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TL,WR: Why do we use signal energy as a measurement of an approximation's error, and not something "simpler" like the absolute value of the error?


Background: I'm reading through my textbook for my introduction to signals class, and they have just gotten to a point where they are discussing Fourier series representations of signals.

Specifically, they're talking about the square wave, and how it needs to be decomposed into an infinite number of harmonically related complex exponentials

$$x(t) = \sum_{k = -\infty}^{+\infty} a_k \cdot e^{j k \omega_0 t}$$ and how to get an approximation error, they want to use a finite number to represent it

$$e_N (t) = x(t) - x_{N} (t) = x(t) - \sum_{k = -N}^{+N} a_k \cdot e^{j k \omega_0 t}$$

Then they say,

In order to determine how good any particular approximation is, we need to specify a quantitative measure of the size of the approximation error. The criterion that we will use is the energy in the error over one period. $$E_N = \int_{T} | e_N (t) |^2 dt$$

Why did they choose to use this as the measurement of the error, instead of, say, the absolute value of the error? Integrals are "hard" (by which I mean more time-consuming) math, and I'd like to hear a justification from anyone who works with this stuff closely. I suppose doing it this way might reveal places where the error in the signal spikes, which could be useful information, but that's the only thing that pops to mind immediately.

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  • \$\begingroup\$ The absolute value (or magnitude) bars in the expression probably aren't needed. Squaring achieves the desired result. Also, this isn't "energy" in the usual sense of physics. This is more a "sum of the squared error" and if memory serves is used to develop the idea of a correlation coefficient that is based upon setting the derivative to zero (to minimize this error result.) The use of a squared term provides a unique solution. Had they used sum of the absolute value, there would be many solutions and no way to choose among them. I'd have to develop the math for you, though, to prove this. \$\endgroup\$
    – jonk
    Feb 10, 2018 at 19:25

2 Answers 2

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Using all the area under a data bit signal (integrating) and its energy gives the maximum signal when compared to the noise error. Less performing methods use peak signal and then peak noise may apply but gives worse SNR and higher BER. Good demodulators may be harder than center peak sample but integrating the entire signal energy over the symbol gives higher SNR and lower BER.

When you get into ADC specs they use about 6 different ways to express error for quantifying ADC dynamic performance.

SINAD (signal-to-noise-and-distortion ratio),
ENOB (effective number of bits),
SNR (signal-to-noise ratio),
THD (total harmonic distortion),
THD + N (total harmonic distortion plus noise),
SFDR (spurious free dynamic range).

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We very likely do our calculations in frequency domain because Fourier series and transforms are handy. Unfortunately they do not present signals exactly, there can be large large error spikes if we transform our Fourier series back to the time domain. But mathematicians have shown that those error spikes have less and less energy, if we add more frequency components into the account. The limit is zero energy error with all terms, but there still can be some zero length error spikes. Mathematicians would say that Fourier series converge only energy-wise.

In addition: Error power and energy are less complex than the absolute value of the error and its average. That's because absolute error hasn't continuous derivative, there's allways a sharp angle at zero crossing. That excludes quite a bunch of powerful math methods.

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