Presentation is loading. Please wait.

Presentation is loading. Please wait.

Fundamentals of Digital Transmission

Similar presentations


Presentation on theme: "Fundamentals of Digital Transmission"— Presentation transcript:

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

18 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.

19 Transmission rates of Typical Services
Speech Audio Fax Coloured Image Video

20 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

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

22 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.

23 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 !

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

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

26 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).

27 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.

28 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

29 Same-maximum-power Scenario
4 signal levels 8 signal levels typical noise

30 signal + noise signal noise High SNR noise signal signal + noise Low
t t t signal noise signal + noise Low SNR t t t Average Signal Power SNR = Average Noise Power

31 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

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

33 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

34 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)

35 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)}


Download ppt "Fundamentals of Digital Transmission"

Similar presentations


Ads by Google