Digital ransmission ethods S-7.5 Second Exercise Session Hypothesis esting Decision aking Gram-Schmidt method Detection.K.K. Communication Laboratory 5//6 Konstantinos.koufos@tkk.fi
Exercise We assume that under hypothesis H the source output is a constant voltage m. Under hypothesis H the source output is zero. Voltage is corrupted by additive noise. he output is sampled with N samples for each second. Each noise sample is modelled as zero mean Gaussian random variable with variance. he samples are assumed to be independent and identically distributed. It is known by Bayesian theory that the optimal test in such case is a likelihood ratio test compared to a decision threshold. a. Define the sufficient statistic of the test. b. What is the distribution of the sufficient statistic under the two hypotheses? c. Calculate the detection and the false alarm probability of the optimal test. d. What is the total error probability?
Exercise Sol. Simple binary hypothesis tests Only possible source outputs Hypotheses he transition mechanism generates points according to known conditional PDFs Observation space is one dimensional in binary hypothesis tests A cost is assigned for each possible event 4 possible events H generated and H chosen H generated and H chosen False alarm probability H generated and H chosen Detection probability H generated and H chosen iss probability
Exercise Sol. (contd) he source output under the hypotheses is : he probability density of a single sample r i under the hypotheses is : r i : the i th random variable R i : the value of the i th sample he joint pdf of the N samples is : What is implied? R R, R,, R, R i N
Exercise Sol. (contd) Construct the likelihood ratio test of the hypotheses : he common terms are cancelled: Bayesian threshold dependent on a priori prob. & decision costs he calculations are brought to the log-domain : Equivalently : ln Finally : Decision threshold ln ln
Exercise Sol. (contd) a. he sufficient statistic of the test adds the N collected amplitudes and compares the sum with the decision threshold. If the sum is lower than the decision threshold, hypothesis H is favoured. Otherwise the voltage m is assumed to be present. b. Under hypothesis H the sufficient statistic is the sum of N independent Gaussian random variables with zero mean and variance. As a result the sufficient statistic follows normal distribution with zero mean and variance N. Under hypothesis H the sufficient statistic is the sum of N independent Gaussian random variables with mean equal to m and variance. he sufficient statistic follows normal distribution with mean equal to Nm and variance N. HIN!! he sum of independent Gaussian random variables is also a Gaussian Random variable with mean the sum of individual means and variance equal to the sum of individual variances For m = and = we plot the PDF of the sufficient statistic under both the hypotheses and for different amount of samples
Exercise Sol. (contd).5 N = Samples.5 N = 4 Samples.45.4 Sufficient Statistic PDF under H Sufficient Statistic PDF under H.45.4 Sufficient Statistic PDF under H Sufficient Statistic PDF under H Sufficient Statistic PDF.35.3.5..5 Sufficient Statistic PDF.35.3.5..5...5.5 - -8-6 -4-4 6 8 Observation Space L - -8-6 -4-4 6 8 Observation Space L As the number of samples increases the mean separation of the two distributions increases too As the number of samples increases the variance of both distributions increases. he probability of decision error is reduced as the number of samples increases How to calculate the probability of decision error?
Exercise Sol. (contd) c. False alarm probability : Decide that the voltage is present when it is actually absent Q z z exp x exp z z erfc dx erfc z x dx Q z c. Detection probability : Decide that the voltage is present when it is actually present d. he probability of decision error is written below : N = Samples P F =.4 P D =.76 N = 4 Samples P F =.6 P D =.84
Exercise Consider a DS-SS system where the following Walsh-Hadamard codes of length N = 4 are used for orthogonalization.. Assume BPSK modulation scheme and symbol energy equal to Eb. he noise in the receiver is modelled as Gaussian random variable with zero mean and variance equal to Compare the performance of the following two schemes: Case : For bit + the firts code is used and for bit the second code is used Case : Only the first code with antipodal signalling is used both for bits and
Exercise Sol. One can see that the two waveforms are orthogonal to each other. his means that for the performance of the two schemes we expect a result similar to the one by comparison of BFSK and BPSK. Case : Orthogonal signalling H : A A A A H : A A A A he joint pdf of the 4 samples is : R A R A R3 A R4 A H : exp exp exp exp H R A R A R3 A : exp exp exp R 4 A exp he likelihood ratio of the two hypotheses is :
Exercise Sol. Finally : H R4 R3 ln H 4A he distribution of the test statistic under the hypotheses is : For equally probable hypotheses the probability of error is : H : ~ N A, H : ~ N A, e e H e H e H he probability of error under hypothesis H is : L A A e e H exp dl Q Under antipodal signalling we do not have the terms of the likelihood ratio cancelled. he sufficient statistic of the test contains all the 4 random variables, R,R,R 3 and R 4 After some manipulation the sufficient statistic is found to be : Why the lower integration limit equals?
