10. Signals and Fourier Analysis

Traveling waves with a definite frequency carry energy but no information. They are just there, always have been and always will be. To send information, we must send a nonharmonic signal.

10.1: Signals in Forced Oscillation¶

Pulse on a String¶

10-1

We begin with the following illustrative problem: the transverse oscillations of a semiinfinite string stretched from $x = 0$ to $\infty$, driven at $x = 0$ with some arbitrary transverse signal $f(t)$, and with a boundary condition at infinity that there are no incoming traveling waves. This simple system is shown in Figure 10.1.

There is a slick way to get the answer to this problem that works only for a system with the simple dispersion relation,

The trick is to note that the dispersion relation, 10.1, implies that the system satisfies the wave equation, 6.4, or

It is a mathematical fact (we will discuss the physics of it below) that the general solution to the one-dimensional wave equation, 10.2, is a sum of right-moving and left-moving waves with arbitrary shapes,

where $g$ and $h$ are arbitrary functions. You can check, using the chain rule, that 10.3 satisfies 10.2,

Given this mathematical fact, we can find the functions $g$ and $h$ that solve our particular problem by imposing boundary conditions. The boundary condition at infinity implies

because the $h$ function describes a wave moving in the $-x$ direction. The boundary condition at $x = 0$ implies

which gives

This describes the signal, $f(t)$, propagating down the string at the phase velocity $v$ with no change in shape.

For the simple function

the shape of the string at a sequence of times is shown in Figure 10.2 and animated in program 10-1.

Fourier integrals¶

Let us think about this problem in a more physical way. In the process, we will understand the physics of the general solution, 10.3. This may seem like a strange thing to say in a section entitled, “Fourier integrals.” Nevertheless, we will see that the mathematics of Fourier integrals has a direct and simple physical interpretation.

The idea is to use linearity in a clever way to solve this problem. We can take $f(t)$ apart into its component angular frequencies. We already know how to solve the forced oscillation problem for each angular frequency. We can then take the individual solutions and add them back up again to reconstruct the solution to the full problem. The advantage of this procedure is that it works for any dispersion relation, not just for 10.1.

Because there may be a continuous distribution of frequencies in an arbitrary signal, we cannot just write $f(t)$ as a sum over components, we need a Fourier integral,

The physics of 10.9 is just linearity and time translation invariance. We know that we can choose the normal modes of the free system to have irreducible exponential time dependence, because of time translation invariance. Since the normal modes describe all the possible motions of the system, we know that by taking a suitable linear combination of normal modes, we can find a solution in which the motion of the end of the system is described by the function, $f(t)$. The only subtlety in 10.9 is that we have assumed that the values of $\omega$ that appear in the integral are all real. This is appropriate because a nonzero imaginary part for $\omega$ in $e^{-i \omega t}$ describes a function that goes exponentially to infinity as $t \rightarrow \pm \infty$. Physically, we are never interested in such things. In fact, we are really interested in functions that go to zero as $t \rightarrow \pm \infty$. These are well-described by the integral over real $\omega$, 10.9.

Note that if $f(t)$ is real in 10.9, then

thus

It is actually easier to work with the complex Fourier integral, 10.9, with the irreducible complex exponential time dependence, than with real expansions in terms of $\cos \omega t$ and $\sin \omega t$. But you may also see the real forms in other books. You can always translate from 10.9 by using the Euler identity

For each value of $\omega$, we can write down the solution to the forced oscillation problem, incorporating the boundary condition at $\infty$. Each frequency component of the force produces a wave traveling in the $+x$ direction.

then we can use linearity to construct the solution by adding up the individual traveling waves from 10.13 with the coefficients $C(\omega)$ from 10.9. Thus

where $\omega$ and $k$ are related by the dispersion relation.

Equation 10.14 is true quite generally for any one-dimensional system, for any dispersion relation, but the result is particularly simple for a nondispersive system such as the continuous string with a dispersion relation of the form 10.1. We can use 10.1 in 10.14 by replacing

Note that while $k^{2}$ is determined by the dispersion relation, the sign of $k$, for a given $\omega$, is determined by the boundary condition at infinity. $k$ and $\omega$ must have the same sign, as in 10.15, to describe a wave traveling in the $+x$ direction. Putting 10.15 into 10.14 gives

Comparing this with 10.9 gives 10.7.

Let us try to understand what is happening in words. The Fourier integral, 10.9, expresses the signal as a linear combination of harmonic traveling waves. The relation, 10.15, which follows from the dispersion relation, 10.1, and the boundary condition at $\infty$, implies that each of the infinite harmonic traveling waves moves at the same phase velocity. Therefore, the waves stay in exactly the same relationship to one another as they move, and the signal is never distorted. It just moves with the waves.

