Lecture Notes 11
Fall 2015
SAMPLING THEOREM and ALIASING
The Poisson Summation Formula
We start by recalling definitions about Fourier series and transforms: ˆ F (~k) ≡ Z dαx (2π)α e −i~k·~xF (~x) F (~x) ≡ Z dαk ei~k·~xF (~k)ˆ transforms in α-dimensions
For series, we have (α = 1 as example) f (x) = ∞ X h=−∞ ei2πLxf (n)ˆ ˆ f (n) = L1 Z L/2 −L/2 f (x) e−i2πnL xdx − L 2 6 x 6 L 2
where f is periodic with period L. The Fourier series goes to the Fourier integral as L → ∞, i.e. f (x) = ∞ X n=−1 eiknxf (n)ˆ kn=2πLn↑ −→ Z dn ˆf (n) eiknx = Z dkn L 2π 1 L Z L/2 −L/2 f (x0) e−iknx0dx0eiknx
−−−→ L→∞ Z dk 1 2π Z ∞ −∞ f (x0) eikx0dx0 | {z } f (k) eikx
Now, let’s derive the Poisson summation formula. For an arbitrary function f (x) we define the following function F (x):
F (x) =
∞
X
n=−∞
f (x + m)
By definition F (x) is periodic with period L = 1, then it can be expanded in Fourier series as follows
F (x) =
∞
X
m=−∞
ei 2πm xF (m)ˆ
where the Fourier coefficients are ˆ F (m) = Z 1/2 −1/2 F (x) e−i 2πm xdx Then: F (x) = ∞ X m=−∞ ei 2πm x Z 1/2 −1/2 F (x) e−i 2πm xdx = ∞ X m=−∞ ei 2πm x Z 1/2 −1/2 ∞ X n=−∞ f (x + n) e−i 2πm xdx
since e−i2πmx = e−i2πm(x+n), and changing variable to y ≡ x + n we have:
F (x) = ∞ X m=−∞ ei2πmx ∞ X n=−∞ Z 1/2+m −1/2+m f (y) e−i2πmydy | {z } =R∞
−∞f (y) e−i2πmydy=2π ˆf (2πm)
⇒ F (x) = ∞ X n=−∞ f (x + n) = 2π ∞ X m=−∞ ei2πmxf (2πm)ˆ
Evaluating this @ x = 0 yields the Poisson summation formula, ∞ X n=−∞ f (n) = ∞ X m=−∞ Z ∞ −∞ dx f (x) ei2πmx = 2π ∞ X m=−∞ ˆ f (2πm)
First, let’s understand this in simple terms. On the LHS, we have sum of f (m), in real space, over all equally-spaced points:
Since we are summing over all points, a wave longer than λ = 1 will average out, so only k > k1 = 2π/1 can contribute to such sum, as given by the RHS
of the formula, so this makes sense.
In fact, when f (x) varies smoothly and slowly over a large range in x (large compared to unity), the m = 0 gives the dominant term and corresponds to giving the sum in LHS as an integral. All other m’s will give small corrections. For the particular case of a “Gaussian”, f (x) = λ1/4e−πλx2, Poisson summa-tion formula gives
z(λ) = λ1/4 ∞ X n=−∞ e−πλn2 = 1 λ1/4 ∞ X m=−∞ e−πm2λ = z(1/λ) !
So, in this case there is a symmetry λ → 1/λ (duality). Thus, if one knows the small-λ result of the sum, one also knows the large-λ result through this transformation. Note that when λ is small it is easiest to evaluate RHS, keeping a few terms in sum close to m = 0 is enough. On the other hand if λ is large, LHS is easiest, keep a few terms about n = 0. This is a general feature of Poisson summation formula:
a) if f (n) varies slowly, sum in real space takes a while to converge, in Fourier space power is concentrated close to m = 0
b) if f (n) has a lot of power from different waves, that is, ˆf (2πm) varies slowly, then sum in Fourier space does not converge very well, however, in real space all contribution is concentrated near n = 0.
We will have a lot more to say about this when we consider phase transition: basically ideas like this connect low and high-temperature expansions.
Sampling Theorem
For simplicity we will consider a 1D random field Φ(x), extension to more dimensions is straightforward. Typically, one knows Φ(x) only at some points and often one can choose the points, then it becomes an interesting question whether one can recover Φ(x) at any other point by sampling Φ(x) only at a finite number of points.
