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EC 723 Satellite Communication Systems

EC 723 Satellite Communication Systems. Mohamed Khedr http://webmail.aast.edu/~khedr. Syllabus. Tentatively. Interleaving. Convolutional codes are suitable for memoryless channels with random error events. Some errors have bursty nature:

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EC 723 Satellite Communication Systems

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  1. EC 723 Satellite Communication Systems Mohamed Khedr http://webmail.aast.edu/~khedr

  2. Syllabus • Tentatively

  3. Interleaving • Convolutional codes are suitable for memoryless channels with random error events. • Some errors have bursty nature: • Statistical dependence among successive error events (time-correlation) due to the channel memory. • Like errors in multipath fading channels in wireless communications, errors due to the switching noise, … • “Interleaving” makes the channel looks like as a memoryless channel at the decoder.

  4. Interleaving … • Interleaving is done by spreading the coded symbols in time (interleaving) before transmission. • The reverse in done at the receiver by deinterleaving the received sequence. • “Interleaving” makes bursty errors look like random. Hence, Conv. codes can be used. • Types of interleaving: • Block interleaving • Convolutional or cross interleaving

  5. A1 A1 A1 A1 A1 A2 B1 B1 A2 A2 A3 A3 C1 A3 C1 A2 A2 B1 B1 B1 B2 B2 B2 B2 B2 B3 B3 B3 C2 C2 A3 C1 C1 A3 C1 C2 B3 C2 B3 C2 C3 C3 C3 C3 C3 Interleaving … • Consider a code with t=1 and 2 coded bits. • A burst error of length 3 can not be corrected. • Let us use a block interleaver 3X3 2 errors Interleaver Deinterleaver 1 errors 1 errors 1 errors

  6. Trellis diagram for K = 2, k = 2, n = 3 convolutional code.

  7. State diagram for K = 2, k = 2, n = 3 convolutional code.

  8. MULTIPLE ACCESS - 1 • THE PROBLEM:HOW DO WE SHARE ONE TRANSPONDER BETWEEN SEVERAL EARTH STATIONS? f1 f2 Satellite Transponder IT IS AN OPTIMIZATION PROBLEM

  9. MULTIPLE ACCESS - 2 • NEED TO OPTIMIZE • Satellite capacity (revenue issue) • Spectrum utilization (coordination issue) • Interconnectivity (multiple coverage issue) • Flexibility (demand fluctuation issue) • Adaptability (traffic mix issue) • User acceptance (market share issue) • Satellite power • Cost Very, VERY, rarely a simple optimum; nearly always a trade-off exercise

  10. HOW DO YOU SEPARATE USERS? • LABEL THE SIGNAL IN A UNIQUE WAY AT THE TRANSMITTER • UNIQUE FREQUENCY SLOT FDMA • UNIQUE TIME SLOT TDMA • UNIQUE CODE CDMA • RECOGNIZE THE UNIQUE FEATURE OF EACH SIGNAL AT THE RECEIVER

  11. CHANNEL RECOGNITION? • FDMA • BAND PASS FILTER EXTRACTS SIGNAL IN THE CORRECT FREQUENCY SLOT • TDMA • DE-MULTIPLEXER “GRABS” SIGNAL IN THE CORRECT TIME SLOT • CDMA • DE-SPREADER OR DE-HOPPER EXTRACTS SIGNAL WITH THE CORRECT CODE Direct Sequence Frequency-Hopped

  12. Multiple access techniques: FDMA, TDMA, and CDMA. Note that in the direct sequence form of CDMA shown here, all the channels overlap in both time and frequency.

