Superposition of waves whose wavelengths are continuous

Question

I know how to find the resultant waves when finitely or countably many waves are superimposed but how do I find the wave equation when there are infinitely many waves whose wavelength is continuous? Is there are formula of any sort?

Answer 1

I am going to answer using complex exponentials to represent waves instead of sinusoids since the answer is much more natural in that language. We can always tranlate between the two using

\begin{align} e^{+i\omega t} &= \cos(\omega t) + i \sin(\omega t)\\ e^{-i\omega t} &= \cos(\omega t) - i \sin(\omega t)\\ \cos(\omega t) &= \frac{1}{2}\left(e^{+i\omega t} + e^{-i\omega t} \right)\\ \sin(\omega t) &= \frac{1}{2i}\left(e^{+i\omega t} - e^{-i\omega t} \right) \end{align}

I will call $f(t)$ the end result of the superposition of many waves. I'll call a wave at a single frequency $h(\omega)$. We have for each $\omega$

\begin{align} h(\omega) = C_{\omega}e^{+i \omega t} = C(\omega)e^{i\omega t} \end{align}

$h(\omega)$ includes an $e^{i\omega t}$ part which describes the oscillation at frequency $\omega$ as well as a (complex) coefficient $C_{\omega}$ which captures the amplitude and phase of the wave at that frequency.

One Frequency

If we have a wave with one frequency then we have

\begin{align} f(t) &= A_{\omega} \cos(\omega t) + B_{\omega} \sin(\omega t) = \frac{1}{2}\left(A_{\omega} - iB_{\omega}\right) e^{+\omega t} + \frac{1}{2}\left( A_{\omega}+iB_{\omega}\right)e^{-i\omega t}\\ f(t)&= C_{\omega}e^{+i\omega t} + C_{-\omega}e^{-i\omega t} \end{align}

Note that the constraint that $f(t)$ is real (which is the case for any real physical signal we are talking about) imposes the constraint that $C_{-\omega} = C_{\omega}^*$. That is the coefficient for frequency $-\omega$ must be the complex conjugate of the coefficient for frequency $+\omega$. We can relax this constraint if we allow $f(t)$ to be complex. In that case $A_{\omega}$ and $B_{\omega}$ can be complex. I won't say more about the translation between sinusoidal and exponential representations. You can use the rules above to translate what comes below if you insist.

Finite Number of Frequencies

\begin{align} f(t) = \sum_{n=1}^N h(\omega) = \sum_{n = 1}^N C_{\omega_n} e^{+i\omega_n t} \end{align}

You said you were familiar with the case for finite number of waves so I assume you understand this. Since we are working in a complex representation, if we insist that $f(t)$ is real then it should be implicitly understood that for every $\omega$ in the set of frequencies over which we are summing, $\{\omega_1 \ldots \omega_N\}$ it's opposite, $-\omega$, is also in the set.

Countable Number of Frequencies

This is the exact same as above but we sum over a countable rather than finite set.

\begin{align} f(t) = \sum_{n=1}^{\infty} h(\omega) = \sum_{n=1}^{\infty} C_{\omega_n} e^{+i\omega_n t} \end{align}

The constraint on positive and negative frequencies is still in place if $f(t)$ is constrained to be real.

Infinite Number of Frequencies

If there are an infinite number of frequencies then we can no longer perform the countable sum. We must instead use an integral. To reflect this we switch over from the notation $C_{\omega}$ (common for a function whose domain is the integers, or at least countable) to the notation $C(\omega)$ indicated above (common for generic functions). We have

\begin{align} f(t) = \int_{\omega = -\infty}^{+\infty} h(\omega) d\omega = \int_{\omega = -\infty}^{+\infty} C(\omega) e^{+i\omega t} d\omega \end{align}

Note that we sum over positive and negative frequencies. If $f(t)$ is real it can be seen that $C(-\omega) = C(\omega)^*$. Here I've integrated over the whole real line. If you just want to include waves within some interval you can just set C(\omega) to zero outside of the interval you are interested in.

This is exactly the formula for the Inverse Fourier transform of $C(\omega)$ to get the function $f(t)$. Thus the answer to your question "is there a formula of any sort" is yes. It is the Fourier transform and Inverse Fourier Transform formulas. There are many functions for which the Fourier transform pairs are well known.

Note that your question amounted to asking if we know $C(\omega)$ can we calculate $f(t)$. The formula above shows how to do that if you can perform the integral. What you may or may not be surprised to learn is that for the smooth functions which decay away at infinity that we often work with in physics ($L^1$ functions) you can do the opposite. For any function $f(t)$ we can find a function $C(\omega)$ with the property that

\begin{align} f(t) = \int_{\omega = -\infty}^{+\infty} C(\omega) e^{+i\omega t} \end{align}

The $C(\omega)$ is given by

\begin{align} C(\omega) = \frac{1}{2\pi} \int_{t=-\infty}^{+\infty} f(t) e^{-i\omega t} \end{align}

Final Notes

For more info on Fourier Transforms and questions about positive, and negative frequencies and factors of $2\pi$ see Fourier transform standard practice for physics and for more info about the complex representation of waves see Using complex numbers to represent waves

Brief note on units: Above you can see that $C_{\omega}$ has the same units as $f(t)$ for the case of finite or countable sums. However, if you recall that $d\omega$ has units of frequency, you will see that $C(\omega)d\omega$ has the same units as $f(t)$ and therefore $C(\omega)$ has the same units as $f(t)$ per frequency, or $f(t)$ times units of time.