Chapter 6: Fourier Analysis of Continuous Functions

6.A Introduction.

In the introduction to Chapter 3 we posed the question of whether or not an arbitrary function defined over a finite interval of time or space can be represented as the sum of a series of weighted sinusoids. We delayed answering that question temporarily while we dealt with the easier question of whether a sampled function can be represented by a weighted sum of sampled sinusoids. We found that indeed it was possible to fit a set of D data points exactly by a Fourier series model which had D coefficients. In the process, we developed simple formulas which allowed us to calculate these unknown Fourier coefficients for any vector of data values. Then, in Chapter 5 we returned to the original problem and argued by extension of these results that a continuous function defined over a finite interval will have a discrete spectrum with an infinite bandwidth. This is Case 3 illustrated in Fig. 5.1. The task now is to formulate an appropriate model for this case and then determine formulas for calculating the unknown Fourier coefficients.

6.B The Fourier Model.

Given an arbitrary, real-valued function f(x), we wish to represent this function exactly by a Fourier series with an infinite number of terms. Accordingly, the model we seek is just eqn. [3.34] extended to have an infinite number of harmonics. To simplify the notation initially, let us assume that f(x) is defined over the interval (-π,π) so the Fourier series is

[6.1]

The method used previously to determine the unknown Fourier coefficients was to evaluate the inner product of the given discrete function with the basis functions of the model. In the discrete case, the basis functions were the sampled trigonometric functions. Now, in the continuous case, the basis functions are the continuous trigonometric functions. Accordingly, we need to form the inner product of eqn. [6.1] with each harmonic function in turn. For example, to find the kth harmonic coefficient a_k, we form the inner product

[6.2]

Notice that the infinite series on the right side of eqn. [6.2] collapses to a single term because of orthogonality. The inner product coskx •coskx equals L / 2=π according to eqn. [5.7]. Therefore, we may conclude from eqn. [6.2] that the coefficient a_k, is given by the formula

[6.3]

Notice that this formula works also for the constant term a₀ since if k=0, then eqn. [6.3] yields twice the mean value. When this result is inserted into the model of eqn. [6.1], the constant term is just the mean value of the function y(x) as required. A similar line of reasoning leads to an analogous formula for the sine coefficient b_k

[6.4]

To obtain the more general formulas which apply when the function y(x) is defined over any finite interval of length L, we replace x by 2πx/L to get

[6.5]

This last result is the form of Fourier series commonly seen in textbooks because it is general enough to allow analysis of one full period with arbitrary starting point.

The trigonometric model of eqn. [6.1] is useful only for real-valued functions. However, we may extend the model to include complex-valued functions just as we did earlier for discrete functions (see eqn. [4.14]). To do so the Fourier series model is written in terms of complex coefficients as

[6.6]

where we take our inspiration for the definition of the complex coefficients from eqn. [4.12] and eqn. [6.5]:

[6.7]

In this model the continuous complex exponential are the orthogonal basis functions for representing any complex-valued function. Since real-valued functions are just a special case of complex-valued functions, the model of eqn. [6.6] subsumes the model of eqn. [6.1] and so is often preferred in textbooks.

6.C Practicalities of Obtaining the Fourier Coefficients.

The preceding section has demonstrated that there is little conceptual difference between the Fourier analysis of discrete and continuous functions over a finite interval. From a practical standpoint, however, there is a large difference in the mechanics of calculating the Fourier coefficients of the model. Anyone who can do arithmetic can compute the inner products required in the Fourier analysis of discrete functions. But to do Fourier analysis of continuous functions requries the ability to do calculus. If evaluating an integral is thought of as a last resort, then there are several alternative strategies which can be tried first.

• Look it up in a reference book. Spectra of many common waveforms have been determined and answers are in handbooks of mathematical functions (e.g. see Bracewell's pictorial dictionary of Fourier transform pairs )
  
  
• Use symmetry arguments to assert that certain coefficients are zero.
  
  
• Break the given function down into a sum of more elementary functions for which the Fourier series is known. Because of linearity, the spectrum of a sum of functions is equal to the sum of their spectra.
  
  
• Approximate the given function by sampling it D times and numerically calculate the corresponding finite Fourier series.
  
