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We have a DAC capable of being driven with IQ data (thus the bandwidth is $-f_s/2$ to $f_s/2$), and the task is to create an "arbitrary" BPSK waveform at IF. I think that means generating this waveform:

$$ y(t) = e^{j\left( 2 \pi f_c t + \phi(t) \right )} $$

where $y$ is complex (the I and Q portion), $f_c$ is the carrier frequency, $\phi(t)$ is the bit/chip sequence over time (will toggle between $0$ and $\pi$). The next step would be the convert this to the sample domain.

I guess my confusion comes from looking online at IQ BPSK and it usually just has a binary sequence driven directly into I and Q, and then a mixer stage. I'm not sure if that applies to me since this is just an IQ dac, not an IQ mixer.

Can someone point me in the right direction?

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  • $\begingroup$ Since the DAC takes in I and Q data, does it provide I and Q analog outputs or a single (real) output? If a real output with I and Q in, then the DAC would also include an IQ mixer- can you provide more specifics on this DAC (Manuf and PN), as that might clear up the confusion. $\endgroup$
    – Dan Boschen
    Commented May 6, 2023 at 19:22

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Your BPSK symbols don't map to I & Q, they only map to I: 1 + 0j and -1 + 0j. You take that resultant complex baseband signal and multiply by $cos({2{\pi}f_ct/Fs)} $

You can see the answer here for some more detail: QAM modulation problem with python

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  • $\begingroup$ But if Q is always 0j, and I multiply by a real cos function, then I would get a real signal out. That wouldn't work for representing negative frequencies, correct? $\endgroup$
    – Sittin Hawk
    Commented Oct 28, 2022 at 18:16
  • $\begingroup$ Check out these two references: dsplog.com/2008/08/08/negative-frequency dspillustrations.com/pages/posts/misc/… $\endgroup$
    – BigBrownBear00
    Commented Oct 30, 2022 at 11:57
  • $\begingroup$ I think if your signal is real you want to just multiply by cos, but if it's complex baseband, then you multiply by $e^{2{\pi}jf_ct/Fs}$ $\endgroup$
    – BigBrownBear00
    Commented Oct 30, 2022 at 11:59
  • $\begingroup$ A real signal has negative frequencies-- when the signal is real the positive and negative frequencies are complex conjugate symmetric (Hermitian symmetric). When the signal is complex then the positive and negative frequencies are no long Hermitian symmetric and can be independent. $\endgroup$
    – Dan Boschen
    Commented May 6, 2023 at 19:24

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