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Chapter 4 Handout #4

Chapter 4 Handout #4. Dr. Clincy Professor of CS. Today’s Agenda: (1) Lecture 26, 27 and 28, (2) Go over Exam 3, (3) Work on final project. Thursday: Exam 4 will cover all chapters, sections and lectures regarding data communications up to (but NOT including), "block coding".

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Chapter 4 Handout #4

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  1. Chapter 4Handout #4 Dr. Clincy Professor of CS Today’s Agenda: (1) Lecture 26, 27 and 28, (2) Go over Exam 3, (3) Work on final project Thursday: Exam 4 will cover all chapters, sections and lectures regarding data communications up to (but NOT including), "block coding". Exam 5: Exam 5 will cover all chapters, sections and lectures regarding data communications including "block coding“ and afterwards. Lecture 2

  2. Four Data-to-Signal Cases DATASIGNAL D D A DATA SIGNAL A D A Already covered Will cover today Lecture 2

  3. ANALOG-TO-DIGITAL CONVERSION We have seen in Chapter 3 that 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.

  4. Components of PCM encoder PCM – Pulse Code Modulation 1st: analog signal is sampled 2nd: sampled signal is quantized 3rd: quantized values are encoded as bit streams (or codes) Analog signal is sampled every Ts seconds Sample rate is fs = 1/Ts

  5. Three different sampling methods for PCM High-speed switch used – able to retain the shape of the signal Ideal but complex Sample-and-hold method that creates flat-top samples by using a circuit Sampling process also called pulse amplitude modulation (PAM)

  6. Recovery of a sampled sine wave for different sampling rates Nyquist theorem states that the sampling rate must be at least 2 times the highest frequency of the signal Catches the essence of the signal Doesn’t improve the case Doesn’t capture the essence of the signal

  7. Quantization and encoding of a sampled signal actual-amplitude/D actual amplitude • Quantization steps: • Determine Vmin and Vmax • Divide range into L zones, each of height D • D = [Vmin - Vmax]/L • 3. Assign quantized values of 0 to L-1 to midpoint of each zone • 4. Map the sample value to a quantized value Norm. Actual Error between actual and nornalized Quant. value for code Code that represents the voltage level Assume sample amplitudes between -20V and +20V Let L = 8 (levels) – therefore, D = [20 - -20]/8 = 5 Quantization error can contribute to Shannon’s SNR: SNRdB = 6.02nb + 1.76 where nb is bits per sample Bit rate = sampling rate x # of bits per sample = fs x nb

  8. Components of a PCM decoder PCM decoder recovers the original signal Smooths out the staircase signal What is the minimum bandwidth of the filter the digitized signal will need ? Bmin = c x nb x 2 x Banalog x 1/r (nb = # bits per sample) If 1/r=1 and c=1/2, Bmin=nb x Banalog If the data rate and number of signal levels are fixed, minimum bandwidth is Bmin = N / [2 x log2 L]

  9. The process of delta modulation PCM is more complex than Delta modulation PCM finds the amplitude of the signal; delta modulation simply finds the change in the signal from the previous sample Delta modulation doesn’t use codes – bits are sent one after another Positive changes are encoded as 1; negative changes are encoded as 0

  10. TRANSMISSION MODES The transmission of binary data across a link can be accomplished in either parallel or serial mode. In parallel mode, multiple bits are sent with each clock tick. In serial mode, 1 bit is sent with each clock tick. While there is only one way to send parallel data, there are three subclasses of serial transmission: asynch, syn and iso approaches (asynchronous, synchronous, and isochronous.)

  11. Parallel transmission

  12. Serial transmission

  13. Asynchronoustransmission 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. Asynchronous here means “asynchronous at the byte level,” but the bits are still synchronized; their durations are the same.

