Spectral Analysis - determine, from finite set of samples, power in each harmonic of signals. For non-periodic signals, use Power Spectral Density.

Transcription

1 Spectral Estimation Spectral Analysis - determine, from finite set of samples, power in each harmonic of signals. For non-periodic signals, use Power Spectral Density. Bank of Band Pass Filters Spectral Resolution - measure of how close in frequency 2 sinusoids can be before they merge. Spectral quality - measure of how good spectral estimation is in terms of mean and variance: Q = (E[Sxx(f)])2 var[s xx(f)] Itcanbeimprovedbyincreasingthenumberofsamples,M. Both resolution and Quality can t be improved at same time for given M. If resolution is improved by reducing BPF bandwith, then BPF impulse response gets longer so less unbiased samples available for estimation. 30 1

2 Spectral Estimators: Parametric or Non-parametric Non-parametric: require no assumptions about the data except (quasi-) stationarity. Parametric: based on model of data so their application is more restrictive. Advantage: when applicable, they give more accurate spectral estimate and for fixed M, can reduce bias and variance due to a-priori knowledge in model. Power Spectral Density: S xx (ω) is defined as DFT of the auto-correlation, R xx (m) =E[x(k)x(k+m)] S xx (f) = m= R xx (m)e jωmt W einer Khintchine theorem (1) R xx can be got from S xx by finding its Fourier coefficients: 31 R xx (m) = T 2π 2πT 0 S xx (ω)e jωmt dω 2

3 Classic Spectral Analysis Eq (1) requires infinite set R xx (m) tocalculates xx (ω) but only set of M samples available One approach is to form M-point DFT: X M (ω) = M 1 k=0 x(k)e jωkt An estimate of PSD is then formed by dividing energy at each frequency by no. of data points: - Periodogram: S xx (ω) = X M(ω) 2 M 32 3

4 Leakage and Windowing In practice only x(k), 0 <k<n available so x(k) = x(k)w(k), w(k) = 1 for 0 <k<n, else= 0 X(f) =X(f) W (f) = Sidelobes of W (f) cause leakage.5.5 X(α)W (f α)dα Eg: Signal with X(f) =1for f < 0.1, else =0, is windowed by rectangular window of length N=61 W (f) = N 1 k=0 x(k)e j2πfk = 1 e jωn sin(ωn/2) = e jω(n 1)/2 1 e jω sin(ω/2) Width of main lobe ω =4π/61 or f =2/61 << widthx(f) 33 4

5 Power Spectral Density For signal x(t), average power is: 1 T P x = lim x 2 (t)dt = S xx (f)df T 2T T S xx = F [R xx (τ)] = R xx (τ)e j2πfτ dτ Properties of P.S.D: 1. S xx is real & +ve 2. Average power in x(t) =R xx (0) = S xx(f)df 3. For x(t) real, S xx ( f) =S(f) P.S.D of Random Sequences: S xx = R xx (k)e j2πfk, 1 2 <f<1 2 if T s =1 E.g: random binary waveform R xx (τ) =1 τ /T, τ <T, =0elsewhere [ sinπft S xx = F [R xx (τ)] = T πft 34 ] 2 5

6 M.F. Spectral Estimation Ŝ xx = N 1 k= (N 1) ˆR xx (k)e j2πfk, 1 2 <f<1 2 for T s =1 where ˆR xx (k) = 1 N k 1 N k i=0 x(i)x(i k), k=0, 1,...N 1 1. With N data points we can estimate R xx only for k <N. 2. As k N, we use fewer points to obtain ˆR xx. So Variance increases as k N Mean = E[ ˆR xx ]=R xx (k), k < N So estimator is unbiased if k<n. Var[ ˆR xx ]= 1 N k σ4 x for Gaussian process whose variance is σ 2 x. As k N, var[ ˆR xx ]increases. Estimate Ŝxx has 2 main disadvantages: 1. Large computation needed for ˆR xx 2. Variance[Ŝxx] canbelarge. 35 6

