Digital to Digital Conversion III-Basic Data Communication Systems-Lecture Slides, Slides of Digital Systems Design

Download Digital to Digital Conversion III-Basic Data Communication Systems-Lecture Slides and more Slides Digital Systems Design in PDF only on Docsity! 4.2 DIGITAL-TO-DIGITAL CONVERSION The conversion involves three techniques: line coding, block coding, and scrambling. Line coding is always needed; block coding and scrambling may or may not be needed. docsity.com Line coding and decoding Sender Receiver rr rr Digital data Digital data A Digital signal 0101+*+101 0101+++101 eee = > Link 4.3 ® docsity.com 4.6 A signal is carrying data in which one data element is encoded as one signal element ( r = 1). If the bit rate is 100 kbps, what is the average value of the baud rate if c is between 0 and 1? Solution We assume that the average value of c is 1/2 . The baud rate is then Example docsity.com 4.7 The maximum data rate of a channel is Nmax = 2 × B × log2 L (defined by the Nyquist formula). Does this agree with the previous formula for Nmax? Solution A signal with L levels actually can carry log2L bits per level. If each level corresponds to one signal element and we assume the average case (c = 1/2), then we have Example docsity.com e 4.3 Effect of lack of synchronization uv AE 4 . . . - oO oO o - ° a. Sent b, Received 48 docsity.com  Figure 4.5 Unipolar NRZ scheme Amplitude Normalized power 4.11 ® docsity.com Figure 4.6 Polar NRZ-L and NRZ-I schemes NRZ-L NRZ-I OFT ;otoritrtiyo! \ \ \ \ \ I | \ T 1 I re. i | 1 I I 1 1 | I Time 1 \ \ I I I I | | 6 Bandwidth ! \ 6 I b 05 andwi — > I Tv T ? I ? I I Time 0 \ I I I \ I | I 0 1 2 f/N \ I I \ | O No inversion: Next bit is 0 @ Inversion: Next bit is 1 docsity.com 4.13 In NRZ-L the level of the voltage determines the value of the bit. In NRZ-I the inversion or the lack of inversion determines the value of the bit. Note docsity.com 4.16 A system is using NRZ-I to transfer 10-Mbps data. What are the average signal rate and minimum bandwidth? Solution The average signal rate is S = N/2 = 500 kbaud. The minimum bandwidth for this average baud rate is Bmin = S = 500 kHz. Example 4.4 docsity.com Figure 4.7 Polar RZ scheme Amplitude Oy ty Of Oo 7 1 | Bandwidth I I I | I 1 I I | 1 I > 0.5 — | = | Time I 1 I I 1 0 0 1 2 f/N 4.17 ® docsity.com 4.18 Figure 4.8 Polar biphase: Manchester and differential Manchester schemes docsity.com 4.21 In bipolar encoding, we use three levels: positive, zero, and negative. Note docsity.com Figure 4.9 Bipolar schemes: AMI and pseudoternary Amplitude A 0 I | | | | I 1 | | | | 1 ' \ I \ I \ I I AMI ' ; 1 —§> 1 \ \ ; Time I I \ \ 1 I I | | | 1 I I 1 I Pseudoternary : > ! ; Time Bandwidth docsity.com MULTILEVEL TRANSITION SCHEMES 4.23 docsity.com Figure 4.11 Multilevel: 8B6T scheme A 00010001 ; 01010011 4+V 4 0 -V | -0-O0++4 ! -+-+4+0 01010000 +--+0+ docsity.com  Figure 4.12 Multilevel: 4D-PAMS scheme 00011110 1 Gbps A 250 Mbps Wire 1 (125 MBd) I Ll 1 ! 250 Mbps | Wire 2 (125 MBd) \ +27 | ' +127 I 1 250 Mbps “ Wire 3 (125 MBd) -2 Wire 4 (125 MBd) ® docsity.com  Figure 4.13 Multitransition: MLT-3 scheme +V ov v4 a. Typical case +V ov V+ b. Worse case Next bit: 0 Next bit: 1 Next bit: 1 Last Last non-zero non-zero Next bit:0 level:+V level: -V Next bit: 1 Next bit: 0 c. Transition states docsity.com  Figure 4.14 Block coding concept Division of a stream into m-bit groups m bits m bits m bits mB-to-nB substitution n bits n bits n bits Combining n-bit groups into a stream docsity.com 4.32 Figure 4.15 Using block coding 4B/5B with NRZ-I line coding scheme docsity.com .2. 