bug repair for X and Y wrong symbols REVD

Repairs PhL 7.22.13 I have wrongly referred to Y'(z) and Y"(z) as X'(z) and X"(z). I reviewed all the places this happened and I have now installed all the repairs! This was a bug in the originally released FT. The bug existed in Sections 33 and 34 as noted below. I think this is all OK. _________________________________________________________________________ 33. Spectral power density of a Simple Pulse Train In subsections (a) through (d) we deal only with simple pulse trains. Along the way, certain facts are developed which apply to general as well as simple pulse trains. In subsection (e) we gather together these general facts, and then show how quantities of interest can be related to the autocorrelation function. Comment: Some texts use the phrase "power spectral density" (PSD). Although this wins on Google by a ratio of 3 to 1, we still prefer the phrase "spectral power density", and the same for "spectral energy density". (a) Infinite Simple Pulse Train This is the first section in which we use the δ(0) notation of Appendix A which may make the reader feel a bit uncomfortable. In subsection (b) we shall repeat everything for a finite pulse train and then take the limit N→∞ to obtain the same results without using δ(0). In both sections we shall include the Z Transform in passing, but our main work is in the ω variable, not the z variable. From Section 14 (a) we know the spectrum of an infinite pulse train formed from pulses xpulse(t) separated by time T1 , x(t) = !Syntax Error, I xpulse(t - nT1) (14.1) X(ω) = Xpulse(ω) !Syntax Error, I 2π δ(ωT1 - 2πm) . (14.4) We can obtain the same expressions from box (25.4) which summarizes amplitude modulated pulse trains by setting all amplitudes to yn = 1, x(t) = !Syntax Error, I yn xpulse(t -tn) = !Syntax Error, I xpulse(t -tn) (33.1) X(ω) = (1/T1)Xpulse(ω) Y'ω) = Xpulse(ω) Y"(z) (33.2) Y"(z) = Y'ω)/T1 = !Syntax Error, Iyn e-iωnT = !Syntax Error, I e-iωnT . (33.3) Y'(ω) is the Digital Fourier Transform of yn = 1, and Y"(z) is the Z Transform, where z = eiωT . In the last line we then use (13.2) !Syntax Error, Ie±ink = !Syntax Error, I2πδ(k - 2πm) -∞ < k < ∞ (13.2) so that (33.3) becomes Y"(z) = Y'ω)/T1 = !Syntax Error, I e-iωnT = !Syntax Error, I2πδ(ωT1 - 2πm) (33.4) and then (33.2) says X(ω) = (1/T1)Xpulse(ω) Y'ω) = Xpulse(ω) !Syntax Error, I2πδ(ωT1 - 2πm) (33.5) in agreement with (14.4) quoted just above (33.1). To find the power spectrum of a signal x(t), our first task is to compute | X(ω) |2. From (33.2) we get |X(ω)|2 = |Xpulse(ω)|2 (1/T1)2 |Y'ω)|2 = |Xpulse(ω)|2 | Y"(z) |2 . z = eiωT (33.6) We therefore must deal with the following object, using (33.4), | Y"(z) |2 = (1/T1)2 |Y'ω)|2 = [ !Syntax Error, I2πδ(ωT1 - 2πm) ] 2 , (33.7) and we are now faced with the issue of squaring delta functions. Formally these objects don't exist in the realm of distribution theory, but (as discussed in Appendix A) we can deal with them in an ad hoc way which proves to be useful. Consider, [ !Syntax Error, I2πδ(ωT1 - 2πm) ] 2 = !Syntax Error, I2πδ(ωT1 - 