  14. Synchronous transmission 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.

  15. Isochronous Transmission • For realtime audio and video, uneven delays between frames is not acceptable – so synchronous transmission doesn’t work well • The entire stream of bits must be synchronized – this is isochronous transmission • Isochronous transmission guarantees data at a fixed rate

  16. Chapter 5Handout #4 Dr. Clincy Professor of CS Lecture 2

  17. Digital-to-analog conversion Based on the digital data, the Modulator changes characteristics of the “controllable” analog signal (bandpass analog signal) on the transmitter side to represent the digital data Demodulator interprets the analog signal in re-creating the digital data on the receiver side Terminology: “modulating digital data into an analog signal” The analog signal we can control ? Sine Wave, Carrier Signal, Periodic Signal Lecture

  18. Types of digital-to-analog conversion Change amplitude to represent a bit Change frequency to represent a bit Change phase to represent a bit Combination of changing both amplitude and phase to represent a set of bits Lecture

  19. Recall • For digital transmission, bit rate (data rate) and signal rate (baud rate) relationship was • S = N x 1/r where r = # of data elements per signal element and N is the data rate in bps (and S is the signaling or baud rate) • For analog, r = log2L where L is the type of signal (versus level) Lecture

  20. Example An analog signal carries 4 bits per signal element. If 1000 signal elements are sent per second, find the bit rate. Solution In this case, r = 4, S = 1000, and N is unknown. We can find the value of N from Lecture

  21. Example An analog signal has a bit rate of 8000 bps and a baud rate of 1000 baud. How many data elements are carried by each signal element? How many signal elements do we need? Solution In this example, S = 1000, N = 8000, and r and L are unknown. We find first the value of r and then the value of L. Lecture

  22. Binary amplitude shift keying changing the original amplitude Explain this not changing the original amplitude Bandwidth (B) is proportional to the signal rate (S) and depending on the modulation and filtering process, the required bandwidth can range between S to 2S (where middle bandwidth is fc). The value of d relates to the modulation and filtering process B = (1 + d) x S Lecture

  23. Example We have an available bandwidth of 100 kHz which spans from 200 to 300 kHz. What are the carrier frequency and the bit rate if we modulated our data by using ASK with d = 1? Solution The middle of the bandwidth is located at 250 kHz. This means that our carrier frequency can be at fc = 250 kHz. We can use the formula for bandwidth to find the bit rate (with d = 1 and r = 1). Lecture S = N * 1/r

  24. Binary frequency shift keying changing the original frequency Use two different carrier frequencies, f1 and f2, for 0 and 1 Explain this not changing the original frequency Lecture

  25. Example We have an available bandwidth of 100 kHz which spans from 200 to 300 kHz. What should be the carrier frequency and the bit rate if we modulated our data by using FSK with d = 1? Solution The midpoint of the band is at 250 kHz. We choose 2Δf to be 50 kHz; this means The difference (delta) between the two frequencies Lecture

  26. Binary phase shift keying changing the original phase Explain this not changing the original phase Lecture

  27. QPSK and its implementation QPSK – Quadrature Phase Shift Keying Use 2 bits in each signal element – decreases baud rate and bandwidth Uses 4 possible phases (versus 2) 2 composite signals are created Because the 2 signals are using the same bandwidth – each signal has ½ bandwidth Lecture

  28. Example Find the bandwidth for a signal transmitting at 12 Mbps for QPSK. The value of d = 0. Solution For QPSK, 2 bits is carried by one signal element. This means that r = 2. So the signal rate (baud rate) is S = N × (1/r) = 6 Mbaud. With a value of d = 0, we have B = S = 6 MHz. B = (1 + d) x S Lecture

  29. Concept of a constellation diagram Helps define the amplitude and phase of a signal element the amplitude of the 2nd carrier Peak Amplitude Phase This is the amplitude given one carrier Only use 1 carrier and phase is static and 2 amplitude levels Only use 1 carrier and 1 amplitude and 2 phases (0o and 180o) Uses 2 carriers and 1 amplitude and 4 phases (45o, 135o, -45o, -135o) Lecture Binary Phase Shift Keying – uses 2 different phases and the same amplitude

  30. Constellation diagrams for some QAMs QAM – Quadrature Amplitude Modulation For QPSK, we only changed the phase For QAM, we change both the phase and amplitude Has a 0 amplitude and a positive amplitude (with 2 carriers) Has a negative amplitude and a positive amplitude (with 2 carriers) Has 2 positive amplitudes (with 2 carriers) Has 4 negative levels and 4 positive levels (with 2 carriers) Lecture

  31. Four Data-to-Signal Cases DATASIGNAL D D A DATA SIGNAL A D A Already covered Will cover today Lecture 2

  32. ANALOG TO ANALOG Analog-to-analog conversion is the representation of analog information by an analog signal. One may ask why we need to modulate an analog signal; it is already analog. Modulation is needed if the medium is bandpass in nature or if only a bandpass channel is available to us. Bandpass – signal being shifted to a particular range Lowpass – signal that IS NOT shifted to a particular range Lecture

