Vector Concept of Signals

Topics to be Covered Vector space concept Representation of signals using vectors Representation of noise using vectors Representation of modulation schemes using vector space concept 2

Signal Space Digitally modulated signal consists of selection / transmission of one out of a finite set of elements. The signals space concept is similar to vector space. Concept of vector space can be applied to represent digital modulated signals. It allows convenient visualization, mathematical modeling and comparison of various modulation schemes Energy

Orthogonal Space of Basis Functions An N-dim orthogonal space consisting of linearly independent basis functions Basis functions are denoted by Basis functions must satisfy the following conditions: When Kj are nonzero, the signal space is called Orthogonal; When Kj=1, the signal space is called Orthonormal.

Orthogonal Space of Basis Functions Any finite set of waveforms can be expressed as a linear combination of N orthogonal waveforms Also called Generalized Fourier Series

Energy of a Signal For orthonormal functions

Topics to be Covered Matched filter Correlator receivers Probability of error 7

OPTIMUM RECEIVER FOR BINARY DIGITAL MODULATION SCHEMES The function of a receiver in a digital communication system is to distinguish between all the M transmitted signals sm(t), in the presence of noise The performance of the receiver is usually measured in terms of the probability of error and the receiver is said to be optimum if it yields the minimum probability of error m=1,2,…,M 4

Received Signal + Noise Vector The shape of the waveforms depends on the type of the modulation used .The output of the modulator passes through a bandpass channel Hc(f) which for purposes of analysis is assumed to be an ideal channel with adequate bandwidth so the signal passes through without suffering any distortion other then propagation delay .the channel nosie n(t) is assumed to be a zero mean stationary, Gaussian random process with a known power spectral .(density Gn(f 8

Demodulation of Digital Signals

Additive white Gaussian noise (AWGN) with zero mean and variance N0 /2 Matched Filter Detection of signal in presence of additive noise Receiver knows what pulse shape it is looking for Channel memory ignored (assumed compensated by other means, e.g. channel equalizer in receiver) s(t) Transmitted signal w(t) x(t) h(t) y(t) t = T y(T) Matched filter T is the symbol period Additive white Gaussian noise (AWGN) with zero mean and variance N0 /2

Matched Filter Derivation Design of matched filter Maximizes signal power i.e. power at t = T Minimizes noise i.e. power of noise Combined design criteria

Matched Filter Derivation PSDs are related by Noise power spectrum SW(f) AWGN Matched filter

Matched Filter Derivation Find h(t) that maximizes pulse peak SNR

Matched Filter Derivation Schwartz’s inequality For functions: upper bound reached if

Matched Filter Derivation For real s(t)

Matched Filter Given transmitted signal s(t) of duration T, matched filter is given by hopt(t) = k s(T-t) for any k Duration and shape of impulse response of the optimal filter is determined by pulse shape s(t) hopt(t) is scaled, time-reversed, and shifted version of s(t) Optimal filter maximizes peak pulse SNR SNR Does not depend on pulse shape s(t) Directly proportional to signal energy (energy per bit) E Inversely proportional to power spectral density of noise

Matched Filter Receiver

Matched filter Example Received SNR is maximized at time T Matched Filter: optimal receive filter for maximized example: transmit filter receive filter (matched)

Correlation Filter At t=T At t=T, matched filter becomes a correlator

Correlator and Matched Filter Outputs for a Sine Wave Input

Matched Filter Receiver

Matched filter Example Received SNR is maximized at time T Matched Filter: optimal receive filter for maximized example: transmit filter receive filter (matched)

Correlation Filter At t=T At t=T, matched filter becomes a correlator

Correlator and Matched Filter Outputs for a Sine Wave Input

Channel Model AWGN channel model: Signal vector is deterministic. Elements of noise vector are i.i.d (independent and identically distributed) Gaussian random variables with zero-mean and variance . The noise vector pdf is The elements of observed vector are independent Gaussian random variables. Its pdf is

Correlation Receiver for BPSK/BASK One basis function, so one branch The received signal is input to a correlator The correlator output is compared with a threshold to decide which symbol was transmitted. Decision Circuit Compare r with threshold. Sample at T

Probability of Error BPSK/BASK Output of the correlator: a1(T) is the correlator output when s1(t) is sent a2(T) is the correlator output when s2(t) is sent n0(T) is the correlator output when AWGN is present at the input Hypotheses: H1: s1 is assumed to have been transmitted H2: s2 is assumed to have been transmitted 29

Probability of Error: BPSK/BASK 30

Probability of Error: BPSK/BASK 31

Probability of Error BPSK/BASK 32

Probability of Error BPSK/BASK Where Q(x) is the Complementary Error Function Now And for BPSK 33

Coherent detection of BFSK For correct detection, the carriers must be orthogonal.

