Sampled Data and Aliasing

4.6.6 Sampled Data and Aliasing

The ideas of Sec. 4.3.5 can be applied to the derivation of the celebrated sampling theo- rem . This is an interpolation formula that expresses a function that is band-limited to the frequency interval [−W, W]. Such a function has a Fourier representation of the form (see also Strang [339, p. 325])

f (z) =

ˆ f (k)e ikz dk,

|ˆ f (k) | ≤ M, (4.6.38)

−W

in terms of its values at all integer points. Theorem 4.6.7 (Shannon’s Sampling Theorem).

Let the function be band-limited to the frequency interval [−W, W]. Then

Proof. We shall sketch a derivation of this for W = π. (Strang [339] gives an entirely different derivation, based on Fourier analysis.) We first note that (4.6.38) shows that f (z) is analytic for all z. Then we consider the same Cauchy integral as many times before,

1 M(u)f (z)

M(z)(z

− u)

Here M(z) = sin πz, which vanishes at all integer points, and D n is the open rectangle with vertices at ±(n + 1/2) ± bi. By the residue theorem, we obtain after a short calculation

I n (u) sin π(j − u)

n M(u)f (j ) f (j )

= f (u) +

= f (u) −

π(j − u) Set z = x + iy. Note that

j =−n M (j )(j − u)

j =−n

M(e |πy|

−|πy| )

|f (z)| ≤

Me −ky dk

−e

, |M(z)| ≥ e |πy| .

These inequalities, applied for y = b, allow us to let b → ∞ (2b is the height of the symmetric rectangular contour). Then it can be shown that I n (u) → 0 as n → ∞, which

establishes the sampling theorem for W = π. The general result is then obtained by “regula

de tri,” 161 but it is sometimes hard to get it right. Note the similarity of (4.6.39) to Lagrange’s interpolation formula. Like Lagrange’s,

it is a so-called cardinal interpolation formula. As W z/π tends to an integer m, all terms except one on the right-hand side become zero; for j = m the term becomes f (mπ/W).

161 This rule from Euclid’s fifth book of Elementa tells us how, given three out of four proportional quantities a/b = c/d, one determines the fourth. This name is used in elementary mathematics, e.g., in Germany and

Sweden, but does not seem to be known under the same name in English-speaking countries.

498 Chapter 4. Interpolation and Approximation Let f be a function which is zero outside the interval [0, L]. Its Fourier transform is

then given by

g(ω) =

f (x)e −2πiωx dx.

We want to approximate g(ω) using values of f (x) sampled at intervals 4x,

0 < j < N − 1, L = N4x. The integral (4.6.40) can be approximated by

f j = f (j4x),

L N −1

g(ω) ≈

f j e −2πiωj4x .

N j =0

Since only N values of f j are used as input and we want the computed values to be linearly independent, we cannot approximate g(ω) at more than N points. The wave of lowest frequency associated with the interval [0, L] is ω = 1/L = 1/(N4x), since then [0, L] corresponds to one full period of the wave. We therefore choose points ω k = k4ω, k = 0 : N in the frequency space such that the following reciprocity relations hold:

(4.6.42) With this choice it holds that W = N4ω = 1/4x,

LW = N,

4x4ω = 1/N.

L = N4x = 1/4ω. (4.6.43) Noting that (j4x)(k4ω) = jk/N we get from the trapezoidal approximation

