ec 723 satellite communication systems n.
Download
Skip this Video
Loading SlideShow in 5 Seconds..
EC 723 Satellite Communication Systems PowerPoint Presentation
Download Presentation
EC 723 Satellite Communication Systems

Loading in 2 Seconds...

play fullscreen
1 / 87

EC 723 Satellite Communication Systems - PowerPoint PPT Presentation


  • 185 Views
  • Uploaded on

EC 723 Satellite Communication Systems. Mohamed Khedr http://webmail.aast.edu/~khedr. Syllabus. Tentatively. Agenda. Modulation Concept Analog Communication Digital Communication Digital Modulation Schemes. EFFECT OF FILTERING - 1. Fig. 5.8 in text. EFFECT OF FILTERING - 2.

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

EC 723 Satellite Communication Systems


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
    Presentation Transcript
    1. EC 723 Satellite Communication Systems Mohamed Khedr http://webmail.aast.edu/~khedr

    2. Syllabus • Tentatively

    3. Agenda • Modulation Concept • Analog Communication • Digital Communication • Digital Modulation Schemes

    4. EFFECT OF FILTERING - 1 Fig. 5.8 in text

    5. EFFECT OF FILTERING - 2 • Rectangular pulses (i.e. infinite rise and fall times of the pulse edges) need an infinite bandwidth to retain the rectangular shape • Communications systems are always band-limited, so • send a SHAPED PULSE • Attempt to MATCH the filter to the spectrum of the energy transmitted Before FILTERS, let’s look at Inter-Symbol Interference

    6. INTER-SYMBOL INTERFERENCE • Sending pulses through a band-limited channel causes “smearing” of the pulse in time • “Smearing” causes the tail of one pulse to extend into the next (later) pulse period • Parts of two pulses existing in the same pulse period causes Inter-Symbol Interference (ISI) • ISI reduces the amplitude of the wanted pulse and reduces noise immunity Example of ISI

    7. ISI - contd. - 1 Form Couch, Fig. 3-23

    8. ISI - contd. - 2 • To avoid ISI, you can SHAPE the pulse so that there is zero energy in adjacent pulses • Use NRZ; pulse lasts the full bit period • Use Polar Signaling (+V & -V); average value is zero if equal number of 1’s and 0’s • Communications links are usually AC coupled so you should avoid a DC voltage component • Then use a NYQUIST filter Nyquist Filter???

    9. NYQUIST FILTER - 1 • Bit Period is Tb • Sampling of the signal is usually at intervals of Tb • Thus, if we could generate pulses that are at a one-time maximum at t = Tb and zero at each succeeding interval of Tb (i.e. t = 2Tb, 3Tb, ….. , NTb then we would have no ISI • This is called a NYQUIST filter

    10. NYQUIST FILTER - 2 Sampling instant is CRITICAL Impulse at this point t 0 Tb 2Tb 3Tb 4Tb

    11. NYQUIST FILTER - 3 NOTE: At each sampling interval, there is only one pulse contribution - the others being at zero level Fig. 5.9 in text

    12. NYQUIST FILTER - 4 • Arranging to sample at EXACTLY the right instant is the “Zero ISI” technique, first proposed by Nyquist in 1928 • Networks which produce the required time waveforms are called “Nyquist Filters”. None exist in practice, but you can get reasonably close

    13. NYQUIST FILTER - 5 • Noise into receiver must be held to a minimum • Place half of Nyquist filter at transmit end of link, half at receive end, so that the individual filter transfer function H(f) is given byVr(f)NYQUIST = H(f) H(f)Filter is a “Square Root Raised Cosine Filter” H(f)matches pulse characteristic, hence it is called a “matched filter” Matched Filter

    14. MATCHED FILTER - 1 f f Roll-off factor =  = (f / f0) where f0 = 6 dB bandwidth B = absolute bandwidth (here shown for  = 0.5) and B = f + f0 f1 = start of ‘roll-off’ of the filter characteristic 6 dB f1 f0 B

    15. MATCHED FILTER - 2 • A Raised Cosine Filter gives a Matched Filter response • The “Roll-Off Factor”, , determines bandwidth of Raised Cosine Low Pass Filter (LPF) • Gives zero ISI when the output is sampled at correct time, with sampling rate of Rb (i.e. at a sampling interval of Tb) BUT how much bandwidth is required for a given transmission rate???

