May be this is a stupid question, so pardon me. For QPSK what is typically the frequency between the I, Q carriers, and the base band frequencies ? I just want to verify if the Nyquist criteria is satisfied, that is the I, Q carrier frequencies are atleast >= 2x the baseband frequency.Thanks in advance for your help.
Question about QPSK
Started by ●May 27, 2011
Reply by ●May 27, 20112011-05-27
On 05/27/2011 04:55 AM, Daku wrote:> May be this is a stupid question, so pardon > me. For QPSK what is typically the frequency > between the I, Q carriers, and the base band > frequencies ? I just want to verify if the Nyquist > criteria is satisfied, that is the I, Q carrier > frequencies are atleast>= 2x the baseband > frequency.Thanks in advance for your help.Your question borders on the meaningless. A real QPSK signal does not have I and Q carriers -- it is a one dimensional function in time (or a one dimensional vector of samples, if you've sampled it). It has _a_ carrier. Before you sample this signal, the Nyquist criteria does not apply, because the Nyquist criteria applies to sampling. What I think you are perhaps driving at, though, is wondering what the _one_ carrier frequency needs to be in QPSK. If you really want the signal to be transmitted cleanly, the answer is that the carrier needs to be more than half the bandwidth of any significant energy in the signal, lest the low-side images fold over 0Hz and interfere with each other. So (e.g.), if the signal is 20kHz wide and symmetrical, you'd need to use at least a 10kHz carrier. But even that doesn't apply all that strongly, because your system can suffer from a significant amount of this image frequency folding (don't ask me how much, I'd have to do the math -- you're welcome to it) and still work just fine. HTH -- Tim Wescott Wescott Design Services http://www.wescottdesign.com Do you need to implement control loops in software? "Applied Control Theory for Embedded Systems" was written for you. See details at http://www.wescottdesign.com/actfes/actfes.html
Reply by ●May 28, 20112011-05-28
Thanks. Maybe my original post was not very clear, but your example below is very concrete. Let us suppose that I am trying to send voice data (20.0KHz bandwidth) with QPSK. In that case what would the carrier frequency be ? And yes, as there is no direct sampling in QPSK, Nyquist criterion does not apply. However, since the symbols are made up of 1s and 0s, the data being sent has to sampled and digitized, at first. But I am assuming that this done beforehand, and I am given a stream of symbols. On May 27, 1:05 pm, Tim Wescott <t...@seemywebsite.com> wrote:> Your question borders on the meaningless. A real QPSK signal does not > have I and Q carriers -- it is a one dimensional function in time (or a > one dimensional vector of samples, if you've sampled it). It has _a_ > carrier. > > Before you sample this signal, the Nyquist criteria does not apply, > because the Nyquist criteria applies to sampling. > > What I think you are perhaps driving at, though, is wondering what the > _one_ carrier frequency needs to be in QPSK. If you really want the > signal to be transmitted cleanly, the answer is that the carrier needs > to be more than half the bandwidth of any significant energy in the > signal, lest the low-side images fold over 0Hz and interfere with each > other. So (e.g.), if the signal is 20kHz wide and symmetrical, you'd > need to use at least a 10kHz carrier. > > But even that doesn't apply all that strongly, because your system can > suffer from a significant amount of this image frequency folding (don't > ask me how much, I'd have to do the math -- you're welcome to it) and > still work just fine. > > HTH > > -- > > Tim Wescott > Wescott Design Serviceshttp://www.wescottdesign.com > > Do you need to implement control loops in software? > "Applied Control Theory for Embedded Systems" was written for you. > See details athttp://www.wescottdesign.com/actfes/actfes.html
Reply by ●May 28, 20112011-05-28
On 05/28/2011 04:41 AM, Daku wrote: (top posting fixed)> > On May 27, 1:05 pm, Tim Wescott<t...@seemywebsite.com> wrote: > >> Your question borders on the meaningless. A real QPSK signal does not >> have I and Q carriers -- it is a one dimensional function in time (or a >> one dimensional vector of samples, if you've sampled it). It has _a_ >> carrier. >> >> Before you sample this signal, the Nyquist criteria does not apply, >> because the Nyquist criteria applies to sampling. >> >> What I think you are perhaps driving at, though, is wondering what the >> _one_ carrier frequency needs to be in QPSK. If you really want the >> signal to be transmitted cleanly, the answer is that the carrier needs >> to be more than half the bandwidth of any significant energy in the >> signal, lest the low-side images fold over 0Hz and interfere