  
• Use theorems to derive new Fourier series from old ones without doing a lot of extra work.
  
  
• Use the brute force method and do the calculus. An efficient strategy is to leave the harmonic number k as a variable so that a formula is produced which determines all of the Fourier coefficients at once.

An example is worked below which uses a combination of the symmetry method and the brute force method. The problem is to find the Fourier series model which fits the continuous function shown in Fig. 6.1. Although the function is only defined over the interval (-π,π), the Fourier series will also fit the periodic function obtained by replicating y(x) to make a "square wave". We begin by evaluating eqn. [6.5] for the given function and observing that the integrand has odd symmetry, therefore all of the cosine coefficients equal zero,

[6.8]

To obtain the sine coefficients we evalute eqn. [6.4] and note that since the integrand is even, the integral simplifies to

[6.9]

Evaluating the first few harmonics we see that

[6.10]

which suggests that the generic formula for the coefficients is

[6.11]

Substituting these coefficients back into the model of eqn. [6.1] we get

[6.12]

Thus we have the fourier series model for a square wave in "sine phase". A similar formula may be determined for the square wave in "cosine phase".

6.D Theorems

1. Linearity

A common way of describing Fourier analysis is that it is a linear operation. There are two general properties of any linear operation which have to do with scaling and adding. In the present context, these properties are:

1a.

if f(x) has the Fourier coefficients ak and bk, then scaling f(x) by a constant s to produce the new function g(x)=s•f(x) will also scale the Fourier coefficients by s. That is, the coefficients for g(x) will be s•ak , s•bk. This theorem is easily verified by substituting the new function into the coefficient generating functions in eqn. [6.5].

1b.

if f(x) has the Fourier coefficients ak and bk, and if g(x) has the Fourier coefficients ak and bk, then the function f(x) + g(x) will have Fourier coefficients ak+ αk and bk +βk. This theorem follows from the fact that the integrals in eqn. [6.5] are themselves linear operations, so if the integrand is the sum of two functions, the integral can be broken into the sum of two integrals, each of which corresponds to the fourier coefficients of the component functions f or g.

2. Shift theorem

If the origin of the time or space reference frame shifts by an amount x´, then the effect is to induce a phase shift in the spectrum of a function. This result is easy to demonstrate algebraically if the original function f(x) is given by the Fourier series in polar form as

[6.13]

Next we obtain g(x) from f(x) by shifting the origin by the amount x´. This is achieved mathematically by substituting the quantity x-x´ for x to yield

[6.14]

In words, this equation says that to evaluate g(x) we subtract the amount x´ from x and submit the result to the function y. The result is that the magnitudes of the Fourier coefficients are unaffected but the phase of each harmonic term is increased by the amount kx´. An example is shown in Fig. 6.2 in which a function y(x) has two harmonic components. Notice that a shift of amount π/2 shifts the phase of the fundamental by 1/4 cycle, whereas the effect on the second harmonic is a phase shift of 1/2 cycle.

The above result is also easily proven algebraically when the spectrum of f(x) is represented by the complex Fourier series

[6.15]

Let g(x) = f(x-x´) be the shifted version of f(x) and substitute x-x´ for x in eqn. [6.15] to get the spectrum of g(x)

[6.16]

The new Fourier coefficient is thus seen to be the old coefficient times e^ikx´. But we know from Chapter 2 that multiplication by the unit phasor e^-iθ has the effect of rotating the given phasor by the angle θ, which is to say the phase is shifted by amount θ. Notice that the amount of phase shift is directly proportional to the harmonic number k and to the amount of displacement x´.

Although more cumbersome algebraically, the Cartesian form of this theorem may provide some insight to the student. If f(x) is given by the Fourier series

[6.17]

and g(x) is obtained from f(x) by shifting the origin by x´ to give

[6.18]

In order to re-write this last equation as a standard Fourier series and thus reveal the new Fourier coefficients,

[6.19]

we apply the trigonometrical identities of eqn. [2.7]

cos(α + δ) = cosαcosδ - sinαsinδ sin(α + δ) = cosαsinδ + sinαcosδ

[2.7]

To see how the result is going to turn out, consider the k-th harmonic term

[6.20]

Thus the new Fourier coefficients are

[6.21]

which are recognized as rotated versions of the phasors (a_k,b_k). In agreement with the solution obtained in polar form above, the amount of rotation is kx.

