Embed Size (px)
Citation preview
Pulse Code Modulation ES442 – Spring 2017
Lecture 9
1
Analog signal
Pulse Amplitude Modulation
Pulse Width Modulation
Pulse Position Modulation
Pulse Code Modulation (3-bit coding)
2
1. Digital is more robust than analog to noise and interference†
2. Digital is more viable to using regenerative repeaters
3. Digital hardware more flexible by using microprocessors and VLSI
4. Can be coded to yield extremely low error rates with error correction
5. Easier to multiplex several digital signals than analog signals
6. Digital is more efficient in trading off SNR for bandwidth
7. Digital signals are easily encrypted for security purposes
8. Digital signal storage is easier, cheaper and more efficient
9. Reproduction of digital data is more reliable without deterioration
10. Cost is coming down in digital systems faster than in analog systems and DSP algorithms are growing in power and flexibility
Reading: Lathi & Ding; Section 6.2.1 on pages 321 and 322.
† Analog signals vary continuously and their value is affected by all levels of noise.
Advantages of Digital Over Analog For Communications
3
Next Topic – Pulse Code Modulation
Pulse-code modulation (PCM) is used to digitally represent sampled analog signals. It is the standard form of digital audio in computers, CDs, digital telephony and other digital audio applications. The amplitude of the analog signal is sampled at uniform intervals and each sample is quantized to its nearest value within a predetermined range of digital levels.
Four-bit coding (16 discrete levels)
Analog to Digital Conversion Process (ADC)
4
Note: “Discrete time” corresponds to the timing of the sampling.
Sample Quantize Encode Analog Signal Captured
Sampled Data Values
Quantized Sampled
Data
Digital Signal
Analog signal is continuous
in time & amplitude
Discrete time values:
few amplitudes from analog
signal
Now have discrete Values in
both time & amplitude
Now have the digital
data which is the final
result
Sampling selects the data points we use to create the
digital data
Quantizing chooses the amplitude
values used to encode
Encoding assigns binary
numbers to those
amplitude values
time
am
plit
ud
e
0100101101011001 1110101010101000 0100011000100011 0101001111010101 1110110111010001
time
amp
litu
de
time amp
litu
de
Three Step Process
5
Second Step – Quantization I
16 levels – 4 bits
=
The process of assigning quantization levels
Lathi & Ding Figure 1.4
p. 4
Natural Binary Pulse Code (Example)
6
To communicate sampled values, we send a sequence of bits that represents the quantized value.
For 16 quantization levels, 4 bits are required.
PCM can use a binary representation of value.
The PSTN uses PCM
Figure 1.5 (page 8) of Lathi & Ding
7
Second Step – Quantization II
We start with a sampled signal {call it m(t)} and now we want to quantize it. The quantized amplitude is limited to a range, say from –mp to +mp. (Note: the range of m(t) may extend beyond (-mp, mp) in some cases.) Divide the range (-mp, mp) into L uniformly spaced intervals. The number intervals is L and the separation between quantized levels is The kth sample point of m(t) is designated as m(kTS) and is assigned a value equal to the midpoint between two adjacent levels. Define: m(kTS) = kth sample’s value, and m(kTS) = kth quantized sample’s value. Then the quantization error q(kTS) is equal to m(kTS) - m(kTS)
Read Section 6.2.2 of Lathi & Ding: Quantizing evaluation of parameter q(t).
2 pm
L
pp. 322-324
8
Error Generated by Quantization
Quantization fluctuation or “noise”
9
Second Step – Quantization III
The quantized levels are separated by The maximum error for any sample point’s quantized value is at most ½. It is shown in Lathi and Ding that the “time average” mean square error from quantization is Let Nq equal q2. Nq is proportional to the fluctuation of the error signal. This is sometimes quantization noise. This means that m(t) = m(t) + q(t) The signal (message) power S0 is proportional to the square of m(t), thus
2 pm
L
22
2
23 12
pmq
L
20 ( )S m t
10
Second Step – Quantization IV
We want a measure of the quality of received signal (that is, the ratio of the strength of the received signal S0 relative to the strength of the error Nq due to quantization). This is given by the ratio
2 220
22
2
( ) ( )3
3q pp
S m t m tL
N mm
L
Conclusion: To keep the quantization error small relative to the message signal level, use smaller quantization steps .
11
The Dilemma of Strong Signals versus Weak Signals
Strong Signal Weak Signal
Note different encoding levels on each side.
