finite N version of stat pulse trains

Repeat Statistical Stuff with Finite N everywhere. PhL 3.26.05 14. The Spectrum of a Pulse Train A pulse train consists of a set of pulses of shape xpulse(t) separated by time T1. We assume that xpulse(t) is some "reasonable" (non-pathological) function. x(t) =!Syntax Error, I xpulse(t - tn) tn = nT1 pulse train (14.1) Let Xpulse(ω) be the spectrum of xpulse(t). This xpulse(t) does not really have to be a "pulse", but it is convenient to think of it as such. We imagine that xpulse(t) is a function that is somewhat localized in the region of t=0, and vanishes for very large positive and negative time. The half-width of the pulse can be larger than T1 as discussed below, so pulses can overlap. Thus, our xpulse(t) is not itself periodic, and we thus expect it to have a continuous spectrum. For example, from (9.2) we already know the spectrum of a single square wave pulse of height A and width τ centered at t=0: Xpulse(ω) = (Aτ) sinc(ωτ/2) . (14.2) Now consider a second pulse which is a copy of our original pulse, but which is translated T1 units to the right in time. From (12.1), we know the spectrum of this second pulse: X(ωsecond pulse) = Xpulse(ω) e-iωT. Now construct the infinite periodic wave by superposing pulses at t = 0, ±T1, ±2T1, ... . We get, X(ω) = Xpulse(ω) !Syntax Error, I eiωnT (14.3) We quote now a result from Appendix A !Syntax Error, I eink = 2π { } = 2π δ5(k,N) (A.25) Thus (14.3) becomes, setting k = ωT1 X(ω) = Xpulse(ω) 2π δ5(ωT1,N) (14.4) STOP HERE and go to (3.31) below and do edits there In getting to the next step, we use the property of delta functions δ(ax) = δ(x)/a, and we define ω1 ≡ 2π/T1 = the fundamental angular frequency of our periodic waveform. The result is: red flag? X(ω) = Xpulse(ω) !Syntax Error, I δ( - m ) . (14.5) Notice that the delta function argument is dimensionless. This is a nice form in theory, but in practice we want ω inside the delta function to be isolated so that, when (14.5) is inserted into a dω integration, we don't pick up any new constants. Thus, we extract the ω1 = 2π/T1 factor, keeping the 2π part right with δ(): X(ω) = !Syntax Error, I (1/T1)Xpulse(ω) 2πδ( ω - mω1) . (14.6) Thus, we have a set of evenly spaced delta function spikes which occur at these frequencies: ωm = mω1 m = 0,1,2,3...... (14.7) In formulas like (14.6) , one should keep in mind that the variable ω appearing inside Xpulse(ω) can at any time be replaced with ωm ≡ mω1 due to the presence of the delta function. In general, one has, f(ω) δ(ω - a) = f(a) δ(ω - a) . Both sides of this equation are zero when ω ≠ a, and at ω = a, f(a) = f(ω). Because the quantity (1/T1)Xpulse(ω) is going to occur very frequently in the following discussion, we shall now define a more compact notation for it as follows: c(ω) ≡(1/T1)Xpulse(ω) . (14.8) Thus, c(ω) is nothing more than our (continuous) pulse spectrum divided by the fundamental period T1. We can then rewrite (14.6) as follows: X(ω) = !Syntax Error, I c(ω) 2π δ(ω - mω1) = !Syntax Error, I c(ωm) 2π δ(ω - mω1) . (14.9) As was just noted above, we can harmlessly replace ω with ωm = mω1 inside c(ω) in (14.8). This leads us to define a set of numbers as follows cm ≡ c(ωm ) = c(mω1) (14.10) These numbers are just the values that the function c(ω) takes at our delta spike frequencies. We arrive then at our final form for the spectrum of a pulse train, X(ω) = !Syntax Error, Icm 2π δ(ω - mω1) . (14.11) Now we are ready to summarize all these results: Fourier Integral Transform of a Pulse Train (14.12) 1. Let xpulse(t) be any reasonable pulse. Construct a pulse train x(t) with spacing T1: x(t) = !Syntax Error, Ixpulse(t - nT1) . (14.1) By its construction, x(t) is periodic with period T1, which we can write formally as: x(t + nT1) = x(t) . n = any integer If x(t) is a known periodic function of period T1, a candidate for xpulse(t) is x(t) over any one period. 