  33. Types of analog-to-analog modulation Lecture

  34. Amplitude modulation Vary the amplitude of the carrier signal to mimic the changing voltage levels (amplitude) of the modulating signal result The total bandwidth required for AM can be determined from the bandwidth of the audio signal: BAM = 2B. Lecture

  35. Frequency modulation Vary the frequency of the carrier signal to mimic the changes in voltage level (amplitude) of the modulating signal result The total bandwidth required for FM can be determined from the bandwidth of the audio signal: BFM = 2(1 + β)B. Would be given Lecture

  36. Phase modulation Vary the phase of the carrier signal to mimic the changes in voltage level (amplitude) of the modulating signal This illustrates the signal starting at different phases The total bandwidth required for PM can be determined from the bandwidth and maximum amplitude of the modulating signal: BPM = 2(1 + β)B. Lecture

  37. Chapter 6: Bandwidth Utilization:Multiplexing and Spreading Lecture

  38. Multiplexing & Spreading (Physical Layer Issues) Up to this point, you have learned about translating “data” into a “signal” – so that the “signal” can travel across the transport It would be very efficient use of the transport’s bandwidth if multiple signals could travel on the transport at the same time ? Also, it would be great if we could protect against eavesdropping That efficiency can be achieved by multiplexing; privacy and anti-jamming can be achieved by spreading. Lecture

  39. SPREAD SPECTRUM In spread spectrum (SS), we combine signals from different sources to fit into a larger bandwidth, but our goals are to prevent eavesdropping and jamming. To achieve these goals, spread spectrum techniques add redundancy. Typically used for wireless applications – privacy outweighs efficiency in this case Frequency Hopping Spread Spectrum (FHSS)Direct Sequence Spread Spectrum Synchronous (DSSS) Lecture

  40. Frequency selection in FHSS Lecture

  41. DSSS – Direct Sequence Spread Spectrum • Each bit sent by the Tx is replaced with a set of bits called a “chip code” • For the time it takes to send the original single bit, it now will take more time to send the chip code • Therefore, the data rate must be N times the original data rate, where N is the # of bits of the chip code • Also, the bandwidth for the chip code should N times greater than the original bit stream’s BW Example of original bits being transmitted as 6-bit chip codes Lecture

  42. DSSS using polar NRZ encoding Lecture

  43. Multiplexing Lecture

  44. Dividing a link into channels – Multiplexing in general Explain this Categories of multiplexing Will also cover Statistical Time-Division Multiplexing Lecture

  45. Frequency-division multiplexing Divide the link’s bandwidth into separate channels (guardband separating each channel) Recall from the Modulation Lectures that – being able to modulate around different “carrier frequencies” was important to being able to adjust the modulated signal into a particular “band” (bandpass signal) On the MULTIPLEXING SIDE Resultant modulated signals are combined into a single composite signal Signals modulate different carrier frequencies (based on amplitude in this case) Lecture

  46. FDM demultiplexing example On the DEMULTIPLEXING SIDE Lecture

  47. Example Assume that a voice channel occupies a bandwidth of 4 kHz. We need to combine three voice channels into a link with a bandwidth of 12 kHz, from 20 to 32 kHz. Show the configuration, using the frequency domain. Assume there are no guard bands. Solution We shift (modulate) each of the three voice channels to different bandwidth. We use the 20- to 24-kHz bandwidth for the first channel, the 24- to 28-kHz bandwidth for the second channel, and the 28- to 32-kHz bandwidth for the third one. Then we combine them into a single composite signal. Lecture

  48. Wavelength-division multiplexing Same as FDM but instead of electrical type signals – muxing optical signals (light signals) Lecture

  49. Time Division Multiplexing (TDM) All networking devices work off clock ticks (explain) Do “tap” analogy Explain this Lecture

  50. Synchronous time-division multiplexing Given n connections needing to be muxed, each frame is divided into n parts (for each slot) Also notice that the time duration before muxing is 1/3 of the time duration after muxing In this case, each frame is divided into 3 time slots For synchronous TDM, the Tx and Rx must be in synch for the Rx to “pull out” of the frame the correct set of data (called interleaving) For synchronous TDM, the data rate of the output link must be n times the data rate of the connection to guarantee the flow of data In keeping the mux and demux in synch, synch bits (framing bits) are added at the beginning of each frame Lecture

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