7 Periodogram Another estimate for R xx is ˆR xx (k) = 1 N N k 1 i=0 x(i)x(i + k), k =0, 1,...N 1 = N k N ˆR xx (k) Define d(n) =1,n=0...N 1, =0, else ˆR xx (k) = 1 N n= [d(n)x(n)][d(n + k)x(n + k)] Ŝ xx (f) =F [ ˆRxx (k)] = ˆR xx (k)e j2πkf n= = [ 1 N n= n= [d(n)x(n)][d(n + k)x(n + k)]]e j2πkf = 1 N [ [d(n)x(n)e j2πnf ][ [d(n + k)x(n + k)e j2π(k+n)f ] n= n= = 1 N X(f)X(f) = 1 N X(f) 2 - Periodogram 36 7

8 Bias of Periodogram: E[Ŝ xx (f)] = E[ ˆR xx (k)e j2πkf ]= E[ ˆRxx (k)]e j2πkf n= n= = n= 1 N [ n= [d(n)x(n)][d(n + k)x(n + k)]]e j2πkf = R xx (k)e j2πfk [ d(n)d(n + k)]/n n= n= = w N (k)r xx (k)e j2πfk,w N (k) = d(n)d(n+k)]/n =1 k /N, k < N n= n= = F [w N (k).r xx (k)] = 1/2 1/2 S xx (α)w N (f α)dα where W N (f) =F [w N (k)] = 1 N [ sin(πfn) sinπf ] 2 Variance of Periogram: var[ˆŝ xx (f)] = Sxx(f)[1 2 + [ sin(πfn) ] 2 ] N Sxx 2 0 sinπf Quality Factor of Periodogram: Q P = [E[Ŝ xx (f)]] 2 var[ˆŝ xx (f)] = S2 xx Sxx 2 =1 37 8

9 Variance of Periodogram (contd).. For zero mean, white Gaussian sequence with variance σ 2 : var[ŝ xx (f)] = σ 4 var[ŝ normalised standard error, ɛ = xx(f)] S xx = σ2 σ 2 i.e var[ˆŝ xx (f)] does NOT 0asN Trade off between Bias and Variance: Ŝ xx (f) =F [ ˆRxx (k)] so examine ˆRxx (k) which estimates R xx (k) for k = 0, 1,...M. When M<<Nwegetgoodestimates of R xx (k) for k =0, 1,...M.. But bias of Ŝ xx (f) is larger since R xx is truncated for k>m. As M N, Bias of Ŝ xx (f) decreases (less truncation of ˆR xx (k)) but var[ ˆˆRxx (k)] increases when k N as fewer points used in estimator. 38 9

10 Smoothing of Spectral Estimates 1. Bartlett Method - Averaging Periodograms: take N(>> 1) samples, x(0)...x(n 1), divide into n sections, each of N/n points, form n different Ŝ xx (f) and average the n different estimators to form averaged spectral estimators S xx (f) = 1 n n Ŝ xx (f) k If Ŝ xx (f) k are independent, then var[ S xx (f)] will be reduced by factor n. But since fewer points used for Ŝ xx (f) k, W N/n (f) will be n times wider than W N (f) so bias will be larger. var[ S xx (f)] = 1 n 2 n k=1 k=1 var[ˆŝ xx (f) k ]= 1 n var[ˆŝ xx (f)] Q B = [E[ S xx (f)]] 2 var[ S xx (f)] = n, N Freq resolution of Bartlett estimate f =.9/M, M = N/n 39 Q B = n = N/M = N.9/ f =1.1N f 10