The simplest case is uniform sampling, where one knows Φ(x) at equally spaced points1, separated by some distance L. In this case the discrete
version Φd(x) of Φ(x) can be written as:
Φd(x) = ∞ X n=−∞ Φ(nL) δD(x − nL) = ∞ X n=−∞ Φ(x) δD(x − nL) ⇒ Φd(x) = Φ(x) ∞ X n=−∞ δD(x − nL)
The Fourier transform of Φd(x) can be written as
ˆ Φd(k) = 1 2π Z dx e−ikxΦd(x) = Φ(nL) ∞ X n=−∞ 1 2πe −iknL
or, more interestingly, using the last expression for Φd(x) and the convolution
theorem, ˆ Φd(k) = ˆΦ(k) ⊗ FT " ∞ X n=−∞ δD(x − nL) # (k) ⊗ : convolution Now, FT [δD(x − nL)](k)= 1 2π Z dx e−ikxδD(x − nL) = 1 2πe −iknL ↓Poisson summ. ⇒ ∞ X n=−∞ 1 2πe −iknL = 2π ∞ X m=−∞ ˆ f (2πm) = ∞ X m=−∞ δD(kL − 2πm) ↑f (x)≡1 2πe −ikLx⇒ ˆf (q)= 1 2πδD(kL−q) = 1 L ∞ X m=−∞ δD k − 2π L m
1Note that L here is the grid spacing, not to be confused with the definition in the
Then: ˆ Φd(k) = ˆΦ(k) ⊗ 1 L ∞ X m=−∞ δD k −2π L m ⇒ ˆΦd(k) = 1 L ∞ X m=−∞ ˆ Φ k − 2π L m
This says that by sampling Φ(x) at intervals L is equivalent to making its Fourier transform 2πL periodic by summing over all its translation ˆΦ k −2πLm. Now, consider the sampling theorem: if ˆΦ(k) vanishes for frequencies higher than the Nyquist frequency kNy = π/L, ˆΦ(k) = 0 for |k| > π/L, then one
can recover Φ(x) everywhere by convolving Φd(x) with a top-hat filter ( or
low-pass filter) WTH ∝ 1 |k| 6 kNy (WTH(k) = 0 otherwise). The idea is
very simple, consider a field that has no frequencies larger than Nyquist (for some L)
Then, by formula above, when sampled uniformly at intervals L, we have:
filter WTH(k):
WTH(k) =
(
L |k| 6 kNy = π/L
0 otherwise which has a real space representation,
WTH(x) = Z ∞ −∞ dk WTH(k) eikx = L Z π/L −π/L dk eikx = L eikx ix π/L −π/L= 2L x sin πx L ⇒ WTH(x) = 2π sin(πx/L) πx/L = 2πj0(πx/L) This looks like, in real and Fourier space:
In real space this is a convolution, then
Φ(x) = ∞ X n=−∞ Φ(nL) j0 hπ L(x − nL) i (Sampling theorem) In our convention, c(x) = Z dk eikxA(k)B(k0) =Z dx0 2πA(x 0)B(x − x0) ≡ (A ⊗ B) (x)
That is, if Φ(x) presents no high-frequency variations, one can use a finite sampling and then interpolate to reproduce Φ(x) at all points.
The lesson from the sampling theorem is then as follows. If we are given a function or random field Φ(x) that is bandwidth limited, that is its Fourier transform vanishes for k > kmax, we can recover the full function
from the sampled function as long as we sample it at a frequency ks= 2kmax
with m = ±1 contributions so it remains unaffected and we want to get rid of m 6= 0 contributions to ˆΦd(k) so we can recover the true ˆΦ(k). Looking
at the interplay between m = 0 and m = 1 terms, to achieve the former we must pick the sampling L just so that kmax = 2πL − kmax so m = 0, 1
just touch each other (at zero amplitude), which sets the sampling frequency ks = 2π/L = 2kmax. To satisfy the latter, we must multiply by a low-pass
filter that eliminates the images with m 6= 0. Intuitively, the condition on the sampling frequency makes sense: given the maximum frequency wave present in our signal, we want to sample it twice per period to catch at least a peak and a trough.
Aliasing
Now, consider what happens when one does not sample the signal well enough, i.e. when ˆΦ(k) does not vanish from frequencies higher than the Nyquist of our sampling (in some cases there is power all the way up to k → ∞, so no sampling will give us perfect reconstruction). We are in the following situation:
Then, we see that the power beyond the Nyquist of our sampling has been “aliased” (transfered) back to the frequency range −π/L 6 k 6 π/L.
It is said that high frequency is aliased to low frequency. Thus high-frequency components (that are not well sampled because of our low Nyquist frequency) get spuriously aliased to low frequencies. Obviously if we use the top-hat filter trick it will not work, because we recover the wrong spectrum for ˆΦ(k):
If one has access to the full signal before sampling it (like when converting an analog signal to digital), the best way to remove aliasing is to first filter it with a top-hat for the Nyquist range, and then sampling it. This will give us back (by the interpolation method) all the correct Fourier coefficients in the range we can access. That’s why analog to digital converters first low-pass filter the signal before sampling. And how antialiasing fonts work in computer displays.
How can a high frequency wave give a low frequency signal under poor sam-pling? There is a simple example. Suppose we have a low frequency wave φ1 = eik1x and a high frequency one φ2 = eik2x with k2 > k1. Depending on
the relationship of k2 to k1 one can have a situation where both waves give
exactly the same sampling waves φ1d = φ2d! This will happen when
φ1d(nL) = eik1nL = φ2d(nL) = eik2nL n = 0, 1, · · · ⇒ ei(k2−k1)nL = 1 n = 0, 1, · · · ⇒ (k2− k1)L = 2πm m = 0, 1, · · · (k2 > k1) in other words: k2− k1 = 2π L m
Thus, if the high frequency wave differs from the low frequency one by a multiple of the frequency width given by twice the Nyquist (i.e. from −π/L to π/L), they will give the same sampled waves. Another way of saying this is that the high frequency k2 has been “aliased back” into the low frequency
mode k1 = k2− (2π/L) m.
Another example of aliasing is the well-known “wagon-wheel” effect, where the wagon wheel seems to be moving in opposite direction from the wagon itself (this happens in old movies where sampling was not enough to satisfy
the sampling theorem). To see how this could happen, consider the similar case of a clock. If you were to only look at the clock every 50 minutes then the minute hand would appear to rotate anticlockwise (10 min back each time you look). The hour hand, however, would still rotate in the correct direction as you have satisfied the sampling theorem for it.