with each >> other. So (e.g.), if the signal is 20kHz wide and symmetrical, you'd >> need to use at least a 10kHz carrier. >> >> But even that doesn't apply all that strongly, because your system can >> suffer from a significant amount of this image frequency folding (don't >> ask me how much, I'd have to do the math -- you're welcome to it) and >> still work just fine. >> >> HTH > > Thanks. Maybe my original post was not very clear, > but your example below(above)> is very concrete. Let us > suppose that I am trying to send voice data (20.0KHz > bandwidth) with QPSK. In that case what would the > carrier frequency be ? And yes, as there is no direct > sampling in QPSK, Nyquist criterion does not apply. > However, since the symbols are made up of 1s and > 0s, the data being sent has to sampled and digitized, > at first. But I am assuming that this done beforehand, > and I am given a stream of symbols.I guess that in the case of QPSK the bandwidth vs. carrier number is kind of like the Pirate's Code -- it's more of a guideline, really. My knee-jerk reaction is to say 10kHz -- but depending on how you're defining bandwidth, and on the details of the signal, it could be much more or much less. I guess the real point is -- can you make the signal unambiguous? If so -- that's enough. Case in point: Manchester encoding is (if you hold your mouth right) a form of MSK, which is, in turn, a form of QPSK. Nearly half of the signal power is above the "carrier frequency" of 3/4 the baud rate, and if you're talking about bandwidth out to 20dB down the bandwidth is well above the baud rate. Yet Manchester encoding works quite well, and there's no reason that you couldn't make a nice 'optimal' encoder for it. So -- I think your question isn't "How should I use bandwidth to set minimum carrier frequency", but "should I use bandwidth to set minimum carrier frequency". And I think the answer is "no". -- Tim Wescott Wescott Design Services http://www.wescottdesign.com Do you need to implement control loops in software? "Applied Control Theory for Embedded Systems" was written for you. See details at http://www.wescottdesign.com/actfes/actfes.html
Reply by ●May 28, 20112011-05-28
On May 27, 12:05=A0pm, Tim Wescott <t...@seemywebsite.com> wrote: ,,,,,,,,>..........=A0If you really want the > signal to be transmitted cleanly, the answer is that the carrier needs > to be more than half the bandwidth of any significant energy in the > signal, lest the low-side images fold over 0Hz and interfere with each > other. =A0So (e.g.), if the signal is 20kHz wide and symmetrical, you'd > need to use at least a 10kHz carrier.Does "20kHz wide and symmetrical" mean a lowpass signal whose energy is mostly contained between -10 kHz to +10 kHz, or (say) a signal whose energy is mostly contained in 20 Hz - 20.020 kHz (and the corresponding negative frequency band) -- a typical spec for hi-fi music -- and symmetric about 10.010 kHz (not a typical spec for a music signal, much less the voice signal suggested by the OP)? In any case, if the OP's clarification that "I am trying to send voice data (20.0KHz bandwidth) with QPSK" means that the baseband signal is at 10 0r 20 kilobaud, I would hesitate about recommending a 10 kHz QPSK carrier frequency even as a minimum possible. Dilip Sarwate
Reply by ●May 28, 20112011-05-28
On Sat, 28 May 2011 04:41:57 -0700 (PDT), Daku <dakupoto@gmail.com> wrote:>Thanks. Maybe my original post was not very clear, >but your example below is very concrete. Let us >suppose that I am trying to send voice data (20.0KHz >bandwidth) with QPSK. In that case what would the >carrier frequency be ?The "carrier" frequency for QPSK is just the center frequency of the signal bandwidth. If you generate the signal at baseband, i.e., zero frequency, by just adding two (perhaps filtered) NRZ sequences together into a complex pair, then the carrier frequency is zero. If you then mix it up with an oscillator to some other frequency, the new frequency becomes the carrier frequency. An easy way to see the "carrier" in a QPSK signal is to set both the I and Q streams to constants, e.g., a constant 1 in the I channel and a constant 1 in the Q chanel. You'll see a spike at the "carrier" frequenc with this data, and at baseband that spike is at DC. Mix it up or down with an oscillator and the "carrier" spike moves to the new carrier frequency. No magic at all.> And yes, as there is no direct >sampling in QPSK, Nyquist criterion does not apply.The Nyquist criterion applies to PSK signals just like it does to anything else.>However, since the symbols are made up of 1s and >0s, the data being sent has to sampled and digitized, >at first. But I am assuming that this done beforehand, >and I am given a stream of symbols.Eric Jacobsen http://www.ericjacobsen.org http://www.dsprelated.com/blogs-1//Eric_Jacobsen.php