11 2. Welch Method: Average Modified Periodograms Two modifications to Bartlett method: 1. Overlap data segments, x i (k) = x(k + id), k = 0, 1...M 1; i = 0, 1...L 1 2. Data Segments windowed prior to computing P gram to give modified P gram: S xx (f) = 1 MU M 1 2 x i (k)w(k)e j2πkf k=0 Where U is normalisation factor for power in window, U = 1 M Welch estimate is average of these periodograms Mean: = 1/2 1/2 S W xx(f) = 1 L L 1 i=0 S xx (f) i E[Sxx W (f)] = 1 L 1 E[ L S xx (f) i ]=E[ S xx (f) i ] i=0 S xx (α)w (f α)dα, W (f) = 1 MU M 1 2 w(k)e j2πkf k=0 M 1 k=0 w2 (k) As N and M, E[S W xx(f)] S xx (f) 40 11

12 Variance of Welch Estimate var[sxx W (f)] = 1 L 1 L 1 L 2 E[ S xx (f) i Sxx (f) i ] {E[Sxx W (f)]}2 i=0 j=0 For no overlap (D=M, L=n): var[sxx W (f)] = 1 L var[ S xx (f) i ] 1 L S2 xx (f) Q W = L = N/M For 50% overlap (D=M/2, L=2n) with triangular window: var[s W xx (f)] = 9 8L S2 xx (f) Q W = 8L 9 = 16N 9M Spectral width of the triangular window at the 3dB points is f =1.28/M Q W =0.78N f for no overlap 41 =1.39N f for 50% overlap + triangular window 12

13 Blackman Tukey Estimate Window autocorrelation sequence, take Fourier tfm: S B xx (f) = M 1 m=(m 1) r xx (m)w(m)e j2πmf = 1/2 1/2 Ŝ xx (α)w M (f α)dα Effect of windowing r xx (m) istosmooths xx, thus decreasing variance but reducing resolution. 1/2 E[Sxx B (f)] = E[Ŝ xx (α)]w M (f α)dα 1/2 = 1/2 1/2 1/2 1/2 S xx (θ)w B (α θ)w M (f α)dαdθ 1/2 1/2 S xx (θ)w M (f θ)dθ if M << N var[s B xx (f)] 1 N 1/2 1/2 Sxx 2 (α)w M 2 (f α)dα Sxx(f)[ 2 1 N if W M (f) narrow c.f. S xx (f) 42 1/2 1/2 W 2 M (f θ)dθ] 13

14 Multiplying by w(k) and taking Fourier tfm: E[ S xx (f)] = 1/2 1/2 S xx (α)w M (f α)dα, W M = F [w(k)] Rectangular: w(k) =1for k M, =0for k >M W M (f) =sin[(2m +1)πf]/sinπf Bartlett: w(k) = 1 k /M for k M, = 0for k > M W M (f) =[1/M ][sin(mπf)/sinπf] 2 Blackman-Tukey: w(k) =.5(1 + cos(πk/m)) for k M, = 0for k >M W M (f) =.25[D M (2πf π/m)+d M (2πf + π/m)] + D M (2πf) 43 14

15 Quality of B-T Estimate Mean, E[S BT xx (f)] 1/2 1/2 S xx(θ)w M (f θ)dθ S xx (f) asn,ie asymptotically unbiased var[s B xx(f)] S 2 xx(f)[ 1 N M 1 m=(m 1) w 2 (m)] = Sxx 2 (f)2m N = Sxx(f) 2 2M 3N for rect. window for triang. window Q BT =1.5N/M for triangular window As window length is 2M 1, f =1.28/2M =.64/M Q BT = N f =2.34N f Estimate Bartlett Welch(50%overlap) Blackman Tukey QualityF actor 1.11N f 1.39N f 2.34N f 44 15

