08/21/03

Mathematical Transforms

A scalar time function, f ( t ) is a simple way to represent information. For example, it might represent the audio from one stereo channel of a CD, or a video signal for display on a screen. From one perspective, the information in f ( t ) is formatted as a time signal. We often find it convenient to transform the information into a different format. The situation parallels having the information in a large book in electronic file format. The electronic file format is great for editing or conducting a word search of the contents, but many people prefer the information in printed format for browsing it quickly and for reading it carefully. It's not that one format is best. Some formats are better for one purpose, others for another.

One popular technique for reformatting the information represented by time functions is with an integral transform:

F ( θ )   =   ∫   a   b k ( θ ,   t )   f ( t )   d t

where

F ( θ ) is the integral transform of f ( t )

k ( θ ,   t ) is the kernal of a particular integral transform

a and b are limits of integration that, together with k ( θ ,   t ) , define this particular integral transform

θ is the independent variable in the transformed function, F .

To perform the transform, we simply multiply the function f ( t ) by the kernal and then integrate from a to b .

It's easy to see that the kernal, k ( θ ,   t ) , must depend on a variable in addition to t . Otherwise, we get a single constant number when we carry out the integral, and a single number, such as 32.461 , clearly contains less information than a time function, such as 3.2 sin ⁡ ( 46 t ) , which includes an infinity of numbers.

Even with the second variable in the kernal, however, most choices for a , b , and k ( θ ,   t ) result in the loss of some of the information originally contained in f ( t ) when the integral is performed. Only those choices for a , b , and k ( θ ,   t ) for which F ( θ ) contains all of the information originally contained in f ( t ) are termed integral transforms . Said another way, an integral transform is invertible : we can reconstruct f ( t ) from F ( θ ) . Reconstruction is impossible, of course, if any information is lost during the integration.

The Laplace transform is an example of an integral transform widely used in science and engineering:

F ( s )     =   ∫   0   ∞ e − s t   f ( t )   d t

To see one of the reasons for the utility of the Laplace transform, let's calculate the Laplace transform of a function g ( t ) , which represents the derivative of some function x ( t ) :

g ( t )     ≡   d x ( t ) d t

From the transform integral, we see that

G ( s )     =   ∫   0   ∞ e − s t   g ( t )     d t   =     ∫   0   ∞ e − s t   d x ( t ) d t     d t

Integrating by parts, we find:

G ( s )     =   e − s t x ( t )   | 0 ∞   −           ∫   0   ∞ ( − s )   e − s t   x ( t )     d t

G ( s )     =   lim ⁡   t   →   ∞   e − s t x ( t )     −     x ( 0 )       +     s ∫   0   ∞ e − s t   x ( t )     d t

For most functions x ( t ) encountered in practice, the limit term goes to zero. For example, if x ( t ) does not become infinite as t approaches infinity, then the exponential, which approaches zero as t approaches infinity, forces the entire term to zero. Thus, this term would approach zero if x ( t ) were, for instance, a sinusoid. In any case for which the first term vanishes, we have

G ( s )     =     s ∫   0   ∞ e − s t   x ( t )     d t     −     x ( 0 )    

or

G ( s )     =     s X ( s )     −     x ( 0 )    

where

X ( s )     =   ∫   0   ∞ e − s t   x ( t )     d t

is just the Laplace transform of x ( t ) .

Note the following key result: when we Laplace transform x ( t ) so that the information it contains is represented as X ( s ) instead, we can find the corresponding derivative simply by multiplying X ( s ) by s , a simple algebraic operation, rather than by differentiating x ( t ) , an operation that can be arduous if x ( t ) is a complicated time function. Stated differently, differentiation in the time domain corresponds to multiplication by s in the Laplace domain. Of course, if we wish to view the derivative as a time function, we must invert the transform to get back to the time domain. More about the inversion process a little later.

The term x ( 0 ) , which reflects the fact that the Laplace transform begins at time zero, shows that the Laplace transform easily includes the initial conditions that often must be imposed on variables in practical problems. In the circuit that recharges the capacitor that powers a flash unit for a camera, for example, the initial voltage on the capacitor when the recharging process begins is important in calculating the time necessary for recharge. Using Laplace transforms to solve such a problem includes the initial voltage on the capacitor in a natural way during the analysis. Other transforms, such as the Fourier transform considered later, begin at t = − ∞ . Including initial conditions on variables is not so straightforward with such transforms, although they certainly have other virtues of their own.