(a) Linear encoding (b) With non-linear encoding Companding
12 http://www.slideshare.net/91pratham/unit-ipcmvsh
Use Compression and Expansion → Companding
Restoration Compression
m(t) m(t) m(t) (m)
13
Companding Laws
A-Law Companding (Europe) -Law Companding (North America)
Lathi & Ding Section 6.2.3 pp. 325-328
1log 1 0 1
log (1 )e
e p p
m my for
m m
10
1 log
11 log 1
1 log
e p p
e
e p p
A m my for
A m m A
AmA my for
A m A m
Input (m/mp) Input (m/mp)
Ou
tpu
t (
y/y m
ax)
Ou
tpu
t (
y/y m
ax)
14
Flattening of the S/N Ratio Using the -Law
Lathi & Ding Figure 6.18
p. 328
For optimal S/N ratio in North America = 255 is used. An approximately constant S/N ratio is the most desirable.
(8 bits)
0
q
S
N
Relative signal power m2(t) , (dB)
15
Transmission Bandwidth
In binary PCM, we have a group of n bits corresponding to L levels with n bits. Thus, L = 2n or n = log2(L)
Signal m(t) is band-limited to B Hz which requires 2B samples per second. For 2nB elements of information, we must transfer 2nB bits/second. Thus, the minimum bandwidth BT needed to transmit 2nB bits/second is BT = nB Hz Practically speaking, usually we choose the transmission bandwidth to be a little higher than the minimum bandwidth required.
16
Example 6.2 (Lathi & Ding, page 329)
Problem: A band-limited signal m(t) of 3 kHz bandwidth is sampled at rate of 33⅓ % higher than the Nyquist rate. The maximum allowable error in the sample amplitude (i.e., the maximum quantization error) is 0.5% of the peak amplitude mp. Assume binary encoding. Find the minimum bandwidth of the channel to transmit the encoded binary signal. Solution: The Nyquist rate is RN = 2 x 3000 Hz = 6000 Hz (samples/second), but the actual rate is 33⅓ % higher, so that is 6000 Hz + (⅓ x 6000) = 8000 Hz. The quantization step is and the maximum quantization error is plus/minus /2. Hence, we can write For binary coding, L, must be a power of two; therefore, knowing that L = 27 = 128 and 28 = 256, we must choose n = 8 to guarantee better than a 0.5% error.
0.5200
2 100
p
p
mm L
L
17
Example 6.2 Continued (Lathi & Ding, page 329)
Solution (continued): Having chosen n = 8 to guarantee 0.5% error, to find the bandwidth required we note that
Total number of bits per second = 8 bits 8000 Hz
= 64,000 bits/second
C
However, we know we can transmit 2 bits/Hz of bandwidth¶, so it requires a bandwidth BT of BT = C/2 = 32,000 Hz = 32 kHz If 24 such signals are multiplexed on a single line (known as a T1 Line in the Telephone system, then CT1 = 24 x 64 kb/s = 1.536 Mb/s, and the bandwidth is 768 KHz
¶A maximum of 2B independent elements of information per second can be transmitted, error-free, over a noiseless channel of bandwidth B Hz.
18
Exponential Increase of the Output SNR (S/N ratio)
2
02
2( )3 2
q p
nS m t
N m
220
2
( )3
q p
S m tL
N m
We start with the SNR (signal-to-noise ratio) equation from slide 10 above:
The number of levels L can be expressed as L2 = 22n where n = log2(L) and is The number of bits to generate L levels. The SNR can now be expressed as
Using the expression for bandwidth, BT = nB, then we arrive at Taking the logarithm gives
2
02
2 /( )3 2 T
q p
B BS m t
N m
2
0 010 10 2
( )10 log 10log 3 2 log 10 2 6 dB
q q pdB
S S m tn n
N N m
19
SNR Example
Given a full sinusoidal modulating signal m(t) of amplitude Am into a load resistance R = 1 ohm, find the signal-to-quantization noise ratio (sometimes called SNR):
Setting mmax = Am
2
maxand set2m
ave m
AP m A
20
2 2max
2 2 2( ) 3 33 (2) (2) (2)
2ave
q p
n n nS m t P
N m m
01010log 1.76 6 dB
q
Sn
N
L n SNR
32 5 31.8 dB
64 6 37.8 dB
128 7 43.8 dB
256 8 49.8 dB
20
Bell System’s T1 Carrier System (1962)
The T-carrier is a member of the group of carrier systems developed by AT&T Bell Laboratories for digital transmission of multiplexed telephone calls using Pulse Code Modulation and Time Division Multiplexing. The first version, the Transmission System 1 (T1), was introduced in 1962 in the Bell System, and could transmit up to 24 telephone calls simultaneously over a single transmission line consisting of copper wire.