16 Parametric Spectral Estimation Model-free methods: simple, efficiently computed via FFT. - but need long data for good frequency response and suffer from spectral leakage. - Basic Limitation: r xx =0form N from S xx (f) = N 1 m= (N 1) r xx (k)e j2πfm Model-based methods don t need this assumption so avoid spectral leakage and give better resolution with less data if model fits well. Based on modelling data x(k) as o/p of H(z) = B(z) q A(z) = i=0 b iz i 1+ p i=1 a...(1) i iz x(k) = q b i w(k i) i=0 p a i x(k i)...(2) Then PSD is S xx (f) = H(f) 2 S ww (f). Assume w(k) is white noise with r ww (m) =σw 2 δ(m). Then S xx (f) =σw H(f) 2 2 = σw 2 B(f) A(f) 2...(3) i=1 To find S xx (f): 1. From x(k), estimate {a i } and {b i } of model 2. From these estimates, compute S xx (f) from(3)

17 Types of Model x(k) generated by pole-zero model (1) is called an ARMA process of order (p,q). If q =0,H(z) =b 0 /A(z) and x(k) is an AR process. If A(z) =1, H(z) =B(z) and x(k) is a MA process. Any ARMA, AR, or MA process can be represented by AR or MA model. AR Spectral Analysis p S ee (f) =σe 2 = H 1 (f) 2 S xx (f), H 1 =1 a i z i z=e j2πft The whitening filter H 1 (z) is a linear predictor: i=

18 a i are chosen to minimise the MSE, E[e 2 (k)]whichleadtoacnequations: Φ xx a = φ xx where Φ xx = E[x(k 1)x T (k 1), φ xx = E[x(k 1)x T (k 1)] Min MSE, σ 2 e = E[x2 (k)] a T φ xx To perform AR analysis directly on the data, use sum of squared error cost fn: ρ = k [x(k) a T x(k 1)] 2 Set of a which minimise ρ is provided by the deterministic form of ACN eqn: R x a = r x Choosing the limits for k makes asumptions about x(k) before x(0) and after x(n-1),which effects the accuracy of a. Eg assume k starts at 0. Then e(0) = x(0) a T x( 1) = x(0) which gives no information about a. Yet e(0) is weighted equally with errors from the middle of x(k). Similarly e(1) = x(1) a T x(0) = x(1) a 1 x(0) which gives information only about a 1. It is only when e(p) is considered that no assumptions about the data have to be made

19 AR Windows Errors at the end of the data cause similar effect for possible windowing variations which lead to four possible estimates of a: Lower Upper Limit N+p 1 N 1 Limit autocorrelation pre windowed k=0 k=p post windowed covariance Despite its inaccuracy, auto-correlation form is popular in real-time systems as R xx is Toeplitz. The minimum sq error ρ is used to get an estimate, ˆσ 2 e of the MSE by dividing ρ by the no. of errors. For covariance form: The AR(p) spectral estimate is ˆσ 2 e = N 1 k=p x2 (k) r T x a N 1 p 48 S xx (f) = ˆσ 2 e 1+ p i=1 a ie jiωt 2 19

20 AR Model Order If p is too low, spectrum is very smoothed and can t seperate close peaks If p is too high, spectrum may have spurious peaks and is inefficient σ 2 e decreases as the order of AR model increases. Rate of decrease can be monitored and pchosenwhenrateofdecreasebecomes slow but this is imprecise. Other methods for selecting p: 1.Final Prediction Error: order is selected to minimise FPE(p) =ˆσ 2 e(n + p +1)/(N p 1) 2. Akaike Information Criterion: order is selected to minimise AIC(p) =lnˆσ 2 e +2p/N 3. Criterion Autoregressive Transfer: order is selected to minimise 49 CAT(p) =(1/N ) p ((N j)/ˆσ ej 2 ) ((N p)/ˆσ2 e ) j=1 20

21 MA Model S xx (z) =σw 2 H(z)H(z 1 )=σw 2 B(z)B(z 1 ) A(z)A(z 1 ) = D(z) σ2 w C(z) q m= q d mz m = σw 2 p m= p c mz m where d m = q m k=0 b k b k+m, m q When a 1 = a 2 =...a p =0itcanbeshownthat and PSD for MA(q) process is: r xx (m) ={ σ2 w d m m q 0 m >q Sxx MA (f) = q m= q = BT modified PERIODOGRAM r xx (m)e 2jπfm i.e we dont need to solve for b i to get Sxx MA (f) which is identical to nonparametric PSD estimate. Since r xx (m) =0for m >q, the order of MA model may be determined by noting if r xx (m) 0 for large lags. If this is not the case, MA model will give poor frequency resolution