The nonharmonic signal is called a “wave packet.” As we have seen, it can be taken apart into harmonic waves, by means of the Fourier integral, 10.9.

10.2: Dispersive Media and Group Velocity¶

For any other dispersion relation, the signal changes shape as it propagates, because the various harmonic components travel at different velocities. Eventually, the various pieces of the signal get out of phase and the signal is dispersed. That is why such a medium is called “dispersive.” This is the origin of the name “dispersion relation.”

Group Velocity¶

10-2

If you are clever, you can send signals in a dispersive medium. The trick is to send the signal not directly as the function, $f(t)$, but as a modulation of a harmonic signal, of the form

where $f_{s}(t)$ is the signal. Very often, you want to do this anyway, because the important frequencies in your signal may not match the frequencies of the waves with which you want to send the signal. An example is AM radio transmission, in which the signal is derived from sound with a typical frequency of a few hundred cycles per second (Hz), but it is carried as a modulation of the amplitude of an electromagnetic radio wave, with a frequency of a few million cycles per second.^[1]

You can get a sense of what is going to happen in this case by considering the sum of two traveling waves with different frequencies and wave numbers,

where

for

The sum can be written as a product of cosines, as

Because of 10.20, the first factor varies slowly in $x$ and $t$ compared to the second. The result can be thought of as a harmonic wave with frequency $\omega_{0}$ with a slowly varying amplitude proportional to the first factor. The space dependence of 10.21 is shown in Figure 10.3.

You should think of the first factor in 10.21 as the signal. The second factor is called the “carrier wave.” Then 10.21 describes a signal that moves with velocity

while the smaller waves associated with the second factor move with velocity

These two velocities will not be the same, in general. If 10.20 is satisfied, then (as we will show in more detail below) $v_{0}$ will be roughly the phase velocity. In the limit, as $k_{+}-k_{-}=2 k_{s}$ becomes very small, 10.22 becomes a derivative

This is called the “group velocity.” It measures the speed at which the signal can actually be sent.

The time dependence of 10.21 is animated in program 10-2. Note the way that the carrier waves move through the signal. In this animation, the group velocity is smaller than the phase velocity, so the carrier waves appear at the back of each pulse of the signal and move through to the front.

Let us see how this works in general for interesting signals, $f(t)$. Suppose that for some range of frequencies near some frequency $\omega_{0}$, the dispersion relation is slowly varying. Then we can take it to be approximately linear by expanding $\omega (k)$ in a Taylor series about $k_{0}$ and keeping only the first two terms. That is

and the higher order terms are negligible for a range of frequencies

where $\Delta \omega$ is a constant that depends on $\omega_{0}$ and the details on the higher order terms. Then you can send a signal of the form

(a complex form of 10.17, above) where $f(t)$ satisfies 10.9 with

This describes a signal that has a carrier wave with frequency $\omega_{0}$, modulated by the interesting part of the signal, $f(t)$, that acts like a time-varying amplitude for the carrier wave, $e^{-i \omega_{0} t}$. The strategy of sending a signal as a varying amplitude on a carrier wave is called amplitude modulation.

Usually, the higher order terms in 10.25 are negligible only if $\Delta \omega \ll \omega_{0}$. If we neglect them, we can write 10.25 as

where $a$ and $b$ are constants we can determine from 10.25,

and $v$ is the group velocity

For the signal 10.28

Thus 10.14 becomes

but then 10.29 gives

The modulation $f(t)$ travels without change of shape at the group velocity $v$ given by 10.32, as long as we can ignore the higher order term in the dispersion relation. The phase velocity

has nothing to do with the transmission of information, but notice that because of the extra $e^{i b x}$ in 10.35, the carrier wave travels at the phase velocity.

You can see the difference between phase velocity and group velocity in your pool or bathtub by making a wave packet consisting of several shorter waves.

10.3: Bandwidth, Fidelity, and Uncertainty¶

The relation 10.9 can be inverted to give $C(\omega)$ in terms of $f(t)$ as follows

This is the “inverse Fourier transform.” It is very important because it allows us to go back and forth between the signal and the distribution of frequencies that it contains. We will get this result in two ways: first, with a fancy argument that we will use again and explain in more detail in chapter 13; next, by going back to the Fourier series, discussed in chapter 6 for waves on a finite string, and taking the limit as the length of the string goes to infinity.

