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High-resolution Interferometric Diagnostics for Ultrashort Pulses

High-resolution Interferometric Diagnostics for Ultrashort Pulses

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3.4 Sampling and uniqueness of solutionssimply in terms of the unknown phase at the sampled points, one requires that <strong>for</strong> every combinationof n and k , ω n + Ω k must be a sampled frequency ω m <strong>for</strong> some m — that is, the frequenciesconnected by every equation must be sampled frequencies. Otherwise, expressing φ(ω n + Ω k ) in(3.6) would require interpolation, complicating the expressions. The shears must there<strong>for</strong>e be aninteger multiple of the sampling rate, so that one may write Ω k = C k Ω. Imposing this restrictionon the shears permits the use of simple linear difference equations, of the <strong>for</strong>mφ n+Ck − φ n = Γ k ,n (3.7)<strong>for</strong> all n and k where φ n = φ(ω n ) and Γ k ,n = Γ k (ω n ). One also imposes φ 0 = 0 as a boundarycondition since the absolute phase is not defined.I now consider the uniqueness of the solution to (3.7). Although not used in practice, a simplesolution method is to start at the boundary condition ω 0 , and to use superposition to “visit” othersampling points. One may determine which of the other sampling points can be visited by applyingseries of jumps, <strong>for</strong>wards or backwards, of the different shear sizes. If one applies a 1 jumps ofshear Ω 1 , a 2 jumps of shear Ω 2 , and so on (the {a k } may be positive or negative) then one will havemoved a total displacement ofω − ω 0 = a 1 Ω 1 + a 2 Ω 2 + ... (3.8)There<strong>for</strong>e, one can reach frequency ω n if and only if there exist integer solutions a 1 ,a 2 ,... toa 1 C 1 + a 2 C 2 + ...= n. (3.9)By Bézout’s Identity [287], this occurs if and only if n is an integer multiple of the greatest commondivisor of {C k }. If this condition is not satisfied, then there is no way that any linear combination ofthe difference equations can refer ω n to ω 0 , and hence φ n is not determined by the experimentaldata. There<strong>for</strong>e, a unique solution <strong>for</strong> all n requires that the {C k } be relatively prime. Examplesof the two cases are illustrated in Fig. 3.2. I proceed assuming the shears are relatively prime, inwhich case one has N unknowns and approximately NM equations, where M is the number of71

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