new Appendix F Aug 8

Appendix F: The Spectrum and Power Density for Repeated-Sequence Pulse Trains Overview: For infinite pulse trains composed of repeats of some length-P subsequence: (a) computes X(ω) in (F.12) (b) computes P(ω) in (F.23) in terms of | YP"(z) |2. (c) computes < P(ω)> for an ensemble of pulse trains which respect the special condition <ym* yn> = α for m ≠ n <ym* yn> = β for m = n s < |m-n| (F.25) The result for <P(ω)> is stated in (F.33) in several different forms. (d) takes the P→∞ limit of the section (c) result for <P(ω)> (e) computes P(ω) for a single pulse train which respects the special condition <ymyn>1 = α for m ≠ n + NP N = any integer <ymyn>1 = β for m = n + NP (F.43) where <ymyn>1 is a horizontal average across the single sequence (autocorrelation). The result for P(ω) is stated in (F.52). It is noted that the results for < P(ω)> of (c) and P(ω) of (e) are exactly the same in terms of their respectively defined α and β constants. It is then shown that these two sets of constants are the same. (f) restates the overall result as the Fact (F.54). A graphical representation is drawn for the spectrum in general, and then for a box pulse. It is shown that the MLS sequence is a candidate for application of (F.54). (g) treats the P = 2 repeated subsequence A,B using the general formulas of (a) and (b) We start with this collection of equations: Y"(z) = !Syntax Error, Iyn e-iωnT Z Transform of yn (24.2) (F.1) | Y"(z) |2 = !Syntax Error, I !Syntax Error, I ym* yn eiω(m-n)T (F.2) X(ω) = Xpulse(ω) Y"(z) (25.3) (F.3) P(ω) = Ppulse(ω) | Y"(z) |2 (34.14) (F.4) z = eiωT (24.1) ω1 ≡ 2π/T1 In (F.4) T is the duration of the infinite pulse train, as in (33.22). The pulse train amplitudes are the yn. Since sequence ym is composed of subsequences of length P that repeat, we have this periodicity property of the yn ym+IP = ym for any integer I (F.5) (a) Calculation of X(ω) for a Pulse Train with a Repeated Sequence Consider first the sum in (F.1). Let n = IP + n' and write this sum as !Syntax Error, Iyn e-iωnT = !Syntax Error, I !Syntax Error, I yIP+n' e-iω(n'+IP)T = !Syntax Error, I e-iωIPT !Syntax Error, I yn' e-iωn'T = (!Syntax Error, I e-iωIPT) (!Syntax Error, I yn e-iωnT) . (F.6) We see that the sum factors into the product of two sums. The first sum we evaluate using !Syntax Error, I e-ink = !Syntax Error, I2πδ(k - 2πm) . -∞ < k < ∞ (A.31) Setting k = ωPT1 we find (!Syntax Error, I e-iωIPT) = !Syntax Error, I2πδ(ωPT1 - 2πm) . (F.7) The second sum we give the name YP"(z) which is the Z transform of the subsequence { y0, y1.....yP-1}. YP"(z) ≡ !Syntax Error, I yn e-iωnT . (F.8) Thus we have shown that Y"(z) = !Syntax Error, Iyn e-iωnT = YP"(z) !Syntax Error, I2πδ(ωPT1 - 2πm) , (F.9) and then from (F.3). X(ω) = Xpulse(ω) Y"(z) = Xpulse(ω) YP"(z) !Syntax Error, I2πδ(ωPT1 - 2πm) . (F.10) It is convenient to write 2πδ(ωPT1 - 2πm) = 2π(PT1)-1δ(ω - 2πm/PT1) = (1/P) ω1 δ(ω - mω1/P) (F.11) and then X(ω) = Xpulse(ω) ω1 (1/P) YP"(z) !Syntax Error, I δ(ω - mω1/P) (F.12) where YP"(z) ≡ !Syntax Error, I yn e-iωnT . (F.8) X(ω) is the Fourier Transform Spectrum of an infinite pulse train composed of a repeating P-length subsequence. Since the pulse