2. Define c(ω) to be the Fourier Integral transform of the pulse, scaled by 1/T1: c(ω) ≡(1/T1)Xpulse(ω) = (1/T1) !Syntax Error, I dt xpulse(t) e-iωt . (14.8) and (1.1) 3. Then the Fourier Integral transform of the Pulse Train is as follows: X(ω) = !Syntax Error, I c(ω) 2π δ(ω - mω1) = !Syntax Error, I cm 2π δ(ω - mω1) (14.9) where cm = c(ωm ), ωm = mω1, ω1 = 2π/T1. 4. These cm are the same cm which appear in the next section. That is, they the complex Fourier Series coefficients. Item 3 is our main result. It says that the Fourier Transform spectrum of an infinite sequence of pulses is a sum of equally-spaced delta function spikes whose coefficients are given by the continuous spectrum of the central pulse evaluated at the spike frequencies ω = mω1. The pulse spectrum c(ω) (1/T1)Xpulse(ω) is normally thought of as the "coefficient envelope", while the equally spaced delta function spikes are the "lines". In this sample symbolic drawing of a spectrum, the infinitely-high delta function spikes of the spectrum are represented by finite vertical red line segments whose heights are the coefficients cm. The red lines are the spectrum, and they track the envelope c(ω). It may happen that certain cm vanish, meaning that such lines are not present. The item 3 sum includes the DC line m=0 having ω0 = 0. Unless c0 happens to vanish, the pulse train has a DC component. As noted in Section 10 above, Xpulse(0) is the area under xpulse(t). Only if this area is zero, do we get c0 = c(0) (1/T1) Xpulse(0) = 0. Whereas the Fourier Transform spectrum of a single pulse (localized, non-periodic) is continuous in ω, that of an infinite sequence of pulses is entirely discrete and has no continuous portions. This conforms with the well-known fact that the spectrum of any periodic function is discrete. In fact, we have just proven this to be so. Note on xpulse(t). In our summary box above, we say that if x(t) is some known periodic function, one can take as a candidate for xpulse(t) the function x(t) restricted to any one period. In this case, the dt integration endpoints for the projection Xpulse(ω) only cover that selected period. If we select the period centered at t=0, then item 2 in the above summary box becomes perhaps more familiar: c(ω) (1/T1)Xpulse(ω) = (1/T1) !Syntax Error, Idt x(t) e-iωt (14.13) What is perhaps less obvious is that there are really many different candidates for xpulse(t) that result in the same x(t) pulse train. These other choices for xpulse(t) are pulses which slop over into more than one period T1. When a pulse train is formed with such pulses, the pulses overlap. To see how this might work, think of a pulse which has a nice gaussian shape and goes about half way into each neighboring T1 interval. Draw some of these, then add them up to make the sum curve x(t). In this case, for your candidate xpulse(t), you can use either the gaussian, which overlaps into several intervals, or you can use one interval's worth of the sum curve x(t) (shown as the darker curve) We have tried to keep our formulas completely general to allow for pulse trains formed from pulses which overlap into more than one period. This note applies to all the analysis and summary boxes which follow. The resulting spectrum is of course the same no matter which xpulse(t) is chosen. To summarize, we can write c(ω) in two equivalent ways c(ω) (1/T1)!Syntax Error, Idt x(t) e-iωt = (1/T1)!Syntax Error, Idt xpulse(t) e-iωt (14.14) If we evaluate (14.14) at the discrete spike frequencies ω = mω1, we get cm = (1/T1)!Syntax Error, Idt x(t) e-imωt = (1/T1)!Syntax Error, Idt xpulse(t) e-imωt (14.15) Now since e-imω(t+T) = e-imωt e-imωT = e-imωt e-im2π = e-imωt, function e-imωt is periodic