Error Probability BPSK and BFSK BPSK and BFSK with coherent detection: “0” “1” “0” “1” BPSK BFSK

Assignment#4 Due Before start of next class on Saturday 23 April, 2011 Q 1: Show that the probability of error for coherently detected BFSK is given by Q 2: Derive the value of decision threshold if the binary transmitted symbols in BPSK are not equiprobable. Hint: See Sec. 3.2 and App B.2.

Quiz#4 Q : (a) A BASK signal transmits 2000 bps. If the energy per bit is 4 Joules and the noise PSD is 1 W/Hz, how many erroneous bits will be received in 1 second? (b) Also sketch the PDFs of the received signal for both the possible transmitted signals and indicate the area that corresponds to the probability of correct decision.

Coherent Detection of MPSK Information resides in the phase, which needs to be detected “110” “000” “001” “011” “010” “101” “111” “100” 8-PSK Compute Choose smallest Decision variable

Coherent Detection of M-QAM (Amplitude Phase Shift Keying) 0000 0001 0011 0010 1000 1001 1011 1010 1100 1101 1111 1110 0100 0101 0111 0110 16-QAM

Signal space dimension How many basis functions does it take to express a signal? It depends on the dimensionality of the signal Some need just 1 some need an infinite number. The number of dimensions is N and is always less than the number of signals in the set N<=M ©2000 Bijan Mobasseri

Example: Fourier series Remember Fouirer series? A signal was expanded as a linear sum of sines and cosines of different frequencies. Sounds familiar? Sines and cosines are the basis functions and are in fact orthogonal to each other ©2000 Bijan Mobasseri

Interpretation s1(t) is now condensed into just two numbers. We can “reconstruct” s1(t) like this s1(t)=(1)1(t)+(-0.5)2(t) Another way of looking at it is this 2 1 1 -0.5 ©2000 Bijan Mobasseri

Signal constellation Finding individual components of each signal along the two dimensions gets us the constellation 2 s4 s2 1 -0.5 0.5 -0.5 s1 s3 ©2000 Bijan Mobasseri

Energy in simple language What we just saw says that the energy of a signal is simply the square of the length of its corresponding constellation vector 2 E=9+4=13 3 ©2000 Bijan Mobasseri

Constrained energy signals Let’s say you are under peak energy Ep constraint in your application. Just make sure all your signals are inside a circle of radius sqrt(Ep ) ©2000 Bijan Mobasseri

Correlation of two signals A very desirable situation in is to have signals that are mutually orthogonal. How do we test this? Find the angle between them transpose s1 s2  ©2000 Bijan Mobasseri

Find the angle between s1 and s2 Given that s1=(1,2)T and s2=(2,1)T, what is the angle between the two? ©2000 Bijan Mobasseri

Distance between two signals The closer signals are together the more chances of detection error. Here is how we can find their separation 2 1 1 2 ©2000 Bijan Mobasseri

Constellation building using correlator banks We can decompose the signal into its components as follows s1 1 s2 N components s(t) 2 sN N ©2000 Bijan Mobasseri

Detection in the constellation space Received signal is put through the filter bank below and mapped to a point s1 1 s2 s(t) 2 components mapped to a single point sN N ©2000 Bijan Mobasseri

Constellation recovery in noise Assume signal is contaminated with noise. All N components will also be affected. The original position of si(t) will be disturbed ©2000 Bijan Mobasseri

Actual example Here is a 16-level constellation which is reconstructed in the presence of noise ©2000 Bijan Mobasseri

Detection in signal space One of the M allowable signals is transmitted, processed through the bank of correlators and mapped onto constellation question is based on what we see , what was the transmitted signal? received signal which of the four did it come from ©2000 Bijan Mobasseri

Minimum distance decision rule It can be shown that the optimum decision, in the sense of lowest BER, is to pick the signal that is closest to the received vector. This is called maximum likelihood decision making this is the most likely transmitted signal received ©2000 Bijan Mobasseri

Defining decision regions An easy detection method, is to compute “decision regions” offline. Here are a few examples decide s2 decide s1 decide s1 s2 s1 decide s1 measurement s2 s1 s1 s4 s3 decide s2 decide s3 decide s4 ©2000 Bijan Mobasseri

More formally... Partition the decision space into M decision regions Zi, i=1,…,M. Let X be the measurement vector extracted from the received signal. Then if XZi si was transmitted ©2000 Bijan Mobasseri

P{error|si}=P{X does not lie in Zi|si was transmitted} Error probability we can write an expression for error like this P{error|si}=P{X does not lie in Zi|si was transmitted} Generally ©2000 Bijan Mobasseri

Example: BPSK (Binary Phase Shift Keying) BPSK is a well known digital modulation obtained by carrier modulating a polar NRZ signal. The rule is 1: s1=Acos(2πfct) 0:s2= - Acos(2πfct) 1’s and 0’s are identified by 180 degree phase reversal at bit transitions ©2000 Bijan Mobasseri

Signal space for BPSK Look at s1 and s2. What is the basis function for them? Both signals can be uniquely written as a scalar multiple of a cosine. So a single cosine is the sole basis function. We have a 1-D constellation cos(2pifct) -A A ©2000 Bijan Mobasseri

