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I know that when we have a message signal

$$m(t) = \cos(\omega t)$$

when it gets squared, its bandwidth gets doubled, because

$$\cos^2(\omega t) = \frac{1}{2}+\frac{1}{2}\cos(2\omega t)$$

But is this true to every possible signal m(t)? What if I raise it to an arbitrary power n?

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    \$\begingroup\$ In a few seconds I could think of a signal where the bandwidth gets lower when you square it. \$\endgroup\$
    – pipe
    May 3, 2020 at 20:01
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    \$\begingroup\$ Your initial signal has 0 bandwidth, and the result of squaring gives \$2\omega\$ bandwidth. So that's much more than a factor of n increase by squaring. \$\endgroup\$
    – The Photon
    May 3, 2020 at 20:04
  • \$\begingroup\$ There is no change in bandwidth with a sinewave signal. What happens is the signal that was at a given frequency is now at twice the frequency. That is why a squaring circuit is often used a a frequency doubler. With more complicated signals, more frequencies that just the doubled one are created due to the cross products generated by the squaring operation. \$\endgroup\$
    – Barry
    May 3, 2020 at 22:09
  • \$\begingroup\$ I think we should give the OP the benefit of the doubt regarding the question. Of course a pure sinewave has zero BW. But what if it is a modulated sinewave? What happens to the BW if two copies of the signal are passed to the inputs of an analog multiplier or other mixer? \$\endgroup\$
    – user57037
    May 4, 2020 at 3:53
  • \$\begingroup\$ To the OP, in order for the bandwidth to be greater than zero, some form of modulation must be introduced. For example w could be a function of time: w(t). You can think of w as the carrier frequency. Your trig identity has demonstrated that the carrier frequency is doubled when the signal is squared. But that is not the same as doubling the bandwidth. \$\endgroup\$
    – user57037
    May 4, 2020 at 3:57

2 Answers 2

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If you raise your time-domain signal to the power n, then your bandwidth does increase, but not necessarily by a factor of n. It will depend upon how you want to define bandwidth.

This is because multiplication in the time domain amounts to convolution in the frequency domain. Let's take the example of n=2. If we denote the Fourier transform operator to be \$F\$, then

$$ F[y^2(t)]= \int Y(\omega')Y(\omega-\omega' )d\omega' $$

You will have non-zero spectral content whenever \$Y\$ overlaps with a shifted version of itself. Therefore your signal has spectral content covering a region twice as wide as that of the original signal, but the 3 dB point may not necessarily double. So it depends a bit on how you want to define bandwidth. For a Gaussian time domain pulse, I believe the 3-dB bandwidth increase will go like \$n^{1/2}\$. The graphic below might help illustrate the point:

enter image description here

(picture borrowed for illustrative purposes from here)

Each time you multiply again by \$y(t)\$ in the time domain, you convolve again with \$Y(\omega)\$, increasing the bandwidth yet again.

The other effect of raising the time-domain signal to a power can be described as generating \$\textit{harmonics}\$ of the original signal. This is best visualized with a signal that has somewhat local support near its center frequency rather than extending from DC on upward (i.e. a 'narrow band' signal). In that case, the spectral content at negative frequency and the spectral content at positive frequency produce, in the convolution, content centered at the sum and difference frequencies (of the center frequency). Each of those 'harmonics' will have a bandwidth larger than the original bandwidth due to the spreading introduced by the convolution.

You could legitimately view these harmonics as an increase in the bandwidth well beyond the the effect described so far because you have spectral content all the way out to the harmonics. So I suppose it depends a bit on what is important to you, and what you will do with the signal after raising it to the nth power. If you intend to filter out harmonics, then you should look at raising to the nth power as increasing the bandwidth by some factor like \$n^{1/2}\$, depending on how you define bandwidth. If you aren't going to filter, then you should probably consider the bandwidth to be approximately n times the center frequency that your original signal contained for a 'narrow band' signal due to the generation of harmonics.

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  • \$\begingroup\$ How about a square wave. If I raise it to any even power, the bandwidth drops to zero because it becomes DC. \$\endgroup\$
    – user57037
    May 3, 2020 at 23:12
  • \$\begingroup\$ @mkeith Great comment. I assume you mean a bipolar square wave. I think this is a pathological case of the idealization, in that it relies on the fact that the value at the transition time is undefined. Is it the positive value or negative? Give it an arbitrarily small, non-zero transition width, and you don't get the result of a dc signal. \$\endgroup\$
    – rpm2718
    May 3, 2020 at 23:17
  • \$\begingroup\$ OK, let's say I have a modulated signal. I feed two copies to a mixer. Normally that is not done. Normally the mixer would be fed a pure sine wave and a modulated sine wave. But if I do that, is the bandwidth double? If I mix it back down to baseband, what is the effect on the baseband signal? Is it frequency doubled (pitch shifted by one octave?) \$\endgroup\$
    – user57037
    May 3, 2020 at 23:31
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    \$\begingroup\$ @mkeith We would really need to get into the details of the signals you are talking about...narrow band high frequency, or broad band? I don't think we can do that in comments. However, the analysis I gave is general enough to describe that situation completely. \$\endgroup\$
    – rpm2718
    May 3, 2020 at 23:35
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    \$\begingroup\$ @mkeith Yes, it is at the heart of the OP's question, and I think I answered that question. One could take the analysis I laid out and apply it to the situation you describe to give a precise, quantitative answer. You and I just can't go back and forth enough in the comments section to do that. \$\endgroup\$
    – rpm2718
    May 3, 2020 at 23:41
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The bandwidth of your original signal is 0, because the support in spectrum is infinitely thin. (Check for yourself: draw the spectrum of a cosine – it's two lines, at \$\pm\omega\$. The bandwidth of a line is infinitesimally small.)

Same applies to your squared signal.

So, you're coming from a wrong statement, in my world; your material might have a less strict definition of bandwidth, though, being defined as the furthest distance between occupied positive frequencies.

But, that definition won't work for long, if you start actually dealing with signals, because you'll need to thing about signals that actually do have a non-zero bandwidth, so I'll stick with my definition (also, because it makes sense, physically).

To answer your question:

No, the bandwidth doesn't get multiplied by \$n\$. In fact, I find it harder finding an example where that actually happens than one where that doesn't happen.

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    \$\begingroup\$ What about the sudden appearance of DC content. Does that extend the bandwidth to 2F when the signal is squared. But then when cubed, the DC gets removed. Dilemma or just dodgy thinking? \$\endgroup\$
    – Andy aka
    May 3, 2020 at 21:35
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    \$\begingroup\$ I think maybe to give the OP the benefit of the doubt we could inject the assumption that the sine wave is modulated so that it occupies non-zero bandwith. In the case of a pure sine wave, since the initial bandwidth is zero, the fact that the bandwidth after doubling is also zero does not disprove the notion. X=2X when X = 0. \$\endgroup\$
    – user57037
    May 3, 2020 at 23:42
  • \$\begingroup\$ @Andyaka neither – cubing the signal, i.e. introducing odd-order intermodulation shifts energy from 0 and -2F and 2F to -F,F and +-3F \$\endgroup\$ May 4, 2020 at 5:59

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