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This signal uses Offset QPSK with a half sine pulse shaping filter for the purpose of getting a constant signal envelop. Does anyone know the exact bandwidth of this signal after being pulse shaped, i.e. the nul to nul bandwidth of the main lobe?

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  • $\begingroup$ There's different PHYs for 802.15.4. Which one are you referring to? $\endgroup$ – Marcus Müller Aug 18 '18 at 0:33
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    $\begingroup$ A half sine pulse shaping filter results in a null to null bandwidth that is 3/T where T is the symbol duration. You can see this by taking the FT of a half sine pulse. (In comparison the null the null bandwidth for a rectangular pulse is 2/T). $\endgroup$ – Dan Boschen Aug 18 '18 at 1:08
  • $\begingroup$ I was referring to the 2.4 GHz part of 802.15.4. 3/T is a very inefficient way of using the spectrum!!. so if i have an unfiltered BPSK with a nul to nul of 2/T , filtering it would increase the BW??!! and why would I do that to start with? since the unfiltered BPSK or QPSK have a constant envelop too? because of the side lobes perhaps? $\endgroup$ – Hatem Tawfik Aug 18 '18 at 5:03
  • $\begingroup$ What i also do not understand is that many websites and documents talk about a BW of 2 MHz for 802.15.4 including the spectrum at 2.4 GHz, I know that the OQPSK symbol rate is 1 MHz . so where does the notion of a BW of only 2 MHz come from? Also I read that using OQPSK with half sine filter is like MSK, and MSK does not have such a high BW! Could you please help me with this confusion? Thank you $\endgroup$ – Hatem Tawfik Aug 18 '18 at 5:10
  • $\begingroup$ I am not intimately familiar with 802.15.4 but the 2MHz you saw may be referring to the channel allocation. MSK and OQPSK have the same spectrum and the -3 dB BW would be 1.2 MHz for a 1 MHz symbol rate. @QasimChaudhari wrote a nice tutorial here you can read for further details on the relationship between MSK and O-QPSK: dsprelated.com/showarticle/1016.php $\endgroup$ – Dan Boschen Aug 18 '18 at 11:47
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I found this link that details the spectrum for 802.15.4: https://www.semanticscholar.org/paper/IEEE-802.15.4-PHY-analysis%3A-Power-spectrum-and-Gupta-Wilson/2892da39a7ecec1945595d46950438a5bb8777af

And pasted the plot below.

802.15.4 spectrum

This is consistent with what I would expect for a half-sine pulse-shaped pulse for a symbol rate of 1 MHz (and that would apply to BPSK, QPSK, OQPSK etc where the power spectrum for a random data sequence is the Fourier Tranform squared of the base pulse that is transmitted.). In Offset-QPSK the I and Q channels can be viewed as two independent (in data content) BPSK transmissions, in quadrature, offset in time by half a symbol. Thus each produces an independent but identical power spectrum that adds in power.

Below I compare the Fourier Transforms for a rectangular pulse of 1 us duration vs the half-sine windowed pulse of the same duration. I normalized the power of the two to be the same. Although the pulse shape increases the width of the main lobe, there is a clear spectral improvement by doing the half-sine pulse shape as demonstrated in this plot (Notably, the power spectrum of the rectangular pulse rolls of at a rate of $1/f^2$, or -20dB/decade, while the power spectrum of the half-sine pulse rolls of at a rate of $1/f^4$ in spectrum).

Fourier Transform of rectangular pulse (Red) vs half-sine pulse (Blue): rect pulse vs half sine

This clearly is not as good as that which can be achieved with more spectral efficient pulse shaping such as raised-cosine shapes but offers the significant advantage to low power systems of being a constant envelope modulation when used in an O-QPSK modulation. In this case we are able to drive the power amplifier (PA, which typically dominates power dissipation in the transceiver) into saturation without introducing AM/PM distortion. In saturation the PA can operate at a much higher efficiency (Tx power radiated at a given DC power in).

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  • $\begingroup$ +1 .. but why do we not have AM/PM ? $\endgroup$ – Ahmad Bazzi Dec 19 '18 at 10:40
  • $\begingroup$ @AhmadBazzi "Constant Envelope" means that there is no AM component. AM/PM is the translation from amplitude variations to phase variations given that characteristic of power amplifiers introducing phase shift at different output operating points especially as the operates at or near saturation where it is most efficient. In this case the amplitude is constant, so the PA runs at a constant output power level, so no additional phase variation is introduced. Does that sufficiently answer your question? $\endgroup$ – Dan Boschen Dec 19 '18 at 11:59
  • $\begingroup$ Hi Dan, thanks for the effort. Yeah, I see what you're saying, thanks once again ;) but you know the AM/PM gives the phase characteristics as a function of input amplitude (hence power level) and not output power level. Or is it because under "constant envelope" assumption, we have constant AM, and therefore constant amp/phase at output? $\endgroup$ – Ahmad Bazzi Dec 20 '18 at 12:33
  • $\begingroup$ Yes! I agree- it would be better if I said "In this case the amplitude of the envelope is constant, so the PA runs at a constant INPUT power level." Thanks! Good point, that is actually an important distinction. $\endgroup$ – Dan Boschen Dec 20 '18 at 12:38
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I think that your confusion comes from your definition of bandwidth as the mainlobe width of the pulse. In this case it's always the rectangular pulse that wins because it has the narrowest mainlobe. You get the same with windowing for spectral estimation: a rectangular window has the best resolution due to its narrow mainlobe, but its sidelobes are large and decay at a slow rate, so you'll have a lot of leakage.

Coming back to pulse shapes, the rectangular pulse has very slowly decaying sidelobes, which makes its effective bandwidth larger than the bandwidth of methods that use smoother pulses. You can better look at the $99$% power-containment bandwidth, and with that definition of bandwidth the rectangular pulse will come out much worse than the half-sinusoidal pulse used in MSK.

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  • $\begingroup$ would you be so kind to send me the matlab code corresponding to the plot? $\endgroup$ – Hatem Tawfik Aug 18 '18 at 22:21

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