The fancy argument goes like this. It is very reasonable that the integral in 10.37 is proportional to $C(\omega)$ because if we insert 10.9 and rearrange the order of integration, we get

The $t$ integral averages to zero unless $\omega = \omega^{\prime}$. Thus the $\omega^{\prime}$ integral is simply proportional to $C(\omega)$ times a constant factor. The factor of $1 / 2 \pi$ can be obtained by doing some integrals explicitly. For example, if

for $\Gamma>0$ then, as we will show explicitly in 10.49-10.56, 10.37 yields

which can, in turn, be put back in 10.9 to give 10.39. For $t = 0$, the integral can be done by the trigonometric substitution $\omega \rightarrow \Gamma \tan \theta$:

To get the inverse Fourier transform, 10.37, as the limit of a Fourier series, it is convenient to use a slightly different boundary condition from those we discussed in chapter 6, fixed ends and free ends. Instead, let us consider a string stretched from $x=-\pi \ell$ to $x=\pi \ell$, in which we assume that the displacement of the string from equilibrium at $x=\pi \ell$ is the same as the displacement at $x=-\pi \ell$,²

The requirement, 10.42, is called “periodic boundary conditions,” because it implies that the function $\psi$ that describes the displacement of the string is periodic in $x$ with period $2 \pi \ell$. The normal modes of the infinite system that satisfy 10.42 are

for integer $n$, because changing $x$ by $2 \pi \ell$ in 10.43 just changes the phase of the exponential by $2 \pi$. Thus if $\psi(x)$ is an arbitrary function satisfying $\psi(-\pi \ell)=\psi(\pi \ell)$, we should be able to expand it in the normal modes of 10.43,

Likewise, for a function $f(t)$, satisfying $f(-\pi T)=f(\pi T)$ for some large time $T$, we expect to be able to expand it as follows

where we have changed the sign in the exponential to agree with 10.9. We will show that as $T \rightarrow \infty$, this becomes equivalent to 10.9.

Equation 10.44 is the analog of 6.8 for the boundary condition, 10.42. The sum runs from $-\infty$ to $\infty$ rather than 0 to $\infty$ because the modes in 10.43 are different for $n$ and $-n$. For this Fourier series, the inverse is

where we have used the identity

Now suppose that $f(t)$ goes to 0 for large $|t|$ (note that this is consistent with the periodic boundary condition 10.42) fast enough so that the integral in 10.46 is well defined as $T \rightarrow \infty$ for all $m$. Then because of the factor of $1/T$ in 10.47, the $c_{n}$ all go to zero like $1/T$. Thus we should multiply $c_{n}$ by $T$ to get something finite in the limit. Comparing 10.45 with 10.9, we see that we should take $\omega$ to be $n/T$.

Thus the relation, 10.45, is an analog of the Fourier integral, 10.9 where the correspondence is

In the limit, $T \rightarrow \infty$, the sum becomes an integral over $\omega$.

Multiplying both sides of 10.46 by $T$, and making the substitution of 10.48 gives 10.37.

Solvable Example¶

For practice in dealing with integration of complex functions, we will do the integration that leads to 10.40 in gory detail, with all the steps.

First we get rid of the absolute value —

and write the second integral as an integral from 0 to $\infty$ —

but we know how to differentiate even complex exponentials (see the discussion of 3.108), so we can write

Thus

or, using the fundamental theorem of integral calculus,

This function of $\omega$ is called a “pole.” While the function is perfectly well behaved for real $\omega$, it blows up for $\omega=-i \Gamma$, which is called the position of the pole in the complex plane. Now we just have to add the complex conjugate to get

which is 10.40. We already checked, in 10.41, that the factor of $1 / 2 \pi$ makes sense.

The pair 10.39-10.40 illustrates a very general fact about signals and their associated frequency spectra. In Figure 10.4 we plot $f(t)$ for $\Gamma=0.5$ and $\Gamma=2$ and in Figure 10.5, we plot $C(\omega)$ for the same values of $\Gamma$. Notice that as $\Gamma$ increases, the signal becomes more sharply peaked near $t = 0$ but the frequency spectrum spreads out. And conversely if ¡ is small so that $C(\omega)$ is sharply peaked near $\omega = 0$, then $f(t)$ is spread out in time. This complementary behavior is general. To resolve short times, you need a broad spectrum of frequencies.

Broad Generalities¶

We can state this fact very generally using a precise mathematical definition of the spread of the signal in time and the spread of the spectrum in frequency.

We will define the intensity of the signal to be proportional to $|f(t)|^{2}$. Then, we can define the average value of any function $g(t)$ weighted with the signal’s intensity as follows

This weights $g(t)$ most when the signal is most intense.