22 ARMA Model Mpy Eq (2) by x (k m) and take expected values: q p E[x(k)x (k m)] = b n E[w(k n)x (k m)] a n E[x(k n)x (k m)] n=0 n=1 p q r xx (m) = a n r xx (m n)+ b n r wx (m n) n=1 n=0 r wx = { σ2 W h( m) m 0 0 m > 0 p n=1 a nr xx (m n) m>q r xx (m) = p n=1 a nr xx (m n)+σw 2 q m n=0 b n+mh(m) 0 m q rxx( m) m > 0 Coeffs a n for m>qfound as in AR case. Then A(z) =1+ p n=1 a nz n, which can filter x(k) togivev(k) =x(k)+ p n=1 a nx(k n) Then apply MA method to v(k) to obtain: S MA vv (f) = q m= q r vv (m)e 2jπfm 53 Sxx ARMA Svv MA (f) = (f) 1+ p n=1 a ne 2jπfm 2 22

23 Maximum Likelihood Spectral Estimation Model Spectrum Analyser as tunable filter. To find power at frequency, ω, tune filter to ω and measure o/p power. Filter must meet 2 criteria: 1. If signal consists of nothing but sinusoid at freq, ω, o/p power = i/p power. 2. Any signal at other freqs must make the min possible contribution to o/p power (Min Variance). Filter o/p, y(n) = N 1 k=0 a kx(n k) and assume i/p consists of complex sinusoid of freq, ω, and other components: x(n) =Ae jωn + p(n) MV design criteria may be written as: 1. If p(n) =0,theny(n) =x(n) 2. Otherwise, o/p power of the filter is to be a minimum

24 Consider 1. It means that Divide by Ae jωn : Use Vectors: N 1 k=0 N 1 k=0 a k Ae jω(n k) = Ae jωn a k Ae jωk =1...(3) a =[a 0,a 1,...a N 1 ] T Then (3) becomes e a =1 Consider 2. Variance, σ 2 = E[y 2 (n)] e =[1,e jω,e 2jω...e j(n 1)ω ] T N 1 N 1 = E[ a i x(n i) a k x(n k)] i=0 k=0 Interchanging order of summations and expected values: σ 2 = N 1 i=0 N 1 k=0 a i a ke[x(n i)x(n k)] 54b = N 1 i=0 N 1 k=0 a i a kr i k = a Ra 24

25 Need to min σ 2 = a Ra subject to e a =1 Using Lagrange s method F (a) =a Ra λe a Use constraint to find λ: df da =[2a R λe ] T =0 a =.5λR 1 e e a =.5λe R 1 e =1 λ =2/(e R 1 e) a = R 1 e e R 1 e = R 1 e e R 1 e For these a, σ 2 is estimate of spectral power at ω σ 2 = a Ra = e R 1 RR 1 e (e R 1 e) 2 = 1 e R 1 e = P ML(ω) e R 1 e evaluated by multiplying elements along any diagonal of R 1 by the same e jiω e jkω = e j(k i)ω. If n = k-i, then e R 1 e = N 1 n= (N 1) ejnω ρ n, ρ n = sum of all elements of R 1 along diagonal n. This can be computed with DFT, so we don t have to evaluate 1 e R 1 e for every freq of interest. 54c 25

26 Maximum Entropy Spectral Est Traditional methods not suited to finite data Windowing assumes data outside window is 0 - an affront to common sense [Burg] Burg s ME method starts with known data and gets everything needed from that. Entropy is measure of uncertainty. Since we want to leave signal free outside window, we maximise entropy of power spectrum. Start with expression of entropy of spectrum and find spectrum that maximises this, subject to condition that spectrum be consistent with known autocorrelations

