Pulse Code Modulation (PCM)

Example

0.5 1 1.5 2 2.5 30

0.5

1

1.5

2

2.5

3

3.5

m(t)

m(t)

Solution

In the previous example we have 8-quantization levels (to encode with 3-bits) from code number 0 to 7 (from binary code 000 to 111)

Sample value: 2.8 3.2 2.7 1.1 Nearest level: 3 3 2.5 1Code number: 6 6 5 2Binary code: 110 110 101 010

PCM Transmitter Block Diagram

L

r=nfs

n

n:

PCM transmitter

n

PCM transmission path

The most important feature of PCM systems is the ability to control the effects of distortion and noise. A PCM signal may be reconstructed from the distorted and noisy input by means of regenerative repeaters placed sufficiently close to each other along the transmission route

Transmission Bandwidth of a PCM Wave

• Each encoded message sample is represented by a n-digit code word.Consequently, the signal bit rate (r) becomes

r = n fs fs ≥ 2 W, Where:W= The message B.W. , fs= Sampling rate

• The transmission bandwidth (BT) required by the PCM wave is:

nWnfrBsT

21

21

Main feature of pulse-code modulation

• If the regenerators are well placed then they cancel the effect of channel distortion and noise.

• In this case the only source of distortion and noise is the quantization error

Time-Division Multiplexing(TDM)

Ts

Tx

For exampleWe multiplex two signals g1(t), and g2(t)

TDM: - Multiple data streams are sent in different time in single data link/medium - Data rate of the link must be larger than a sum of the multiple streams - Data streams take turn to transmit in a short interval - widely used in digital communication networks

A

B

C

A

B

C

MUX D

EMUX

… A1 B1 C1 A2 B2 C2 …

TDM: Concept of Framing and Synchronization

DefineTs = sampling interval (period)fs = sampling frequency (rate)Tx = time spacing between adjacent samples in the TDM signal waveform

• Consider two cases: Equally sampled ( N of input signals ) Tx = Ts / N N : number of samples Non equally sampled Signals have non equal bandwidths

Minimum Bandwidth

• The resultant signal could be considered as a new one sampled with fx . To get the B.W. of the TDM-PAM signal:

• The used system must have a minimum B.W. equals to fx / 2 in order to pass the TDM-PAM signal.

• The B.W = K fx (½) < K < 1

Equally sampled signals• Example-2

Two signals each band-limited to W = 3 KHz N = 2 signals• Ts = 1 / ( 2W) = ( 6000)-1 = 166.7 μ sec.

• Tx = Ts / 2 = 83.3 μ sec.

• B ≥ ( 1 / 2Tx ) ≥ 6 KHz

• N.B. PAM is as efficient in conserving bandwidth as SSB when all the signals have the same bandwidth.

• Example-3If we have 20 signals each band-limited to 3.3

KHz sampled with rate 8 KHz and PAM-TDM is used. Find the min. clock freq. and min. B.W.

Solution:Clock frequency fx=8K20=160 KHzMin. B.W.=fx/2 =80 KHzIf PCM-TDM is used using 3-bits/sample:The bit rate = 8K 203 = 480 Kb/s

Non-equally sampled signals

• For signals having non equal bandwidths we do non equal sampling

• To conserve B.W. PAM time multiplexed systems usually group input signals of comparable bandwidths

• Example three signals band limited to W , W and 2W

Equally Sampled:• fs = 4 W

• Ts = 1 / 4 W

• Tx = 1 / 12 W

• fx = 12 W

• B = fx/2 = 6 W > SSB B.W.

number Of

samplesfs

B.W Signal

1 2W W S1

1 2W W S2

1 4W 2W S3

3 1 2 3

Non equally sampled• Ts = 1 / 2W• N = 4• Tx = 1/(8W)• B = 4W• = SSB B.W.

number of

samplesfs B.W Signal

1 2W W S1

1 2W W S2

2 4W 2W S3

1 3 2 3 1

1.544Mb/s

T1

44.736 Mb/s

T3

PCM

Mux

12

24

M12

Mux

12

4

6.312 Mb/s

T2 M23

Mux

12

7

M34

Mux

1

2

6

274.176 Mb/s

North American Digital Hierarchy

EXAMPLE OF NON EQUALLY SAMPLING T1– DIGITAL SYSTEM

T4

Multiplexing PCM signals

• 24 – analog signals each band-limited to 3.3 K sampled at a rate of 8 K. Frame time=125 s

• At 6000 ft, repeaters are used• Each sample is encoded by 8-bits• No. of levels = 256 level• No. of bits /frame = (24 8) + 1F bit = 193 bitsWhere: 1F is one frame synchronization bit• Bit rate for T1 channel =(193 8K)=1.544 Mb/s

T1 TDM format for one frame.

Multiplexing T1-Lines

• 4-T1 lines are multiplexed to generate one T2 line.

• M12 multiplexer adds 17 bits for synch.

• Number of bits/frame = (193 4) + 17 = 789 bits/frame• Bit rate for T2 =789 8K = 6.312 Mb/s

Multiplexing T2-Lines

• 7-T2 lines are multiplexed to generate one T3 line.

• M23 multiplexer adds 69 bits for synch.

• Number of bits/frame = (789 7) + 69 = 5592 bits/frame• Bit rate for T3 = 5592 8K = 44.736 Mb/s

Multiplexing T3-Lines

• 6-T3 lines are multiplexed to generate one T4 line.

• M34 multiplexer adds 720 bits for synch.

• Number of bits/frame = (5592 6) + 720 = 34272 bits/frame• Bit rate for T4 = 34272 8K = 274.176 Mb/s

Line Coding

• How to represent the ‘1’ and ‘0’

• Each line code has its advantage and disadvantage

• A good line code should have the following requirements:

Requirements of line code:

• Transmission bandwidth : as small as possible• Error detection and correction• Avoid D.C. ( due to the A.C. coupling used at the

repeaters• Adequate timing content: It should be possible to

extract timing or clock information from the signal

• (long sequence of zeroes could cause errors in timing extraction for certain line codes)

Line Codes

1

1

1

1

1

1

0

0

0

Examples• On/Off (unipolar)– “1” send p(t), “0” nothing – Return to zero (RZ)– Non-Return to Zero (NRZ)

• Polar (bipolar)– “1” send p(t), “0” send -p(t) Bipolar (RZ) Bipolar (NRZ)

1 1 1 0 0 1 1

1 1 1 0 0 1 1

1 1 1 0 0 1 1 1 1 1 0 0 1 1

• Bipolar Alternate Mark Inversion (AMI):

– “0” has no pulse– “1” changes the sign of the waveform p(t)

1 1 1 0 0 1 1

• Bi-phase Codes (Manchester)

• More complex waveforms can “split” the phase of the two signals

1 1 1 0 0 1 1