For example, $\langle t\rangle$ is the average time, that is the time value around which the signal is most intense. Then

measures the mean-square deviation from the average time, so it is a measure of the spread of the signal.

We can define the average value of a function of $\omega$ in an analogous way by integrating over the intensity of the frequency spectrum. But here is the trick. Because of 10.9 and 10.37, we can go back and forth between $f(t)$ and $C(\omega)$ at will. They carry the same information. We ought to be able to calculate averages of functions of $\omega$ by using an integral over $t$. And sure enough, we can. Consider the integral

This shows that multiplying $C(\omega)$ by $\omega$ is equivalent to differentiating the corresponding $f(t)$ and multiplying by $i$.

Thus we can calculate $\langle\omega\rangle$ as

and

$\Delta \omega$ is a measure of the spread of the frequency spectrum, or the “bandwidth.”

Now we can state and prove the following result:

One important consequence of this theorem is that for a given bandwidth, $\Delta \omega$, the spread in time of the signal cannot be arbitrarily small, but is bounded by

The smaller the minimum possible value of $\Delta t$ you can send, the higher the “fidelity” you can achieve. Smaller $\Delta t$ means that you can send signals with sharper details. But 10.63 means that the smaller the bandwidth, the larger the minimum $\Delta t$, and the lower the fidelity.

To prove 10.62 consider the function³

which depends on the entirely free parameter $\kappa$. Now look at the ratio

This ratio is obviously positive, because the integrands of both the numerator and the denominator are positive. What we will do is choose $\kappa$ cleverly, so that the fact that the ratio is positive tells us something interesting.

First, we will simplify 10.65. In the terms in 10.65 that involve derivatives of $f(t)^{*}$, we can integrate by parts (and throw away the boundary terms because we assume $f(t)$ goes to zero at infinity) so that the derivatives act on $f(t)$. Then 10.65 becomes

All other terms cancel. But

Thus the last term in 10.66 is just $\kappa$, and 10.65 becomes

10.68 is clearly greater than or equal to zero for any value of $\kappa$, because it is a ratio of positive integrals. To get the most information from the fact that it is positive, we should choose $\kappa$ so that 10.65 (=10.68) is as small as possible. In other words, we should find the value of $\kappa$ that minimizes 10.68. If we differentiate 10.68 and set the result to zero, we find

We can now plug this back into 10.68 to find the minimum, which is still greater than or equal to zero. It is

which immediately yields 10.62.

Equation 10.62 appears in many places in physics. A simple example is bandwidth in AM radio transmissions. A typical commercial AM station broadcasts in a band of frequency about 5000 cycles/s (5 kc) on either side of the carrier wave frequency. Thus

and they cannot send signals that separate times less than a few $\times 10^{-5}$ seconds apart. This is good enough for talk and acceptable for some music.

A famous example of 10.62 comes from quantum mechanics. There is a completely analogous relation between the spatial spread of a wave packet, $\Delta x$, and the spread of $k$ values required to produce it, $\Delta k$:

In quantum mechanics, the momentum of a particle is related to the $k$ value of the wave that describes it by

where $\hbar$ is Planck’s constant $h$ divided by $2 \pi$. Thus 10.72 implies

This is the mathematical statement of the fact that the position and momentum of a particle cannot be specified simultaneously. This is Heisenberg’s uncertainty relation.

²A example of a physical system with this kind of boundary condition would be a string stretched around a frictionless cylinder with radius $\ell$ and (therefore) circumference $2 \pi \ell$. Then 10.42 would be true because $x=-\pi \ell$ describes the same point on the string as $x=\pi \ell$.

³This is a trick borrowed from a similar analysis that leads to the Heisenberg uncertainty principle in quantum mechanics. Don’t worry if it is not obvious to you where it comes from. The important thing is the result.

10.4: Scattering of Wave Packets¶

In a real scattering experiment, we are interested not in an incoming harmonic wave that has always existed and will always exist. Instead we are interested in an incoming wave packet that is limited in time. In this section, we discuss two examples of scattering of wave packets.

Scattering from a Boundary¶

10-3

We begin with the easier of the two examples. Consider the scattering of a wave packet from the boundary between two semi-infinite dispersionless strings both with tension $T$ and different densities, $\rho_{I}$ and $\rho_{II}$, as shown in Figure 9.1. The dispersion relations are:

where $v_{I}$ and $v_{II}$ are the phase velocities in the two regions.

Specifically, we assume that the boundary condition at $-\infty$ is that there is an incoming wave,

in region $I$, but no incoming wave in region $II$, and we wish to find the outgoing waves, the reflected wave in region $I$ and the transmitted wave in region $II$.

