0
$\begingroup$

I saw details indicating "total energy of a Fourier-transformable signal equals the total area under the curve of squared amplitude spectrum of this signal."

I have no issue with that. But I have found pretty much no discussions about the consequence of applying the Rayleigh energy equation to the Fourier transform of sinusoidal signal - which of course has infinite energy. There are certainly discussions indicating that the energy equation can only be applied to finite energy signals, and therefore is not applicable to the Fourier transform of a sinusoidal time domain signal.

So - what is the bottom line? Does this mean that if one attempts to apply the energy equation to the Fourier transform of a sinusoid, then the result will be a finite value or finite number - and it will be a meaningless number, right?

That is - when applied to finite energy signals, the energy equation returns meaningful results. But when applied to infinite energy signals like a sinusoid, then it returns a value, but it will be a meaningless result, right?

$\endgroup$
2
  • 1
    $\begingroup$ When Rayleigh'e energy equation is mis-applied to a sinusoidal equation, one side "equals" infinity while the other side cannot even begin to be defined because the Fourier transform of the sinusoid does not exist in the conventional sense of the term. So yes, whatever you believe to be the "value" "returned" by Rayleigh's energy equation is pretty meaningless. $\endgroup$ – Dilip Sarwate Mar 13 '19 at 13:35
  • $\begingroup$ Thanks Dilip. Makes sense. It will mean that the energy equation just does not work for 'impulses' having finite area. Eg. cos(wo.t) transforms to two impulses, each impulse with area of 'pi' in the angular frequency domain. So we cannot simply square the areas and sum them. Eg. pi^2 + pi^2 = 2.(pi^2). The Rayleigh energy expression must be applied to 'magnitude' (not the area). And impulse functions have infinite magnitude. Definitely makes sense now. Thanks very much. $\endgroup$ – Kenny Mar 13 '19 at 21:32
1
$\begingroup$

An (infinite) sinusoidal signal does not really have a Fourier transform, since as you say it is not a finite-energy signal. It can be represented as a Fourier series, and there is a useful way to sort of write it as a "pseudo-Fourier transform" with Diract delta distributions, as well. However, if you try to take the magnitude of that spectrum and square it, it is not well-defined, since you would need the squared-magnitude of the Dirac deltas.

$\endgroup$
1
  • $\begingroup$ Thanks very much mateC. I can now see (after what you taught) that the energy formula applies to the amplitudes of the Fourier Transform components. So impulses have infinite amplitude, so the energy formula actually works for sinusoids too (as in predicting infinite energy). The mistake is to wrongly assume that the energy formula can be applied to the 'area' associated with the impulse. Thanks again! $\endgroup$ – Kenny Mar 14 '19 at 12:21
1
$\begingroup$

Rayleigh's theorem says that if $x(t)$ is a _square-integrable signal (meaning that $\int_{-\infty}^\infty |x(t)|^2 \,\mathrm dt$, which is called the energy of the signal, is finite) then $$\int_{-\infty}^\infty |x(t)|^2 \,\mathrm dt = \int_{-\infty}^\infty |X(f)|^2 \,\mathrm df \tag{1}$$ where $X(f)$ is the Fourier transform $$X(f) = \int_{-\infty}^\infty x(t)\exp(-j2\pi ft) \,\mathrm dt\tag{2}$$ of $x(t)$. The finite-energy assumption suffices to assure us that $X(f)$ is properly defined, that is, the integral in $(2)$ exists (doesn't diverge or otherwise fail to converge), and everything is hunky-dory.

Now, if we perversely insist on using Raylegh's theorem on signals such as $\cos(t), -\infty < t < \infty$ which are not square-integrable, we have the problem of dealing with the left side of $(1)$ since the integral there is \begin{align} \int_{-\infty}^\infty |x(t)|^2 \,\mathrm dt &= \lim_{T_1\to -\infty}\lim_{T_2\to \infty}\int_{T_1}^{T_2}|x(t)|^2 \,\mathrm dt\\ &= \lim_{T_1\to -\infty}\lim_{T_2\to \infty}\int_{T_1}^{T_2}\cos^2(t) \,\mathrm dt\\ &= \lim_{T_1\to -\infty}\lim_{T_2\to \infty}\int_{T_1}^{T_2}\frac{1+\cos(2t)+1}{2} \,\mathrm dt\\ &=\lim_{T_1\to -\infty}\lim_{T_2\to \infty} \left.\frac t2+\frac{\sin(2t)}{4} \right|_{T_1}^{T_2}\\ &= \lim_{T_1\to -\infty}\lim_{T_2\to \infty} \frac{T_2-T_1}{2} + \frac{\sin(2T_2)-\sin(2T_1)}{4} \end{align} in which the inner limit doesn't exist since the limitand diverges to $\infty$ as $T_2 \to \infty$. So, at best we can say with a straight face (after all, we are DSP engineers, not mathematicians!) that the left side of $(1)$ has "value" $\infty$. But matters are even worse for the right side of $(1)$ because $X(f)$ is even properly defined! Note that \begin{align}X\left(\frac{1}{2\pi}\right) &= \int_{-\infty}^\infty x(t)\exp(-jt) \,\mathrm dt\\ &= \int_{-\infty}^\infty \frac{\exp(jt)+\exp(-jt)}{2}\exp(-jt) \,\mathrm dt\\ &= \int_{-\infty}^\infty \frac{1+\exp(-2jt)}{2} \,\mathrm dt\\ \end{align} which diverges to $\infty$ (as does the integral for $X\left(\frac{1}{2\pi}\right)$) while for $f \neq \frac{1}{2\pi}$, $$X(f) = \int_{-\infty}^\infty \frac{\exp(jt)+\exp(-jt)}{2}\exp(-j2\pi ft) \,\mathrm dt$$ doesn't converge in that the limit $$\lim_{T_1\to -\infty}\lim_{T_2\to \infty}\int_{T_1}^{T_2}\frac{\exp(jt)+\exp(-jt)}{2}\exp(-j2\pi ft) \,\mathrm dt$$ doesn't exist because the integral oscillates between two fixed values regardless of how large $T_1$ or $T_2$ gets. In short, we can't even get started on the right side of $(1)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.