27 Entropy: Let x be discrete random variable and p(x i ) be probability that x takes value x i. Entropy of x is H(x) = x i p(x i )logp(x i ) E.g if x can take only 2 values x 0 and x 1 then: where p = p(x 0 ) H(x) = plogp (1 p)log(1 p) H(x) has the following properties: 1. H(x)is continuous fn of p(x i ) 2. H(x)is non-negative and is 0 only for cases where p(x i ) = 0 for all the values of x i but one. 3. H(x) is max when all values equally likely 4. H(x) increases with N, if N can take on N different values, all equally likely 5. H(x) is additive in following sense: suppose we partition possible values of x into 2 sets A and B. Then finding the value of x can be divided into 2 steps: determine whether x is in A or B; determine actual value of x given that it is known to be in A (or B). The entropies of these 2 steps will sum to H(x) 56 27

28 If x is cont random variable with pdf = f(x), then entropy is H(x) = f(x)logf(x)dx... (2) Base of log gives unit of entropy. For discrete r.v., base = 2 and unit is bit For cont. r.v., use natural logs and unit is nat -forc.r.vwithstddevσ, pdf with max entropy is Gaussian For zero-mean Gaussian pdf with std dev σ f(x) = 1 2πσ e (x2 /2σ 2 ) From (2): H(x) =ln( 2πeσ) nats For n-dimensional, 0-mean, mulitivariate Gaussian pdf with covariance matrix C: f(x) =[(2π) n/2 C ] 1/2 e.5(xt C 1 x) H (x) =ln( (2πe) n C )nats...(3) Avoid problem of non-convergence of H(x) as signal length increases by using entropy rate: H 1 (x) = lim n n ln( (2πe) n C ) =ln( 2πe)+ 1 2 lim ln( C )/n n If signal is stationary, C is Toeplitz and: lim ln( C )/n = 1 W lns(f)df n 2W W 57 H (x) =ln( 2πe)+ 1 W lns(f)df... (4) 4W W 28

29 Application of ME to Spectral Analysis Assume we know r n out to max lag p, none beyond p. If we knew all r n,we could find PSD from: S(f) = 1 2W n= r n e j2πfnt, T =1/(2W ) But few known r n are not enough; there are an infinite number number of spectra, depending on unknown r n. ME Method: Find S(f) which maximises H (x) subject to r n = 1 W S(f)e j2πfnt df, 2W W n p... (5) So (4) becomes: H (x) =ln( 2πe)+ 1 W 4W W 1 ln[ 2W n= r n e j2πfnt ]df H (x) = 1 W 1 [ r n 4W W 2W n= r n e j2πfnt ] 1 e j2πfnt df = 1 W 4W W 1 S(f) e j2πfnt df Constraint (5) means derivatives = 0 only w.r.t. r n, n > p But these derivatives are Fourier coeffs of 1/S(f). So, ME criterion means that Fourier expansion of 1/S(f) must have finite no. of terms. So S(f) must be of form: 58 S(f) =[ p n= p q n e j2πfnt ] 1... (6) 29

30 To find coefficients {q n } : Write (6) as z-tfm: S(f) =S D (z) z=e j2πfnt, wheres D (z) =1/Q(z) S D (z) =P/[A(z)A (z 1 )], wherep =1/q 0 and A(z) = If R(z) is z-tfm of r n, then (5) becomes: R(z) =P [A(z)A (z 1 )] p a i z i i=0 R(z)A(z) =P/A (z 1 )... (7) As P/A(z) is z-tfm of impulse response h(n) of casual system, P/A (z 1 )is z-tfm of h ( n). So: r n a n = h ( n) p a i r n i = h ( n)... (8) i= p Equation (8) to be satisfied for n=0 to p (to match r n out to r p ) But if h(n) is causal, h ( n) = 0 for +ve n. Hence p a i r n i = i= p { h (0), n =0 0, 1 n p...(9) These equations are similar to ACNE for linear predictor. Once we get {a} we can get power spectum: use P S(f) = A(z) 2... (9a) z=e j2πfnt and obtain S ME (f) from the reciprocal of the absolute-value-squared Fourier tfm of {a 0,a 1...a p, 0,...0} 59 30

