group delay

Pondering Group Delay PhL 3.4.13 Here I was trying to figure this out. The web just gives the formula for group delay but does not say why it has that name, nor what group is being delayed, etc. I had to roll my own as usual. It now appears as section 21(b) in spectral doc. U(ω) = G(ω) F(ω) u(t) = (1/2π) !Syntax Error, Idω G(ω) F(ω) e+iωt Now suppose F(ω) is for some pulse, U(ω) = G(ω) Xpulse(ω) u(t) = (1/2π) !Syntax Error, Idω G(ω) Xpulse(ω) e+iωt Suppose G(ω) = |G(ω)| e-φ(ω) u(t) = (1/2π) !Syntax Error, Idω |G(ω)| e-iφ(ω) Xpulse(ω) e+iωt Plan A. Let's try f(t) = δ(t) so that Xpulse(ω) = 1 from (8.2). Then u(t) = (1/2π) !Syntax Error, Idω |G(ω)| e-iφ(ω) e+iωt Let's assume that most of the contribution comes from the region near ω = 0. Does this assumption have a name. That name is not "stationary phase" because that is the limit of an integral for some large x. I don't see how dφ/dω is going to "come in". It must be a parts integration. Plan B. Web says it applies to a "narrow band" signal, so my δ is the wrong thing to work with which has a very wide band spectrum. So let's assume that Xpulse(ω) has a narrow band around some ω2. In the extreme case we could try Xpulse(ω) = δ(ω-ω2) and then above we get u(t) = (1/2π) !Syntax Error, Idω |G(ω)| e-iφ(ω) Xpulse(ω) e+iωt = (1/2π) !Syntax Error, Idω |G(ω)| e-iφ(ω) δ(ω-ω2) e+iωt = (1/2π) |G(ω2)| e-iφ(ω2) e+iω2t This is no good because Xpulse(t) = 1 and you have no spatial localization. I have no idea right now where that derivative thing is coming from! Hard to get info on this, so I should incorporate it when I learn how it works. I know I have done this before. Plan C. Suppose we make the thing less narrow, so that it has a spectrum from ω2-a to ω2+ a. Then we get roughly u(t) = (1/2π) |G(ω2)| !Syntax Error, Idω e-iφ(ω) Xpulse(ω) e+iωt Plan D. Suppose xpulse(t) has max at t = 0. Where does u(t) have its max u(t) = (1/2π) !Syntax Error, Idω |G(ω)| e-iφ(ω) Xpulse(ω) e+iωt du/dt = (1/2π) !Syntax Error, Idω |G(ω)| e-iφ(ω) Xpulse(ω) dte+iωt = (1/2π) !Syntax Error, Idω |G(ω)| e-iφ(ω) Xpulse(ω) iω e+iωt = 0 That does nothing for me. Every website just quotes the definition of group delay and gives no derivation or explanation! Plan E. Wiki suggests using this function xpulse(t) = a(t)cos(ω2t+θ) Again they quote a result but don't derive it/ I will try this instead xpulse(t) = a(t)exp(iω2t) What is the spectrum of such a pulse" X(ω) = !Syntax Error, Idt a(t)exp(iω2t) e-iωt ≈ a(0) !Syntax Error, Idt xp(iω2t) e-iωt ≈ a(0) δ(ω-ω2) Put this in to get u(t) = (1/2π) !Syntax Error, Idω |G(ω)| e-iφ(ω) Xpulse(ω) e+iωt = (1/2π) !Syntax Error, Idω |G(ω)| e-iφ(ω) a(0) δ(ω-ω2) e+iωt ≈ (1/2π) |G(ω2)| e-iφ(ω2) a(0) e+iω2t Nothing is happening for me here! This question is hard to get answered! Plan F. Let's try it for the simplest possible case treated in Lam page 9. He uses G(ω) = e-ikω/ωc in range (-ωc, ωc) This is a narrow passband filter, it is G = 0 elsewhere! Then I would get u(t) = (1/2π) !Syntax Error, Idω e-ikω/ωc Xpulse(ω) e+iωt We will use x(t) = δ(t) with Xpulse= 1 and talk "impulse response". Then we have u(t) = (1/2π) !Syntax Error, Idω e-ikω/ωc e+iωt = (1/2π) !Syntax Error, Idω e-i(k/ωc-t)ω = (1/2π) !Syntax Error, Idω cos [(k/ωc-t)ω] = (1/π) !Syntax Error, Idω cos [(k/ωc-t)ω] Maple gives this integral as = ωc [ sin(ωct)cos(k) - cos(tωc)sin(k)]/(ωct - k) = ωc [ sin(ωct - k)] /(ωct - k) = ωc sinc(ωct - k) The original δ(t) had peak at t = 0, this thing has peak at t = k/ωc . This result agrees more or less with Pam page 10. We see a time delay here of Δt = k/ωc and this is then the "group delay". Meanwhile, the filter was e-ikω/ωc = e-iφ(ω) so it had φ(ω) = kω/ωc and dφ/dω = k/ωc and in this case things work our right. So there is your "n=2" case. Now make it a little more complicated. replace e-ikω/ωc by e-iφ(ω). Then we get u(t) = (1/2π) !Syntax Error, Idω e-iφ(ω) e+iωt Now maybe claim that ωc is small, so ω is always near ωc in this integral. We then can do (1/2π) !Syntax Error, Idω e-i[φ(0)+φ'(0)ω] e+iωt = e-iφ(0) !Syntax Error, Idω e-iφ'(0)ω e+iωt where now φ'(0) plays the role of k/ωc before. We will then end up with Td = φ'(0) and this then shows that the group delay is approximately Td = φ'(0) . The narrower the frequency band and the smoother φ(ω) is, the more this is true. Plan G. Let's now try a slight generalization. Take G(ω) = |G(ω)|e-iφ(ω) in a narrow passband (ω2-a, ω2+a), G = 0 elsewhere. x(t) = δ(t) as before Then we get u(t) = (1/2π) !Syntax Error, Idω |G(ω)| e-iφ(ω) Xpulse(ω) e+iωt In the narrow band we approximate with first two terms of a Taylor series φ(ω) ≈ φ(ω2) + φ'(ω2)ω Then get u(t) ≈ (1/2π) |G(ω2)| Xpulse(ω2) !Syntax Error, Idω e-iφ(ω2) e-iφ'(ω2)ω e+iωt = (1/2π) |G(ω2)| Xpulse(ω2) e-iφ(ω2) !Syntax Error, I dω e+iω[t-φ'(ω2)] Now set φ'(ω2) = b. We then have this integral !Syntax Error, I dω e+iω[t-b] = 2!Syntax Error, I dω cos[(t-b)ω] = 2 (t-b)-1 sin[(t-b)(ω2-a)] = 2 (ω2-a) sin[ (ω2-a) (t-b)] / [ (ω2-a) (t-b)] = 2 (ω2-a) sinc[ (ω2-a) (t-b)] The sinc function has its peak at t = b, so the group delay is b = φ'(ω2) , there you are! In total we get u(t) ≈ (1/π) (ω2-a) |G(ω2)| Xpulse(ω2) e-iφ(ω2) sinc[ (ω2-a) (t-b)] OK, I have now written this up as part (b) of Section 21. It is OK.