train is periodic, the spectrum is entirely discrete with lines at ωm = (m/P)ω1. (F.13) (b) Calculation of P(ω) for a Pulse Train with a Repeated Sequence In Section 35 [ see below (35.26)' ] it was noted that P(ω) can be calculated either by the Autocorrelation approach or the Double Sum approach. Here we shall take the latter approach based on (F.2) above. We can reorganize the double sum in (F.2) into a quadruple sum by defining : n = IP + n' m = JP + m' . Then the double sum above becomes, !Syntax Error, I !Syntax Error, I ym* yn eiω(m-n)T = !Syntax Error, I !Syntax Error, I !Syntax Error, I !Syntax Error, I (yJP + m')* (yIP +n') eiωP(I-J)T eiω(m'-n')T = !Syntax Error, I !Syntax Error, I !Syntax Error, I !Syntax Error, I (ym')* (yn') eiωP(I-J)T eiω(m'-n')T , (F.14) where in the last line we have used the periodicity (F.5) of the yn. The following illustration shows how, in this reorganization, we first sum over a square grid patch with n' and m', and then we sum over an array of those patches with I and J. Fig F.1 In order to regulate things, we shall assume that the I and J sums range from -N to N rather than from -∞ to ∞. This means we are assuming that the sequence is (2N+1) repeated periods in length and not infinite. Then of course we can say T = (2N+1)PT1 . (F.15) As usual, we keep N finite as long as possible, and then take N→ ∞ in the end. We now rewrite (F.2) by removing the primes from summation indices and reordering the factors | Y"(z) |2 = !Syntax Error, I !Syntax Error, I ym* yn eiω(m-n)T → = [ !Syntax Error, I !Syntax Error, I eiωP(I-J)T ] [ !Syntax Error, I !Syntax Error, I ym* yn eiω(m-n)T ] = [ !Syntax Error, I !Syntax Error, I eiωP(I-J)T ] | YP"(z) |2 (F.16) As happened in (F.6) with a single sum, our double sum factors into a product of two double sums. The second double sum we recognize from (F.8) as | YP"(z) |2. The first double sum is [ !Syntax Error, I !Syntax Error, I eiωP(I-J)T ] = | !Syntax Error, I eiωPIT | 2 (F.17) Then apply (A.30) !Syntax Error, I eink = 2π δ5(k,N) = 2π { } (A.30) with k = ωPT1 to get [ !Syntax Error, I !Syntax Error, I eiωP(I-J)T ] = | 2π δ5(ωPT1,N) | 2 = ( 2π δ5(ωPT1,N) ) 2 . (F.18) Then from (A.20), δ6(k,N) ≡ = , (A.20) we can write the first factor of (F.16) as [ !Syntax Error, I !Syntax Error, I eiωP(I-J)T ] = (2N+1) 2π δ6(ωPT1,N) = (1/P) 2π δ6(ωPT1,N) (F.19) where T is from (F.15). At this point we have, looking at (F.16) and (F.19), | Y"(z) |2 = [ !Syntax Error, I !Syntax Error, I eiωP(I-J)T ] [ !Syntax Error, I !Syntax Error, I ym* yn eiω(m-n)T ] = [ (1/P) 2π δ6(ωPT1,N) ] | YP"(z) |2 . (F.20) Now finally we take N→∞ using (A.21) limN→∞ δ6(k,N) = !Syntax Error, Iδ(k-2πm) (A.21) or limN→∞ δ6(ωPT1,N) = !Syntax Error, Iδ(ωPT1-2πm) (F.21) with this result | Y"(z) |2 = [ (1/P) 2π!Syntax Error, Iδ(ωPT1-2πm) ] | YP"(z) |2 or | Y"(z) | = (1/P) !Syntax Error, I 2π δ(ωPT1-2πm) | YP"(z) |2 Using (F.11) we can rewrite this as | Y"(z) | = ω1 (1/P)2 !Syntax Error, I δ(ω - mω1/P) | YP"(z) |2 . (F.22) Installing this into (F.4) then gives P(ω) = Ppulse(ω) ω1 (1/P)2 !Syntax Error, I δ(ω - mω1/P) | YP"(z) |2 (F.23) where YP"(z) ≡ !Syntax Error, I yn e-iωnT . (F.8) P(ω) is the Spectral Power Density of an infinite pulse train composed of a repeating P-length subsequence. The power density is discrete with lines at ωm = (m/P)ω1, the same lines observed in the spectrum X(ω) of (F.12). (c) Calculation for an Ensemble of such Pulse Trains subject to Certain Conditions We now imagine an ensemble of P-length subsequences si. We then create a corresponding ensemble of infinite length sequences Si according to Si = {.......si, si, si, si, ...