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My understanding is that if a single tone at frequency F is inputted to the frequency doubler, one should observe at its output in the spectrum the same tone but now at a frequency 2F.

However, what happens if this signal is no longer a tone? To make it simpler and to minimize the impact of possible modulation, let's consider this signal now being a wide-band Gaussian noise with a bandwidth B and centered at F. I am wondering when this noise signal encounters the frequency doubler will the doubler make any practical impact on the bandwidth of such a signal.

Practically, can we expect that at the output of the doubler we will now see spectrum of the wideband white noise (with the possible realistic modulation due to the doubler spectrum) centered at 2F and with the same bandwidth B, or with bandwidth 2B, or something else?

Edit (with some additional details about the setup): What is meant here as a frequency doubler/multiplier is a type of PiN diode, such as Cobham DH267. The carrier frequency of the wideband noise signal at the input to the doubler, F, is 6GHz, and bandwidth B of the wideband noise signal is 300 MHz.

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  • $\begingroup$ The term "frequency doubler" is too broad -- to double the frequency of an input sinusoid, the circuit must include a nonlinear element. If you allow that nonlinear element to be anything, then you could get any answer. If you have a specific example of a frequency doubler you're looking at, then edit your question with a schematic or a reference and ask about that. $\endgroup$
    – TimWescott
    Apr 6 at 14:58
  • $\begingroup$ Thanks for the feedback. I added in the edit some extra details about the doubler based on a PiN diode. $\endgroup$
    – Akhaim
    Apr 6 at 16:04
  • $\begingroup$ Could you provide a schematic. Specifically, how is your diode connected, are there any passive components (including striplines, cavities, etc.), is the thing followed by a filter, etc. A diode, by itself, is not a doubler, but it certainly can be a component of one. $\endgroup$
    – TimWescott
    Apr 6 at 19:44

2 Answers 2

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The result of frequency doubling is dependent on the waveform and doubling approach. I completed a simulation showing the result of doubling a Gaussian noise filtered to 300 MHz BW centered at 6 GHz.

I moved the original spectrum at 6 GHz (in orange below), to be superimposed with the resulting spectrum at 12 GHz which is in blue. The doubler in this case was done by multiplying the signal with itself (squaring), where the result is intuitive when you consider the product in time to be a convolution of the spectrums.

spectrum of result

I repeated the experiment with an absolute-value non-linearity (which also has a strong second harmonic to be used as a doubler) with the comparative result shown below:

spectrum of absolute value

The absolute value process results in many more harmonics, resulting in more intermodulation products in vicinity to the doubled spectrum. This explains the additional smearing we see of the noise in the lower plot. The point here is the actual doubling process as well as the waveform used must be considered in evaluating its effect on the resulting spectrum at 2x the carrier frequency.

The characteristics of the resulting doubled waveform depend on the waveform being doubled and the doubling approach used. For example, if the waveform being doubled had amplitude modulated components only (no phase modulation), the doubled output would have a strong component right at the 2F frequency with much reduced modulated energy (this is a carrier recovery technique):

AM only doubled

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depends on what you mean with "frequency doubler".

  • If it's just a mixer that mixes its input with itself, yes, you get all intermodulation products between components in the signal. That can double the bandwidth; for special signals, however, intermodulation products might cancel. It depends on the signal, and what the bandwidth of said mixer is.
  • If it's, for example, a PLL that trains an oscillator to oscillate at twice the input frequency, then the bandwidth around the input carrier will look like noise to the PLL; you might get a single stable output frequency, you might get a very noisy output, the PLL might not lock at all, it might lock to a subharmonic or harmonic of the input instead; if the bandwidth of your input signal falls into the loop bandwidth of the PLL, you would potentially even get a frequency shifted version of your input of twice the bandwidth.
    That depends on what your actual input signal is and how your output oscillation synthesizer works.
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    $\begingroup$ Given how small and inexpensive processors are, it could be designed so that when it encounters noise on the input it plays Orson Well's War Of The Worlds. Who knows? $\endgroup$
    – TimWescott
    Apr 6 at 14:57
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    $\begingroup$ I'm trying to think of what a simple diode-filter frequency doubler would do with wide-band Gaussian. For wide-enough Gaussian noise, it'd probably not respond at all (because the voltage would never light up the diode). At some point you'd just generate a signal out of the diode that would completely "fill" the bandwidth of the filter. Etc. -- each doubler topology would have its own response. $\endgroup$
    – TimWescott
    Apr 6 at 15:00
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    $\begingroup$ @TimWescott I like the idea. "In this paper, we present an adaptive filter that is advantageous for high-speed wireless communications subject to War Of The Worlds-Noise and its phase equivalent" $\endgroup$ Apr 6 at 15:00
  • $\begingroup$ April 1 2024 is nearly a year away... $\endgroup$
    – TimWescott
    Apr 6 at 15:01
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    $\begingroup$ We note with small angle phase noise that the bandwidth due to the phase modulation doesn't increase when we multiply by N (as long as we remain small angle) but the side bands increase by 20Log10(N)-- sideband level relative to carrier is the phase and the offset is the modulation rate. Multiplying does not change the modulation rate, but it does increase the peak phase deviation by N (easier to see this on a phasor IQ plot). $\endgroup$ Apr 6 at 15:26

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