Eb= A2Tb/2 --->A=sqrt(2Eb/Tb) Bringing in Eb We want each bit to have an energy Eb. Bits in BPSK are RF pulses of amplitude A and duration Tb. Their energy is A2Tb/2 . Therefore Eb= A2Tb/2 --->A=sqrt(2Eb/Tb) We can write the two bits as follows ©2000 Bijan Mobasseri

BPSK basis function As a 1-D signal, there is one basis function. We also know that basis functions must have unit energy. Using a normalization factor E=1 ©2000 Bijan Mobasseri

Formulating BER BPSK constellation looks like this -√Eb √Eb received if noise is negative enough, it will push X to the left of the boundary, deciding 0 instead X|1=[√Eb+n,n] noise -√Eb √Eb noise transmitted ©2000 Bijan Mobasseri

Finding BER Let’s rewrite BER But n is gaussian with mean 0 and variance No/2 -sqrt(Eb) ©2000 Bijan Mobasseri

BER for BPSK Using the trick to find the area under a gaussian density(after normalization with respect to variance) BER=Q[(2Eb/No)0.5] ©2000 Bijan Mobasseri

BPSK Example Data is transmitted at Rb=106 b/s. Noise PSD is 10-6 and pulses are rectangular with amplitude 0.2 volt. What is the BER? First we need energy per bit, Eb. 1’s and 0’s are sent by ©2000 Bijan Mobasseri

Solving for Eb Since bit rate is 106, bit length must be 1/Rb=10-6 Therefore, Eb=20x10-6=20 w-sec ©2000 Bijan Mobasseri

Solving for BER Noise PSD is No/2 =10-6. We know for BPSK BER=0.5erfc[(Eb/No)0.5] What we have is then Finish this using erf tables ©2000 Bijan Mobasseri

M-ary signaling Binary communications sends one of only 2 levels; 0 or 1 There is another way: combine several bits into symbols 1 0 1 1 0 1 1 0 1 1 1 0 0 1 1 Combining two bits at a time gives rise to 4 symbols; a 4-ary signaling ©2000 Bijan Mobasseri

QPSK constellation 00 01 45o √E 10 11 Basis functions S=[0.7 √E,- 0.7 √E] ©2000 Bijan Mobasseri

QPSK decision regions 00 01 10 11 Decision regions re color-coded ©2000 Bijan Mobasseri

QPSK error rate Symbol error rate for QPSK is given by This brings up the distinction between symbol error and bit error. They are not the same! ©2000 Bijan Mobasseri

Symbol error Symbol error occurs when received vector is assigned to the wrong partition in the constellation When s1 is mistaken for s2, 00 is mistaken for 11 s2 s1 11 00 ©2000 Bijan Mobasseri

Symbol error vs. bit error When a symbol error occurs, we might suffer more than one bit error such as mistaking 00 for 11. It is however unlikely to have more than one bit error when a symbol error occurs 00 10 10 11 10 Sym.error=1/10 Bit error=1/20 00 11 10 11 10 10 symbols = 20 bits ©2000 Bijan Mobasseri

QPSK vs. BPSK Let’s compare the two based on BER and bandwidth BER Bandwidth BPSK QPSK BPSK QPSK Rb Rb/2 EQUAL ©2000 Bijan Mobasseri

M-phase PSK (MPSK) If you combine 3 bits into one symbol, we have to realize 23=8 states. We can accomplish this with a single RF pulse taking 8 different phases 45o apart ©2000 Bijan Mobasseri

8-PSK constellation Distribute 8 phasors uniformly around a circle of radius √E 45o Decision region ©2000 Bijan Mobasseri

Quadrature Amplitude Modulation (QAM) MPSK was a phase modulation scheme. All amplitudes are the same QAM is described by a constellation consisting of combination of phase and amplitudes The rule governing bits-to-symbols are the same, i.e. n bits are mapped to M=2n symbols ©2000 Bijan Mobasseri

16-QAM constellation using Gray coding 16-QAM has the following constellation Note gray coding where adjacent symbols differ by only 1 bit 0000 0001 0011 0010 1000 1001 1011 1010 1100 1101 1111 1110 0100 0101 0111 0110 ©2000 Bijan Mobasseri

Vector representation of 16-QAM There are 16 vectors, each defined by a pair of coordinates. The following 4x4 matrix describes the 16-QAM constellation ©2000 Bijan Mobasseri

What is energy per symbol in QAM? We had no trouble defining energy per symbol E for MPSK. For QAM, there is no single symbol energy. There are many We therefore need to define average symbol energy Eavg ©2000 Bijan Mobasseri

Eavg for 16-QAM Using the [ai,bi] matrix and using E=ai^2+bi^2 we get one energy per signal Eavg=10 ©2000 Bijan Mobasseri

Relation between Eb/N0 & S/N With R=Rb In M-ary schemes In most cases e.g. FSK

Time Division Multiplexing