fourier series method Sec 34

This is the Title PhL 3.26.05 Note that page numbering is turned on in this template. (c) Pulse Trains with Repeated Amplitude Sequences Suppose a pulse train has period 2T1 during which time the pulse Consider a pulse train composed of some general pulse xpulse(t) whose amplitudes are repeated sequences of A,B. We let each pulse have width T1 so the repeating sequence is 2T1 long. Then we think of these two pulses as forming a single longer pulse xPulse of length 2T1. Then xPulse(t) = A xpulse(t) 0 < t < T1 xPulse(t) = B xpulse(t-T1) T1 < t < 2T1 We then apply our Fourier Series results of box (15.12) using XPulse for the pulse and 2T1 for the period. x(t) = !Syntax Error, IxPULSE(t - n2T1) Cm = (1/2T1) !Syntax Error, I dt xPULSE(t) e-imωt/2 = (1/2T1) A !Syntax Error, I dt xpulse(t) e-imωt/2 + (1/2T1) B !Syntax Error, I dt xpulse(t-T1) e-imωt/2 = (1/2T1) A !Syntax Error, I dt xpulse(t) e-imωt/2 + (1/2T1) B !Syntax Error, I dt' xpulse(t') e-imω(t'+T1)/2 = (1/2T1) A !Syntax Error, I dt xpulse(t) e-imωt/2 + (1/2T1) Be-imω(T/2) !Syntax Error, I dt' xpulse(t') e-imωt'/2 = (1/2T1) [ A + B e-imω(T/2 ] !Syntax Error, I dt xpulse(t) e-imωt/2 = (1/2) [ A + B e-imω(T/2 ](1/T1) !Syntax Error, I dt xpulse(t) e-imωt/2 = (1/2) [ A + B e-imω(T/2 ]cm/2 Thus we find that Cm = (1/2) [ A + B e-imω(T/2) ]cm/2 where the cn go with xpulse(t) having period T1. But ω1T1/2= π and then e-imω(T/2) = e-imπ = (-1)m So we then have Cm = (1/2) [ A + B(-1)m ]cm/2 = (1/T1)Xpulse(mω1/2) (1/2) [ A + B(-1)m ] Our spectrum is then for the 2T1 period pulse train is then X(ω) = !Syntax Error, I C(ω) 2π δ(ω - mω1/2) = !Syntax Error, I Cm 2π δ(ω - mω1/2) = (1/2) [ A + B(-1)m ]cm/2 2π δ(ω - mω1/2) But of course cm/2 = c(mω1/2) = (1/T1)Xpulse(mω1/2) so we seem then to get X(ω) = !Syntax Error, I (1/2) (1/T1)Xpulse(mω1/2) [ A + B(-1)m ] 2π δ(ω - mω1/2) which compare to my Section 34 result X(ω) = Xpulse(ω) (1/2) ω1!Syntax Error, I [ A + B(-1)m ] δ(ω-mω1/2) (34.18) They agree!! So here is some support for my result. Next, from (33.29) we get P(ω) = !Syntax Error, I |Cm|2 δ(ω - mω1/2) = !Syntax Error, I |(1/T1)Xpulse(mω1/2) (1/2) [ A + B(-1)m ] |2 δ(ω - mω1/2) = (1/T1)2 (1/4) !Syntax Error, I| Xpulse(mω1/2)|2 |[ A + B(-1)m ] |2 δ(ω - mω1/2) = (1/T1)2 | Xpulse(ω)|2 (1/4) !Syntax Error, I |[ A + B(-1)m ] |2 δ(ω - mω1/2) = (1/T1)Ppulse(ω) (1/4) !Syntax Error, I |[ A + B(-1)m ] |2 2π δ(ω - mω1/2) and this agrees as well. We have a small problem here. 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) X'ω) = Xpulse(ω) X"(z) (34.6) |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-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. Therefore X"(z) = [ A + B e-iωT] 2πδ5(2ωT1,N/2) (34.16) Using the Appendix A result, limN→∞ δ5(k,N) = !Syntax Error, Iδ(k-2πm) (A.19) we obtain the N→∞ limit for our spectrum X"(z) = [ A + B e-iωT] 2π !Syntax Error, Iδ(2ωT1-2πm) = (1/2)[ A + B e-iωT] (1/T1) 2π !Syntax