1.544 Mbit/s data rates
193 bit frame – 122 sec/frame
21
T1 Carrier – Time Division Multiplexing
Lathi & Ding Figure 6.20
p. 333
22
Comparison of T-Carrier (North America) and E-Carrier (Europe)
Carrier Level T-Carrier Data Rates E-Carrier Data Rates
Zero-level 64 kbits/s (DS-0)
64 kbits/s
First-level 1.544 Mbits/s (DS-1)
T1 – 24 channels
2.048 Mbits/s (E1) 32 user channels
Second-level 6.312 Mbits/s (DS-2)
T2 – 96 channels
8.448 Mbits/s (E2) 128 channels
Third-level 44.736 Mbits/s (DS3)
T3 – 672 channels
34.368 Mbits/s (E3) 512 channels
Fourth-level 274.176 Mbits/s (DS4)
T4 – 4032 channels
139.264 Mbits/s (E4) 2048 channels
Fifth-level 400.352 Mbits/s (DS5)
T5 – 5760 channels
565.148 Mbits/s (E5) 8192 channels
23
Worked PCM Example
We are given a signal m(t) = 2cos(2250 t) as the signal input.
(a) Find the SNR with 8-bit PCM.
For 8-bit encoding, L = 2n where n = 8, therefore, the number of levels = 256. The amplitude Am of the sinusoidal waveform means that mp = 2 volts. The total signal swing possible (- mp to + mp) will be 2mp = 4 volts, therefore, the average signal power is Pave = [(Am)2/2] = [22/2] = 2 watts. (See slide 19)
The interval = [2mp/L] = 4 volts/256 levels = 1.625 10-2 volt. (See slide 19)
Now we can find the SNR (signal-to-quantized noise ratio) (See side 18) Using for the quantization noise Nq = [()2/12], and taking Pave = 2 W, the SNR is given by
22 2
10
2 2412 98,304
(1.5625 10 )
10 log 98,304 49.93 dB
ave
q q
dB
PS
N N
SNR
24
Worked Example for PCM (continued)
We are given a signal m(t) = 2cos(2250 t) as the signal input.
(b) If the minimum SNR is to be at least 36 dB, how many bits n are needed to encode the signal (i.e., find n)? Other parameters such as signal power remain the same as in part (a) on previous slide.
Note that 36 dB is numerically equivalent to 3,981 [about 4,000]. Remembering that the interval is = [2mp /L] and 2mp = 4 volts. Therefore, we can determine the number of levels L, and then n. The lowest integer number of bits n that will give at least 31.5 levels is n = 5 because 25 = 32 levels. So the answer is 5 bits.
2
2 2
2 4 43,981 ; 0.001005 and 0.0317 volt
3981
pm
2 431.5
0.0317
pmL
25
Differential Pulse Code Modulation (DPCM)
PCM is not really efficient because it generates so many bits taking up a lot of bandwidth. Can we improve on this? YES. Suppose we have a slowly varying signal m(t), then we exploit this by using the difference in two adjacent samples. This will form the basis of differential pulse code modulation (DPCM). Let m[k] be the kth sample of signal m(t). Then we can express the difference between adjacent samples as d[k] = m[k] – m[k-1] Instead of transmitting m[k], we instead transmit d[k].
Lathi & Ding, Section 6.5, pp. 341-344
26
Differential Pulse Code Modulation (continued)
At the receiver knowing d[k] and the previous value of m[k-1] allows us to construct the value of m[k].
How do we benefit from doing this?
The difference of successive samples almost always is much smaller than the full range of the sample values of m(t) (full range covers -mp to +mp). We use this fact to improve upon the efficiency of PCM by requiring fewer bits.
In addition, we make use of the estimate of m[k], denoted by m[k]. We use previous sample values of m(t) to make this estimate. Suppose m[k] is the estimate of the kth sample, then the difference d[k] is given by d[k] = m[k] – m[k] and it is the difference d[k] that is transmitted.
Lathi & Ding, Section 6.5, pp. 341-344
27
Differential Pulse Code Modulation (continued)
Receiver Concept:
At the receiver we determine the estimate m[k] from previous sample values, and then generate m[k] by adding the received d[k] values to the estimate m[k]. Thus the reconstruction of the samples is done iteratively.
