where do image spectra

This is the Title PhL 3.26.05 Note that page numbering is turned on in this template. 29. Where do Image Spectra come from? Consider the amplitude-modulated pulse train and its Fourier Integral spectrum from summary box (25.4). The pulse shape is xpulse(t) and the pulse spacing is T1. w(t) = !Syntax Error, I yn xpulse(t -tn) (25.1) (29.1) W(ω) = (1/T1)Xpulse(ω) Y'ω) . (25.3) (29.2) In the last equation, Y'(ω) is the Digital Fourier Transform of the sequence of yn values which provide the modulation of the pulses, Y'ω) ≡ T1!Syntax Error, I yn e-iωt projection = transform (22.2) (29.3) We assume now that the samples yn are of some continuous signal y(t) which has Fourier Integral spectrum Y(ω). Then the relationship between Y'(ω) and Y(ω) is given by Y'(ω) =!Syntax Error, IY(ω- mω1) = [ Y(ω) + !Syntax Error, IY(ω- mω1) ] . (23.1) (29.4) We noted that Y'(ω) therefore has some sort of central region of shape Y(ω) and an infinite set of image spectra going off to the left and right of this central region, spaced by ω1. A priori, we don't know exactly what Y(ω) looks like because we don't know y(t), we only know its sample values yn. In order to have something to work with, we have to assume some shape for y(t). This shape, whatever it might be, is completely independent of shape of xpulse(t) in (29.1). For purposes of discussion, then, let us assume that the signal y(t), whose samples are yn, is a stair-step function which just tracks the yn values as in this example, But this is an amplitude-modulated finite pulse train with x'pulse(t) = a box of height A = 1 and width τ = T1. The Fourier Integral Transform spectrum Y(ω) associated with just the box centered at t = 0 we know from (9.2) is y0(T1) sinc(ωT1/2). The spectrum of the next box is, according to the time translation rule (12.1), y1(T1) sinc(ωT1/2) e-iωT and by superposition we find that Y(ω) = (T1) sinc(ωT1/2) Σn yn e-inωT = (T1) sinc[π(ω/ω1)] Σn yn e-i2π(nω/ω) . = sinc[π(ω/ω1)] Y'(ω) (29.5) in agreement with (25.3) for the spectrum of an amplitude-modulated pulse train with x'pulse = box. As a function of ω, the sum part Y'(ω) of Y(ω) is a periodic function of period ω1 repeating out to ω = ± ∞, but the sinc function tames the spectrum so that Y(ω) is localized to a region around ω = 0 of half-width ω1. We shall now explore the above set of equations for some simple pulse train examples. Example 1: Even the simplest case is interesting. The yn sequence is taken as a unit impulse scaled by y0 Then |Y(ω)| = y0(T1) |sinc(ωT1/2)| which we plot for T1= 1 so ω1 = 2π, and ω in (-40,40), Meanwhile, we know from (29.3) that Y'ω) ≡ y0T1. In this example, how exactly does the image spectrum equation Y'(ω) = [ Y(ω) + Σm≠0Y(ω- mω1) ] work out? First, suppose we plot this Y'(ω) sum limiting the sum range to m = -100 to 100 , using Y(ω) = y0(T1) sinc(ωT1/2) for Y(ω) : It appears that the shifted sinc functions are adding up to produce the constant Y'ω) ≡ T1 = 1. In terms of the math, this must mean that Y'(ω) = !Syntax Error, IY(ω- mω1) = yo T1!Syntax Error, I sinc[π(ω/ω1-m)] = yoT1 (29.6) which implies the following unusual sum rule, valid for any real x : !Syntax Error, I sinc[π(x-m)] = 1 . (29.7) This result is sometimes quoted with x = 0 as Σm=-∞∞ sinc(πm) = 1, but it is in fact valid for any x. We