2πm) !Syntax Error, I2πδ(ωT1 - 2πn) = !Syntax Error, I !Syntax Error, I2πδ(ωT1 - 2πm) 2πδ(ωT1 - 2πn) . Looking at the product of the two delta functions, there can be no contribution to the double sum unless m = n, so we continue = !Syntax Error, I 2πδ(ωT1 - 2πm) 2πδ(0) = [2πδ(0)] !Syntax Error, I 2πδ(ωT1 - 2πm) so that [ !Syntax Error, I2πδ(ωT1 - 2πm) ] 2 = [2πδ(0)] !Syntax Error, I 2πδ(ωT1 - 2πm) . (33.8) The object δ(0) is formally undefined, but in Appendix A we ascribe the meaning that 2πδ(0) = 2N+1 in the limit that N→ ∞ and we can always "undo the limit" when necessary. We shall firm up this idea in section (b) directly below. So we have shown then that | Y"(z) |2 = (1/T1)2 |Y'ω)|2 = [2πδ(0)] !Syntax Error, I 2πδ(ωT1 - 2πm) (33.9) or = (1/T1)2 = !Syntax Error, I 2πδ(ωT1 - 2πm) . (33.10) Then from (33.6) = |Xpulse(ω)|2 = |Xpulse(ω)|2 (1/T1)2 = |Xpulse(ω)|2 !Syntax Error, I 2πδ(ωT1 - 2πm) . (33.11) We shall now repeat the above set of steps for a finite pulse train. (b) Finite Simple Pulse Train Our finite pulse train always has pulses ranging from n = -N to N instead of from n = -∞ to ∞. We start off exactly as in the previous section but with limited sums, x(t) = !Syntax Error, I yn xpulse(t -tn) = !Syntax Error, I xpulse(t -tn) (33.12) X(ω) = (1/T1)Xpulse(ω) Y'ω) = Xpulse(ω) Y"(z) (33.13) Y"(z) = Y'ω)/T1 = !Syntax Error, Iyn e-iωnT = !Syntax Error, I e-iωnT . (33.14) In the last line we then use (13.3), !Syntax Error, I eink = 2π { } ≡ 2π δ5(k,N) -∞ < k < ∞ (13.3) where δ5 is a periodic delta function model discussed in Appendix A (b). Equation (33.14) becomes Y"(z) = Y'ω)/T1 = !Syntax Error, I e-iωnT = 2π δ5(ωT1,N) (33.15) and then equation (33.13) says X(ω) = Xpulse(ω) 2π δ5(ωT1,N) . (33.16) This δ5 is periodic with period 2π and has identical peaks separated by 2π. For finite N, these are peaks of finite width and height. Squaring, we find | X(ω) |2 = | Xpulse(ω) |2 [2π δ5(ωT1,N)]2 . (33.17) Recalling the definition of the δ6 delta function model from Appendix A (A.20), 2π δ6(k,N) ≡ (A.20) we obtain = | Xpulse(ω) |2 2π δ6(ωT1,N) (33.18) and this is the finite pulse train result. We can then take the limit N→∞ and make use of (A.21), limN→∞ δ6(ωT1,N) = !Syntax Error, Iδ(ωT1 - 2πm) (A.21) to find that limN→∞ [] = | Xpulse(ω) |2 !Syntax Error, I2π δ(ωT1 - 2πm) (33.19) and this replicates (33.11) with the promised connection 2πδ(0) = limN→∞ (2N+1). It is useful now to provide some side-by-side comparisons of results : Y"(z) = Y'ω)/T1 = !Syntax Error, I e-iωnT = !Syntax Error, I2π δ(ωT1 - 2πm) infinite (33.4) Y"(z) = Y'ω)/T1 = !Syntax Error, I e-iωnT = 2π δ5(ωT1,N) finite (33.15) X(ω) = Xpulse(ω) !Syntax Error, I2π δ(ωT1 - 2πm) infinite (33.5) X(ω) = Xpulse(ω) 2π δ5(ωT1,N) . finite (33.16) = |Xpulse(ω)|2 !Syntax Error, I 2π δ(ωT1 - 2πm) infinite (33.11) = | Xpulse(ω) |2 2π δ6(ωT1,N) finite (33.18) One can interpret as the value of |X(ω)|2 per pulse in an infinite pulse train. ________________________________________________________________ Things are OK again until we get to Section 34 (b) so I will repair it below: ________________________________________________________________ (b) Spectral power density for a General Pulse Train In Section 33 (d) we dealt with simple pulse trains. Here we consider the more general amplitude modulated pulse train. In all equations, one can replace !Syntax Error, I by !Syntax Error, Ito adapt the equation to a finite pulse train instead of an infinite one. We start then with x(t) = !Syntax Error, I yn xpulse(t -tn) (34.5) X(ω) = (1/T1)Xpulse(ω) Y'ω) = Xpulse(ω) Y"(z) (34.6) Y"(z) = Y'ω)/T1 = !Syntax Error, Iyn e-iωnT . (34.7) To find the frequency-domain power spectrum of a signal x(t), our first task is to compute | X(ω) |2. From (34.6) we get |X(ω)|2 = |Xpulse(ω)|2 (1/T1)2 |Y'ω)|2 = |Xpulse(ω)|2 | Y"(z) |2 z = eiωT (34.8) We therefore must deal with the following object, using (34.7), | Y"(z) |2 = (1/T1)2 |Y'ω)|2 = | !Syntax Error, Iyn e-iωnT | 2 = !Syntax Error, Iyn e-iωnT !Syntax Error, Iym* e+iωmT = !Syntax Error, I !Syntax Error, I ym* yn eiω(m-n)T (34.9) so that | X(ω) |2 = | Xpulse(ω) |2 !Syntax Error, I !Syntax Error, I ym* yn eiω(m-n)T . (34.10) From box (34.4) we then have E(ω) ≡ |X(ω)|2/2π = (1/2π) | Xpulse(ω) |2 !Syntax Error, I !Syntax Error, I ym* yn eiω(m-n)T (34.11) P(ω) ≡ = (1/2πT) | Xpulse(ω) |2!Syntax Error, I !Syntax Error, I ym* yn eiω(m-n)T (34.12) which we can write as (again using box (34.4) results) E(ω) = T1 Ppulse(ω) !Syntax Error, I !Syntax Error, I ym* yn eiω(m-n)T (34.13) P(ω) = T1 Ppulse(ω) (1/T) !Syntax Error, I !Syntax Error, I ym* yn eiω(m-n)T (34.14) P = !Syntax Error, Idω P(ω) . (33.31) (34.15) Not knowing details of the yn there is not much else we can do in these expressions. (c) Pulse Trains with Repeated Sequences Consider a pulse train composed of some general pulse shape xpulse(t) whose amplitudes are repeated sequences of A,B. We shall compute the spectrum X(ω) and spectral power density P(ω) by two different methods. The first method is more or less by brute force, and it reveals a potential pitfall in using the δ(0) notation and shows a clean way to avoid the pitfall. The second method, much simpler, is to use the Fourier Series results in box (15.12) applied to the repeating sequence. We then state X(ω) and P(ω) for a few special cases including various square waves. At the very end we generalize the results to any repeated sequence A,B,C... . Method 1: Brute Force Approach Our starting point is (34.8) with (34.7), where we assume N is large and later we will take N→∞ : X(ω) = (1/T1)Xpulse(ω) Y'ω) = Xpulse(ω) Y"(z) (34.6) |X(ω)|2 = |Xpulse(ω)|2 (1/T1)2 |Y'ω)|2 = |Xpulse(ω)|2 | Y"(z) |2 z = eiωT (34.8) Y"(z) = Y'ω)/T1 = !Syntax Error, Iyn e-iωnT. (34.7) The main problem is to compute Y"(z) and then square it. We have !Syntax Error, Iyn e-iωnT = A !Syntax Error, I e-iωnT + B !Syntax Error, I e-iωnT . Now process the sums as follows, where !Syntax Error, I e-iωnT = !Syntax Error, I e-iω(2m)T where we used n = 2m !Syntax Error, I e-iωnT = !Syntax Error, Ie-iω(2m+1)T where we used n = 2m + 1 . We assume N is very large, so we regard (N±1)/2 ≈ N/2 . We then find Y"(z) = Y'ω)/T1 = !Syntax