Dan, you're right on track to the answer. Let s = exp(iw), so that the transfer function is 1/(1 - exp(iw)) = 1/(1 - s). It's easy to see that f(w) = (i/2)cot(w/2) + 1/2, f(w) exp(iw) = exp(iw)/(1 - exp(iw)) = (i/2)cot(w/2) - 1/2. In any engineering application, we can replace the step function response of 1 in the n-th term by exp(-ne), with e > 0, and at the end of the calculation, let e go to zero. This is the same as letting s approach the unit circle from inside. When w is not a multiple of 2 pi, the limit is just the function above. But we must be careful in defining f(w) at w = 2 n pi. This is important because we may need to use f(w) in further calculations. By the residue theorem, (1/(2 pi i)) int(ds/(1 - s)) = 0, the path being once CCW keeping inside the unit circle. We must define f(w) so that this remains true when e --> 0, i.e. on the unit circle. Change the integration variable to w and integrate over a 2 pi period. 0 = (1/(2 pi i)) int(i exp(iw) dw f(w)) = (i/(4 pi)) int(cot(w/2) dw) - 1/2. The integral goes through the pole of cot. We must choose a specific method to define this integral. It is customary to take the principal value, i.e. exclude the interval (-eps, +eps), do the integral, and let eps --> 0. But then the cot integral is zero. So f(w) must have in addition a compensating Dirac-delta function of the form pi delta(w - 2 n pi) (summed over all integers n). So the transfer function of the step response filter is F(w) = P[1/(1 - exp(iw))] + pi sum(delta(w - 2 n pi), in which P denotes that the principal value is to be used Now one can use the correspondence in both directions: X(w) = sum(x[n] exp(i n w)), x[n] = (1/(2 pi)) int(X(w) exp(-i n w) dw). Gene --- <dasimov@earthlink.net> wrote:
Fred wrote:
<< . . . The "delta" is not a Kronecker delta, but an infinite spike whose integral is nonzero (here pi): this is an example of a "distribution", a generalisation of the classical notion of "function" which permits such limiting cases to be discussed compactly. . . .
Thanks.
(A slip of the fingers; I meant Dirac delta. Yes, indeed, a linear functional on the vector space of functions, given by delta(f) = f(0). I'm only half as confused as I seem to be.)
I'm almost sure you're right that the author is "by convention" setting 1/(1-exp(iw)) to 0 where w = 2k*pi, and then adding explicitly the "singular part", a periodic delta function. ---------------------------------------------------------------------
I'm thinking, f(w) = 1/(1-exp(iw)) = -2i exp(-iw/2) / sin(w).
So maybe somehow integral_a^b(f(w)) = (1/2)*(2 pi*i) * (Res(f(w)) at w=0) can be used to express the "singular part" of f(w) ???
--Dan
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