In the Laplace transform domain, a result parallel to that for differentiation is that integration corresponds to division by s . In the Laplace transform domain, both integration and differentiation correspond to algebraic operations and, therefore, calculus simplifies to algebra . The substitution of algebra for calculus and the ease of including initial conditions on variables are most attractive features of the Laplace transform, and account for much of its use.

Historically, the eccentric electrical engineer Oliver Heaviside first replaced calculus by algebra, in about 1890. Heaviside, who first understood that circumglobal radio communication meant that the earth must be surrounded by a layer of charged particles that we now call the ionosphere, who first wrote the Maxwell's equations of electromagnetic theory in their modern vector form, who first discovered how to send signals over long cables without muddling the signals, and who (but not first) chose to sponge off relatives and friends rather than work for a living and then treated those who helped him like dirt, this Oliver Heaviside developed what he called operational calculus , in which he denoted differentiation by the letter p , and integration by the term 1 / p . He then treated these letters as though they were algebraic variables. Mathematically, it appeared to be utter nonsense, for p , unlike ordinary algebraic variables, could not take on numerical values. Heaviside could not explain rigorously what he was doing, but those who would discount his ideas as lunacy faced the problem that he always got the right answer, and got it faster than the critics could. The arguments between Heaviside and his critics became quite bitter in this dispute, as well as in countless others with the mathematical, scientific, and engineering "establishment" that an outsider like Heaviside seemed to relish and hate at the same time. Of course, Heaviside's arrogant attitude did not encourage dispassionate discussions with those who disagreed with him. Ultimately, most people (but not Heaviside) came to understand Heaviside's operational calculus in terms of the Laplace transform, which therefore gave Heaviside's approach (of differentiating through multiplying by something and integrating through dividing by the same thing) a mathematically rigorous basis.

Now, how can we invert Laplace transforms to time functions? After all, operating in the Laplace transform domain may be convenient for avoiding some drudgeries of calculus, but our intuitions have much more experience in envisioning the information in functions of time than that in functions of s .

In practice, the inversion of Laplace transforms usually relies on published tables of transform pairs of time functions, f ( t ) , and their corresponding Lapace transforms, F ( s ) . The following table shows a few lines from such a table of Laplace transforms:

Using tables with perhaps hundreds of entries and using the properties (mainly linearity) of the integral that defines the Laplace transform, we can invert most of the transforms that we encounter in practice. Suppose, as a simple example, that through calculations in the Laplace domain we find that

X ( s )     =     a s 2 ​   +   ω 2 ​     =     a ω   ω s 2 ​   +   ω 2 ​

From the table, we find the corresponding representation as a time function to be

x ( t )     =     a ω   sin ⁡ ( ω   t )

In principle, we can calculate f ( t ) directly from F ( s ) with the Laplace transform inversion integral , as well:

f ( t )     =   1 2 π   j ∫   a   −   j ∞   a   +   j ∞ F ( s )     e s t d s

where

j   =   − 1

Thus, we have the Laplace transform pair of integrals:

F ( s )     =   ∫   0   ∞ e − s t   f ( t )     d t

f ( t )     =   1 2 π   j ∫   a   −   j ∞   a   +   j ∞ F ( s )     e s t d s

Writing this pair of integrals emphasizes the point that in transforming f ( t ) to F ( s ) with the first integral, we have lost no information, since we can recover f ( t ) from F ( s ) using the second integral.

In answer to the question "Why bother?", we say that we can accomplish some things, such as calculus, more easily in the Laplace transform domain than in the time domain. The situation is somewhat analogous to asking why make a round trip to New York City. The answer, of course, is that we can do some things in New York City, such as see a Broadway play, more easily than in Starkville.

We now turn our attention to an integral transform that is perhaps even more widely used than the Laplace transform, the Fourier transform pair :

F ( ω )     =   ∫   − ∞   ∞ e − j ω   t   f ( t )     d t

f ( t )     =     1 2 π ∫   − ∞   ∞ e j ω   t   F ( ω )     d ω

Mainly because they both use an exponential function as a kernal, the Fourier transform and the Laplace transform show a number of similarities. In the Fourier transform domain, for example, it turns out that calculus also becomes algebra: differentiation corresponds to multiplication by j ω and integration corresponds to division by j ω .

For our present purposes, however, we want to see that the Fourier transform justifies our thinking of practical signals in terms of their sinusoidal frequency components. That is, in thinking of audio signals as consisting of sinusoidal frequency components with frequencies that range from 20Hz to 20,000Hz , for example.