with period T1. Since x(t) in the first integral in (14.15) is also assumed periodic with period T1, the integration can be over any interval of width T1, so one usually takes this interval to be (0,T1). Thus, cm = (1/T1) !Syntax Error, Idt x(t) e-imωt = (1/T1)!Syntax Error, Idt xpulse(t) e-imωt (14.16) Amplitude Modulated Pulse Train Notes W(ω) = Xpulse(ω) !Syntax Error, I yn e+iωt Chapter 6: Statistical Pulse Trains 32. The Autocorrelation Function This is a very simple idea which is often made to seem complicated. Starti with a reasonable function x(t). Define the autocorrelation function of x(t) as follows: rx(t) ≡ !Syntax Error, I dt' x(t') x(t' + t) (32.1) The integrand is the function evaluated at time t' times the same function evaluated at later time t'+t. Some texts might define the above with an extra overall "normalizing" constant factor; we leave it as shown. A simple reflection property follows from the above definition (use t" = t' + t ): rx(-t) = rx(t) (32.2) Thus, in (32.1) we could put either +t or -t in the last parentheses. Although we have not yet mentioned statistics and randomness, one could easily imagine the following situation. Suppose x(t') is some sort of random function ("noise") that takes values in the range -1 to 1. It seems likely that for a value of the separation t that is larger than some small value, you might get rx(t) = 0. The vague argument would be that there is no "correlation" between x(t') and x(t'+t), so the product of these two functions ought be be pretty random, and a sum of random numbers in the range -1 to 1 ought to be zero. Even in this case, we can see that the result is not zero if t = 0, since we are then summing a positive quantity. In fact, rx(0) is the area under x(t)2. (a) Autocorrelation function for a Square Pulse. Before going any further, let us compute the autocorrelation function for some simple case we are familiar with. A good candidate is x(t) = a square pulse of width τ and amplitude A. As in (9.1), xpulse(t) = A [ θ(t + τ/2) - θ(t - τ/2) ] . (9.1) One can easily do the above integral (32.1) to get the answer, but it is very obvious what the answer is. We are multiplying a box times a box shifted by t. Where they overlap, the integrand is A2. The boxes only overlap if the absolute value of shift tis less than the width τ of the pulse. If this is so, the size of the overlap is τ-|t|. If |t| is larger than τ, there is no overlap, so the integral is 0. Thus, rpulse(t)  A2(τ-|t|) θ(τ-|t|) = (A2τ) [ 1 - |t|/τ ] θ(τ-|t|) (32.3) (b) Energy in a finite signal x(t) The total energy in a finite duration signal x(t) can be computed in either the t-domain or the ω-domain using using Parseval's formula (10.5), to which we add 1/R to each side. !Syntax Error, Idt |x(t)|2/R = !Syntax Error, Idω (32.4) If we think of x(t) as the voltage across a resistor R, then dr x2(t)/R is the energy delivered to the resistor in time dt, and the integral on the left is the total energy in signal x(t). The dimensions on the right are, looking at (1.1), dω = sec-1 (volt-sec)2/ohms = (volt2/ohms)sec = watts-sec = joules = energy (32.5) Setting R = 1Ω, we can write this as !Syntax Error, Idt p(t) = !Syntax Error, Idω P(ω) (32.6) p(t) ≡ |x(t)|2 = energy density in t-domain p(t)dt = energy in dt P(ω) ≡ |X(ω)|2/2π = energy density in the ω-domain P(ω)dω = energy in dω We shall refer to P(ω) as the spectral energy density of signal x(t) whose Fourier Transform is X(ω). The isolated single pulse xpulse(t) has a corresponding Ppulse(ω) . (b) The Spectral Density Connection: Wiener-Khintchine Changing to t" = -t', we can trivially rewrite the definition (32.1) as follows: rx(t)  !Syntax Error, Idt"x(t - t") x(-t") . // energy units (32.7) This has the standard convolution equation form (3.1) where we select b(t) = x(t) ↔ B(ω) = X(ω) c(t) = x(-t) ↔ C(ω) = X(ω) = X(ω)* // from (7.1) and (7.2) Thus, the diagonalized frequency domain form (3.6) is Rx(ω) = |X(ω)|2 (32.8) Dividing by 2π we find that P(ω) = (1/2π) Rx(ω) (32.9) This is the key result. It says that the Fourier integral transform of the autocorrelation function of x(t) is 2π times the spectral energy density of x(t). This seemingly simple result gets the elaborate label of a Wiener-Khintchine Relation, see Bennett and Davey, Data Transmission, p 334. Note also from (32.1) that rx(t) evaluated at t = 0 gives the total energy in signal x(t), rx(0) = !Syntax Error, Idt x2(t) ≡ Ex = total energy in signal x(t) (32.10) For us, the significance of (32.9) is that we can "inject statistics" into a computation of the autocorrelation function, and then we will know the power spectrum of our statistical signal from (32.9). All we have to do is Fourier transform the autocorrelation function rx(t). Examples will follow. (c) Verification of Wiener-Khintchine for a Square Pulse. In (32.3) we have the autocorrelation function for a square pulse. One can insert this into the Fourier transform (1.1) to compute Rx(ω). Rx(ω) = !Syntax Error, I dt (A2τ) [ 1 - | t | / τ ] e-iωt = 2(A2τ) !Syntax Error, Idt (1-t/τ) cos(ωt) = 2(A2τ) (1-cos(ωτ))/(ω2τ) = 4(A2τ) sin2(ωτ/2)/(ω2τ) = (A2τ2) sin2(ωτ/2)/(ωτ/2)2 = (Aτ)2 [ sinc(ωτ/2) ]2 (32.10) and this is recognized from (9.2) to be |X(ω)|2 for the square pulse, in agreement with (32.8) . ok to here 33. Autocorrelation Function for a Pulse Train (b) Spectral Energy Density of a Finite Pulse Train Recall these equations from the end of Section 14, x(t) = !Syntax Error, I xpulse(t - nT1) (14.1) X(ω) = Xpulse(ω) 2π δ5(ωT1,N) (14.4) where δ5 is a certain delta function model described in Appendix A. We then have | X(ω) |2 = | Xpulse (ω) |2 [2π δ5(ωT1,N)]2 (33.11) Recalling the definition of the δ6 delta function model from Appendix A, δ6(k,N) ≡ (A.20) we obtain = | Xpulse (ω) |2 δ6(ωT1,N) (33.12) But the left side is P1(ω) as defined in (33.6) so we obtain P1(ω) = | Xpulse (ω) |2 δ6(ωT1,N) (33.13) for a finite pulse train. Recalling now that limN→∞ δ6(k,N) = !Syntax Error, Iδ(k-2πm) (A.21) we obtain for an infinite pulse train P1(ω) = | Xpulse (ω) |2 !Syntax Error, Iδ(ωT1-2πm) = !Syntax Error, I | Xpulse (mω1) |2δ(ωT1-2πm) and this agrees with (33.7) derived in the previous section. P1(ω) = | Xpulse (ω) |2 δ6(ωT1,N) so that P(ω) = = | Xpulse (ω) |2 2π [ δ5(ωT1,N)]2 (33.3) We now define P1(ω) ≡ P(ω) (33.6) which we can interpret as the spectral energy density of an average pulse in the pulse train, or as the average spectral energy density of the pulse train overall, it is the same either way. Then (33.3) reads, P1(ω) = | Xpulse (ω) |2 2π [ δ5(ωT1,N)]2 (33.7) Example: We can estimate the sharpness of the delta function line away from the limit. Suppose we run a signal into our HP Spectrum Analyzer, and suppose it consumes 1 msec for its computation of each spectral "bin". If the signal under test is a 1 GHz pulse train, then N = 106 pulses of the pulse train are analyzed for each bin. As shown in Appendix A, the peaks of (33.4) all have the same shape since the function is periodic, and the width of a peak is roughly Δk = 2π/N. Applying this to the δ in (33.7), the width of one of the δ lines is then Δ(ωT1) = 2π/N so that Δω = ω1/N. Our line width is thus ∆ω = ω1/ N or ∆f = f1/N . (33.8) In this case, the line width ∆f would be 109/106 = 1 kHz. This is just an application of the uncertainty principle which says ∆t ∆f ≈ 1. Here, ∆t is the time of measurement (the 1 msec), and ∆f is the line width (the 1 KHz). The frequency is uncertain by amount ∆f, so the line has width Δf. (b) Autocorrelation function for a Pulse Train We can compute the autocorrelation function for pulse train x(t) directly in