4B/5B mapping codes Data Sequence Encoded Sequence Control Sequence Encoded Sequence 0000 11110 Q (Quiet) 00000 0001 O1L001 I dle) 11111 0010 10100 H (Halt) 00100 OO1L 10101 J (Start delimiter) 11000 0100 OL010 K (Start delimiter) 10001 O101 O1011 T (End delimiter) OL1OL 0110 O1110 S (Set) 11001 O11 OLLI R (Reset) OO111 1000 10010 1001 10011 1010 10110 1011 10111 1100 11010 1101 11011 1110 11100 1111 11101 docsity.com 4.33  Figure 4.17 8B/10B block encoding 8B/10B encoder 5B/6B encoding = encoding 10-bit block 8-bit block ® docsity.com e 4.18 AMI used with scrambling Violated digital signal Receiver Modified AMI encoding 4.37 docsity.com 4.38 Figure 4.19 Two cases of B8ZS scrambling technique docsity.com 4.41 HDB3 substitutes four consecutive zeros with 000V or B00V depending on the number of nonzero pulses after the last substitution. Note docsity.com 4.42 4-2 ANALOG-TO-DIGITAL CONVERSION A digital signal is superior to an analog signal. The tendency today is to change an analog signal to digital data. In this section we describe two techniques, pulse code modulation and delta modulation. docsity.com Figure 4.21 Components of PCM encoder Quantized signal PCM encoder Analog signal Quantizing PAM signal T1***1100 Digital data docsity.com 4.46 Figure 4.23 Nyquist sampling rate for low-pass and bandpass signals docsity.com 4.47 For an intuitive example of the Nyquist theorem, let us sample a simple sine wave at three sampling rates: fs = 4f (2 times the Nyquist rate), fs = 2f (Nyquist rate), and fs = f (one-half the Nyquist rate). Figure 4.24 shows the sampling and the subsequent recovery of the signal. It can be seen that sampling at the Nyquist rate can create a good approximation of the original sine wave (part a). Oversampling in part b can also create the same approximation, but it is redundant and unnecessary. Sampling below the Nyquist rate (part c) does not produce a signal that looks like the original sine wave. Example 4.6 docsity.com 4.48 Figure 4.24 Recovery of a sampled sine wave for different sampling rates docsity.com 4.51 An example related to Example 4.7 is the seemingly backward rotation of the wheels of a forward-moving car in a movie. This can be explained by under-sampling. A movie is filmed at 24 frames per second. If a wheel is rotating more than 12 times per second, the under- sampling creates the impression of a backward rotation. Example 4.8 docsity.com 4.52 Telephone companies digitize voice by assuming a maximum frequency of 4000 Hz. The sampling rate therefore is 8000 samples per second. Example 4.9 docsity.com 4.53 A complex low-pass signal has a bandwidth of 200 kHz. What is the minimum sampling rate for this signal? Solution The bandwidth of a low-pass signal is between 0 and f, where f is the maximum frequency in the signal. Therefore, we can sample this signal at 2 times the highest frequency (200 kHz). The sampling rate is therefore 400,000 samples per second. Example 4.10 docsity.com 4.56 What is the SNRdB in the example of Figure 4.26? Solution We can use the formula to find the quantization. We have eight levels and 3 bits per sample, so SNRdB = 6.02(3) + 1.76 = 19.82 dB Increasing the number of levels increases the SNR. Example 4.12 docsity.com 4.57 A telephone subscriber line must have an SNRdB above 40. What is the minimum number of bits per sample? Solution We can calculate the number of bits as Example 4.13 Telephone companies usually assign 7 or 8 bits per sample. docsity.com 4.58 We want to digitize the human voice. What is the bit rate, assuming 8 bits per sample? Solution The human voice normally contains frequencies from 0 to 4000 Hz. So the sampling rate and bit rate are calculated as follows: Example 4.14 docsity.com e 4.28 The process of delta modulation Amplitude Generated Time binary data 4.61 docsity.com  Figure 4.29 Delta modulation components DM modulator Digital data Analog signal Delay unit 4.62 ® docsity.com  Figure 4.30 Delta demodulation components DM demodulator 11°°*1100 6 \ Digital data Analog signal Delay unit 1.63 docsity.com  Figure 4.32 Parallel transmission ) {| \ - POO OF +- OD wf rd 4 N\ ~ \, hm @ 4.66 docsity.com  Figure 4.33 Serial transmission 0 1 1 0 0 O 0 0 1 0 ae docsity.com 4.68 In asynchronous transmission, we send 1 start bit (0) at the beginning and 1 or more stop bits (1s) at the end of each byte. There may be a gap between each byte. Note docsity.com 4.71 In synchronous transmission, we send bits one after another without start or stop bits or gaps. It is the responsibility of the receiver to group the bits. Note docsity.com Figure 4.35 Synchronous transmission Direction of flow Frame docsity.com