I have a rather elementary frequency upconversion/mixer question, perhaps so simple that it does not warrant a real question but rather just a reference to a website/tutorial (that I have not been able to find). If that is the case, feel free to comment that and we can resolve the question that way.

My question is as follows. Say I send in a 10ns duration square pulse at an intermediate frequency (IF) of 10 MHz (ideal, for the sake of the question? If that leads to issues we can make it Gaussian), along with an local oscillator (LO) frequency at 4 GHz, into a mixer . The mixer has an appropriate IF bandwidth (something broadband), and the LO an appropriate power.

Now, the result of this, as I understand it, should be a 10ns square pulse of a 4.01 GHz signal. But how was the mixer able to 'tell' that it had a 10 MHz signal coming in, if it only received 1/10th of a period of that signal? In the sense that 10 MHz has a period of 100ns, and the signal was only 'on' for 10ns.

How is it able to (passively) reconstruct the IF, only being able to 'sample' it for less than a full period? Is this indeed possible? And what kind of limitations does one run into here? What if it is a 1ns signal?

At this point I've asked my question. I can speculate a bit more/elaborate on my thought process, but feel free to skip that if you prefer. As far as I can guess, what goes on as something to do with the bandwidth of the mixer, and also the sampling rate of the instrument/entity producing the 10ns pulse to begin with. That is to say, if I have a 1 GS/s instrument, that 10ns pulse will have 10 points. That should certainly lead to a different effect than sending in a single point, or 100.

That makes me think it is related to how many cycles one needs to perform a Fourier transform. In a way, what we are doing is sending in some $\sin{(2\pi f_{\rm LO} t)}$ multiplied by some square impulse, which needs to be reconstructed. That again either brings me to the sampling rate of the instrument, or the bandwidth of the mixer.. Somehow also the fact that a 'square' 10ns pulse produces frequency components far in excess of 10 MHz should be factored in, I suppose.

Would it be possible to write down what goes on here mathematically? I suppose it must be. Maybe that makes it easier to spot the limitations and distortions that arise in various cases.

  • $\begingroup$ Please explain the abbreviations (IF, LO) before you use them, so that the question is understandable for a broader audience. $\endgroup$ Commented Mar 3, 2022 at 10:10
  • $\begingroup$ I've added that, thanks $\endgroup$
    – user129412
    Commented Mar 3, 2022 at 10:17

1 Answer 1


a 'square' 10ns pulse produces frequency components far in excess of 10 MHz should be factored in, I suppose.

Yes. A 50 ns square pulse (50% duty cycle) at 10 MHz would also have components exceeding 10 MHz. The only 10 MHz waveform that doesn't have components at higher frequencies is a sine wave.

A 10 MHz signal with 10% duty cycle includes a 10 MHz sine wave component, and components at higher frequencies. If your whole system has bandwidth of 10 MHz, you will reconstruct a 10 MHz sine wave. If the bandwidth is higher, the output will more resemble the input 10 ns pulses.

how many cycles one needs to perform a Fourier transform.

If you'd like to learn the frequency spectrum of a waveform (e.g. a square wave with 10% duty cycle) you need just one cycle (providing you know the cycle length). If you want to do it using digital Fourier transform (rather than analytically), the higher sampling frequency (the more points per cycle), the larger bandwidth the transform will output.


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