the time domain using definition (32.1), and we shall do this later with the addition of statistics. Here instead we inverse Fourier transform both sides of (33.3), P(ω) = = | Xpulse (ω) |2 2π [ δ5(ωT1,N)]2 (33.3) Using (1.2) apply !Syntax Error, Idω e+iωt to both sides to get !Syntax Error, Idω e+iωt = !Syntax Error, Idω e+iωt!Syntax Error, I|Xpulse(mω)|2 2π [ δ5(ωT1,N)]2 From (32.8) we know that |X(ω)|2 = Rx(ω), so from (1.2) the left side is just rx(t). On the right side the delta function pins the phasor ω to a specific value and the result is = (1/T1) !Syntax Error, I|Xpulse(mω1)|2 eimωt (33.9) Undoing the limit, = (1/T1) !Syntax Error, I|Xpulse(mω1)|2 eimωt (33.10) In this undone limit, we define the duration of the pulse train to be T ≡ (2N+1)T1 (33.11) Dividing both sides of (33.10) by T1 then gives = (1/T1)2 !Syntax Error, I|Xpulse(mω1)|2 eimωt (33.12) But we know that (1/T1) Xpulse(mω1) = c(mω1) = cm, the complex Fourier coefficient of (14.13). So we can rewrite the above result one more time to get a simple result for the autocorrelation function of x(t), = !Syntax Error, I|cm|2 eimωt (33.13) Finally, we can set t=0 to get, = !Syntax Error, I|cm| 2 // average power (33.14) Recall from (32.10) that rx(0) is the total energy in the pulse train, rx(0)/T is the average power carried by the pulse train (delivered to a 1Ω load). Some texts add (1/T) in their definition of rx(t), but we have chosen not to do that. So, the average power in a pulse train is the sum of the squared magnitudes of the complex Fourier Series coefficients. We know that c-m = cm* when x(t) is real, so we can reflect the negative sum over to the positive side and gain a factor of 2. We also know that 2cm = am - ibm from (15.4), so this gives us several other ways to write the above sum = |c0|2 + 2 !Syntax Error, I|cm|2 = (1/4) |a0|2 + (1/2) !Syntax Error, I{ |am|2 +|bm|2 } (33.15) The magnitude-squares of the Fourier Series coefficients give a power decomposition of a pulse train x(t). 34. Statistical Pulse Trains Suppose we have, in place of the infinite pulse train of (33.1) with all coefficients unity, a new pulse train where the coefficients are completely general numbers, presumably real. Later we might want to have each coefficient be either a 1 or a 0. For now, we stay general, so here is our amplitude moduldate pulse train, x(t) = !Syntax Error, I an xpulse(t - nT1) . (34.1) Now, lets imagine that we have a "statistical ensemble" of such pulse trains. Each such pulse train gets a superscript i as its label. Then we can say, xi(t) = !Syntax Error, Iain xpulse(t - nT1) . (34.2) We now define the operation < > to indicate the average of some quantity over a large statistical ensemble of systems. Assume there are ' I ' systems in the ensemble. We then have, <rx(t)> = (1/I) !Syntax Error, I!Syntax Error, I dt' xi(t') xi(t' + t) . (34.3) If we now insert (34.2) twice into (33.3), <rx(t)> = (1/I) !Syntax Error, I!Syntax Error, I dt' { !Syntax Error, Iaim xpulse(t' - mT1)}{ !Syntax Error, Iain xpulse(t' + t - nT1)} =!Syntax Error, I !Syntax Error, I { (1/I) !Syntax Error, I aim ain } !Syntax Error, I dt' xpulse(t' - mT1) xpulse(t' + t - nT1) Following our statistical ensemble averaging rule, we define <aman> ≡ (1/I) !Syntax Error, Iaim ain (34.4) so that <rx(t)> = !Syntax Error, I !Syntax Error, I <aman> !Syntax Error, I dt' xpulse(t' - mT1) xpulse(t' + t - nT1) (34.5) Now replace integration variable t' by t" = t' - mT1 to evalaute the integral appearing above, !Syntax Error, I dt" xpulse(t") xpulse(t" + { t + (m-n)T1 }) = rpulse[t + (m-n)T1] (34.6) and we end up with <rx(t)> = !Syntax Error, I !Syntax Error, I<aman> rpulse[t + (m-n)T1)] (34.7) This expresses the autocorrelation function of the pulse train as a sum, with statistically averaged coefficients, over the autocorrelation function of the pulse used to build the pulse train. We wish to Fourier transform (34.7) to the ω domain using (1.1), so apply !Syntax Error, Idt e-iωt to both sides, !Syntax Error, Idt e-iωt <rx(t)> = !Syntax Error, I !Syntax Error, I<aman> !Syntax Error, Idt e-iωt rpulse[t + (m-n)T1)] On the left move the integration inside the statistical sum implied by <...