1. Fundamentals of Digital Transmission
Chapter 7
Fundamentals of Digital Transmission

2. Baseband Transmission (Line codes)
ON-OFF or
Unipolar
(NRZ)
Non-Return-to-Zero
Polar
(NRZ)

3. Performance Criteria of Line Codes
Zero DC value
Inherent Bit-Synchronization
Rich in transitions
Average Transmitted Power
For a given Bit Error Rate (BER)
Spectral Efficiency (Bandwidth)
Inversely proportional to pulse width.

4. Comparison Between On-Off and Polar
Zero DC value:
Polar is better.
Bandwidth:
Comparable
Power:
BER is proportional to the difference between the two levels
For the same difference between the two levels, Polar consumes half the power of on-off scheme.
Bit Synchronization:
Both are poor (think of long sequence of same bit)

5. More Line Codes On-Off RZ Bi-Polar Better synch., at extra bandwidth
at same
bandwidth

6. More Line Codes Polar RZ Manchester (Bi-Phase) Perfect synch 3 levels

7. Spectra of Some Line Codes

8. Pulse Shaping
The line codes presented above have been demonstrated using (rectangular) pulses.
There are two problems in transmitting such pulses:
They require infinite bandwidth.
When transmitted over bandlimited channels become time unlimited on the other side, and spread over adjacent symbols, resulting in Inter-Symbol-Interference (ISI).

9. Nyquist-Criterion for Zero ISI
Use a pulse that has the following characteristics
One such pulse is the sinc function.

10. The Sinc Pulse t p(t) P(f)1 Tb
2Tb t
3Tb
4Tb
5Tb
6Tb
-6Tb
-5Tb
-4Tb
-3Tb
-2Tb
-Tb
P(f)
Note that such pulse has a bandwidth of Rb/2 Hz.
Therefore, the minimum channel bandwidth
required for transmitting pulses at a rate
of Rb pulses/sec is Rb/2 Hz f
-1/(2Tb)
1/(2Tb)

11. Zero ISI

12. More on Pulse Shaping
The sinc pulse has the minimum bandwidth among pulses satisfying Nyquist criterion.
However, the sinc pulse is not fast decaying;
Misalignment in sampling results in significant ISI.
Requires long delays for realization.
There is a set of pulses that satisfy the Nyquist criterion and decay at a faster rate. However, they require bandwidth more than Rb/2.

13. Raised-Cosine Pulses
where b is 2Rb and x is the excess bandwidth. It defines how much bandwidth required above the minimum bandwidth of a sinc pulse, where

14. Spectrum of Raised-Cosine Pulses

15. Extremes of Raised-Cosine Spectra

16. Raised-Cosine Pulses

17. Bandwidth Requirement of Passband Transmission
Passband transmission requires double the bandwidth of baseband transmission.
Therefore, the minimum bandwidth required to transmit Rb pulses/sec using carrier modulation is Rb Hz.

18. Transmission rates of Typical Services
Speech
Audio
Fax
Coloured Image
Video

19. Speech (PCM) B = 3.4 kHz Rs = 8000 samples/sec
Encoding = 8 bits/sample
Transmission rate = 64 kbps
Required bandwidth (passband) = 64 kHz
One hour of speech = 64000x3600 = Mb

20. Audio B = 16-24 kHz Rs = 44 000 samples/sec Encoding = 16 bits/sample
Stereo type = 2 channels
Transmission rate = 1.4 Mbps

21. Fax Resolution 200x100 pixels/square inch 1 bit/pixel (white or black)
A4 Paper size = 8x12 inch
Total size = 1.92 Mb = 240 KB
Over a basic telephone channel (3.4 kHz, baseband) it takes around 4.7 minutes to send one page.

22. Colour Image (still pictures)
Resolution 400x400 pixels/inch square
8 bits/pixel
3 colours/photo
A 8x10 inch picture is represented by Mb = 38.4 MB !

23. Video (moving pictures)
Size of still pictures
15 frames/sec
307 Mb/frame x 15 frames/sec = 4605 Mbps =4.6 Gbps !!

24. Solutions Compression M-ary communication reduces data size
Expands channel ability to carry information

25. M-ary Transmission In the binary case one pulse carries one bit.
Let each pulse carry (represent) m bits.
Bit rate becomes m multiples of pulse rate
We need to generate 2m different pulses.
They can be generated based on:
Multiple Amplitudes (baseband and passband)
Multiple Phases (passband)
Multiple frequencies (passband)
Some combination (Amplitude and Phase).

26. Signal Constellation
Signal constellation is a convenient way of representing transmitted pulses.
Each pulse is represented by a point in a 2-dimensional space.
The square of the distance to the origin represents the pulse energy.
The received signals form clouds around the transmitted pulses.
A received points is decoded to the closest pulse point.

27. Multiple Amplitudes (PAM)1 00 10 11 01
000
100
110
010
011
111
101
001
8 “levels”
3 bits / pulse
3×B bits per second
2 “levels”
1 bits / pulse
B bits per second
4 “levels”
2 bits / pulse
2×B bits per second

28. Same-maximum-power Scenario
4 signal levels
8 signal levels
typical noise

29. signal + noise signal noise High SNR noise signal signal + noise Lowt t t
signal
noise
signal + noise
Low
SNR t t t
Average Signal Power
SNR =
Average Noise Power

30. Same-BER Scenario Average power for binary case: ½ A2 + ½ A2 = A2
Average power for 4-ary case: ¼ (9 A2 + A2 + A2 + 9 A2 ) = 5 A2

31. Carrier Modulation of Digital Signals
Information 1 +1 -1 T 2T 3T 4T 5T 6T
Amplitude
Shift
Keying
Frequency
Phase t

32. Spectrum

33. TDM for Digital

34. Digital Hierarchy

35. Multiple Phases (MPSK)
2 bits / pulse
2×B bits per second
8 “phases”
3 bits / pulse
3×B bits per second

36. Quadrature Amplitude Modulation (QAM)Ak Bk
4 “levels”or pulses
2 bits / pulse
2xB bits per second
QAM Bk
16 QAM Ak
16 “levels” or pulses
4 bits / pulse
4xB bits per second

37. The Modulation Process of QAM
Modulate cos(wct) and sin (wct) by multiplying them by Ak and Bk respectively: Ak x
cos(wc t)
Yi(t) = Ak cos(wc t) Bk
sin(wc t)
Yq(t) = Bk sin(wc t) +
Y(t)

38. QAM Demodulation x x Y(t) LPF Ak 2cos(wc t)
2cos2(wct)+2Bk cos(wct)sin(wct)
= Ak {1 + cos(2wct)}+Bk {0 + sin(2wct)} x
LPF Bk
2sin(wc t)
2Bk sin2(wct)+2Ak cos(wct)sin(wct)
= Bk {1 - cos(2wct)}+Ak {0 + sin(2wct)}