Error, Iyn e-iωnT = [ A + B e-iωT]!Syntax Error, I e-iω(2m)T . We now use (13.3), !Syntax Error, I eink = 2π { } ≡ 2π δ5(k,N) , -∞ < k < ∞ (13.3) to write !Syntax Error, I e-iω(2m)T = 2π δ5(2ωT1,N/2) where δ5 and δ6 to come are explained in Appendix A (b). Therefore, Y"(z) = [ A + B e-iωT] 2πδ5(2ωT1,N/2) . (34.16) Using this Appendix A result, limN→∞ δ5(k,N) = !Syntax Error, Iδ(k-2πm) (A.19) we obtain the N→∞ limit for our spectrum Y"(z) = [ A + B e-iωT] 2π!Syntax Error, Iδ(2ωT1-2πm) = (1/2)[ A + B e-iωT] (1/T1) 2π !Syntax Error, I δ(ω - mω1/2) = (1/2) ω1 !Syntax Error, I[ A + B (-1)m] δ(ω-mω1/2) (34.17) and correspondingly X(ω) = Xpulse(ω) (1/2) ω1!Syntax Error, I [ A + B(-1)m ] δ(ω - mω1/2) . (34.18) There is a certain logic to the [ A + B e-iωT] factor. If we set A = K and B = 0 we get one result, and if we set A = 0 and B = K we get the same result multiplied by e-iωnT . The second pulse train is just the first pulse train shifted T1 units to the right, and this adds phase e-iωnT as in (12.1). If we were to square (34.18) and use our usual 2πδ(0) = 2N+1 association, we get a result that is off by a factor of 2. The reason is that our pre-limit sums are going from -N/2 to N/2, so we would get the right answer if we were to adjust and say 2πδ(0) = N+1. Rather than make an arm-waving argument to this effect, it is safer to continue along with our pre-limit expressions, having paused to take the limit for the spectrum X(ω) as in (34.18). So, backing off again from limit, we square (34.16) to get |Y"(z)|2 = |A + Be-iωT|2 [2π δ5(2ωT1,N/2)]2 . Then from (A.20) applied with N → N/2 δ6(k,N/2) ≡ (A.20) we get |Y"(z)|2 = |A + Be-iωT|2 [2π δ5(2ωT1,N/2)]2 or = |A + Be-iωT|2 { } = |A + Be-iωT|2 δ6(2ωT1,N/2) . Now for large N we ignore the difference between N and N + 1 and so on, so we divide both sides by 2 to get, = (1/2) |A + Be-iωT|2 δ6(2ωT1,N/2) Notice that a very important factor of 1/2 appears on the right in the last step. We now insert the squared pulse spectrum to get = |Xpulse(ω)|2 = |Xpulse(ω)|2 (1/2)|A + Be-iωT|2 δ6(2ωT1,N/2) . If we divide both sides by T1 the left side is where T is the length of the pulse train and this in turn equals P(ω), all as shown in box (34.4). So for large N we have shown that P(ω) = |Xpulse(ω)|2 (1/T1)(1/2) |A + Be-iωT|2 δ6(2ωT1,N/2) = Ppulse(ω)(1/2) |A + Be-iωT|2 2π δ6(2ωT1,N/2) . (34.19) Now at last we take the limit N→∞ and use limN→∞ δ6(k,N) = !Syntax Error, Iδ(k-2πm) (A.21) to get our desired infinite pulse train result P(ω) = Ppulse(ω)(1/2) |A + Be-iωT|2 !Syntax Error, I2π δ(2ωT1 - 2πm) = Ppulse(ω)(1/4) |A + Be-iωT|2 (1/T1)!Syntax Error, I2π δ(ω - mω1/2) = Ppulse(ω)(1/4) (1/T1)!Syntax Error, I |A + B(-1)m |2 2π δ(ω - mω1/2) = Ppulse(ω)(1/4) ω1!Syntax Error, I { |A|2 + |B|2 + 2Re(AB)(-1)m } δ(ω - mω1/2) . (34.20) Summarizing the key results: Fig 34.1 X(ω) = Xpulse(ω) (1/2) ω1!Syntax Error, I [ A + B(-1)m ] δ(ω - mω1/2) . (34.18) P(ω) = Ppulse(ω) (1/4) ω1!Syntax Error, I { |A|2 + |B|2 + 2Re(AB)(-1)m } δ(ω - mω1/2) (34.20) _______________________________________________________________________________- .