum sampling frequency is 2×3.4 = 6.8 ksample/sec. actually, using sampling frequency of 8 ksamole/sec. Bit rate:
R= fs (samples/s)×n(bits/sample) = 8 k sample/s ×8 bits/sample = 64 kbps 6
Your site here
Bandwidth of PCM signals The bandwidth of (serial) binary PCM waveforms depends on:
The bit rate The waveform pulse shape used to represent the data For rectangular pulse, the first null bandwidth is
B PCM = R = nf s
Example. the result for the case of the minimum sampling: Number of quantizer levels, M
Length of the PCM, n(bit)
Bandwidth of PCM signal (the first null bandwidth)
2
1
2B
4
2
4B
8
3
6B
256
8
16B
……
7
Your site here
Effects of noise Two main effects produce noise or distortion:
Bit errors in the recovered PCM signal . (channel noise, improper channel filtering, ISI etc. ) Quantizing noise that is caused by the M-step quantized at PCM transmitter
8
Your site here
Effects of noise
x(t)
PCM transmitter (A/D conversion) ~ x x n
sampler
n
quantizer
encoder
0101110……
channel
0101010……
~ x(t)
9
Low-pass filter
decoder
Your site here
Effects of noise Under certain assumptions, the ratio of the recover analog peak signal power to the total average noise power is: 2 3M S = 1 + 4 ( M 2 − 1) Pe N pk out
The ratio of the average signal power to the average noise power is 2 S M = 1 + 4 ( M 2 − 1) Pe N out 10
Your site here
Effects of noise If Pe=0 (no ISI), the peak SNR resulting form only quantizing errors is S 2 = 3M N pk out
The average SNR due only to S 2 quantizing error is= M N out
6-dB rule S = 6.02n + α N dB 11
α=4.77 for the peak SNR, α=0 for the average SNR. Your site here
Effects of noise This equation points out the significant performance characteristic for PCM:
An additional 6-dB improvement in SNR is obtained for each bit added to the PCM word. Assumptions: ① No bit errors ② the input signal level is large enough to range over a significant number of quantizing levels 12
Your site here
Performance
13
Your site here
Example SA3-2 (page 219) PCM signal bandwidth and SNR In a communications-quality audio system, an analog voice-frequency (VF) signal with a bandwidth of 3200 Hz is converted into a PCM signal by sampling at 7000 samples/s and by using a uniform quantizer with 64 steps. The PCM binary data are transmitted over a noisy channel to a receiver that has a bit error rate (BER) of 10-4. What is the null bandwidth of the PCM signal if a polar line code is used? What is the average SNR of the recovered analog signal at the14 receiving end? Your site here
Nonuniform Quantizing Characteristic voice analog signal
Nonuniform amplitude distribution The granular quantizing noise will be a serious problem if uniform quantizing is used.
Solution: nonuiform quantizing Nouniform Quantizing: a variable step size is used
15
Your site here
Nonuniform Quantizing Method:
passing the analog signal through a compression (nonlinear) amplifier and then into a PCM circuit that uses uniform quantizer.
Analog Signal
A Compression (nonlinear) Amplifier
Nonuniform PCM (uniform quantized) Quantizing signal
µ-Law and A-Low 16
Your site here
Nonuniform Quantizing µ-Law
w2 (t ) =
ln (1 + µ w1 (t ) ) ln (1 + µ )
1.0
1.0
0.8
0.8 Compression quantizer characteristic
0.6 0.4
0 ≤ w1 (t ) ≤ 1 μ=225
μ=100
0.6
Uniform quantizer characteristic
0.4
0.2
μ=0 μ=1
0.2
μ=5
0 0
0.2
0.4
0.6
0.8
1.0
0
0.2
0.4
0.6
0.8
(a) M=8 Quantizer Characteristic 17
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1.0
Nonuniform Quantizing A-Law
A w1 (t ) , ln (1 + µ ) w2 (t ) = 1 + ln (A w1 (t ) ) , 1 + ln A 1.0
0
0.2
0.4
0.6
1 0 ≤ w1 (t ) ≤ A 1 ≤ w1 (t ) ≤ 1 A
0.8
1.0
0.8 0.6 A=100
0.4
A=87.6 A=5
0.2 A=1 0
0
0.2
A=2
0.418
0.6
0.8
1.0 Your site here
Compression characteristics In practice, the smooth nonlinear characteristics of μ-Law and A-Low are approximated by piecewise linear chords 19
Your site here
Output SNR Following 6-dB law: S = 6 .02 n + α N dB where
α = 4 .77 − 20 log (V / x rms )
Uniform quantizing
α = 4.77 − 20 log[ln(1 + µ )]
μ-Law companding
α = 4.77 − 20 log[1 + ln A]
A-Law companding
20
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Comparison of output SNR
21
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