31 Linear Prediction and Burg s Method Similarity of equation (9) to ACNE for Linear predictor can be used to develop Burg s Method. Recursive solution: a 0 (0) = 1 where e 0 = r 0 a n+1 (j) =a n (j)+k n+1 a n (n +1 j)... (10)a E n+1 = E n (1 Kn+1) 2...(10)b K n+1 = 1 E n n a n (i)r n+1 i... (11) i=0 So far we have assumed we know r n to find {a} But real r n may not be known; how to find {k} and {a}? Burg s Method: Suppose we need a way of minimizing E n that 1. is consistent with recursion (10) 2. gives us {k} directly, 3. does not rely on r n There is such a way based on recurrence relations among individual forward and backward prediction errors for order-p linear predictor: f p (n) =f p 1 (n)+k p b p 1 (n)... (12a) 60 b p (n) =b p 1 (n 1) + K p f p 1 (n 1)... (12b) 31

32 On each recursion step n, use Eq (12) to find value of K n that minimises sum of mean-squared forward and backward prediction errors E over sequence of length N: where = E = N 1 1 f 2 2(N m) n(j)+b 2 n(j +1) j=n N 1 1 [f n 1 (j)+k n b n 1 (j)] 2 +[b n 1 (j)+k n f n 1 (j)] 2 2(N m) j=n N 1 E 1 = K n [f 2 K n (N m) n 1(j)+b 2 n 1(j)] + 2f n 1 (j)b n 1 (j) j=n Setting this derivative to 0 gives: K n = 2P/Q...(13) P = Q = N 1 j=n N 1 j=n f n 1 (j)b n 1 (j) f 2 n 1 (j)+b2 n 1 (j) (13) is expression for K n, using errors known from previous step and doesn t need r. Having found new K n, we can find new {a} from (10) and new prediction errors from (12) To start the process, we observe that for n = 0, predicted values are all 0, so errors are simply the values of the sequence. Burg recursion then proceeds as follows: 61 32

33 Burg s Recursion: 1. For n = 0: 2. For n=1 to p: f 0 (j) =y(j),b 0 (j) =y(j 1),j =1 to N, a 0 (0) = 1 (a) K n = 2P/Q from (13) (b) a n (n) =K n and a n (0) = 1 (c) From (10a), for i=1 to n-1 (d) From (12), for j = n to n-1 a n (i) =a n 1 (i)+k n a n 1 (n i) f n (j) =f n 1 (j)+k n b n 1 (j) b n (j) =b n 1 (j 1) + K n f n 1 (j 1) 3. Using {a}, obtain S(f) from (9a) Step 3 is carried out by finding the reciprocal of the squared magnitude of the DFT of {a 0,a 1,...a p, 0...0} 62 33

34 The Burg method dosen t need the calculation of {r} but if they are required, they can be got as a by-product of the Burg recursion. To see how this can be done, refer to Eq (11) rewritten as: K n = 1 n 1 a n 1 (i)r n i E n 1 i=0 n 1 r n = E n 1 K n a n 1 (i)r n i i=1 = K n n i=1 n 1 a n 1 (n i)r n i a n 1 (i)r n i i=1 n = a n (i)r n i i=1 So we can include computation of r into Burg recursion by adding to step 1: N r 0 = y 2 (j) j=1 andtostep2: r n = 63 n a n (i)r n i i=1 34

35 Performance of ME Spectral Estimate E.g for 3 sinusoids at 764, 955 and 1019 Hz 64 35

36 64a 36

37 64b 37