Lathi & Ding, Section 6.5, pp. 341-344
28
Digression on Signal Prediction
Starting with a Taylor series (time step is Ts),
2 2 3 3
2 3
( ) ( ( ) ( ( )[ ] ( ) ...
2! 3!( )
[ ] ( ) for small
S SS S
S S S
dm t T d m t T d m tm t T m t T
dt dt dtdm t
m t T m t T Tdt
We denote the kth sample of m(t) by m[k], that is, m[kTS] = m[k], and m[kTS TS] = m[k 1], and so on. This is a first-order predictor.
In handling the derivatives, we write Thus, So we get an approximation of the (k+1)th sample, m[k+1], from the two prior samples.
( ) ( )
( ) S S SS
S
m kT m kT Tdm kT
dt T
[ ] [ 1][ 1] [ ]
[ 1] 2 [ ] [ 1]
S
S
m k m km k m k T
T
m k m k m k
29
Signal Prediction (continued)
But we can do better than this. In general, The set of {ai} are the prediction coefficients. This is the predicted value of m[k]. It is an Nth order predictor. Note that the input consists of the weighted previous samples m[k-1], m[k-2],etc. We say that input m[k] gives output m[k]. For first-order prediction, m[k] = m[k-1]. The next slide shows how to implement this prediction of m[k].
1 2[ ] [ 1] [ 2] . . . [ ] [ ]Nm k a m k a m k a m k N m k
30
Linear Predictor Implemented With Transversal Filter
Lathi & Ding Figure 6.27
p. 343
Delay TS
Delay TS
Delay TS
a1
Input m[k]
a2 a3
Delay TS
. . . Delay TS
aN
Output m[k]
Transversal filter is a tapped delay line (with required weights {ai} )
1 2[ ] [ 1] [ 2] . . . [ ]Nm k a m k a m k a m k N
31
DPCM Transmitter
[ ] [ ] [ ] and is quantized to yield,
[ ] [ ] [ ] where [ ] is the quantization error
q
q
d k m k m k
d k d k q k q k
The predictor output m[k] is fed back to the input so the predictor input mq[k] is given by
[ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ]q q q qm k m k d k m k d k d k m k q k
This shows that mq[k] is the quantized version of m[k].
Lathi & Ding Figure 6.28(a)
p. 344
Quantizer
Predictor
Input m[k]
mq[k]
+
+
+
mq[k]
d[k] Output
dq[k]
32
DPCM Receiver
Predictor
Input dq[k]
+
mq[k]
Output mq[k]
+ Lathi & Ding Figure 6.28(a)
p. 344
The receiver’s output (which is the predictor’s input) is also the same, mq[k] = m[k] + q[k]. Hence, we are able to receive the desired signal m[k] plus the quantization noise, q[k]. It is important to note that from the difference signal d[k] is much smaller that the noise associated with m[k].
33
DPCM SNR Improvement
How much better is DPCM with regard to SNR? To determine this, define mp and dp as the peak amplitudes of m(t) and d(t), respectively. Assuming the same number of steps L for both, then the quantization step in DPCM is reduced in magnitude by dp/mp. The quantization noise is proportional to ()2 – the quantization noise power is reduced by a factor (dp/mp)2 and the SNR is therefore increased by (mp/dp)2. Maintaining the same SNR, the number of bits can be reduced.
Example: The AT&T telephone system sometimes operates at 32 kbits/s (or even 24 kbits/s) when using DPCM. [The telephone system was initially designed to use a 64 kbits/s data rate.]
34
Adaptive Differential PCM
Adaptive differential PCM (ADPCM) can further improve upon DPCM by Incorporating an adaptive quantizer (variable ) at encoding.
The quantized prediction error dq[k] is a good measure of the predicted error size – it can be used to change which minimizes dq[k]. When dq[k] fluctuates around large positive or negative values, the prediction error is large and needs to increase, but when dq[k] fluctuates around zero (small values), then needs to decrease.
Example: An 8-bit PCM sequence can be encoded into a 4-bit ADPCM sequence at the same sampling rate. This reduces the channel bandwidth by one-half with no loss in quality.
Adaptive Quantizer
nth order Predictor
+
m[k]
To Channel
dq[k]
35
Adaptive Differential PCM Output Example
time
36
Next Topic is Delta Modulation