3. Scaling theorem

If the scale of the time or space reference frame changes by the factor s , then the effect is to inversely scale the frequency axis of the spectrum of the function. This result is easy to demonstrate algebraically if f(x) is given by the Fourier series

[6.22]

We now create the new function g(x) from f(x) by scaling the x-axis by the factor s. It follows that we need only to substitute the quantity sx for x to yield

[6.23]

In words, this equation says that to evaluate g(x) we multiply variable x by the constant s and submit the result to the function y. This result shows that the new frequency of each harmonic is now s times the old frequency. Another way of saying the same thing is that the spectrum has been stretched by the factor 1/s. A graphical example for the case s=2 is shown in Fig. 6.3.

This result may be easier to grasp if the frequency spectrum is conceived in terms of harmonic number rather than physical frequency. Since harmonic number does not depend upon the length L of the interval, the harmonic spectrum will be exactly the same before and after scaling the x-axis. However, to convert harmonic numbers to physical frequency requires knowledge of the fundamental period. If the observation period is compressed by the scaling factor s, then the fundamental frequency is correspondingly larger and thus the frequency spectrum will be expanded by the factor s.

4. Differentiation theorem

If f(x) is given by the Fourier series

[6.24]

and a new function g(x) is created by differentiating f(x) with respect to x, then the model for g(x) is

[6.25]

This shows that the new a_k equals k times the old b_k and that the new b_k equals -k times the old a_k . The reason the coefficients are scaled by the harmonic number is that the rate of change of higher harmonics is greater so the derivative must be greater. For example, the function y(x) in Fig. 6.1 has the Fourier coefficients b₁=2, b₂=-1. According to this theorem, the derivative of y(x) should have the Fourier coefficients a₁=2, a₂=-2. This is verified by differentiating y(x) directly to get 2cosx - 2cos2x.

The above result may be interpreted as a rotation and scaling of the spectrum in the complex plane. To see this, let f(x) be represented by the Fourier series

[6.26]

and a new function g(x) is created by differentiating f(x) with respect to x, then the model for g(x) is

[6.27]

In words, the effect of differentiating a function is to rotate each phasor c_k by 90° (the factor i) and to scale all of the Fourier coefficients by the harmonic number k.

5. Integration theorem

If f(x) is given by the Fourier series

[6.28]

and a new function g(x) is created by integrating f(x) with respect to x, then the model for g(x) is

[6.29]

The last step in eqn. [6.29] is possible since the integral of a sum of terms is equal to the sum of each separate integral. Evaluating these integrals we get

[6.30]

The variable C in eqn. [6.30] is a constant of integration which absorbs the cos(kπ) terms and the a₀ term. Notice that if the mean of the original function f(x) is not zero, then a linear term shows up after integration. Thus, the result is not necessarily a proper Fourier series unless a₀=0. This result shows that the new a_k equals -1/k times the old b_k and that the new b_k equals 1/k times the old a_k . The reason the coefficients are scaled inversely by the harmonic number is that the area under a sinusoid grows smaller as the frequency grows larger, hence the integral is less.

In a manner similar to that shown above in section 4 above, eqn. [6.30] may be interpreted as a rotation of the spectrum in the complex plane. Thus, the effect of integrating a function is to rotate each phasor c_k by -90° and to scale all of the Fourier coefficients by 1/k, the inverse of the harmonic number.

An example of the use of the integration theorem is shown in Fig. 6.4 in which the square wave of Fig. 6.1 is integrated to produce a triangle wave. By inspection we see that the mean value of g(x) is -π/2. To get the other Fourier coefficients we apply the theorem to the Fourier series of f(x) found above in eqn. [6.12]

[6.11]

Since the old a-coefficients are zero, the new b-coefficients are zero. The new a-coefficients are equal to -1/k times the old b-coefficients. Therefore, we conclude that the new function g(x) has the Fourier coefficients

[6.31]

in which case the Fourier series for g(x) is

	[6.32]