}. This just an arbitrary ensemble, not a random ensemble or any other special kind ensemble. Then we apply the ensemble average <..> to (F.23) to get (assuming now that the yn are real ), <P(ω)> = Ppulse(ω) ω1 !Syntax Error, I δ(ω - ω1m/P) !Syntax Error, I !Syntax Error, I <ymyn> eiω(m-n)T (F.24) At this point, suppose it happens that <ymyn> = α for m ≠ n <ymyn> = β for m = n (F.25) where α and β are independent of the indices shown. Notice in the (F.24) sum that max(n-m) = P-1, so we don't have to worry about these indices differing by an integral multiple of P. We have now restricted our interest to the sequence {y0,y1, ...yP-1}. Comment: In Section 35 we showed in (35.11) that if a finite pulse train source has "stationarity", then <ymyn> = f(n-m) for m ≠ n <ymyn> = f(0) for m = n (35.11) This is not sufficient to meet the condition (F.25). If in addition we assume that Yn and Ym are independent, we find from (35.15) that f(n-m) = . // stationarity and independence assumed (35.15) This does meet condition (F.25), but this assumption is more than we want to assume, so we just leave condition (F.25) as stated. Notice that if (F.25) is met, then (35.11) is valid so we are in the stationary realm. The general picture might be illustrated by this Venn diagram, So assuming (F.25) we can write the double sum in (F.24) as, using z = eiωT, !Syntax Error, I !Syntax Error, I <ymyn> zm-n = α !Syntax Error, I !Syntax Error, I zm-n + β !Syntax Error, I 1 = α !Syntax Error, I !Syntax Error, I zm-n + β P (F.26) To evaluate the double sum, we write it as !Syntax Error, I !Syntax Error, I zm-n = !Syntax Error, I[ !Syntax Error, I zm-n – !Syntax Error, I zm-n ] = !Syntax Error, I[ !Syntax Error, I zm-n – 1 ] = !Syntax Error, I !Syntax Error, I zm-n - P = | !Syntax Error, I zm |2 – P = | |2 – P . (F.27) With z = eiωT we can have Maple evaluate | |2 using k = ωT1 (F.28) so we find that | |2 = . (F.29) Now recall from (A.20) that δ6(k,N) ≡ . (A.20) Setting N = (P-1)/2 we find P = 2N+1 and N+1/2 = P/2 so, 2πδ6(k, ) = . (A.20) and thus | |2 = = P 2πδ6(k, ) . (F.30) Therefore from (F.27), !Syntax Error, I !Syntax Error, I zm-n = 2π P δ6(k, ) - P = P [2πδ6(k, ) - 1 ] (F.31) and from (F.26) !Syntax Error, I !Syntax Error, I <ymyn> zm-n = α P [2πδ6(k, ) - 1 ] + β P = P [ (β-α) + 2π α δ6(k, ) ] (F.32) and then our resulting power density (F.24) becomes <P(ω)> = Ppulse(ω) ω1 !Syntax Error, I δ(ω - ω1m/P) [ (β-α) + α 2πδ6(ωT1, ) ] (F.33a) where 2π δ6(ωT1, ) =   . Evaluating at the delta function hit values ω = ω1m/P we find 2π δ6 = = for N = any integer . To get alternate forms for (F.33a), we first express each term as a separate sum, <P(ω)> = Ppulse(ω) ω1 [(β-α)!Syntax Error, Iδ(ω - ω1m/P) + α!Syntax Error, Iδ(ω - ω1m/P) 2πδ6(ωT1, ) ] . Since only those