> to get <Rx(ω)> . On the right, replace integration variable t by t' = t + (m-n)T1 so that e-iωt = e-iωt' eiω(m-n)T . The result is <Rx(ω)> = Rpulse(ω) !Syntax Error, I !Syntax Error, I <aman> eiω(m-n)T (34.8) We now use the relation (32.8) to replace each R(ω) with it's corresponding spectral density. Thus, <|X(ω)|2 > = |Xpulse(ω)|2 !Syntax Error, I !Syntax Error, I <aman> eiω(m-n)T (34.9) Divide both sides by 2π and use (32.6) to write this as < P(ω) > = Ppulse(ω) !Syntax Error, I !Syntax Error, I <aman> eiω(m-n)T (34.10) This result describes the spectral energy density of a statistical pulse train in terms of the spectral energy density of the pulse we select to build the train, and in terms of the statistically averaged coefficients. The integral !Syntax Error, Idω Ppulse(ω) is the total energy in one of the pulses from which the pulse train is constructed, while !Syntax Error, Idω < P(ω)> is the total energy in the average pulse train of a statistical ensemble of pulse trains. If N = ∞, then this integral is infinite as well. If we now make some statistical assumptions about the coefficients, we will arrive at the spectrum of a statistical pulse train. This is exactly what we want to know. We already see from (34.9) a principle fact. The spectral density of the pulse acts as an "envelope" function, regardless of what happens with the double summation business. More on this below. ok to here 35. Dealing with the statistical coefficient averages By our definition (34.4) above, we have this average over our statistical ensemble of pulse trains: <aman> ≡ (1/I) !Syntax Error, Iaimain . (35.1) A very simple case to consider is this: am = A probability p am = B probability 1-p (35.2) Remember that index m labels a particular pulse in the pulse train. We can now state the averaged coefficients as follows, first for two different pulses, then for the same pulse: n ≠ m <aman> = [pp] AA + [p(1-p)] AB + [(1-p)p]BA + [(1-p)(1-p)]BB ≡ α n = m <am2> = [p]AA + [(1-p)] BB ≡ β (35.3) In the square brackets [..] we indicate the probability of some case occurring, and this is multiplied by the value that the quantity in question takes in that case. The reader must now stop reading and stare at the above equations until they make complete sense. Notice that the second line is quite distinct from the first line. In the double summation in (34.9), there are both "diagonal" terms where m = n, and off diagonal terms where m≠n. These groupings must be treated separately according to the above. We have defined new symbols α and β to emphasize that, for the situations shown, there is no longer any dependence on the n and m indices for these quantities. Finite Pulse train with Random Coefficients We repeat the previous section for a finite random pulse train. Using the above forms for the coefficient averages, (34.13) becomes <|X(ω)|2> = |Xpulse(ω)|2 { α!Syntax Error, I !Syntax Error, I [ eiω(m-n)T] +β!Syntax Error, I !Syntax Error, I [1] } (35.9) To evaluate the first double sum, we write it as !Syntax Error, I !Syntax Error, I [ eiω(m-n)T] = !Syntax Error, I { !Syntax Error, I [ eiω(m-n)T] – 1 } = ( !Syntax Error, I e+iωnT ) (!Syntax Error, Ie-iωmT ) – !Syntax Error, I1 = |!Syntax Error, I e+iωnT |2 - (2N+1) From Appendix A we have !Syntax Error, I eink = 2π { } = 2πδ5(k,N) so the first double sum in (35.9) becomes !Syntax Error, I !Syntax Error, I [ eiω(m-n)T] = [2πδ5(ωT1,N)]2 - (2N+1) . (35.10) The