Consider we have created a QPSK signal with a carrier frequency 100kHz. Before applying matched filter or demodulation we need to take samples from the received signal. So what is the minimum sampling rate we must have? Consider that somehow we have perfect synchronization and there is no drift of the carrier.

Any help would be appreciated.

I have asked this question (for more attention) at stackexchange .

  • $\begingroup$ Sampling rate depends on the signal bandwidth, not the carrier frequency. What is the bandwidth of the signal? $\endgroup$
    – BlackMath
    Jun 20, 2019 at 3:39
  • $\begingroup$ actually, what is the bit rate? $\endgroup$ Jun 20, 2019 at 4:51

1 Answer 1


As best as I can tell from the terminology, QPSK a.k.a. 4-PSK or 4-QAM can convey twice the number of bits of information as BPSK in the same bandwidth.

Given a one-sided bandwidth of $B$ and assuming sufficient S/N ratio so that no bits are corrupted, BPSK can transfer bits at a rate of $2B$. That means, given the same one-sided bandwidth, QPSK can transfer bits at a rate of $4B$.

But the sample rate must be at least $2B$. If the receiver is perfectly synchronized to the transmitter, there would be one sample per bit with no bit error, if the transmitter sends a bipolar $\operatorname{sinc}(\cdot)$ for each bit.

But for QPSK, each sample is a complex value with a real part and imaginary part, each part containing one bit sample if this is 4-PSK.

So if you use quadrature demodulation to bump the IF or RF down to baseband $i(t)$ and $q(t)$, and sample those signals, you will need one complex-valued sample per pair of bits.

  • $\begingroup$ geee, i never got points for a digital communications question. i have never worked in this, so i consider myself a bit pendantic. but i thought about it, particularly about OBPSK which i think is very kewl. $\endgroup$ Jun 20, 2019 at 7:09
  • $\begingroup$ You definitely deserve more points for your digital comms answers :) $\endgroup$ Jun 20, 2019 at 7:59
  • $\begingroup$ thanks, Marcus. it's only what happens when someone like me (DSP but no digital comms since grad school during the Reagan era) comes upon, for the first time, OQPSK and realizes that this naturally takes the even and odd indexed bits in natural order and there is no need for a contrived "dibit" paring operation. i really think of this in terms of a DSP problem. and i still think that markers like 0011001100110011 or 0110011001100110 should be used for frame sync since they cause the extreme low and extreme high instantaneous frequencies and can be more grossly detected. $\endgroup$ Jun 20, 2019 at 19:56
  • $\begingroup$ It's always more than refreshing when someone with a slightly different background looks at techniques. One stands to learn quite a bit. To mention why OQPSK is so commonly used: It looks like FSK; to be specific, it looks like a specific MSK, i.e. an FSK with the shortest possible symbol period that still allows detection of whether you're doing a positive or a negative frequency :) GMSK (gaussian minimum shift keying) is that variant, and it's easy to produce thanks to its equality to OQPSK, and it's robust thanks to its detectability as FSK. $\endgroup$ Jun 20, 2019 at 21:20
  • $\begingroup$ well, i asked about OQPSK at the EE.SE, about what were the currently established protocols. i think my idea of using the above bit patterns, one as channel idle and the other as the start frame, might be a better idea than the existing protocols. and they should do something like the old SDLC did to prevent spurious frame sync in case either of those two words appear accidently in the actual data. but taking a serial stream of bits, routing the even bits to $i[n]$ and the odd bits to $q[n]$ is just so natural for OQPSK. no need to group two bits together in a symbol. $\endgroup$ Jun 20, 2019 at 21:59

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