Exercise Sol. H 3 4 4 8A R R R R ln H he distribution of the test statistic under the hypotheses becomes : H : ~ N A, 4 H : ~ N A, 4 By following the same procedure as previously : L A 4A e e H exp dl Q 4 4
Exercise 3 A digital transmission system in an AWGN channel is power limited but not bandwidth limited. he receiver power is pw and the one-sided noise power spectral density is - W/Hz. Using orthogonal signalling where the number of symbols is an integer power of two: a. Determine the minimum to transmit bps if the BER requirement equals -6 b. Calculate how many times must the signal bandwidth be increased from the value in a. when the bit rate is doubled but the BER requirement remains unchanged.
Exercise 3 Sol. For orthogonal signalling the upper bound for the BEP is : E Prx Prx log ( ) BEP Q Q Q No NoRs NoRb For the input values : Prx NoR b BEP Q 7 log ( ) For different values of : 4 BEP Q log ( ) Q 7.83 6 4 BEP Q log ( ) Q 7.74 8 8 BEP Q log ( ) 4Q 3 8.64
Exercise 3 Sol. (contd) When the bit rate is doubled we can determine the required -value in the same way : 4 8 BEP Q 5log ( ) 4Q 5.5 5 6 BEP Q 5log ( ) 8Q 3. 6 3 BEP Q 5log ( ) 6Q 5 4.59 7 64 BEP Q 5log ( ) 3Q 3 6.9 he required bandwidth is proportional to the number of symbol values. hus the increase of the bandwidth is 64/8 = 8-fold.
Exercise 4 Using the Gram-Schmidt procedure find an orthogonal basis functions set for the three signals depicted in the following figure: s (t) s (t) s 3 (t) 3 t t 3 t -4 Express the three signals with respect to the orthogonal basis functions set.
Exercise 4 Sol. Gram-Schmidt orthogonalization procedure permits to: Represent each of energy signals as a linear combination of N orthonormal basis functions where N i N j ij j s t s t, t i,,, where Orhtogonalization Algorithm: Define the first basis function as: Define the other basis functions as: In practice: he energy of the first signal is: herefore the first basis function is: i,,, sij si t j t dt, j,,, N E s t t E t i and gi t, i,3, N i g dt 4 t t dt s t, if t E, otherwise t j, i t dt, if i j if i j where git sit sij jt i j
Calculate: s s t t dt 4 dt 4 Apply the formula: 4, t g t s t s t, otherwise Similarly: Exercise 4 Sol. (contd) t, if t, otherwise 3 3 and 3 3 s s t t dt dt 3 3 s s t t dt 3 dt 3 3, t3 g3 t s3 t s3 t s3 t, otherwise 3t, if t3, otherwise he signals can be represented as: 4 4 3 3 3 s t t s t t t s t t t t 3 3 Remark: he sum of the squares of the coefficients equals the signal energy
Exercise 5 Consider the following signal set: E cos fcti t si t 4 otherwise a. What is the dimensionality of the signal set? b. Find an orthogonal basis functions set for the above signal set. c. Using the formula: N j s t s t, i,,3, 4 i ij j c. find the coefficients sij i,,3, 4 c. Sketch the signal set within the orthogonal basis functions space. f c n c
Break the cosine into sums: Exercise 5 Sol. E E E si t cos fcti cos fct cos i sin fct sin i, t 4 4 4 Linear combination of orthogonal function!! he orthonormal functions are: t cos fct t sin fct Let s calculate s(t): t E E E c c c s t cos f t cos f t cos sin f t sin 4 4 4 E E cos c sin c s t f t f t s t E cos fct E sin fct s s
Exercise 5 Sol. (contd) he signal coordinates in the basis functions orthogonal system are: s s 3 t t : E, E s t :, E : E, E s4 t : E, s 4 (t) s 3 (t) (t) s (t) s (t) (t)
Exercise 6 hree messages m, m, m3 are to be transmitted over an AWGN channel with power spectral density he messages are: s t t otherwise N t st s3t t otherwise a. What is the dimensionality of the signal space? b. Find an appropriate basis for the signal space. c. Draw the signal constellation for this problem d. Derive and sketch the optimal decision regions R, R, R3 e. Which of the three messages is more vulnerable to errors and why? P error m transmitted, i,,3, is larger? In other words which of i
Exercise 6 Sol. a. he dimensionality of the signal space is N= since s 3 (t) is antipodal to s (t) b. A simple set of othogonal functions can be time-shifted rectangular pulses (t) t) / / c. he signal coordinates in the basis functions system are: s s s 3 :, :, :, s3 s s
x x x 3 x 3 P m P m P m P m P m P m P m P m x x x x x x x x x x Exercise 6 Sol.. (contd( contd) d. he borders of the decision areas are calclated as: 3,,,,,,,, 3 x 3 x 3 x 3 x P m P m P m P m P m P m P m P m x x x x x x x x x xm xm xm xm 3 3,,,,,,,, x! x! xm xm xm xm x x x 3 x 3 P m P m P m P m P m P m P m P m x x x x x x x x x 3,,,,,,,, x! x xm xm xm xm s3 s s Equiprobable alphabet set and same signals energy
