PULSE Code MODULATIon

ES442 – Spring 2017

Lecture 9

Analog signal

Pulse Amplitude Modulation

Pulse Width Modulation

Pulse Position

Modulation

Pulse Code Modulation

1
(3-bit coding)

2
 Advantages of Digital Over Analog For Communications

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
†
Analog signals vary continuously and their value is affected by all levels of noise.

Reading: Lathi & Ding; Section 6.2.1 on pages 321 and 322.
 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)
Three Step Process
amplitude

amplitude
amplitude


   0100101101011001
time time time 1110101010101000
  0100011000100011

  0101001111010101
   
 1110110111010001


  
Sample Quantize Encode
Analog Digital
Signal Captured Quantized Signal
Sampled Data Sampled
Sampling Values Quantizing Data Encoding
selects the chooses the assigns binary
data points amplitude numbers to
Discrete time values: few amplitudes
Nowfrom
haveanalog
discrete Values in both time those
& amplitude
we use to values used
Analog signal is continuous in time & amplitude
signal amplitude
create the to encode
values Now have the digital
digital
data which is the final result
data

Note: “Discrete time” corresponds to the timing of the sampling.

 Second Step – Quantization I
The process of assigning quantization levels

Lathi & Ding

Figure 1.4
p. 4

= 

16 levels – 4 bits
 Natural Binary Pulse Code (Example)

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
 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
2mp
 
L
th
The k 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).

7
 Second Step – Quantization
pp. 322-324II

8
 Error Generated by Quantization



Quantization fluctuation or “noise”

8
 Second Step – Quantization
III
2mp
The quantized levels are separated  
by L
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

q 
2
mp2  
3L
2 
2
12

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

S0  m2(t)
9
 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).
S m 2 (t)  2 
2 m (t)
0   3L
This is given by the ratio  m2p  m2 
  p 
q 
N  3L2 

Conclusion:
To keep the quantization error small relative to the message signal level,
use smaller quantization steps .

10
 The Dilemma of Strong Signals versus Weak Signals

Strong Signal Weak Signal

(a) Linear encoding (b) With non-linear encoding

Companding
Note different encoding levels on each side.

11
 Use Compression and Expansion → Companding

Compression Restoration
(m) m(t)
m(t) m(t)

http://www.slideshare.net/91pratham/unit-ipcmvsh 12
 Companding Laws

A-Law Companding (Europe) -Law Companding (North America)

Output (y/ymax)

Output (y/ymax)
Input (m/mp) Input (m/mp)

A  m    for 0  m m 1   
y1 A y 1 loge 1  m m
loge A  mp  log (1   )  m fo 0m 1
p r
A 
 Am   1 m e  p  p
y  1  loge m   for  1
1  log A  p A m
 e p

 
Lathi & Ding
Section 6.2.3
pp. 325-328
13
 Flattening of the S/N Ratio Using the -Law

For optimal S/N ratio in North America  = 255 is used.

An approximately constant S/N ratio is the most desirable.

 S 
 N0  Lathi & Ding
 q Figure 6.18
p. 328
(8
bits)

Relative signal power m2(t) , (dB)

14
 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.

15
 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
p
  mp
0.5
2  L 
100m

16
  L  200
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.

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 C = 8 bits  8000 Hz

= 64,000 bits/second

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.
17
 Exponential Increase of the Output SNR (S/N ratio)

We start with the SNR (signal-to-noise ratio) equation from slide 10 above:
S  m 2 (t)  2
0 
3  m2  L
Nq  p 

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
S  m2(t)  2n
0
3
 m2   
2
Nq  p 

Using the expression for bandwidth, BT = nB, then we arrive at

S  m 2 (t) 
2B /B
2
0
Nq  3
T

 m
2

Taking the logarithm gives p 
S    m (t) 
S 
 10
log  10 3 2  2n log 10  2     6n dB
 0 log0 10  
10
   2 

18
 Exponential Increase of the Output SNR (S/N ratio)
Nq dB  Nq   mp  

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):
2
A
Pav m and mmax  Am
e 2 set
Setting mmax =
Am
S  m 2 (t)  2n  3P  2n  3  2n
N0  3 m2 (2)  m2ave  (2)   2 (2)
   max 
q  p 

L n SNR

10  S  1.76  6n dB 32 5 31.8 dB

log 0
64 6 37.8 dB
10 N
 
 q 128 7 43.8 dB
256 8 49.8 dB

19
 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.