    16. BANDWIDTH REQUIRED - 1 • Bandwidth required depends on whether the signal is at BASEBAND or at PASSBAND • Bandwidth needed to send baseband digital signal using a Nyquist LPF isBandwidth = (1/2)Rb(1 + ) • Bandwidth needed to send passband digital signal using a Nyquist Bandpass filter isbandwidth = Rb(1 + ) NOTE: It is the Symbol Rate that is key to bandwidth, not the Bit Rate

    17. BANDWIDTH REQUIRED - 2 • SYMBOL RATE is the number of digital symbols sent per second • BIT RATE is the number of digital bits sent per second • Different modulation schemes will “pack” different numbers of Bits in a single Symbol • BPSK has 1 bit per symbol • QPSK has 2 bits per symbol

    18. BANDWIDTH REQUIRED - 3 • OCCUPIED BANDWIDTH, B, for a signal is given by B = Rs ( 1 +  ) where Rs is the symbol rate and  is the filter roll-off factor • NOISE BANDWIDTH, BN, for a channel will not be affected by the roll-off factor of filter. Thus BN = Rs

    19. BANDWIDTH EXAMPLE - 1 • GIVEN: • Bit rate 512 kbit/s • QPSK modulation • Filter roll-off, , is  = 0.3 • FIND: Occupied Bandwidth, B, and Noise Bandwidth, BN • SOLUTION:Symbol Rate = Rs = (1/2)  (512  103) = 256  103 2 bits per symbol Number of bits/s

    20. BANDWIDTH EXAMPLE - 2 • Occupied Bandwidth, B, isB = Rs (1 +  ) = 256  103 ( 1 + 0.3) = 332.8 kHz • Noise Bandwidth, BN, isBN = Rs = 256 kHz • Now what happens if you have FEC? Example with FEC

    21. BANDWIDTH EXAMPLE - 3 • SAME Example, but 1/2-rate FEC is now used • SOLUTION Symbol Rate, Rs= (1/2)  (2)  (512  103) = 512  103 symbols/sOccupied Bandwidth, B, isB = Rs ( 1 +  ) = 665.6 kHz 2 bits per symbol 1/2-rate FEC used Number of bits/s

    22. BANDWIDTH EXAMPLE - 3 • Noise Bandwidth, BN, isBN = Rs = 512  103 = 512 kHz • Summary: • High Modulation Index  More Bandwidth Efficient • FEC (Block or Convolutional)  Increases bandwidth required

    23. Digital Modulations Functional model of passband data transmission system

    24. Digital Modulations • In digital communications, the modulating signal is a binary or M-ary data. • The carrier is usually a sinusoidal wave. • Change in Amplitude: Amplitude-Shift-Keying (ASK) • Change in Frequency: Frequency-Shift-Keying (FSK) • Change in Phase: Phase-Shift-Keying (PSK) • Hybrid changes (more than one parameter). Ex. Phase and Amplitude change: Quadrature Amplitude Modulation (QAM)

    25. Binary Modulations – Basic Types These two have constant envelope (important for amplitude sensitive channels)

    26. Coherent and Non-coherent Detection • Coherent Detection (most PSK, some FSK): • Exact replicas of the possible arriving signals are available at the receiver. • This means knowledge of the phase reference (phased-locked). • Detection by cross-correlating the received signal with each one of the replicas, and then making a decision based on comparisons with pre-selected thresholds. • Non-coherent Detection (some FSK, DPSK): • Knowledge of the carrier’s wave phase not required. • Less complexity. • Inferior error performance.

    27. Design Trade-offs • Primary resources: • Transmitted Power. • Channel Bandwidth. • Design goals: • Maximum data rate. • Minimum probability of symbol error. • Minimum transmitted power. • Minimum channel bandiwdth. • Maximum resistance to interfering signals. • Minimum circuit complexity.