Exercise 6 Sol.. (contd( contd) e. Symbol error probability calculation for m, P e m P c m P x x m Pe m exp " " Pe m exp " " Pe m exp x t t dt Note : Q x Q x " " dt dx Pe m Q Q Q N N N Q z z exp x exp z z erfc dx erfc z x dx Q z
e. For symbol m the error probability is:!, P e m P c m P x x x m Exercise 6 Sol.. (contd( contd) x y Pe m exp dx exp dy " " 4" 4" " Pe m exp exp " 4" t u " dt " " du Pe m Q Q Q Q Q N N N N N y x x ~ N,"
Exercise 7 Consider a set of signal waveforms s m t, m, t all of which have the same energy E. he signal waveforms are separated in frequency: Es c sm t cos f t % f t, m,,,, t a. For which frequencies df the above signal set is orthogonal? b. Assume that Sm(t) are orthogonal and define a new set of waveforms as: s# t s t s t, m, t m m k k Show that the signal waveforms have equal energy given by: E# c. Show that the signals are now correlated with correlation coefficient: $ mn s# m t s# n t dt E# E
Exercise 7 Sol. he frequency separation df determines the degree to which we can disciminate among the possible transmitted signals. Es cos cos & s t s t dt f t m% f t f t n% f t dt mn m n c c Es Es sin m n % f & mn cos m n% f t dt cos 4 fct m n% f t dt m n % f f c We look for neighboring frequency separation m-n= he signal waveforms are orthogonal when the separation between neighboring frequencies is a multiple of / he minimum frequency separation for rthogonality between neighboring frequencies is: df = /
he energy of the signal waveform Exercise 7 Sol. (contd) s t m# is E# sm# t dt sm t sk t dt sm t dt sk t sn t dt sm t sk t dt k k n k E E# E E E E E he multiple is if kn Less energy Less dimensions he correlation coefficient is given by: $ $ $ s# ts# tdt s t s t s t s t dt E# E# mn m n m k n l k l s ts tdt s t s tdt s t s tdt s t s tdt E# l k k l mn m n m l n k k l mn E E E E E E # # Evaluate both results for =
Appendix In binary Bayesian hypotheses testing problems it is decided that hypothesis is present in a way to minimize the Bayesian risk: R C p Pr say H H is true C p Pr say H H is true C p Pr say H H is true C p Pr say H H is true Given the observation vector, show that the division of the observation space minimizing the Bayesian risk is the following test: say H say H if if r r (!' ( ' p ( R r p r H H r H R H ' p C C p C C Assume that under hypothesis H the generated signal is absent and the received signal is: x t where w(t) stands for one dimensional Gaussian noise with zero mean and variance. Under hypothesis H the generated signal is present and the received signal is: w t x t w t s t where s(t) stands for constant DC level A. For equal probable hypothesis assume unity amplitude and noise variance and evaluate the error probability with respect to the value of the decision threshold. For which value of the decision threshold the error probability becomes minimal?
Appendix Simple binary hypothesis tests Only possible source outputs according to a known PDF In general the generated hypothesis is N-dimensional he transition mechanism generates points according to known conditional PDFs he received points have N coordinates 4 possible events H generated and H chosen H generated and H chosen H generated and H chosen H generated and H chosen A cost is assigned for each possible event he mean value of the cost on average must be as small as possible (Bayes Criterion) Problem Formulation: How to use the a priori knowledge for hypotheses generation and probabilistic transition mechanism in order to divide optimally the observation space.
Appendix Write an expression for the mean value of the cost including the decision regions and the transition probabilities: Use that: End up with: fixed cost Note!! C>C & C>C he terms inside the integral are positive Choose Z in a way to minimize the integral
Appendix o minimize the integral the observation space points that make it positive must be assigned to Z. Say H when: r H R ) r HR P C C p H P C C p H Equivalently: pr H R H H P C C ) p R H P C C r H For one dimensional binary hypothesis testing the observation space is linear Pr Prdet P P Pr say H H is true PPr say H H istrue e P P Pr false alarm P miss e Pe P Pr false alarm P ection & & A Pe erfc erfc " "