  13. MULTIPLE ACCESS • If the proportion of the resource (frequency, time, code) is allocated in advance, it is called PRE-ASSIGNED MULTIPLE ACCESS or FIXED MULTIPLE ACCESS • If the proportion of the resource is allocated in response to traffic conditions in a dynamic manner it is called DEMAND ASSIGNED MULTIPLE ACCESS - DAMA

  14. FDMA

  15. FDMA • SHARE THE FREQUENCY • TIME IS COMMON TO ALL SIGNALS • DEVELOP A FREQUENCY PLAN FROM USER CAPACITY REQUESTS • TRANSPONDER LOADING PLAN USED TO MINIMIZE IM PRODUCTS TRANSPONDER LOADING PLAN

  16. FDMA TRANSPONDER LOADING PLAN One large and four small digital signals Four medium-sized FM signals Available transponder bandwidth typically 27 to 72 MHz IMPORTANT TO CALCULATE INTERMODULATION PRODUCTS

  17. INTERMODULATION • INTERMODULATION • WHEN TWO, OR MORE, SIGNALS ARE PRESENT IN A CHANNEL, THE SIGNALS CAN “MIX” TOGETHER TO FORM SOME UNWANTED PRODUCTS • WITH THREE SIGNALS, 1, 2 AND 3, PRESENT IN A CHANNEL, IM PRODUCTS CAN BE SECOND-ORDER, THIRD-ORDER, FOURTH-ORDER, ETC. ORDER OF IM PRODUCTS

  18. IM PRODUCT ORDER • Second-order is 1 + 2, 2 + 3, 1 + 3 • Third-order is 1 + 2 + 3, 21 - 2, 22 - 1.. • Usually, only the odd-order IM products fall within the passband of the channel • Amplitude reduces as order rises • Only third-order IM products are usually important 3-IM products very sensitive to small signal changes. Hence, IM ‘noise’ can change sharply with output amplifier back-off

  19. IM EXAMPLE • There are two 10 MHz signals at 6.01 GHz and 6.02 GHz centered in a 72 MHz transponder • 2-IM product is at 12.03 GHz • 3-IM products are at [2(6.01) - 6.02] = 6.00 and [2(6.02) - 6.01] = 6.03 GHz 3-IM products

  20. FDMA LIMITATIONS • Intermods cause C/N to fall • Back-Off is needed to reduce IM • Parts of band cannot be used because of IM • Transponder power is shared amongst carriers • Power balancing must be done carefully

  21. FDMA Techniques

  22. TDMA

  23. TDMA • SHARE THE TIME • FREQUENCY IS COMMON TO ALL SIGNALS • DEVELOP A BURST TIME PLAN FROM USER CAPACITY REQUESTS • LARGE SYSTEM BURST TIME PLANS CAN BE COMPLICATED AND DIFFICULT TO CHANGE BURST TIME PLAN

  24. BURST TIME PLAN #1 #2 #3 #N time Frame Time for Burst Time Plan USERS OCCUPY A SET PORTION OF THE FRAME ACCORDING TO THE BURST TIME PLAN NOTE: (1) GUARD TIMES BETWEEN BURSTS (2) LENGTH OF BURST  BANDWIDTH ALLOCATED

  25. TDMA - 1 • THE CONCEPT: • Each earth station transmits IN SEQUENCE • Transmission bursts from many earth stations arrive at the satelliteIN AN ORDERLY FASHION and IN THE CORRECT ORDER

  26. TDMA - 2 NOTE: Correct timing accomplished using Reference Transmission

  27. TDMA - 3 FRAME “Traffic Burst” Pre-amble in each traffic burst provides synchronization, signaling information (e/s tx, e/s rx, etc.), and data “Pre-amble”

  28. TDMA - 4 • Timing obtained by • organizing TDMA transmission into frames • each e/s transmits once per frame such that its burst begins to leave the satellite at a specified time interval before (or after) the start of a reference burst • Minimum frame length is 125 s • 125 s  1 voice channel sampled at 8 kHz

  29. TDMA - 5 • Reference burst(s) and pre-amble bits are system overhead and earn no revenue • Traffic bits earn the revenue • Need to minimize system overhead • Complicated system trade-off with number of voice (or data) channels, transmission bit rate, number of bursts, etc.