Error, Iδ(ω-mω1/2) (34.17) and correspondingly X(ω) = (1/2) Xpulse(ω) [ A + B e-iωT] ω1!Syntax Error, Iδ(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 |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) 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) 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) (34.20) Meanwhile, |A + Be-iωT|2 = |A|2 + |B|2 + 2|A||B| cos(ωT1) so here is the final result: P(ω) = Ppulse(ω) (1/4)(1/T1) { |A|2 + |B|2 + 2|A||B| cos(ωT1)} !Syntax Error, I 2π δ(ω - mω1/2) (34.21) When cos(ωT1) is brought inside the sum, it becomes cos(T1mω1/2) = cos(mπ) = (-1)m-1, so the result can also be written P(ω) = Ppulse(ω) (1/4)(1/T1) !Syntax Error, I { |A|2 + |B|2 – 2|A||B| (-1)m} 2π δ(ω - mω1/2) (34.22) Special case 1: Suppose A = 1 and B = -1. Then { |A|2 + |B|2 – 2|A||B| (-1)m} = 2{1 - (-1)m} This vanishes for even m and equals 4 for odd m, so we then get P(ω) = Ppulse(ω) ω1!Syntax Error, I 2π δ(ω - mω1/2) = Ppulse(ω)!Syntax Error, I 2π δ(x - m/2) where x = ω/ω1 (34.23) Meanwhile, the spectrum from (34.18) becomes X(ω) = (1/2) Xpulse(ω) [ 1 – e-iωT] ω1!Syntax Error, Iδ(ω-mω1/2) We can write (1/2) [ 1 – e-iωT] = i e-iωT/2 sin(ωT1/2) Since ωT1/2 → mπ/2 under that action of the delta function, we have sin(mπ/2) which vanishes for even m and equals (-1)m-1 for odd m. The phasor becomes e-iωT/2 = e-imπ/2 = (i)-m so we get X(ω) = Xpulse(ω) !Syntax Error, I i (i)-m(-1)m-1 ω1 δ(ω-mω1/2) = Xpulse(ω) !Syntax Error, I (i)m-1 ω1 δ(ω-mω1/2) = Xpulse(ω) !Syntax Error, I (-1)(m-1)/2 ω1 δ(ω-mω1/2) (34.24) Square wave pulse train with peak-to-peak = 2 units: Specializing further to the case of a square pulse, we can write from (9.2) and box (34.4) Xpulse(ω) = T1 sinc(ωT1/2) = T1 sin(ωT1/2)/ (ωT1/2) Ppulse(ω) = = (1/2π) T1 sinc2(ωT1/2) = (1/ω1) sin2(ωT1/2)/ (ωT1/2)2 When these are evaluated at ω = mω1/2 we get ωT1/2 = mπ/2 and we know sin(mπ/2) = (-1)m-1, so Xpulse(ω) = T1 (-1)m-1/ (mπ/2) = (2/π) T1 (-1)m-1/ m Ppulse(ω) = (1/ω1)T1 / (mπ/2)2 = (2/π)2 (1/ω1)T1/ m2 Then we find for this square wave pulse train P(ω) = Ppulse(ω) ω1!Syntax Error, I 2π δ(ω - mω1/2) = = (2/π)2 T1 !Syntax Error, I (1/ m2) 2π δ(ω - mω1/2) X(ω) = Xpulse(ω) !Syntax Error, I (-1)(m-1)/2 ω1 δ(ω-mω1/2) = !Syntax Error, I (-1)(m-1)/2 (2/π) T1 (-1)m-1/ m ω1 δ(ω-mω1/2) = (2/π) T1 ω1 !Syntax Error, I (-1)(m-1)/2 (-1)m-1/ m δ(ω-mω1/2) = (2/π) T1 ω1 !Syntax Error, I (-1)(m-1)/2 (1 / m) δ(ω-mω1/2) We shall now challenge this result for various special cases. Special Case 0. If A = 1 and B = -1 we get { |A|2 + |B|2 + 2|A||B| cos(ωT1)} = 2(1-cos(ωT1)) = 4 sin2(ωT1/2) so then P(ω) = Ppulse(ω) (1/ω1) sin2(ωT1/2) !Syntax Error, I δ(ω - mω1/2) (34.21) and If we convert this to dimensionless frequency x = ω/ω1 it becomes P(ω) = Ppulse(ω) !Syntax Error, I sin2(mπ/2)δ(x - [m/2]) (34.21) All even m terms vanish so this becomes P(ω) = Ppulse(ω) !Syntax Error, I δ(x - [m/2]) (34.21) 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)