Electronic – How to we interpret the negative side of a double-sided frequency spectrum

frequencyspectrumwireless

We know that a frequency spectrum includes both positive and negative frequencies. However, how can we explain the negative frequency physically?

For example, in a wireless communication system, suppose the carrier frequency is 2.4GHz. The power carried by this frequency is, say 10uW. In a double-sided spectrum, this carrier frequency can be represented by two tones, one at f = 2.4GHz, another at f = -2.4GHz. Each of these tones carries a power of 5uW.

What I understand is that, the rotations of these two tones are in opposite directions. However, in the carrier signal, what is rotating? Is it related to the propagation of the EM waves?

Could anyone offer some insights?

Thanks.

Best Answer

It is useful to think of observable (real-valued) waves as the sum of two complex waves:

  • it doesn't require a nonlinear operation to drop the imaginary part, like \$\cos x=\Re(e^{ix})\$ does.
  • it doesn't require an (arbitrary) definition whether an observed sine wave has increasing or decreasing phase (because \$\sin x=-\sin -x\$).
  • it explains mixing behaviour

The latter is the most important point for practical purposes: if you mix two sine waves by multiplying them, you end up with two sines, one with the sum of the frequencies, one with the difference, and a phase shift:

$$ (\sin x)(\sin y)=\frac{\cos(x-y)-\cos(x+y)}{2} $$

This makes a lot more sense if you substitute

$$ \begin{eqnarray} \sin u&=&\frac{e^{-iu}-e^{iu}}{2}i\\ \cos u&=&\frac{e^{-iu}+e^{iu}}{2} \end{eqnarray} $$

to get

$$ \begin{eqnarray} (\frac{e^{ix}-e^{-ix}}{2}i)(\frac{e^{iy}-e^{-iy}}{2}i)&=&-\frac{(e^{ix}-e^{-ix})(e^{iy}-e^{-iy})}{4}\\ &=&-\frac{e^{ix}e^{iy}-e^{-ix}e^{iy}-e^{ix}e^{-iy}+e^{-ix}e^{-iy}}{4}\\ &=&-\frac{e^{i(x+y)}-e^{i(-x+y)}-e^{i(x-y)}+e^{i(-x-y)}}{4}\\ &=&-\frac{e^{i(x+y)}-e^{-i(x-y)}-e^{i(x-y)}+e^{-i(x+y)}}{4}\\ &=&-\frac{e^{i(x+y)}+e^{-i(x+y)}-e^{-i(x-y)}-e^{i(x-y)}}{4}\\ &=&-\frac{(e^{i(x+y)}+e^{-i(x+y)})-(e^{-i(x-y)}+e^{i(x-y)})}{4}\\ &=&-\frac{\cos(x+y)-\cos(x-y)}{2}\\ &=&\frac{\cos(x-y)-\cos(x+y)}{2} \end{eqnarray} $$

That way it is also way more intuitive how the spectrum of the multiplication of two signals is the convolution of the signal's spectra: the input signals have two peaks each, and the output signal has four.

This system also remains consistent when you operate on complex-valued signals, e.g. digital baseband, where it is possible to mix with a complex-valued sine and get a frequency shift without moving half the signal's energy to a mirror image.

The mixing behaviour is observable in the real world, so a model that adequately explains it is useful.