AB example version 2

Section 34 (c) An extended example PhL 3.26.05 This example is fairly general and we shall at the end consider its special cases, since these will be referred to later on. The example is this: The pulse train is composed of some general pulse xpulse(t) and the amplitudes are repeated sequences of A,B. Our starting point is (34.8) with (34.7), where we assume N is large and later we will take N→∞ : |X(ω)|2 = |Xpulse(ω)|2 (1/T1)2 |X'ω)|2 = |Xpulse(ω)|2 | X"(z) |2 z = eiωT (34.8) X"(z) = X'ω)/T1 = !Syntax Error, Iyn e-iωnT. (34.7) The main problem is to compute X"(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 X"(z) = X'ω)/T1 = !Syntax Error, Iyn e-iωnT = [ A + B e-iωnT]!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. Therefore X"(z) = [ A + B e-iωnT]2π δ5(2ωT1,N/2) (34.16) Squaring we get |X"(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 |X"(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) (34.17) Notice that a very important factor of 1/2 appears on the right in the last step. We now add back the squared pulse spectrum to get = |Xpulse(ω)|2 = |Xpulse(ω)|2 (1/2)|A + Be-iωT|2 δ6(2ωT1,N/2) (34.18) 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) (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) = |Xpulse(ω)|2 (1/T1)(1/2) |A + Be-iωT|2 !Syntax Error, Iδ(2ωT1 - 2πm) (34.20) We then make two ancillary calculations |A + Be-iωT|2 = |A|2 + |B|2 + 2|A||B| cos(ωT1) δ(2ωT1 - 2πm) = (2T1)-1 δ(ω - (1/2)[2π/T1]m) = (2T1)-1 δ(ω - mω1/2) ω1 = 2π/T1 Here then is our final result: P(ω) = |Xpulse(ω)|2 (1/4)(1/T12) { |A|2 + |B|2 + 2|A||B| cos(ωT1)} !Syntax Error, I δ(ω - mω1/2) (34.21) We shall now challenge this result for various special cases. Special Case 1. xpulse(t) is a box of width T1 and height 1. From (9.2) we then have Xpulse(ω) = T1 sinc(ωT1/2) so that P(ω) = (1/4) sinc2(ωT1/2) { |A|2 + |B|2 + 2|A||B| cos(ωT1)} !Syntax Error, I δ(ω - mω1/2) (34.22) Special Case 1a: Let A = B = 1. First, let's calculate this another way. If the pulse train is a DC signal x(t) = 1, it must be that P(ω) = kδ(ω) for some k. The total power is then P = !Syntax Error, Idω P(ω) = k. But for a load of R = 1Ω, we know that P will be 1 watt for A = 1 volt, so k = 1 and P(ω) = δ(ω). So how exactly is this going to happen? First { |A|2 + |B|2 + 2|A||B| cos(ωT1)} = 2[1+cos(ωT1)] = 4 cos2(ωT1/2) Evaluated at ω = mω1/2 we get ωT1/2 = mπ/2. So then P(ω) = (1/4) !Syntax Error, Isinc2(mπ/2) {4 cos2(mπ/2)} δ(ω - mω1/2) = δ(ω) For all integer m ≠0 we have a factor 2sin(mπ/2)cos(mπ/2) = sin(mπ) = 0 so there is no contribution. Then for m = 0, sinc = 1 and cos = 1 and the result is P(ω) = δ(ω), as expected. Special Case 1b: Let A = 1 and B = 0. Then P(ω) = (1/4) sinc2(ωT1/2) {1} !Syntax Error, I δ(ω - mω1/2) = (1/4) !Syntax Error, I sinc2(mπ/2) δ(ω - mω1/2) = (1/4)δ(ω) + (1/4) !Syntax Error, I (mπ/2)-2 sin2(mπ/2) δ(ω - mω1/2) = (1/4)δ(ω) + (1/4) !Syntax Error, I (mπ/2)-2 sin2(mπ/2) δ(ω - mω1/2) = (1/4)δ(ω) + (1/4) !Syntax Error, I (mπ/2)-2 δ(ω - mω1/2) = (1/4)δ(ω) + (1/4) (2/π)2 !Syntax Error, I δ(ω - mω1/2) (34.23) For verification, we recall (33.27) that P(ω) = !Syntax Error, I |cm|2 δ(ω - mω1) and we consider the waveform shown in Fig ** with coefficients as in (16.2a). For that waveform we get P(ω) = !Syntax Error, I (1τ/T1)2 sinc2(mπτ/T1) δ(ω - mω1) To get a 50% duty cycle, we set τ = T1/2 to get P(ω) = !Syntax Error, I (1/2)2 sinc2(mπ/2) δ(ω - mω1) But to get our current A = 1 B = 0 waveform, we have to scale T1/2 → T1 in Fig ** . This means we are taking T1→ 2T1 so that means ω1→ ω1/2 and then we get P(ω) = !Syntax Error, I (1/2)2 sinc2(mπ/2) δ(ω - mω1/2) which agrees an early line of our result quoted above. This can be written, by the way, as P(ω) = (1/4) δ(ω) + !Syntax Error, I (1/2)2 sinc2(mπ/2) δ(ω - mω1/2) The origin for t = 0 in the waveform plays no role in P(ω) since the phase shift dies off in |X(ω)|2. The DC component is consistent with an average DC level of 1/2. Special Case 1c: Let A = 1 and B = -1. Then P(ω) = (1/4) sinc2(ωT1/2) { |A|2 + |B|2 + 2|A||B| cos(ωT1)} !Syntax Error, I δ(ω - mω1/2) = (1/4) sinc2(ωT1/2) { 2- 2cos(ωT1)} !Syntax Error, I δ(ω - mω1/2) = sinc2(ωT1/2) sin2(ωT1/2) !Syntax Error, I δ(ω - mω1/2) = !Syntax Error, I sinc2(mπ/2) sin2(mπ/2) δ(ω - mω1/2) = !Syntax Error, I (mπ/2)-2 sin4(mπ/2) δ(ω - mω1/2) = !Syntax Error, I (mπ/2)-2 sin4(mπ/2) δ(ω - mω1/2) = !Syntax Error, I (mπ/2)-2 δ(ω - mω1/2) so the final result is P(ω) = (2/π)2 !Syntax Error, I δ(ω - mω1/2) (34.24) Verification: If we add a DC offset of +1 to our case 1c waveform, we will get P(ω) = δ(ω) + (2/π)2 !Syntax Error, I δ(ω - mω1/2) But this is the same waveform as in Special Case 1b with A = 2 and B = 0, so P(ω) should be 4x the result of Special Case 1b, which in fact it is. Exercises for the Reader: (a) If the repeating amplitude sequence is A,B,C then P(ω) = |Xpulse(ω)|2 (1/T1)(1/3) |A + Be-iωT + C e-i2ωT|2 !Syntax Error, Iδ(3ωT1 - 2πm) (34.25) (b) If the repeating sequence is A0,A1....AM-1 then P(ω) = |Xpulse(ω)|2 (1/T1)(1/M) | !Syntax Error, IAke-ikωT|2 !Syntax Error, Iδ(MωT1 - 2πm) (34.26)