new section 33

New version of Section 33 (a) PhL 3.26.05 33. Energy and Power for a Pulse Train; using the Autocorrelation Function 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. (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 and changing w,W to x,X: x(t) = !Syntax Error, I yn xpulse(t -tn) = !Syntax Error, I xpulse(t -tn) (33.1) X(ω) = (1/T1)Xpulse(ω) X'ω) = Xpulse(ω) X"(z) (33.2) X"(z) = X'ω)/T1 = !Syntax Error, Iyn e-iωnT = !Syntax Error, I e-iωnT . (33.3) X'(ω) is the Digital Fourier Transform of x(t), and X"(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 X"(z) = X'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) X'ω) = Xpulse(ω) !Syntax Error, I2πδ(ωT1 - 2πm) (33.5) in agreement with (14.4) quoted just above (33.1). To find the frequency domain 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 |X'ω)|2 = |Xpulse(ω)|2 | X"(z) |2 z = eiωT (33.6) We therefore must deal with the following object, using (33.4), | X"(z) |2 = (1/T1)2 |X'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. [ !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 | X"(z) |2 = (1/T1)2 |X'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) X'ω) = Xpulse(ω) X"(z) (33.13) X"(z) = X'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 X"(z) = X'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, 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 limN→∞ δ6(ωT1,N) = !Syntax Error, Iδ(ωT1-2πm) (A.21) to find that limN→∞ [] = | Xpulse (ω) |2 !Syntax Error, Iδ(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 X"(z) = X'ω)/T1 = !Syntax Error, I e-iωnT = !Syntax Error, I2πδ(ωT1 - 2πm) infinite (33.4) X"(z) = X'ω)/T1 = !Syntax Error, I e-iωnT = 2π δ5(ωT1,N) finite (33.15) X(ω) = Xpulse(ω) !Syntax Error, I2πδ(ωT1 - 2πm) infinite (33.5) X(ω) = Xpulse(ω) 2π δ5(ωT1,N) . finite (33.16) = |Xpulse(ω)|2 !Syntax Error, I 2πδ(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. (c) Spectral Energy Density and Spectral Power Density in a Simple Pulse Train In this section, everything is in the frequency domain, nothing is in the time domain. Recall from (32.6) that E(ω) ≡ |X(ω)|2/2π = energy density in the ω-domain (joule-sec) (32.6) E(ω)dω = energy in dω (joules) where E(ω) is the spectral energy density of signal x(t) whose Fourier Transform is X(ω). If we divide E(ω) by 2N+1 or 2πδ(0) we obtain the pulse train's average spectral energy density per pulse, which is the same as the energy density of an average pulse in the pulse train. Using our two expressions above for infinite and finite pulse trains, we then find = = |Xpulse(ω)|2 !Syntax Error, I 2πδ(ωT1 - 2πm) = = | Xpulse (ω) |2 2π δ6(ωT1,N) (33.20) If we divide the energy of the average pulse by T1, we obtain power of an average pulse, and this is the same as the average power of the pulse train. Thus, P(ω) ≡ = = |Xpulse(ω)|2 (1/T1)!Syntax Error, I 2πδ(ωT1 - 2πm) P(ω) ≡ = = |Xpulse(ω)|2 (1/T1) 2π δ6(ωT1,N) (33.21) If is perhaps helpful to define T ≡ (33.22) Then the above maybe be restated P(ω) ≡ = = |Xpulse(ω)|2 (1/T1)!Syntax Error, I 2πδ(ωT1 - 2πm) P(ω) ≡ = = |Xpulse(ω)|2 (1/T1) 2π δ6(ωT1,N) (33.23) Meanwhile, our xpulse(t) which lasts only for duration T1 itself has an energy and a power Ppulse(ω) ≡ = (33.24) We can then write (***) in the following compact form P(ω) ≡ Ppulse(ω)!Syntax Error, I 2πδ(ωT1 - 2πm) joules P(ω) ≡ Ppulse(ω) 2π δ6(ωT1,N) joules (33.25) Notice that δ(ωT1 - 2πm) and δ6(ωT1,N) are both dimensionless, so the dimensions in each equation trivially match. A power density P(ω) has dimensions of energy = joules, so that P(ω)dω then has the dimensions of joules/sec = watts, and this is the pulse train power contained in interval dω of the spectrum. For the infinite pulse train, we are always allowed to write 2π δ(ωT1 - 2πm) = (2π/T1) δ(ω - m(2π/T1)) = ω1 δ(ω - mω1) ω1 ≡ 2π/T1 to get P(ω) ≡ Ppulse(ω) ω1!Syntax Error, I δ(ω - mω1) infinite (33.26) which shows more explicitly that the power lines occur at the harmonics ω = mω1 . Recall now these two earlier facts c(ω) ≡(1/T1)Xpulse(ω) (14.14) cm ≡ c(mω1) = (1/T1)Xpulse(mω1) (14.10) For the infinite pulse train we can then write P(ω) ≡ Ppulse(ω) ω1!Syntax Error, I δ(ω - mω1) = !Syntax Error, I δ(ω - mω1) = !Syntax Error, I δ(ω - mω1) = |c(ω)|2 !Syntax Error, I δ(ω - mω1) = !Syntax Error, I |c(mω1)|2 δ(ω - mω1) so that P(ω) = !Syntax Error, I |cm|2 δ(ω - mω1). (33.27) Recall from box (15.12) that Dim(cm) = Dim[x(t)] = volts (say), so |cm|2 = watts into a 1Ω resistor, and since δ(ω - mω1) has dimensions sec, |cm|2 δ(ω - mω1) then has dimensions watt-sec = joules, as befits any P(ω) object. If we assume x(t) is a real pulse train, then X(ω) is Hermitian X(-ω) = [ X(ω)]* by (7.1), which means c-m= (1/T1)Xpulse(-mω) = (1/T1)[Xpulse(mω)]* = cm* so |c-m|2 = |cm|2 x(t) real (33.28) and then we can fold the negative part of the sum in (**) to get P(ω) = !Syntax Error, I |cm|2 δ(ω - mω1) = |c0|2 δ(ω) + 2!Syntax Error, I|cm|2 δ(ω - mω1) (33.29) and then finally, recalling from our Fourier Series box (15.2) that cm = [ am - ibm ]/2 and b0= 0 we get P(ω) = a02 δ(ω) + (1/2) !Syntax Error, I(am2+bm2) δ(ω - mω1) (33.30) (d) Average Power P for a Simple Pulse Train We seek an expression for the average power P in a general pulse train. This is of course a time domain quantity, not a frequency domain quantity. P = [ total energy in pulse train / time duration of pulse train ] = average pulse train power = (1/T)!Syntax Error, Idt |x(t)|2 = !Syntax Error, Idω // from (32.4) with R = 1Ω so P = !Syntax Error, Idω P(ω) . // from (33.23) (33.31) This is certainly reasonable since P(ω) is the average spectral power density of the pulse train. For the special case of a simple pulse train, we found above that P(ω) = !Syntax Error, I |cm|2 δ(ω - mω1) = a02 δ(ω) + (1/2) !Syntax Error, I(am2+bm2) δ(ω - mω1) Therefore the power in a simple pulse train is given by P = !Syntax Error, I |cm|2 = a02 + (1/2) !Syntax Error, I(am2+bm2) (33.32) (e) General Pulse Train results and connection with the Autocorrelation Function Certain results of the previous section apply to general pulse trains and we gather them here: E(ω) ≡ |X(ω)|2/2π = energy density in the ω-domain (joule-sec) (32.6) T ≡ (33.22) P(ω) ≡ = (33.23) Ppulse(ω) ≡ = (33.24) P = !Syntax Error, Idω P(ω) . (33.31) We now bring the autocorrelation function into the discussion. Let x(t) be an arbitrary but real pulse train, and recall that rx(t) ≡ !Syntax Error, I dt' x(t') x(t' + t) (32.1) Then, using T from (33.22), rx(0) ≡ !Syntax Error, I dt' x(t')2 = P T = E = total energy in the pulse train (33.33) so the average pulse train power maybe written in terms of the autocorrelation function evaluated at t = 0, P = rx(0)/T . (33.34) We diagonalized (32.1) treated as a convolution equation to obtain |X(ω)|2 = Rx(ω) which we called the Wiener-Khintchine Relation. Here Rx(ω) is the Fourier Integral Transform of the autocorrelation function rx(t) of x(t). Then from (33.23) that P(ω) = we get P(ω) = Rx(ω)/ (2πT) (33.35) In this way, both P and P(ω) can be expressed in terms of the autocorrelation function. Thus, one approach to finding P and P(ω) for a pulse train is to try and determine rx(t). Here then is a box summarizing all the general pulse train results: Energy and Power Properties of a General Pulse Train (33.36) E(ω) ≡ |X(ω)|2/2π = energy density in the ω-domain (joule-sec) (32.6) T ≡ (33.22) P(ω) ≡ = = Rx(ω)/ (2πT) joules (33.23) and (33.35) Ppulse(ω) ≡ = (33.24) P = !Syntax Error, Idω P(ω) = rx(0)/T watts (33.31) and (33.34) If P(f)df = P(ω)dω = P(ω) 2πdf , then P(f) = 2πP(ω).