have in effect proven the sum rule with the above analysis, but as usual we would like to find verification. We first process the sum as follows !Syntax Error, I sinc[π(x-m)] = !Syntax Error, I sin[π(x-m)]/ [π(x-m)] = (1/π) sin[πx]!Syntax Error, I(-1)m / (x-m). The sum on the right can be further processed, !Syntax Error, I(-1)m / (x-m) = 1/x + [!Syntax Error, I+!Syntax Error, I] (-1)m / (x-m) = 1/x + 2x !Syntax Error, I(-1)m . According to GR 142.3 (page 44), we may replace 1/x + 2x !Syntax Error, I(-1)m = π csc(πx) and then we find that !Syntax Error, I sinc[π(x-m)] = (1/π) sin[πx] π csc(πx) = 1 . Example 2: Here we use the exact sample sequence shown in Figure *** above. We start here with a plot of |Y'(ω)| = |Σn yn e-inωT | of (29.3) [T1 = 1 so ω1= 2π, and ω in (-20,20)], As expected, the pattern of peaks goes on forever in each direction. At ω = 0 the real peak is at 19 since this is the sum of the yn values in the figure. The corresponding Y(ω) of (29.5) then looks like this, which is localized roughly to the range (-ω1,ω1) = (-2π,2π). There are no image spectra in Y(ω). On the other hand, Y'(ω) does have image spectra as the previous figure shows. Now in this example, how does the equation Y'(ω) = [ Y(ω) + Σm≠0Y(ω- mω1) ] work out? Using (29.5) we write Y'(ω) = !Syntax Error, IY(ω- mω1) = !Syntax Error, Isinc[π([ω-mω1]/ω1)] Y'(ω-mω1) = !Syntax Error, Isinc[π([ω-mω1]/ω1)] Y'(ω) // Y'(ω-mω1) = Y'(ω) from (22.3) = Y'(ω) !Syntax Error, I sinc[π([ω-mω1]/ω1)] = Y'(ω) * 1 = Y'(ω) where we used (29.7) that the sinc sum is 1. Therefore, if we implement Y(ω) + Σm≠0Y(ω- mω1) using the second figure for Y(ω), we will obtain exactly the first figure for Y'(ω). where, once again, xpulse(t) is some arbitrary pulse shape. For any physical pulse shape xpulse(t), the spectrum Xpulse(ω) will be localized in some region around ω = 0, and the fact that Y'(ω) has an infinite set of image spectra does not cause W(ω) to have image an infinite set of image spectra. It might be that W(ω) has some finite number of image spectra if Xpulse(ω) is very broad. However, in the idealized limit that xpulse(t) = δ(t), in which case Xpulse(ω) = 1, we find that W(ω) = (1/T1) Y'ω) and in this limiting case (only), W(ω) inherits the full infinite set of image spectra of Y'(ω). This means that, when considered in the frequency domain, our pulse train w(t) = !Syntax Error, I yn xpulse(t - tn) must have infinite energy. Question: Suppose we have an infinite pulse train with all yn = 1 and xpulse= box width T1. Then we have w(t) = K = constant at all time. You would think that the energy in such a pulse (treat at a voltage) would be infinite, sort of K2 * infinite time. Yet is seems that |W(ω)|2 is integrable! Resolution: In this case we have, from above, Y(ω) = (T1) sinc(ωT1/2) Σn=-∞∞ e-inωT = (T1) sinc(ωT1/2) !Syntax Error, I2πδ(ωT1 - 2πm) -∞ < ω < ∞ Although this is tamed a bit by the sinc function, it still contains δ functions and they contain infinite energy, so W(ω) = (1/T1)Xpulse(ω) Y'ω) still has infinite energy due to these δ spikes. Now if we consider only finite pulse trains, then Y(ω) = (T1) sinc(ωT1/2) Σn yn e-inωT has no delta spikes, and the sinc function really does tame it to a certain bandwidth. In this case, we expect finite energy, and that then agrees with a finite set of pulses.