L N −1

g(ω k ) ≈

f j e −2πikj/N

= Lc k , k = 0 : N − 1,

j =0

where c k is the coefficient of the DFT. The frequency ω c = 1/(24x) = W/2 is the so-called Nyquist’s critical frequency. Sampling the wave sin(2πω c x) with sampling interval 4x will sample exactly two points per cycle . It is a remarkable fact that if a function f (x), defined on [−∞, ∞], is band-width limited to frequencies smaller than or equal to ω c , then f (x) is completely determined by its sample values j4x, ∞ ≤ j ≤ ∞; see Shannon’s sampling theorem, our Theorem 4.6.7. If the function is not band-width limited the spectral density outside the critical fre- quency is moved into that range. This is called aliasing. The relationship between the Fourier transform g(ω) and the DFT of a finite sampled representation can be character- ized as follows. Assuming that the reciprocity relations (4.6.42) are satisfied, the DFT of

f j = f (j4x), 0 ≤ j < N, will approximate the periodic aliased function

(4.6.44) where

˜g k = ˜g(k4ω),

0 ≤ j < N,

˜g(ω) = g(ω) + (g(ω + kW ) + g(ω − kW )) , ω ∈ [0, W ]. (4.6.45)

k =1

4.6. Fourier Methods 499 Since by (4.6.43) W = 1/4x, we can increase the frequency range [0, W] covered by

decreasing 4x .

Example 4.6.4.

The function f (x) = e −x , x > 0, f (x) = 0, x < 0, has Fourier transform

(cf. Example 4.6.3). Set f (0) = 1/2, the average of f (−0) and f (+0), which is the value given by the inverse Fourier transform at a discontinuity.

Set N = 32, T = 8, and sample the f in the interval [0, T ] at equidistant points j4t, j = 0 : N − 1. Note that T is so large that the aliased function (4.6.44) is nearly equal to

f . This sampling rate corresponds to 4t = 8/32 = 1/4 and W = 4. The effect of aliasing in the frequency domain is evident. The error is significant for frequencies larger than the critical frequency W/2. To increase the accuracy W can be increased by decreasing the sampling interval 4x; see Figure 4.6.4.

Figure 4.6.4. The real (top) and imaginary (bottom) parts of the Fourier transform (solid line) of e −x and the corresponding DFT (dots) with N = 32, T = 8.

500 Chapter 4. Interpolation and Approximation

ReviewQuestions

4.6.1 Derive the orthogonality properties and coefficient formulas which are fundamental to Fourier analysis, for both the continuous and the discrete case.

4.6.2 Give sufficient conditions for the Fourier series of the function f to be everywhere convergent to f . To what value will the Fourier series converge at a point x = a of

discontinuity of f ?

4.6.3 How can the Fourier expansion be generalized to a function f (x, y) of two variables?

4.6.4 (a) Give two ways in which a real function f defined on the interval [0, π] can be extended to a periodic function.

(b) What disadvantage (for Fourier analysis) is incurred if the periodic continuation has a discontinuity—for example, in its derivative at the end points of 80, π9?

4.6.5 Describe how the behavior of the coefficients of the discrete Fourier expansion can

be modified to improve the approximation of the corresponding continuous Fourier expansion.

4.6.6 Formulate the Fourier integral theorem.

4.6.7 What is meant by aliasing, when approximating the Fourier transform by a discrete transform?

Problems and Computer Exercises

4.6.1 Give a simple characterization of the functions which have a sine expansion con- taining odd terms only.

4.6.2 Let f be an even function, with period 2π, such that

f (x) = π − x,

0 ≤ x ≤ π.

(a) Plot the function y = f (x) for −3π ≤ x ≤ 3π. Expand f in a Fourier series. (b) Use this series to show that 1 + 3 2 −2 +5 −2 +7 −2 +···=π / 8.

(c) Compute the sum 1 + 2 −2 +3 −2 +4 −2 +5 −2 +···. (d) Compute, using (4.6.9), the sum 1 + 3 −4 +5 −4 +7 −4 +···. (e) Differentiate the Fourier series term by term, and compare with the result for the

rectangular wave in Sec. 4.6.1.