193 bit frame – 122 sec/frame

1.544 Mbit/s data rates

20
T1 Carrier – Time Division
Multiplexing

Lathi & Ding

Figure 6.20
p. 333

21
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) 2.048 Mbits/s (E1)

T1 – 24 channels 32 user channels
Second-level 6.312 Mbits/s (DS-2) 8.448 Mbits/s
T2 – 96 channels (E2) 128
channels
Third-level 44.736 Mbits/s (DS3) 34.368 Mbits/s (E3)
T3 – 672 channels 512 channels
Fourth-level 274.176 Mbits/s (DS4) 139.264 Mbits/s
T4 – 4032 channels (E4) 2048
channels
Fifth-level 400.352 Mbits/s (DS5) 565.148 Mbits/s
T5 – 5760 channels (E5)
8192 channels

22
 Worked PCM Example

We are given a signal m(t) = 2cos(2250 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, 2 2
the average signal power is Pave = m
[(A ) /2] = [2 /2] = 2 watts. (See slide 19)
The interval  = [2m
p /L] = 4 volts/256 levels = 1.625  10 volt. (See
-2

slide 19) Now we can find the SNR (signal-to-quantized noise ratio)
(See side 18)
av

Using for the quantization noise = [()2/12], and taking = 2 W, the

Nq P
SNR is given by
 S  P   2  24 
 ave
 122    98, 304
 N   N    (1.5625 102 )2
 q 
23
 Worked PCM Example
q    
SNRdB  10 log 98, 304  49.93 dB
10

24
 Worked Example for PCM (continued)

We are given a signal m(t) = 2cos(2250 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.

3, 981 2m 4
4    0.001005 and   0.0317 volt
 p   2 ; 2 3981

2

Therefore, we can determine the number of levels L, and then n.

2mp 4
L     31.5
0.0317
24
 Worked Example for PCM (continued)
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.

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

25
 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
26
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

 Digression on Signal Prediction
Starting with a Taylor series (time step is Ts),
dm(t)
TS 2 d2(m(t) TS 3 d3(m(t)
m[t  TS ]  m(t)     ...
TS
dt 2! dt2 3! dt3
m[t  T dm(t)
]  m(t)  for small T
T
S S S
dt
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 d
write m(kTS )  m(kTS )  m(kTS  TS )
dt TS
Thus,
T  1]  m[k]   m[k]  m[k  1] 
m[k
S 
 TS 
m[k  1]  2m[k]  m[k  1]

28
 Digression on Signal Prediction
So we get an approximation of the (k+1)th sample, m[k+1], from the two
prior samples.

29
 Signal Prediction (continued)
But we can do better than this. In general,

m[k]  a1m[k  1] a2m[k  . .  aNm[k  N] m[k]

2] . 

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].

29
 Linear Predictor Implemented With Transversal Filter

m[k]  a1m[k  1]  a2m[k  . .  aNm[k  N]

2]  .

Input
m[k] Delay Delay Delay Delay Delay
TS TS TS TS ... TS

Lathi & Ding aN

a1 a2 a3
Figure 6.27
p. 343

Output m[k]

Transversal filter is a tapped delay line (with required weights {ai} )

30
 DPCM Transmitter
Input Outpu
m[k] d[k] t dq[k]
 Quantizer
+
+
Lathi & Ding
Figure 6.28(a) mq[k]
p. 344 + 
mq[k]
Predictor

d[k]  m[k]  mq [k] and is quantized to yield,

dq [k]  d[k]  q[k] where q[k] is the quantization error

The predictor output m[k] is fed back to the input so the predictor input
mq[k] is given by
mq [k]  mq [k]  dq [k]  m[k]  d[k]  dq[k]  m[k]  q[k]

31
 DPCM Transmitter
This shows that mq[k] is the quantized version of m[k].

32
 DPCM Receiver
Input Outpu
dq[k] t
+ mq[k]

Lathi & Ding
Figure 6.28(a) +
p. 344 mq[k]
Predictor

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].

32
 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.

2
The quantization noise is proportional to () – the quantization noise
power is reduced by a factor (d ) and the SNR is therefore increased by
2
/m
(m / )2. p p
d
p p

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.]

33
 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 Adaptive dq[k]
used to change m[k] Quantizer
 which minimizes dq[k]. When
d [k] fluctuates around large positive
+  To
Channel
q
or negative values, the prediction nth order Predictor
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.
34
Adaptive Differential PCM Output Example

time

35
Next Topic is Delta Modulation

36