We can solve this problem without decomposing the wave packet into its harmonic components with a trick that is analogous to that used at the beginning of this chapter to solve the forced oscillation problem, Figure 10.1. The most general solution to the boundary conditions at $\pm \infty$ is

where $g$ and $h$ are arbitrary functions. To actually determine the reflected and transmitted waves, we must impose the boundary conditions at $x = 0$, that the displacement is continuous (because the string doesn’t break) and its x derivative is continuous (because the knot joining the two strings is massless):

and

Using the chain rule in 10.79, we can relate the partial derivatives with respect to $x$ to deriviatives of the functions,

or

Differentiating 10.78, we get

Now for every value of $t$, 10.81 and 10.82 form a pair of simultaneous linear equations that can be solved for $g^{\prime}(t)$ and $h^{\prime}(t)$ in terms of $f^{\prime}(t)$:

Undoing the derivatives, we can write

where $k_{1}$ and $k_{2}$ are constants, independent of $t$. In fact, though, we must have $k_{1} = k_{2}$ to satisfy 10.78, and adding the same constant in both regions is irrelevant, because it just corresponds to our freedom to move the whole string up or down in the transverse direction. Thus we conclude that

and the solution, 10.77, becomes

The same result emerges if we take the incoming wave packet apart into its harmonic components. For each harmonic component, the reflection and transmission components are the same (from 9.16):

When we now put the harmonic components back together to get the scatter and transmitted wave packets, the coefficients, $\rho$ and $\tau$ appear just as overall constants in front of the original pulse, as in 10.86.

This scattering process is animated in program 10-3. Here you can input different values of $v_{I I} / v_{I}$ to see how the reflection and transmission is affected. Notice that $v_{I I} / v_{I}$ very small corresponds to a large impedance ratio, $Z_{I I} / Z_{I}$, which means that the string in region $II$ does not move very much. Then we get a reflected pulse that is just the incoming pulse flipped over below the string. In the extreme limit, $v_{I I} / v_{I} \rightarrow \infty$, the boundary at $x = 0$ acts like a fixed end. $v_{I I} / v_{I}$ very large corresponds to a small impedance ratio, $Z_{I I} / Z_{I}$, in which case the string in region $I$ hardly notices the string in region $II$. In the limit $v_{I I} / v_{I} \rightarrow 0$, the boundary at $x = 0$ acts like a free end.

Mass on a String¶

10-4

A more interesting example of the scattering of wave packets that can be worked out using the mathematics we have already done is the scattering of an incoming wave packet with the shape of 10.39 encountering a mass on a string. Here the dispersion relation is trivial, so the wave packet propagates without change of shape until it “hits” the mass. But then interesting things happen. This time, when we decompose the wave packet into its harmonic components, the reflection and transmission coefficients depend on $\omega$. When we add them

back up again to get the reflected and transmitted wave packets, we will find that the shape has changed. We will work this out in detail. The familiar setup is shown in Figure 10.6.

For an incoming harmonic wave of amplitude $A$, the displacement looks like

The solution for $R$ and $\tau$ was worked out in the last chapter in 9.39-9.45. However, the parameter $\epsilon$ of 9.38 depends on $\omega$. In order to disentangle the frequency dependence of the scattered wave packets, we write $R$ and $\tau$ as

where

is independent of $\omega$ — it depends just on the fixed parameters of the string and the mass. Note that in the notation of 9.38,

Suppose that we have not a harmonic incoming wave, but an incoming pulse:

Now the situation is more interesting. We expect a solution of the form

where $\psi_{\tau}(x+v t)$ is the transmitted wave, traveling in the $+x$ direction, and $\psi_{R}(x+v t)$ is the reflected wave, traveling in the $-x$ direction. To get the reflected and transmitted waves, we will use superposition and take $\psi_{in}$ apart into harmonic components. We can then use to determine the scattering of each of the components, and then can put the pieces back together to get the solution. Thus we start by Fourier transforming $\psi_{in}$:

We know from our discussion of signals that

Now to get the reflected and transmitted pulses, we multiply the components of $\psi_{in}$ by the reflection and transmission amplitudes $R$ and $\tau$ for unit $\psi_{in}$

Now we have to reverse the process and find the Fourier transforms of these to get the reflected and transmitted pulses. This is straightforward, because we can rewrite 10.98 and 10.99 in terms of single poles in $\omega$:

Now we can work backwards in 10.100 and 10.101 to get the Fourier transforms. We know from 10.55 that each term is the Fourier transform of an exponential. It is straight-forward, but tedious, to put them back together. The result is reproduced below (note that we have combined the two terms in each expression proportional to $1 /(2 \Omega-i \omega)$).