engineering mathematics class # 15 fourier series, integrals, and transforms (part 3)

Sheng-Fang Huang

11.7 Fourier IntegralConsider the periodic rectangular wave

ƒL(x) of period 2L > 2 given by

The left part of Fig. 277 shows this function for 2L = 4, 8, … as well as the nonperiodic function ƒ(x), which we obtain from ƒL if we let L → ∞,

Amplitude SpectrumConsider the Fourier coefficients of ƒL as L

increases. Since ƒL is even, bn = 0 for all n. For an,

This sequence of Fourier coefficients is called the amplitude spectrum of ƒL because ｜ an｜ is the maximum amplitude of the ancos (nπx/L).

Fig. 277. Waveforms and amplitude spectra

(See Fig. 277) For increasing L these amplitudes become more and more dense on the positive wn-axis, where wn = nπ/L. For 2L = 4, 8, 16 we have 1, 3, 7 amplitudes

per “half-wave” of the function (2 sin wn)/(Lwn). Hence, for 2L = 2k we have 2k-1 – 1 amplitudes

per half-wave.These amplitudes will eventually be

everywhere dense on the positive wn-axis (and will decrease to zero).

From Fourier Series to Fourier IntegralConsider any periodic function ƒL(x) of period

2L that is represented by a Fourier series

what happens if we let L → ∞?

We should expect an integral (instead of a series) involving cos wx and sin wx with w no longer restricted to integer multiples w = wn = nπ/L of π/L but taking all values.

If we insert an and bn , and denote the variable of integration by v, the Fourier series of ƒL(x) becomes

We now set

Then 1/L = Δw/π, and we may write the Fourier series in the form

(1)

Let L → ∞ and assume that the resulting nonperiodic function

is absolutely integrable on the x-axis; that is, the following limits exist:

1/L → 0, and the value of the first term on the right side of (1) → zero. Also Δw = π/L → dw. The infinite series in (1) becomes an integral from 0 to ∞, which represents ƒ(x), namely,

(3)

If we introduce the notations

(4)

Fourier integral we can write this in the form

(5)

This is called a representation of ƒ(x) by a Fourier integral.

Fourier Integral

Fourier Integral

THEOREM 1

If ƒ(x) is piecewise continuous in every finite interval and has a right-hand derivative and a left-hand derivative at every point and if the integral exists, then ƒ(x) can be represented by a Fourier integral with A and B given by (4). At a point where ƒ(x) is discontinuous the value of the Fourier integral equals the average of the left- and right-hand limits of ƒ(x) at that point.

Applications of Fourier IntegralsExample 2: Single Pulse, Sine Integral

Find the Fourier integral representation of the function

Solution.

Sine IntegralThe case x = 0 is of particular interest. If x

= 0, then (7) gives

(8*)

We see that this integral is the limit of the so-called sine integral

(8)

as u → ∞. The graphs of Si(u) and of the integrand are shown in Fig. 279.

Fig. 279. Sine integral Si(u) and integrand

In the case of the Fourier integral, approximations are obtained by replacing ∞ by numbers a. Hence the integral

(9)

which approximates ƒ(x).

Gibbs PhenomenonWe might expect that these oscillations

disappear as a → ∞. However, with increasing a, they are shifted closer to the points x = ±1.

This unexpected behavior is known as the Gibbs phenomenon.

Fourier Cosine Integral and Fourier Sine Integral

If ƒ(x) is an even function, then B(w) = 0 and

(10)

The Fourier integral (5) then reduces to the Fourier cosine integral

Fourier Cosine Integral and Fourier Sine IntegralIf ƒ(x) is an odd function, then A(w) = 0 and

(12)

The Fourier integral (5) then reduces to the Fourier cosine integral

11.8 Fourier Cosine and Sine TransformsFourier Cosine TransformFor an even function ƒ(x), the Fourier

integral is the Fourier cosine integral(1)

We now set A(w)= , where c suggests “cosine.” Then from (1b), writing v = x, we have

(2)

and from (1a),

(3)

Fourier Sine TransformSimilarly, for an odd function ƒ(x), the

Fourier integral is the Fourier sine integral

(4)

We now set B(w)= , where s suggests “sine.” From (4b), writing v = x, we have

(5)

This is called the Fourier sine transform of ƒ(x). From (4a)

(6)

Fourier Sine TransformEquation (6) is called the inverse Fourier

sine transform of . The process of obtaining from ƒ(x) is also called the Fourier sine transform or the Fourier sine transform method.

Other notations are

and and for the inverses of and , respectively.

Example 1: Fourier Cosine and Fourier Sine Transforms

Find the Fourier cosine and Fourier sine transforms of the function

Solution:

Example 2: Fourier Cosine Transform of the Exponential Function

FindSolution.