m which are multiples of P contribute to the second sum in (F.33a), we may rewrite that second sum as follows, <P(ω)> = Ppulse(ω) ω1 [(β-α)!Syntax Error, Iδ(ω - ω1m/P) + α !Syntax Error, Iδ(ω - ω1N) P ] // m = NP = Ppulse(ω) ω1 [(β-α) !Syntax Error, Iδ(ω - ω1m/P) + α !Syntax Error, Iδ(ω - ω1m) ] (F.33b) = Ppulse(ω) ω1 !Syntax Error, I[ (β-α) δ(ω - ω1m/P) + α δ(ω - ω1m) ] (F.33c) = Ppulse(ω) ω1 [ (β-α) !Syntax Error, Iδ(ω - ω1m/P) + α !Syntax Error, Iδ(ω - ω1m) + { (β-α) /P + α } δ(ω) ] . (F.33d) If xpulse(t) is real, then by (7.5) Ppulse(ω) is an even function of ω, and we can then reflect the negative part of the sum to the positive side to get, = Ppulse(ω) ω1 [ (β-α) !Syntax Error, Iδ(ω - ω1m/P) + 2α !Syntax Error, Iδ(ω - ω1m) + { (β-α) /P + α } δ(ω) ] . (F.33e) In all forms of (F.33) we have: α = <ymyn> for m≠n β = <yn2> . These slightly different forms of <P(ω)> are useful for different purposes. All results are valid for any finite integer P. Since each sequence in the ensemble is periodic with the same period P, the ensemble average spectrum is entirely discrete. The m≠0 sums include positive and negative integers. (d) Limit as P → ∞ of the Ensemble Result We would now like to take the limit of the above as P→∞. Write (F.33b) as <P(ω)> = Ppulse(ω) [(β-α) { !Syntax Error, Iδ(ω - ω1m/P) } + ω1 α !Syntax Error, Iδ(ω - ω1m) ] . (F.34) Then define fP(ω) ≡ !Syntax Error, Iδ(ω - ω1m/P) . (F.35) As P→∞, the spacing of the δ lines becomes closer and closer, while the amplitude of each δ line becomes less and less. Perhaps we can argue that in the limit this becomes some continuous function. In line with the distribution theory approach to symbolic functions noted in Appendix A, suppose we integrate this function from some a to a+ε for small ε, where a is an arbitrary real number, !Syntax Error, I fP(ω) dω = !Syntax Error, I!Syntax Error, I δ(ω - ω1m/P) = !Syntax Error, I Θ(a ≤ω1m/P ≤ a+ε) (F.36) where we use the notation of Appendix A (e) for the Θ function which takes value 1 if the inequality argument is valid, meaning there is a delta hit. If P is a large integer, how many non-zero terms does this Σm have? The inequality argument reads Pa/ω1 ≤ m ≤ Pa/ω1 + Pε/ω1 . Since P is large, we round each term in this equation to the nearest integer, making little error. We select a very small ε first, and then we make sure P is large enough so Pε/ω1 is still a reasonably large integer when rounded. Then we have !Syntax Error, I fP(ω) dω = !Syntax Error, I Θ(a ≤ω1m/P ≤ a+ε) = !Syntax Error, I 1 = ( Pε/ω1) = ε . (F.37) Since we then have (for very large P) that !Syntax Error, I fP(ω) dω = ε for any real a and for ε as small as we like, and since the integral over range ε is proportional to ε, the function fP(ω) is equivalent to the constant function 1. Thus we have shown that, limP→∞ fP(ω) = limP→∞ [ !Syntax Error, Iδ(ω - ω1m/P)] = 1. (F.38) We then obtain this P→∞ limit of (F.34), <P(ω)> = Ppulse(ω) [(β-α) + ω1 α !Syntax Error, Iδ(ω - ω1m) ] (F.39) = Ppulse(ω) [(β-α) + (1/T1) α !Syntax Error, I2π δ(ω - ω1m) ] // ω1 = 2π/T1 = Ppulse(ω) [(β-α) + α !Syntax Error, I2π δ(ωT1 - 2πm) ] . This limit agrees with our result (35.11) for a random ensemble of infinite