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.

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EC 723 Satellite Communication Systems

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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

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

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???

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

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

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

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

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

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

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???

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

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

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

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

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

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

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

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)

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.

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.

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

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

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

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

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

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

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

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

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!

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

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 =>

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

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.

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!