    28. Coherent Binary PSK (BPSK) • Two signals, one representing 0, the other 1. • Each of the two signals represents a single bit of information. • Each signal persists for a single bit period (T) and then may be replaced by either state. • Signal energy (ES) = Bit Energy (Eb), given by: Therefore 

    29. Orthonormal basis representation • Gram-Schmidt Orthogonalization: basis of signals that are both ortogornal between them and normalized to have unit energy. • Allows representation of M energy signals {si(t)} as linear combinations of N orthonormal basis functions, where N<=M. • Ex.: N=2

    30. BPSK representation • Let’s consider the unidimensional base (N=1) where: • Let’s also rewrite the signal amplitudes as a function of their energy:

    31. BPSK representation • Therefore, we can write the signals s1(t) and s2(t) in terms of 1(t): • Which can be graphically represented as:

    32. BPSK Physical Implementation +A -A

    33. Detection of BPSK • Actual BPSK signal isreceived with noise • We assume AWGN inthis class • Other noise properties are possible • AWGN is a good approximation • Other noise models are more complex • Constellation becomes a distribution because of noise variations to signal

    34. Recall Gaussian Distribution Area to the right of this line represents Probability (x>x0) • = mean =standard deviation x0 x  Approximation for large positive values of y  Where: Both Q(.) and erfc(.) functions are integrals widely tabled and available as functions in Excel and calculators

    35. Calculating Error Probability Noise Spectral Density = N0 Noise Variance: BPSK error probability

    36. Bit Error Rate (BER) for BPSK Approximation valid for Eb/No greater than ~4 dB • BER is therefore given by Eb/No (dB) BER 0 0.08 2 0.04 4 0.014 6 0.0027 8 2*10-4 10 4*10-6 10.543 10-6 Note that these calculations are for synchronous detection

    37. Ambiguity Resolution • We haven’t discussed yet how to tell which signal state is a 1 and which a 0 • Because of variations in the signal path, its impossible to tell a priori • Two common approaches resolutions: • Unique Word • Differential Encoding

    38. Unique Word Ambiguity Resolution • A specific, known unique word is sent • The unique word is sent at a known time in the data • The correct signal state is chosen as 1 to correctly decode the unique word • Usually implemented with two detectors - the output of the correct one is simply used • Could lead to problems until a new UW is RX if a phase slip occurs • All bits after slip will be received wrong!

    39. Differential Encoding Ambiguity Resolution • Data is not transmitted directly • Each bit is represented by: 0 => phase shift of p radians 1 => no phase shift in the carrier • This results in ~ doubling the BER since any error will tend to corrupt 2 bits • BER is then Valid for BER<~0.01

    40. Coherent Quaternary PSK (QPSK) • Four signals are used to convey information • This leads to a constellation of:when shown as a phasorreferenced to the signal phase, q • Each of the two states representsa two bits of information Constant Modulus =>

    41. QPSK Constellation Representation • In this case we use the following orthonormal basis: • Which gives, after application of some trigonometric identities, the following constellation representation:

    42. QPSK Constellation

    43. QPSK Waveform

    44. QPSK Physical Implementation Note that the QPSK signal can be seen to be two BPSK signals in phase quadrature

    45. Block diagrams of (a) QPSK transmitter and (b) coherent QPSK receiver.

    46. The two QPSK constellations. Note that they differ by п /4. When going from (1,1) to (-1, -1), the phase is shifted by п. When going from (1, -1) to (1,1), the phase shifts by п /2. Thus, depending on the incoming symbol, transitions from (1,1) can occur to (1,1), (1,-1), (-1, 1), or (-1, -1) or vice versa, leading to phase shifts of 0, ± п /2, or ± п in QPSK. I and Q represent the in-phase and quadrature bits, respectively. Arrows show all possible transitions.

    47. Bit Error Rate (BER) for QPSK • The BER is still the probability of choosing the wrong signal state (symbol now) • Because the signal is Gray coded (00 is next to 01 and 10 for instance but not 11) the BER for QPSK is that for BPSK: • BER (after a lot of derivation) is given by: Approximation valid for Eb/No greater than ~4 dB Note that Eb is here, not Es!

    48. Possible paths for switching between the message points in (a) QPSK and (b) offset QPSK.

    49. Block diagram of the OQPSK modulator.