sequences for which <aman> does not depend on the values of m and n, <P(ω)> = Ppulse(ω) { (β-α) + α!Syntax Error, I2π δ(ωT1- 2πm) } (35.11) In the limit P→∞ , the discrete lines δ(ω - ω1m/P) shown in (F.34) have coalesced into a continuous function. See Fig F.2 below for a graphical view. (e) Calculation of P(ω) for a single P-periodic Pulse Train subject to Certain Conditions We really know ahead of time how this calculation will come out, but we do it nevertheless to convince the reader that the result is valid. After getting the result, we shall comment on why it is the way it is. Recall this expression for P(ω), which incorporates the discrete Wiener-Khintchine relation, P(ω) = Ppulse(ω) | Y"(z) |2 = Ppulse(ω) R"(z) (34.14a) (F.40) where R"(z) is the Z transform of the autocorrelation sequence rs obtained from the yn. Our first task is to find rs, which we defined this way, rs ≡ limN→∞ [!Syntax Error, I yn yn+s ] = <yn yn+s>1 . (32.16) (F.41a) Since yn is periodic as shown in (F.5), rs may be written in this alternate form: rs = !Syntax Error, I yn yn+s = <yn yn+s>1 (F.41b) Proof: For large N, we can replace the sum endpoints by -N = -MP and +N = MP ≈ (M+1)P where M is also large. The sum then has (2M+1)P terms, so rs ≡ limM→∞ [!Syntax Error, I yn yn+s ] . We now let n = IP + n' where n' takes values 0,1..P-1 and I = Int(n/P). Then the single sum above can be written, !Syntax Error, I yn yn+s = !Syntax Error, I !Syntax Error, I yIP+n' yIP+n'+s where this drawing shows how the sum is now a double sum where I denotes segments containing P points, and n' counts the points in each segment, We then process this sum using the periodicity (F.5) to get = !Syntax Error, I !Syntax Error, I yn' yn'+s = (!Syntax Error, I1 ) (!Syntax Error, I yn yn+s ) = (2M+1) (!Syntax Error, I yn yn+s ) . Inserting the expression into rs, we get this alternate form for rs rs = limM→∞ [{ (2M+1) (!Syntax Error, I yn yn+s ) } ] = !Syntax Error, I yn yn+s QED From now on we assume that the ym are real. Back in (F.25) we made the assumption that <ymyn> = α for m ≠ n s < | m-n | <ymyn> = β for m = n (F.25) where <...> were ensemble averages. Here we are going to make a completely different assumption, and this assumption applies to a single sequence: <ymyn>1 = α for m ≠ n s < | m-n | <ymyn>1 = β for m = n (F.42) Comment: If we imagine drawing each sequence of an ensemble as a row in a set of rows ... * * yn * yn+2 ... sequence #1 ... * * y'n * y'n+2 ... sequence #2 ... * * y"n * y"n+2 ... sequence #3 then any ensemble average like <ynyn+2> is a vertical average through the ensemble, whereas an average like <ynyn+2>1 is a horizontal average across one particular sequence row and <ynyn+2>1 never depends on n as (F.41) shows. These two averages are unrelated and (F.42) being true does not imply that (F.25) is true. As an example, one might take 10 infinite sequences each of which satisfies (F.42) and put them down as a set of rows, and then each row is shifted horizontally some amount to cause a 0 to be in column n. For the resulting 10 row ensemble, one would find that <yn2> = 0 for column n, whereas <yn2>1 = β ≠ 0 . This contrived situation, however, would be a violation of "stationarity" as discussed in (35.10), and in fact for a very large ensemble of our infinite pulse trains we will argue below that in fact <ynyn+2> = <ynyn+2>1, but we put that argument on a back burner for the moment and maintain the distinction between <..