MATH 262/CME 372: Applied Fourier Analysis and Winter 2021 Elements of Modern Signal Processing Lecture 6 — January 28, 2021 Prof. Emmanuel Candes Scribe: Carlos A. Sing-Long, Edited by E. Bates
1 Outline
Agenda: Fourier series
1. Aliasing example 2. Discrete convolutions 3. Fourier series 4. Another look at Shannon’s sampling theorem
Last Time: We introduced the Dirac comb and proved the Poisson summation formula, which states that the Fourier transform of a Dirac comb of period T is also a Dirac comb, but with period 2π/T (modulo a mutiplicative factor 2π/T ). We discussed the problem of sampling an analog signal f(t) by measuring its values at time intervals of length T , and we represented the discretized signal fd as a superposition of Dirac delta functions. The Poisson summation formula was instrumental in proving the aliasing formula, which states that the Fourier transform fˆd of the discretized signal is the 2π/T -periodization of the spectrum fˆ of the analog signal. From this calculation, we saw that when f is bandlimited and T is sufficiently large, no aliasing occurs in the periodization. When this is the case, we can recover the original signal from its samples via Shannon’s sampling theorem.
Today: We will first revisit the topic of aliasing by observing it for an actual signal. Then, in the pursuit of further studying discrete signals, we shall introduce Fourier series. Understanding these discrete sums as a special case of the Fourier integrals we know well, we will survey the properties Fourier series have in analogy with Fourier transforms. In particular, the Poisson summation formula will be employed once more to show that complex exponentials again form an orthonormal basis of L2-space. The machinery of this basis will actually provide a second proof of Shannon interpolation.
2 A case of aliasing
To make concrete our study of aliasing, let us consider the very simple signal 3π 1 h i 3π t i −3π ti f(t) = cos t = e 2 + e 2 . 2 2
1 Its Fourier transform is 1 3π 3π fˆ(ω) = δ ω + + δ ω , 2 2 − 2 3π 3π meaning f is bandlimited on 2 , 2 . From last lecture we know that the Nyquist rate is then 3 − 2 (three samples for every two units of time). Suppose we did not know this and instead guessed f were bandlimited on [ π, π]. We would then believe samples were only needed every 1 unit of − time. But this sampling rate is now below Nyquist and thus too slow to prevent aliasing. Let us P see why: The Fourier transform of the discretization fd = n∈Z f(n)δn with T = 1 is X 1 X 3π 3π fˆd(ω) = fˆ(ω 2πk) = δ ω + 2πk + δ ω 2πk . (1) − 2 2 − − 2 − k∈Z k∈Z which is a spike train of spacing π. What we then see in the frequency range [ π, π] is not fˆ, but − rather ˆ 1 π π fd(ω)I ω π =g ˆ(ω) = δ ω + + δ ω , {| | ≤ } 2 2 − 2 | {z } | {z } from k = −1 from k = 1
5⇡ 3⇡ ⇡ ⇡ 3⇡ 5⇡ 2 2 2 2 2 2
k = 2 k =0 k = 1 k =1 k =0 k =2
Figure 1: Shown above is the portion of the Dirac comb fˆd in [ π, π], which has contributions from − five different copies of fˆ, corresponding to k = 0, 1, and 2 in (1). Part of each of the k = 1 copies ± ± ± is aliased into the interval [ 3π , 3π ], giving artifact delta functions (displayed with solid lines) in the − 2 2 supposed frequency band [ π, π]. Such aliasing produces low-frequency artifacts in signal reconstruction. −
This is the Fourier transform of π g(t) = cos t , 2 a cosine wave with frequency three times smaller than that of f. Indeed, if Shannon’s formula is naively applied to fˆd, the signal recovered is
X X 3π X π X f(n) sinc(t) = cos n sinc(t) = cos n sinc(t) = g(n) sinc(t) = g(t) 2 2 n∈Z n∈Z n∈Z n∈Z
2 since g is bandlimited to [ π, π] and f is not. − In other words, by sampling f at too low a rate, we cannot detect the difference between the signal f and the signal g: their values at the integers are identical. Shannon interpolation can only return the unique function that gives those samples and is bandlimited to [ π, π]. In this case, that − function is g(t).
2 2