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We use OQPSK because of envelope of signal but my question is

"In Phase Shift Keying(or QPSK) the amplitude of the transmitted signal is constant then how the envelope will change in OQPSK"??

Matt L.
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Avinash Baldi
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1 Answers1

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Note that a QPSK signal using a band-limited pulse has an envelope that passes through zero every time there is a phase transition of $\pi$. So its envelope is not constant if all symbol transitions are allowed.

Offset QPSK (OQPSK) doesn't have phase transitions of $\pi$. By staggering the $I$ and $Q$ signals by half a symbol interval, the maximum phase transition is $\pi/2$, leading to an approximately constant envelope. Slight droops in the envelope (occurring at phase transitions of $\pi/2$) can be eliminated by hard-limiting.

Note that with an appropriate choice of the transmit pulse, phase transitions in OQPSK can be completely avoided. This results in a modulation scheme with an exactly constant envelope. One example of such a continuous-phase modulation is minimum-shift keying (MSK).

For more information on OQPSK take a look at this question and its answers. Also browse this site for other questions on OQPSK.

Matt L.
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  • To improve this good answer slightly but an important clarification: If I am not mistaken OQPSK could be a constant envelope waveform only if specifically pulse shaped to do so, but OQPSK on it's own is not necessarily constant envelope. If this is accurate it could be more generally stated "leading to a reduced peak to average power ratio". – Dan Boschen Apr 16 '19 at 00:25
  • @DanBoschen: Thanks for your remark. You're right that I was oversimplifying, and I've changed the answer accordingly. – Matt L. Apr 16 '19 at 08:13
  • i wonder if, referring to this rambling of mine, if the pulse shape was $$ p[n] = \operatorname{sinc}\left( \tfrac{n}{2} \right) w[n] $$ and the window $w[n]$ is good and wide so it's very nearly a sinc function of bandwidth that is $\frac14$ of the bit rate, i wonder if that would be a constant envelope? – robert bristow-johnson Apr 16 '19 at 09:24
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    @robertbristow-johnson To be constant envelope and if rotating at a constant rate the pulse shape would be a cosine, thus shifting half a pulse one pulse is a cosine and one is a sine and thus it is clear we stay on the unit circle. This is MSK as Matt stated. The rotating at a constant rate is not a requirement for constant envelope and results in the abrupt transitions when we change direction of rotation, resulting in a higher spectrum than if those direction changes were slowed (such a GMSK). Note that a Sinc function does closely approximate a cosine function..... – Dan Boschen Apr 16 '19 at 10:52
  • at least over half of the main lobe. What I didn't proceed to check but would answer your wondering specifically regarding constant envelope condition is if the summation of all p[n] for n odd together with all j p[n] for n even is equal to 1. – Dan Boschen Apr 16 '19 at 10:57
  • that's what i'm wondering, @DanBoschen . certainly if the summation included only the neighboring two values of $p[n-2m]$, it should be a cosine, which would flip polarity if $a[2m]$ or $a[2m+1]$ toggles the bit instead of keeping the same bit. what i really like about OQPSK is how any of these 4 sequences leave it in DC: 00000000, 11111111, 01010101, 10101010. and it's 00110011 or 01100110 that makes the phasor swing around clockwise or counterclockwise at the most rapid angular speed. – robert bristow-johnson Apr 16 '19 at 18:10
  • actually @DanBoschen , if the data is such that the position of the phasor is constant (that would be ...00000000..., ...11111111..., ...01010101..., or ...10101010....), then it seems to me that the pulse shape would have to be a sinc to yield a constant envelope, not a cosine. i think this is true:

    $$ \sum\limits_{n=-\infty}^{\infty} \operatorname{sinc}(t-n) = 1 \qquad \forall t \in \mathbb{R} $$

    i don't think that would be true for a rectangular-windowed cosine.

     – robert bristow-johnson Apr 16 '19 at 19:34
  • If the pulse is only one symbol duration, then in that case it is simply cosine shaped (with constant angular speed)... it is the partial response signaling cases where I can’t see the result in my head (infinite summations). Note that the DFT itself is such a summation of (aliased) sinc pulses and does add to 1, but in that case the phase of each adjacent bin is 2pi/N – Dan Boschen Apr 17 '19 at 18:55