> and <..>1. Since we require rs for arbitrary s in order to compute R"(z), we extend (F.42) in this manner <ymyn>1 = α for m ≠ n + NP N = any integer <ymyn>1 = β for m = n + NP (F.43) which is to say, for s being any integer whatsoever, <yn yn+s>1 = α s ≠ NP <yn yn+s>1 = β s = NP . (F.44) The reason of course is that {ym} is periodic with period P, ym+NP = ym for any integer N (F.5) so we must have for example <yn yn+P>1 = <yn yn>1 = β. Thus we have arrived at our characterization of the autocorrelation sequence rs for our specific infinite periodic sequence with the assumption (F.44), rs = <yn yn+s>1 = N = any integer (F.45) Our next step is to compute its Z transform R"(z), R"(z) = !Syntax Error, Irn z-n = β !Syntax Error, Iz-n + α !Syntax Error, Iz-n = β !Syntax Error, Iz-n + α ( !Syntax Error, I z-n – !Syntax Error, Iz-n ) = (β-α) !Syntax Error, Iz-n + α !Syntax Error, I z-n . (F.46) The first sum is over n = 0, ±P, ±2P and so on. We can replace summation index n by index N, !Syntax Error, Iz-n = !Syntax Error, I z-NP = !Syntax Error, I z-nP . (F.47) From (24.1) we know that z lies on the unit circle in the z-plane and is related to ω by z = eiωT (24.1) where T1 is the duration of a pulse of the pulse train. We then have !Syntax Error, Iz-n = !Syntax Error, I (eiωT)-nP = !Syntax Error, I e-iωTnP . (F.48) According to (A.31), !Syntax Error, I eink = !Syntax Error, I2πδ(k - 2πm) -∞ < k < ∞ (A.31) so setting k = -ωT1P we get !Syntax Error, Iz-n = !Syntax Error, I2πδ(-ωT1P - 2πm) = !Syntax Error, I2πδ(ωT1P + 2πm) = !Syntax Error, I2πδ(ωT1P - 2πm) (F.49) where we use δ(x) = δ(-x) and in the last step take m→ -m. Meanwhile, our other sum of interest in (F.46) is this one, !Syntax Error, I z-n = !Syntax Error, I (eiωT)-n = !Syntax Error, I e-iωTn which is just the previous sum without the P. Thus, !Syntax Error, I z-n = !Syntax Error, I2πδ(ωT1 - 2πm) . (F.50) Inserting (F.48) and (F.49) into (F.46) gives R"(z) = (β-α) !Syntax Error, Iz-n + α !Syntax Error, I z-n = (β-α) !Syntax Error, I2πδ(ωT1P - 2πm) + α !Syntax Error, I2πδ(ωT1 - 2πm) = (β-α) (T1P)-1 !Syntax Error, I2πδ(ω - 2πm/[T1P]) + α(T1)-1 !Syntax Error, I2πδ(ω - 2πm/T1) = (2π/T1) { (β-α) (1/P) !Syntax Error, Iδ(ω - 2πm/[T1P]) + α !Syntax Error, Iδ(ω - 2πm/T1) } = ω1 { (β-α) (1/P) !Syntax Error, Iδ(ω - mω1/P) + α !Syntax Error, Iδ(ω - mω1) } (F.51) and this concludes our calculation of the Z Transform R"(z) of the autocorrelation sequence rs. It only remains to install this into the Z Transform Wiener-Khintchine relation (34.14a) which says P(ω) = Ppulse(ω) R"(z) (34.14a) so then P(ω) = Ppulse(ω) ω1 { (β-α) !Syntax Error, Iδ(ω - mω1/P) + α !Syntax Error, Iδ(ω - mω1) } = Ppulse(ω) ω1 !Syntax Error, I[ (β-α) δ(ω - ω1m/P) + α δ(ω - ω1m) ] (F.52c) We may compare this to the ensemble result of the section (c) above, <P(ω)> = Ppulse(ω) ω1 !Syntax Error, I[ (β-α) δ(ω - ω1m/P) + α δ(ω - ω1m) ] . (F.33c) The expressions are exactly the same ! However, the