Consider the second (inversion) Fourier integral:

f ( t )       =     1 2 π ∫   − ∞   ∞ e j ω   t   F ( ω )     d ω

The term e j ω   t may seem a little peculiar because it includes an "imaginary" number, j , in the argument of the familiar exponential function. To understand the meaning of this term more clearly, we use a mathematical identity, called Euler's identity :

e j φ   ≡     cos ⁡ ( φ ) ​     +   j sin ⁡ ( φ )

The symbol ≡ distinguishes a mathematical identity , which holds for every value of φ , from a mathematical equation (with the symbol =), which would hold only for particular values of φ . Leonhard Euler proved this somewhat peculiar identity by showing that the power series expansions of each side of the identity agreed term-by-term. But at the moment we're more interested in applying the identity than proving it, so we note that it tells us that

e j ω   t   ≡     cos ⁡ ( ω   t ) ​     +   j sin ⁡ ( ω   t )

so that

f ( t )     =     1 2 π ∫   − ∞   ∞ e j ω   t   F ( ω )     d ω     =   1 2 π ∫   − ∞   ∞ [ cos ⁡ ( ω   t ) ​     +   j sin ⁡ ( ω   t ) ]   F ( ω )     d ω

With Euler's identity, then, this Fourier integral says that any function, f ( t ) , for which the Fourier transform exists (most functions encountered in practice) can be expressed as a sum, or superposition, of sinusoids (sines and cosines) with different frequencies, ω . Although the details are hard to see without further work, it turns out that the function F ( ω ) determines the phase and amplitude of the sinusoids at each frequency. Thus, this Fourier integral expresses f ( t ) as an integral sum, weighted by F ( ω ) / 2 π , of sinusoids with various frequencies. If a particular f ( t ) has no frequency component at a certain frequency ω , then at that frequency, F ( ω ) = 0 .

You may have noted one other peculiarity. The limits on the Fourier integral extend from − ∞ to + ∞ so that the integral sums over negative as well as positive frequencies, ω . What in the world are negative frequencies?

To understand this point better, we again employ Euler's identity. Because the cosine is an even function of its argument and the sine is an odd function of its argument, we can write

cos ⁡ ( − φ )     ≡     cos ⁡ ( φ ) ​

and

sin ⁡ ( − φ )     ≡     − sin ⁡ ( φ ) ​

Because Euler's identity holds for every value of its argument, it must hold, in particular, when the argument is − φ :

e − j φ   ≡     cos ⁡ ( − φ ) ​     +   j sin ⁡ ( − φ ) ≡     cos ⁡ ( φ ) ​     −   j sin ⁡ ( φ )

By adding these two forms of Euler's identity, we see

e j φ   +     e − j φ   ​ ​ ≡     cos ⁡ ( φ ) ​     +   j sin ⁡ ( φ )     +     cos ⁡ ( φ ) ​     −   j sin ⁡ ( φ )       =     2 cos ⁡ ( φ )

or

cos ⁡ ( φ )     ​ ​ ≡     e j φ   +     e − j φ 2

By subtracting the two versions of Euler's identity, it follows similarly that

sin ⁡ ( φ )     ​ ​ ≡     e j φ   −     e − j φ 2 j

These two identities show that we can express both cosine and sine functions in terms of the exponential function.

Why would we want to do that? Well, mainly because exponential functions are usually easier to manipulate (multiply, divide, differentiate, integrate, ...) than sines or cosines. But note particularly the following:

cos ⁡ ( ω   t )   ​   ​ ≡     e j ω   t   +     e − j ω   t 2

sin ⁡ ( ω   t )     ​ ​ ≡     e j ω   t   −     e − j ω   t 2 j

When we express sine or cosine time functions in terms of the exponential function, we need to use both negative , as well as positive frequencies. Thus, when we represent both signs and cosines in the Fourier integral with a single exponential function, we must include negative, as well as positive, frequencies in the second Fourier integral. This peculiarity seems a small price to pay for the compact and elegant form of the Fourier integral written in terms of the exponential function compared to the form the integral takes when we write it in terms of sines and cosines explicitly.

Having said all that, however, we can show that for signals of practical interest, the magnitude of the sinusoidal component at frequency − ω is the same size as the magnitude of the sinusoidal component at frequency + ω , so we usually plot, without loss of any information, the Fourier spectrum of a signal only for positive frequencies.