  30. TDMA - 6 Number of bursts in a frame Number of bits in each pre-amble Transmission bit rate Number of voice channels Frame period For INTELSATR = 120 Mbit/s and TF = 2 ms Bit rate for one voice channel No allowance for guard times

  31. TDMA - 7 • PROBLEM • Delay time to GEO satellite is  120 ms • TDMA Frame length is 125 s to 2 ms • There could be almost 1000 frames on the path to the satellite at any instant in time • Timing is therefore CRUCIAL in a TDMA system

  32. LONG TDMA FRAMES • To reduce overhead, use longer frames • 125 s frame: 1 Word/Frame • 500 s frame: 4 Words/Frame • 2000 s frame: 16 Words/Frame 2000 s = 2 ms = INTELSAT TDMA standard NOTE: 1 Word is an 8-bit sample of digitized speech, a “terrestrial channel”, at 64 kbit/s 8 kHz × 8 bits = 64 kbit/s

  33. TDMA EXAMPLE - 1 • Transponder bandwidth = 36 MHz • Bit rate (QPSK) 60 Mbit/s = 60 bits/s • Four stations share transponder in TDMA using 125 s frames • Pre-amble = 240 bits • Guard time = 1.6 s Assuming no reference burst we have

  34. TDMA EXAMPLE - 2 FRAME= 125 s #1 #2 #3 #4 Guard time 96 bits = 1.6 s First thing to do: draw theTiming Recovery Diagramto get a picture of the way the frame is put together Traffic: N bitslet it = T s Pre-amble 240 bits= 4 s @ 60 bits/ s

  35. TDMA EXAMPLE - 3 • WITH THE TDMA EXAMPLE • (a) What is the transponder capacity in terms of 64 kbit/s speech channels? • (b) How many channels can each earth station transmit? • ANSWER • (a) There are four earth stations transmitting within the 125 s frame, so we have

  36. TDMA EXAMPLE - 4 • 125 s frame gives125 = (44 s) + (41.6 s) + (4T s) Four earth stations, 4 s pre-amble, 1.6 s guard time, T s traffic bits This gives T = (125 - 16 - 6.4)/4 = 25.65 s60 Mbit/s  60 bits/s, thus 25.65 s = 1539 bitsHence channels/earth station = 1539/8 = 192(.375) 8 bits/word for a voice channel

  37. TDMA EXAMPLE - 5 • (a) What is the transponder capacity in terms of 64 kbit/s speech channels?Answer: 768 (64 kbit/s) voice channels • (b) How many channels can each earth station transmit?Answer: 192 (64 kbit/s) voice channels

  38. TDMA EXAMPLE - 6 • What happens in the previous example if we use an INTELSAT 2 ms frame length?2 ms = 2,000 s = 44 + 41.6 + 4TTherefore, T = 494.4 sand, since there are 60 bits/s (60 Mbit/s), we have T  29,664 bits Remember we have 128 bits for a satellite channel

  39. TDMA EXAMPLE - 7 • With 128 bits for a satellite channel we haveNumber of channels/access = 29,664/128 = 231(.75) • Capacity has increased due to less overhead125 s frame  192 channels/access2 ms frame  231 channels/access

  40. TDMA SYNCHRONIZATION • Start-up requires care!! • Need to find accurate range to satellite • Loop-back (send a PN sequence) • Use timing information from the controlling earth station • Distance to satellite varies continuously • Earth station must monitor position of its burst within the frame AT ALL TIMES

  41. Structure of an Intelsat traffic data burst. A satellite channel is a block of sixteen 8-bit samples from one terrestrial speech channel. Other blocks in the traffic burst are used to synchronize the PSK demodulator, the bit clock, and the frame clock in the receiver (CBTR, UW) and to provide communication links between earth stations (TTY, SC, and VOW). CBTR, carrier and bit timing recovery; UW, unique word; TTY, teletype; SC, satellite channel; VOW, voice order wire. Carrier and bit Time recovery Unique word Telegraphy Telephony Order wires Service channel

  42. Unique word correlator. The example shown here has a 6 bit unique word for illustration—practical satellite systems use unique words of 24-48 bits. The bits stream from the receiver output is clocked into the shift register serially. When the contents of the shift register match the stored unique word the output of the summer is a maximum and exceeds the threshold, marking the end of the unique word. This provides a time marker for the remainder of the earth station’s transmission.

  43. TDMA SUMMARY - 1 • ADVANTAGES • No intermodulation products (if the full transponder is occupied) • Saturated transponder operation possible • Good for data • With a flexible Burst Time Plan it will optimize capacity per connection

  44. TDMA SUMMARY - 2 • DISADVANTAGES • Complex • High burst rate • Must stay in synchronization

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