second double sum in (35.9) is just !Syntax Error, I !Syntax Error, I [1] = !Syntax Error, I [1] = (2N+1) and therefore we may write (35.9) as <|X(ω)|2> = |Xpulse(ω)|2{ α [ [2πδ5(ωT1,N)]2 - (2N+1) ] + β (2N+1) } = (2N+1) |Xpulse(ω)|2{ α [- 1] + β (2N+1) } so = |Xpulse(ω)|2 {(β-α) + α } (35.11) The factor multiplying α can be replaced using (A.20), δ6(k,N) ≡ = (A.20) to give = |Xpulse(ω)|2 { (β-α) + α 2π δ6(ωT1,N) } (35.12) From (33.1) and (33.2) we identify the left side with 2π P1(ω). Dividing by 2π then gives <P1(ω)> = Ppulse(ω) [ (β-α) + α 2π δ6(ωT1,N) ] (35.13) which is the finite pulse train version of (35.8). In the limit N → ∞ we know from (A.21) limN→∞ δ6(k,N) = !Syntax Error, Iδ(k-2πm) (A.21) so that (35.13) becomes <P1(ω)> = Ppulse(ω) [ (β-α) + α!Syntax Error, I2π δ(k-2πm) ] which reproduces (35.8). _________________________________________________- = |Xpulse(ω)|2 { (1/4) + (1/4) } (35.7) Now we divide both sides by 2π. The left side then becomes <P(ω)>/(2N+1) = <P1(ω)> as in (33.6). And |Xpulse(ω)|2/2π = Ppulse(ω) so, <P1(ω)> = Ppulse(ω) { α + (β-α) } (35.8) Now take the large N limit and what happens? What is the limit of ?? In Appendix A I consider large N only and look only at the peak at ω = 0 and integrate it from (-a,a), and that is how I show the area under δ5 is 1 for that peak. Now consider this maple code This tells me that limN→∞ ∫-aa = (2π)2 (1/2π) = 2π Therefore I can regard limN→∞ = 2πδ(ωT1) for the central peak only, and more generally = 2π!Syntax Error, I δ(ωT1 – 2πn) Then (35.8) becomes <P1(ω)> = Ppulse(ω) { α + (β-α) } (35.8) = Ppulse(ω) { α 2π!Syntax Error, I δ(ωT1 – 2πn) + (β-α) } And this agrees with my all delta function approach. But now I have a result for finite N as well. Here Ppulse(ω) is the spectral energy denstiy of xpulse(t), and its integral over dω is finite. The left side is <P1(ω)> which is the statistically averaged energy density per pulse for the pulse train, and its integral over dω is also finite. Note that no N's or δ(0)'s appear explicitly in this result! If N = ∞, δ(ωT1- 2πm) is a true delta function, but if N is large but finite, then δ(ωT1- 2πm) is given by (33.4). Spectral Energy Density of a Random Pulse Train (35.9) <P1(ω)> = Ppulse(ω) [(β-α) + α !Syntax Error, I2π δ(ωT1- 2πm) ] // per pulse (35.8) α = [pp] AA + [p(1-p)] AB + [(1-p)p]BA + [(1-p)(1-p)]BB // = <aman> when n≠m (35.3) β = [p]AA + [(1-p)] BB // = <aman> when n=m A pulse has probably p of having amplitude A, and probability 1-p of having amplitude B. T1 = pulse spacing. For coefficients A=1, B=0, α = p2 , β = p, (β - α) = p(1-p), where p = probability of a coded 1. This is a fascinating result for <P1(ω)>. There are two weighted terms: the first term has weight (β - α), the second term has weight α. The first term is just (β-α) times the spectrum of the pulse used to build the pulse train -- it is a continuous function of ω. The second term is the same discrete spectrum we found in (33.3) for the regular pulse train. It's delta functions pick off specific values of |Xpulse(ω)|2 at the lines. Looking back at (35.3), we see that if the pulse train coefficients are A=1 or B=0, then α = p2 and βp, so that (β - α) = p(1-p). Here are a few examples using A = 1 and B = 0. Example 1: Let p=1 with A=1 and B=0. Then α = β = 1. This is the non-statistical pulse train (33.1) where each pulse has a unity coefficient. Now having the experience with handling the delta functions, we can just stare at (33.2) and mentally square both sides and divide by 2π. The resulting 2πδ(0) gets moved to the left side denominator, it is interpreted as (2N+1), so the left side becomes P1(ω), so (33.2) squared reads P1(ω) = !Syntax Error, I Ppulse(mω1) 2π δ(ωT1 - 2πm) But this is exactly what (35.8) says when α = β = 1. Since all pulses are the same, <P1(ω)> = P1(ω). Example 