I am using Xilinx's RFSoC in I/Q mode. The important point is that in the digital space I have real and complex samples. I take the incoming data and multiply it by a complex sinusoid.

In my test setup I am sampling a 125 MHz sine wave and am mixing it with a 5 MHz sinusoid. I take the mixed signal and pass it out a DAC to view it on a spectrum analyzer. What I would like to see is the 125 MHz digitally upconverted to 130 MHz. What I actually see is two tones: 120 MHz and 130 MHz.

Is there a way, without just filtering the 120 MHz sine wave, to upconvert the 125 MHz tone to 130 MHz by mixing?

  • $\begingroup$ I had mis-read your question. Are you positive you are multiplying with a complex exponential and still see a peak at 120 and 130 MHz? This would happen if you had multiplied with a cosine or sine wave but shouldn't happen otherwise. $\endgroup$ – Engineer Apr 5 '20 at 18:56
  • $\begingroup$ I mean you have a 125 MHz sine wave so if you look at all frequencies positive and negative, you should see spikes at $\pm$125. Then multiplying with a true complex exponential with $f_0=5$ MHz should shift the spectrum to the right by 5 MHz so now the two spikes are at -120 and 130 MHz. $\endgroup$ – Engineer Apr 5 '20 at 18:59
  • $\begingroup$ Thank you for the reply. Given your response I am wondering if the problem is actually with the ADC. If the two ADCs are actually sampling at the same time instead of with a 90 degree offset, you would see a pure sine wave in the digital domain. If I mixed the pure sine wave (Instead of a complex sinusoid) with my internally generated complex sinusoid, this is the behavior I would expect, right? $\endgroup$ – annapCoug Apr 5 '20 at 19:17
  • $\begingroup$ Yes, the pure sine wave and complex exponential will give you two peaks as in my answer $\endgroup$ – Engineer Apr 5 '20 at 20:28
  • $\begingroup$ How do you view a complex signal on a spectrum analyzer ? $\endgroup$ – Hilmar Apr 5 '20 at 21:30

You can use the property of Fourier transforms which says that: $x(t)e^{j2\pi f_0t}\leftrightarrow X(f-f_0)$. That is, multiplying the time domain signal with a complex exponential will give you a shift in the frequency domain.

By multiplying by another sine, you get: $x(t)\text{sin}(2\pi f_0 t) \leftrightarrow X(f)*\text{FT}\big(\text{sin}(2\pi f_0 t)\big)=\frac{X(f-f_0)-X(f+f_0)}{2j}$, and so you see an extra peak at $120$ MHz.


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