4.6.3 (a) Show that the function G 1 (t ) = t − 1/2, 0 < t < 1, has the expansion

∞ sin 2nπt

be as in (a) and consider a sequence of functions such that G ′ p +1 (t ) =

G p (t ) , p = 1, 2, . . . . Derive by termwise integration the expansion for the functions

Problems and Computer Exercises 501

G p (t ) , and show that c p −G p (t ) has the same sign as c p . Show also that for p even

4.6.4 (a) Using that sin x is the imaginary part of e ix , prove that

2N (b) Determine a sine polynomial n −1 j =1 b j sin jx, which takes on the value 1 at the

k =1

points x k = πk/n, k = 1 : n − 1. Hint: Use (4.6.26) or recall that the sine polynomial is an odd function. (c) Compare the limiting value for b j as n → ∞ with the result in Example 4.6.1.

4.6.5 (a) Prove the inequality in (4.6.10). (b) Show, under the assumptions on f which hold in (4.6.10), that for k ≥ 1, f can

be approximated by a trigonometric polynomial such that

4.6.6 (a) Fourier approximations to the rectangular wave, f (x) = 1, 0 < x < π, f (x) = −1, −π < x < 0, are shown in Figure 4.6.2 for one, two, five, and ten terms.

Plot and compare the corresponding smoothed approximations when the σ -factors in (4.6.15) have been applied.

(b) The delta function δ(x) is defined as being zero everywhere except between the limits ±ǫ, where ǫ tends to zero. At x = 0 the function goes to infinity in such a

way that the area under the function equals one. The formal Fourier expansion of δ(x) yields

which does not converge anywhere. If the σ -factors in (4.6.15) are applied, we obtain the expansion

1 y m (x) =

m −1 cos(m − 1)x , (4.6.46) which can be considered the trigonometric representation of the delta function. Plot

π 2+ cos x + σ 2 cos 2x + · · · + σ

y m (x) , −6 ≤ x ≤ 6, for several values of m.

4.6.7 (a) Consider a real function with the Fourier expansion

F (φ) =

c n e inφ .

n =−∞

502 Chapter 4. Interpolation and Approximation Show that this rewritten for convergence acceleration with the generalized Euler’s

method is

Hint: Show that c −n = ¯c n . (b) Set c n =a n − ib n , where a n ,b n are real. Show that

1+n ) ? (d) Consider also how to handle a complex function F (φ).

(c) How would you rewrite the Chebyshev series ∞ T n (x)/(

n =0

In the following problems, we do not require any investigation of whether it is permissible to change the order of summations, integrations, differentiations; it is sufficient to treat the problems in a purely formal way.

4.6.8 The partial differential equation ∂u

∂t = ∂x 2 is called the heat equation. Show that the function

sin(2k + 1)x

u(x, t )

e −(2k+1) 2 t

2k + 1

satisfies the differential equation for t > 0, 0 < x < π, with boundary conditions u( 0, t) = u(π, t) = 0 for t > 0, and initial condition u(x, 0) = 1 for 0 < x < π (see Example 4.6.1).

4.6.9 Show that if g(t) is the Fourier transform of f (x), then (a) e 2πkt g(t ) is the Fourier transform of f (x + k). (b) (2πit) k g(t ) is the Fourier transform of f (k) (x) , assuming that f (x) and its derivatives up to the kth order tend to zero as x → ∞.

4.6.10 The correlation of f 1 (x) and f 2 (x) is defined by

Show that if f 1 (x) and f 2 (x) have Fourier transforms g 1 (t ) and g 2 (t ) , respectively,

then the Fourier transform of c(ξ) is h(t) = g 1 (t )g 2 ( −t).

Hint: Compare Theorem 4.6.6.

4.6.11 (a) Work out the details of the proof of the sampling theorem. (b) The formulation of the sampling theorem with a general W in Strang [339] does

not agree with ours in (4.6.39). Who is right?

4.6.12 Suppose that f (z) satisfies the assumptions of our treatment of the sampling theorem for W = π/h. Show by integration term by term that

lim

f (u) du = h lim

f (j h).

→∞ −R

n →∞

j =−n

4.7. The Fast Fourier Transform 503 Hint: Use the classical formula

Full rigor is not necessary.