meaning of the symbols α and β is not the same according to the definitions given above in (F.25) and (F.42). Why are the expressions the same? This goes back to the general discussion of Section 35 (h) where it was shown that, for infinitely long pulse trains, <ymyn>1 = <ymyn> (35.24) and <P(ω)> = P(ω) (35.27) and all pulse trains in the ensemble are statistical pulse trains having the same statistics and having the same P(ω). Thus, our result (F.52c) above had to come out the same as (F.33c). Moreover, the quantities α and β are in fact the same values in the two cases. Since the expressions have the exact same form, we can rewrite (F.52c) in the same alternate ways that (F.33c) was written: P(ω) = Ppulse(ω) ω1 !Syntax Error, Iδ(ω - ω1m/P) [(β-α) + α 2π δ6(ωT1, ) ] (F.52a) = Ppulse(ω) ω1 [(β-α)!Syntax Error, Iδ(ω - ω1m/P) + α !Syntax Error, Iδ(ω - ω1N) P ] // m = NP = Ppulse(ω) ω1 [(β-α) !Syntax Error, Iδ(ω - ω1m/P) + α !Syntax Error, Iδ(ω - ω1m) ] (F.52b) = Ppulse(ω) ω1 !Syntax Error, I[ (β-α) δ(ω - ω1m/P) + α δ(ω - ω1m) ] (F.52c) = Ppulse(ω) ω1 [ (β-α) !Syntax Error, Iδ(ω - ω1m/P) + α !Syntax Error, Iδ(ω - ω1m) + { (β-α) /P + α } δ(ω) ] (F.52d) (f) Summary and an Example: The MLS Sequence Fact : For an infinite statistical sequence made of repeated subsequences of length P : (F.54) if the following is found to be true, <ymyn>1 = α for m ≠ n + NP N = any integer <ymyn>1 = β for m = n + NP (F.43) then the spectral power density is given by : (this is one of several forms shown in (F.52) P(ω) = Ppulse(ω) ω1 !Syntax Error, I[ (β-α) δ(ω - ω1m/P) + α δ(ω - ω1m) ] (F.52c) The parameters α and β can be interpreted as elements of the autocorrelation sequence rs = <yn yn+s>1 = N = any integer (F.45) The form shown in (F.52d) is P(ω) = Ppulse(ω) ω1 [ (β-α) !Syntax Error, Iδ(ω - ω1m/P) + α !Syntax Error, Iδ(ω - ω1m) + { (β-α) /P + α } δ(ω) ] which has the following graphical representation (drawn for P = 4), Fig F.2 Each vertical arrow represents a spectral δ line. The height of the arrow is the value of the red envelope curve times the quantity shown. The red curve Ppulse(ω) ω1 will in general have an infinite extent, an example being Ppulse(ω) ω1 = sinc2(ωT1/2) = sinc2(πω/ω1) for a box shaped pulse as in (9.2). As P → ∞, we showed in section (d) the spectrum (F.52d) becomes, Ppulse(ω) ω1 [ (β-α) + α !Syntax Error, Iδ(ω - ω1m) + α δ(ω) ] Regarding the three images of Fig F.2, we see that the left set of dense arrows coalesces into the continuous function (β-α) Ppulse(ω) ω1 the middle set of arrows stays exactly the same the amplitude of the DC line becomes α Once a particular Ppulse(ω) is specified, some of the spectral lines may be quenched. For the box pulse Ppulse(ω) ω1 = sinc2(ωT1/2) = sinc2(πω/ω1) (36.1) lines are quenched when ω = Nω1 for N = ±1,±2 .... In this case, all the lines of the central image go away and the corresponding lines in the left image also vanish, this being every Pth line in that image: Fig F.3 Example: The MLS Sequence A Maximum Length Sequence (MLS) (Lucht, Polynomial Multipliers... ) which is created by a shift register generator has the property (F.44), specifically, <yn yn+s>1 = α = (1/4)(1 + 1/P) s ≠ NP <yn yn+s>1 = β = (1/2)(1 + 1/P) s = NP (F.44) where P must be one of the special values P = 2k-1 for k = 1,2,3.... Therefore, by Fact (F.54) the spectrum P(ω) of an MLS sequence is given by P(ω) = Ppulse(ω) ω1 !Syntax Error, I[ (β-α) δ(ω - ω1m/P) + α δ(ω - ω1m) ] . (F.52) = [Ppulse(ω) ω1] (1/4)(1 + 1/P) !Syntax Error, I[ δ(ω - ω1m/P) + δ(ω - ω1m) ] . (F.55) (g) Results for an A,B repeated sequence This subject is treated in Section 34 (c) using a "brute force" approach and a "Fourier Series" approach. Here we duplicate the main results of that section by applying our general formulas to a sequence where the repeated subsequence is just P = 2, {y0,y1} = {A,B}, and we allow A and B to be complex. The fact that the results here agree with Section 34 lends some confidence to all three methods of computation. Our general expressions for X(ω) and P(ω) are, X(ω) = Xpulse(ω) ω1 (1/P) YP"(z) !Syntax Error, I δ(ω - mω1/P) (F.12) P(ω) = Ppulse(ω) ω1 (1/P)2 !Syntax Error, I δ(ω - mω1/P) | YP"(z) |2 (F.23) where YP"(z) ≡ !Syntax Error, I yn e-iωnT . (F.8) For a repeated A,B sequence P = 2 we have YP"(z) = A + B e-iωT | YP"(z) |2 = | A + B e-iωT |2 = |A|2 + |B|2 + 2 Re{A*B e-iωT} Then, X(ω)AB = Xpulse(ω) ω1 (1/2) [A + B e-iωT]!Syntax Error, I δ(ω - mω1/2) P(ω)AB = Ppulse(ω) ω1 (1/4) [ |A|2 + |B|2 + 2 Re{A*B e-iωT} ]!Syntax Error, I δ(ω - mω1/2) . We can slide the square-bracketed factors inside the sum and then use ωT1 = (mω1/2)T1 = π m(ω1/2π)T1 = πm => e-iωT = (-1)m and the results simplify to X(ω)AB = Xpulse(ω) ω1 (1/2) !Syntax Error, I [A + B (-1)m] δ(ω - mω1/2) P(ω)AB = Ppulse(ω) ω1 (1/4) !Syntax Error, I [ |A|2 + |B|2 + 2 Re(A*B) (-1)m ] δ(ω - mω1/2) . These results agree with (34.18) and (34.20), see just below Fig 34.1 . For a box pulse of height 1 and width T1 we know that Xpulse(ω) = T1 sinc(ωT1/2) = (1/ω1) 2π sinc(ωT1/2) (9.2) Ppulse(ω) = = (1/ω1) sinc2(ωT1/2) . (33.24) Then for a square wave with alternating A,B amplitudes we get X(ω)AB = 2π sinc(ωT1/2) (1/2) !Syntax Error, I [A + B (-1)m] δ(ω - mω1/2) P(ω)AB = sinc2(ωT1/2) (1/4) !Syntax Error, I [ |A|2 + |B|2 + 2 Re(A*B) (-1)m ] δ(ω - mω1/2) . When the sinc functions are moved inside the sum, ωT1/2 = πm/2, so X(ω)AB = 2π (1/2) !Syntax Error, I [A + B (-1)m] sinc(πm/2) δ(ω - mω1/2) P(ω)AB = (1/4) !Syntax Error, I [ |A|2 + |B|2 + 2 Re(A*B) (-1)m ] sinc2(πm/2)δ(ω - mω1/2) But sinc(πm/2) = (2/πm) sin(πm/2) = 1 for m = 0 = 0 for m even = (2/πm) (-1)(m-1)/2 for m odd so we get X(ω)AB = 2π (1/2) !Syntax Error, I [A – B (-1)m] (2/πm) (-1)(m-1)/2 δ(ω - mω1/2) + 2π (1/2) [A+B] δ(ω) P(ω)AB = (1/4) !Syntax Error, I [ |A|2 + |B|2 + 2 Re(A*B) (-1)m ] (2/πm)2 (-1)(m-1) δ(ω - mω1/2) + ω1 (1/4) [ |A|2 + |B|2 + 2 Re(A*B) ] δ(ω) or X(ω)AB = 2 !Syntax Error, I [A – B] (1/m) (-1)(m-1)/2 δ(ω - mω1/2) + π [A+B] δ(ω) P(ω)AB = (1/π2) !Syntax Error, I [ |A|2 + |B|2 – 2 Re(A*B) ] (1/m)2 δ(ω - mω1/2) + ω1 (1/4) [ |A|2 + |B|2 + 2 Re(A*B) ] δ(ω) For a standard-issue square wave with B = - A we have [A – B] = 2A [ |A|2 + |B|2 – 2 Re(A*B) ] = 4 |A|2 [A + B] = 0 [ |A|2 + |B|2 + 2 Re(A*B) ] = 0 and therefore X(ω)A,-A = 4A !Syntax Error, I (1/m) (-1)(m-1)/2 δ(ω - mω1/2) P(ω)A,-A = 4A2 (1/π2) !Syntax Error, I (1/m)2 δ(ω - mω1/2) in agreement with (34.23).