2: Let p=0 with A=1 and B=0. Then every pulse has B = 0, x(t) ≡ 0, α = β = 0, and both terms in (38.5) vanish so there is zero spectral energy density. Example 3: Let p=1/2 with A=1 and B=0, so α = p2 = 1/4 and β = p = 1/2, This pulse train then has an equal probably of 1's and 0's. We find from (38.5) that <P1(ω)> = Ppulse(ω) [ 1/4 + 1/4 !Syntax Error, I2π δ(ωT1- 2πm) ] (35.10) This shows that the discrete spectrum has been reduced to 1/4 of its full strength, and a continuous spectrum exists with coefficient 1/4 as shown. It would be interesting to try and confirm this last result using a spectrum analyzer for various pulses. 36. A View from the High Ground In this Chapter we introduced some new definitions, such as the autocorrelation function, and the Wiener-Khintchine relation. We used these in our derivation of the statistical average of the spectral density of a pulse train. The whole thing was somewhat of a deception, and here we want to make sure the reader has not missed this point. Our main purpose for introducing autocorrelation and Wiener-Khintchine was to get these well-known terms into our dialog, since they appear in texts. In reality, there is nothing at all new in this Chapter, and we certainly did not need to use the words autocorrelation or Wiener-Khintchine to get our main results. As shown in (32.7), autocorrelation is just a standard convolution of x(t) with x(-t), and Wiener-Khintchine is just the convolution theorem (3.6) applied to this convolution equation. In effect, we could have just defined R(ω) to be |X(ω)|2, given it the new name "autocorrelation", and then forgotten about it. Here for example is a simple derivation of (34.10): Start off with the pulse train "i" with statistical coefficients ain, as in (34.2), xi(t) = !Syntax Error, I ain xpulse(t - nT1) (36.1) Fourier transform both sides in the usual manner, picking up the usual time-shift phase, Xi(ω) = Xpulse(ω) !Syntax Error, I ain e-inωT (36.2) Magnitude-square this thing to get |Xi(ω)|2 = |Xpulse(ω)|2 !Syntax Error, I !Syntax Error, I aimain eiω(m-n)T (36.3) Finally, divide by 2π, average over ensemble index i, identify <|Xi(ω)|2/2π> = <P(ω)> and we get < P(ω) > = Ppulse(ω) !Syntax Error, I !Syntax Error, I <aman> eiω(m-n)T which is (34.10). The dervivation of (35.8) would then proceded as outlined in Section 35. A paradox and its resolution Now while we are here, we can back up and apply our statistical average directly to the spectrum in (36.2). Our result is <X(ω)> = Xpulse(ω) !Syntax Error, I<an> e-inωT (36.4) We could then argue that <an> = [p] A + [1-p] B = p if A=1, B=0 . (36.5) In thise case, we can extract p from the sum, which then collapses to form the usual (13.2) delta functions. The result is then the same as with the regular pulse train (33.2) with an overall factor of p out front, namely, <X(ω)> = p !Syntax Error, IXpulse(mω1) 2π δ( ωT1 - 2πm) (36.6) Thus says that our average spectrum is 100% discrete, there is no continuous part! If p = 1/2, the average spectrum is just 1/2 times our unit amplitude pulse train spectrum (14.4). How can this be true, if we just showed in (35.10) that the average spectral density <|X(ω)|2> has a continuous spectral component? The answer lies in the fact that <ab> ≠ <a><b>, where < > is our averaging operation. Thus <|X(ω)|2> ≠ <X(ω)><X(ω)*> (36.7) There is no reason in the world why the average of a product should be the product of the averages. The HP Spectrum Analyzer measures <|X(ω)|2>, not <X(ω)>. Idea: have Maple generate a long pulse train with p= 1/2, then have it compute the Fourier spectrum X(ω) and see if its spectrum looks discrete. Then maybe try to get it do the energy density and see if it seems to have a discrete and continuous piece. This would be quite interesting. ok to here pass 2