Linearity, Transforms of DerivativesThe Fourier cosine and sine transforms are

linear operations,

(7)

Cosine and Sine Transforms of Derivatives THEOREM 1

Let ƒ(x) be continuous and absolutely integrable on the x-axis, let ƒ'(x) be piecewise continuous on every finite interval, and let let ƒ(x) → 0 as x → ∞. Then

(8)

Formula (8a) with ƒ' instead of ƒ gives (when ƒ', ƒ'' satisfy the respective assumptions for ƒ, ƒ' in Theorem 1)

hence by (8b)

(9a)

Similarly,

(9b)

Example 3: An Application of the Operational Formula (9)Find the Fourier cosine transform (e-ax) of

ƒ(x) = e-ax, where a > 0.Solution.

11.9 Fourier Transform. Discrete and Fast Fourier TransformsThe complex Fourier integral is

(4)

Fourier Transform and Its InverseWriting the exponential function in (4) as a

product of exponential functions, we have

(5)

The expression in brackets is a function of w, is denoted by , and is called the Fourier transform of ƒ; writing v = x, we have

(6)

With this, (5) becomes

(7)

and is called the inverse Fourier transform of .

Another notation for the Fourier transform is

so that

The process of obtaining the Fourier transform

(ƒ) =from a given ƒ is also called the Fourier transform or the Fourier transform method.

Example 1: Fourier TransformFind the Fourier transform of ƒ(x) = 1 if ︱

x ︱ < 1 and ƒ(x) = 0 otherwise.Solution. Using (6) and integrating, we

obtain

As in (3) we have eiw = cos w + i sin w, e-iw = cos w – i sin w, and by subtraction

eiw – e-iw = 2i sin w. Substituting this in the previous formula,

we see that i drops out and we obtain the answer

Example 2: Fourier TransformFind the Fourier transform (e-ax) of ƒ(x) =

e-ax if x > 0 and ƒ(x) = 0 if x < 0; here a > 0.Solution.

Linearity. Fourier Transform of Derivatives

Linearity of the Fourier Transform

THEOREM 2

The Fourier transform is a linear operation; that is, for any functions ƒ(x) and g(x) whose Fourier transforms exist and any constants a and b, the Fourier transform of aƒ + bg exists, and

(8)

Fourier Transform of the Derivative of ƒ(x)

THEOREM 3

Let ƒ(x) be continuous on the x-axis and ƒ(x) → 0 as ︱ x ︱ → ∞ . Furthermore, let ƒ'(x) be absolutely integrable on the x-axis. Then

(9)

ConvolutionThe convolution ƒ * g of functions ƒ and g

is defined by

(11)

Taking the convolution of two functions and then taking the transform of the convolution is the same as multiplying the transforms of these functions (and multiplying them by ):

Convolution Theorem

Convolution Theorem

THEOREM 4

Suppose that ƒ(x) and g(x) are piecewise continuous, bounded, and absolutely integrable on the x-axis. Then

(12)

Convolution TheoremBy taking the inverse Fourier transform

on both sides of (12), writing and as before, and noting that and 1/ in (12) and (7) cancel each other, we obtain

(13)

Discrete Fourier Transform (DFT)The function f(x) is given only in terms of

values at finitely many points.Dealing with sampled values, we can replace

Fourier transform by the so-called discrete Fourier transform.

Let f(x) be periodic with the period 2π. Assume N measurements are taken over the interval 0 ≦ x 2≦ π at regular spaced points

Discrete Fourier Transform (DFT)We now to determine a complex

trigonometric polynomial that interpolates f(x) at the nodes. That is,

Hence, we must determine the coefficients c0, …, cN-1

Multiply by and sum over k from 0 to N-1

Denote […] by r. For n = m, r = e0 = 1. The sum of these terms over k equals N.

Discrete Fourier Transform (DFT)For n≠m we have r ≠1 and by the formula for

a geometric sum:

Because

This shows that the right side of (17) equals cmN. Thus, we

obtain the desired coefficient formula:

Discrete Fourier Transform (DFT)It is practical to drop the factor 1/N from cn

and define the discrete Fourier transform of the given signal to be the vectorwith components

This is the frequency spectrum of the signal.

Fourier MatrixIn vector notation, , where the N×N

Fourier matrix FN=[enk] has the entries [given in (18)]:

Example 4

Let N = 4 measurements (sample values) be given. Then w = e-2πi/N = e-πi/2 = –i and thus wnk = (–i)nk. Let the sample values be, say f = [0 1 4 9]T. Then by (18) and (19),

(20)