PoissonSum_13
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Handout for Physics 214, Winter 2013, giving a proof of the Poisson sum formula as a sum of exponentials equal to a sum of delta functions at multiples of 2π. It expands a sawtooth function in a Fourier series and differentiates it, using a step function to account for the jumps. It then states the common form relating the sum of f(αm) to a sum of its Fourier transform, and cites Lighthill's book on Fourier analysis and generalized functions. It is tied to a Jackson electrodynamics problem.
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Physics 214 Winter 2013
The Poisson sum formula
The Poisson sumformula takes ona number ofdifferent formsinthelit erature.
Here is one useful version,
1
2π∞/summationdisplay
n=−∞einx=∞/summationdisplay
m=−∞δ(x−2πm). (1)
You will use this version of the Poisson sum formula in solving problem 14 .13 of
Jackson.
To prove this formula, consider the following periodic function, defin ed by:
f(x) =1
2−x
2π,0≤x≤2π, (2)
wheref(x+ 2π) =f(x). Note that f(x) is discontinuous at x= 2πm, where
m= 0,±1,±2,.... It follows that one can expand f(x) in a Fourier series:
f(x) =∞/summationdisplay
n=−∞cneinx,
where
cn=1
2π/integraldisplay2π
0e−inxf(x)dx.
Inserting f(x) as given by eq. (2), one easily obtains:
c0= 0, c n=−i
2πn,(n/negationslash= 0).
That is,
f(x) =−i
2π∞/summationdisplay
n=−∞
n/negationslash=0einx
n. (3)
Consider the derivative of f(x), which we denote by f′(x). From its definition
[eq. (1)], f′(x) =−1/(2π) forx/negationslash= 2πm(m= 0,±1,±2,...). Atx= 2πm, the
discontinuity of f(x) can be described by the step function Θ( x). Specifically, in
the vicinity of x= 2πm,
f(x) =−1
2+Θ(x−2πm),forx≃2πm. (4)
That is, f(x) =−1/2 forx= 2πm−ǫandf(x) = 1/2 forx= 2πm+ǫ, where
ǫ >0 is an infinitesimal quantity. Taking the derivative of eq. (4) yields:
f′(x) =δ(x−2πm),forx≃2πm.
We conclude that:
f′(x) =−1
2π+∞/summationdisplay
m=−∞δ(x−2πm). (5)
We can also compute f′(x) by differentiating the Fourier series of f(x) term-by-
term. Using eq. (3), we obtain:
f′(x) =1
2π∞/summationdisplay
n=−∞
n/negationslash=0einx=1
2π/bracketleftBigg
−1+∞/summationdisplay
n=−∞einx/bracketrightBigg
. (6)
Equating eqs. (5) and (6) yields the desired result announced in eq. (1).
Actually, the most common form for the Poisson sum formula is as follo ws.
Given a function f(t) and its Fourier transform,
F(ω)≡/integraldisplay∞
−∞eiωtf(t)dt, (7)
then the Poisson sum formula is given by:
∞/summationdisplay
m=−∞f(αm) =1
α∞/summationdisplay
n=−∞F/parenleftbigg2πn
α/parenrightbigg
, (8)
whereαis any number. One can derive eq. (8) by inserting the integral expr ession
forF[eq. (7)] on the right-hand side of eq. (8), and then performing th e sum over
nusing eq. (1). The resulting integrals are then trivially performed, a nd the left
hand side of eq. (8) is immediately obtained.
For further details, see for example M.J. Lighthill, Introduction to Fourier
Analysis and Generalized Functions , pp. 67–71.