I've found some conflicting information about how the I/Q waveforms look compared to the 4-PAM signals before the modulator.

Here it is shown that the 4-PAM digital signals are filtered, so the analog I/Q components look like the 4-PAM levels:

Relation between Bandwidth and Baud-Rate for 8-PSK

A confirmation for this method can be found in the paper "FPGA implementation of software defined radio model based 16QAM. Naghmash, Majid & Ain, M.F. & Hui, C.Y. (2009)". From this paper, the image that best exemplifies the filtering is this: enter image description here

But then I found this: enter image description here Source: https://www.slideshare.net/aknigin/digital-communicationunit-3

So the output of the balance modulator in the above picture does not look like the filtered 4-PAM. The schematic for the transmitter can be seen at page 76 of the presentation.

Some observations I've made:

-to get the filtered 4-PAM look approximately the same as the digital signal in terms of amplitude, there must be no ISI (intersymbol interference), thus a higher bandwidth must be occupied by the filtered signal than the absolute minimum required. Specifically, the raised cosine filter must have a higher rolloff, which in the case of beta parameter equal to 1 instead of 0 means double occupied bandwidth. (https://en.wikipedia.org/wiki/Raised-cosine_filter)

-in the second case, where the signal does not look like the 4-PAM levels, the bandwidth also appears to have doubled because of the phase shifting. The main lobe bandwidth is doubled compared to the PAM signal.


  1. Are these different methods for the same 16-QAM?
  2. How are these compared in terms of bandwidth efficiency?
  • $\begingroup$ for any modulation where the symbols are 0 if you all average them, the spectrum is completely defined by the pulse shaping filter, and not by the modulation; hence, the slide you show doesn't have much to do with reality; it's not pulse-shaped $\endgroup$ – Marcus Müller Jan 27 at 17:52

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