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In QPSK, after splitting the odd and even bits into the I and Q channels, and then multiplying them with the sine and cosine function, what next am I suppose to do?

Do I rearrange them back into their odd and even positions?

Also, in OQPSK do I make the first bit of my Q array a dummy bit zero to represent the half symbol delay in the Q channel ?

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  • $\begingroup$ Your first question "..what next I am suppose to do?" is answered in great detail here, and no, you do not rearrange the bits in their odd and even positions: you sum the results of the multiplications by the sine and cosine. A QPSK modulator accepts $2$ bits every $T$ seconds, and spits out a high frequency (phase-modulated) signal that lasts for a full $T$ seconds, not a cosine for $T/2$ seconds followed by a sine for the next $T/2$ seconds. With regard to your second question about the dummy bit, yes you insert a dummy bit $\endgroup$ – Dilip Sarwate Jan 24 '14 at 19:56
  • $\begingroup$ what dummy bit? $\endgroup$ – robert bristow-johnson Jan 25 '14 at 5:15
  • $\begingroup$ @robertbristow-johnson If one assumes that an OQPSK signal begins somewhere, say at $t = 0$ instead of being present from $-\infty$ onwards, then during the transmission of the very first data bit, there is no bit in the other branch because it has not entered the QPSK modulator as yet. Some people insert a dummy bit in the other branch so that from $t=0$ onwards there is a QPSK signal with phase being one of the 4 values instead of a BPSK signal for a brief period with phase not being one of the 4 values. $\endgroup$ – Dilip Sarwate Jan 25 '14 at 13:21
  • $\begingroup$ @Dilip, i confess i still don't see it. you turn the machine on, the bits start flowing in, the first two bits are grouped together into a "semi-nibble" or "crumb" or whatever becomes the QPSK symbol. even though you can represent it sorta as a filter (and i have), the pump really does not need to be primed. a "dummy bit" is something you would have to add to an odd number of bits, then you transmit the symbols, then the symbols are recovered in the receiver, then the receiver tosses the dummy bit. but i don't see that here at all. $\endgroup$ – robert bristow-johnson Jan 25 '14 at 15:48
  • $\begingroup$ @robertbristow-johnson In OQPSK with serial input, the first bit interval is $(0,T/2)$ and it modulates the cosine during $(0,T)$. The second bit comes in during $(T/2, T)$ and modulates the sine during $(T/2, 3T/2)$. The third bit comes in during $(T,3T/2)$ and modulates the cosine during $(T, 2T)$, and so on. In the steady state, there are always two different bits modulating the cosine and sine. The question is what modulates the sine during $(0,T/2)$? It is a dummy bit, or the sine is suppressed. $\endgroup$ – Dilip Sarwate Jan 25 '14 at 18:23
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This question seems to be based on several misconceptions.

In a QPSK modulator operating at a rate of $N$ baud, two bits enter the modulator during each $T = N^{-1}$ second interval. If the bits are entering on a single wire (at a rate of $2N$ bits per second, the QPSK modulator first converts this serial input bit stream at $2N$ bits per second into two parallel bit streams at $N$ bits per second. For example, alternate bits are steered to alternate streams so that, say, all the odd-numbered bits are in one stream and all the even-numbered bits are in the other stream.

In some cases, the bits entering the QPSK modulator are already formatted into two parallel bit streams at $N$ bits per second. One important and commonly observed arrangement of this type is when a rate-$\frac12$ convolutional encoder precedes the QPSK modulator, so that information bits enter the encoder serially at rate $N$ bits per second, and the convolutional encoder spits out two parallel bit streams at $N$ bits per second. However, regardless of encoding or not, the bits that the QPSK modulator uses are in two parallel streams that are synchronous (that is, the bit transitions occur at the same times (once every $T$ seconds) and in each bit stream, the bit occupies the entire $T$ second interval.

As the OP's question says, the bits are modulated onto cosine and sine carriers. As discussed in more detail here, if the bits in the two streams are denoted by $b_I$ and $b_Q$, then the corresponding signals, say, over the interval $[0,T)$ can be taken to be $$(-1)^{b_I}\cos(2\pi f_c t) ~~ \text{and}~~ -(-1)^{b_Q}\sin(2\pi f_c t), 0 \leq t < T. \tag{1}$$ (Ignore the negative sign on the $\sin$, it is there to simplify notation in more detailed analyses). The QPSK signal is the sum $$(-1)^{b_I}\cos(2\pi f_c t) -(-1)^{b_Q}\sin(2\pi f_c t), 0 \leq t \leq T. \tag{2}$$ It is not the case, as the OP seems to think that the two signals are separated into even and odd bits, so that $(-1)^{b_I}\cos(2\pi f_c t)$ lasts from $t=0$ to $t=T/2$ and $-(-1)^{b_Q}\sin(2\pi f_c t)$ from $t=T/2$ to $t=T$: the cosine and sine signals last for the full $T$-second duration and the bits $b_I$ and $b_Q$ both affect the signal phase for all $T$ seconds.

Offset QPSK also converts the serial bit stream into two parallel bit streams in which the bit intervals are of duration $T$ on each wire, but the bit transitions are not synchronous but occur with an offset of $T/2$ seconds. Thus, $b_I$ modulates the cosine carrier to produce $$(-1)^{b_I}\cos(2\pi f_c t), 0 \leq t < T$$ but $b_Q$ modulates the sine carrier to produce $$-(-1)^{b_Q}\sin(2\pi f_c t), \frac{T}{2} \leq t < \frac{3T}{2}.$$ The modulator output is still the sum of the two sinusoids, but the sinusoids change phase at different times. In fact, the signal can be deemed to be a $2N$ baud signal since the phase can change once every $T/2 = (2N)^{-1}$ seconds. Each bit still affects the phase for $T$ seconds, but the bit transitions are staggered. In the receiver, integrations (or matched filterings) are carried out over (staggered) $T$-second intervals in the two branches of the receiver, and the decision devices spit out one bit every $T$ seconds but at staggered times. It is easy to multiplex these decisions onto a single wire that thus carries a bit stream at $2N$ bits per second, just as the input to the QPSK modulator (if anyone still remembers that far back in this answer) is a bit stream on a single wire at $2N$ bits per second.

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  • $\begingroup$ BTW, Dilip's "$T$ seconds" covers two sampling periods (like $n$ and $n+1$) in my analysis below (which is still not complete, but i think someone could commit an entire book or chapter of a book on QPSK and maybe a whole section on OQPSK). $\endgroup$ – robert bristow-johnson Jan 25 '14 at 20:12
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sure, in reception of QPSK, you receive bit pairs as they are transmitted, and detangle them in the reverse manner that they were grouped together.

what's really cool about OQPSK is that the order of the bits going in can naturally determine the order of bit changing in OQPSK. in fact, it can be made into a DSP modulation system without the need for grouping bits together in pairs.

define the bitstream $a[n] \in \{0,1\}$, and the bipolar binary signal as:

$$ x[n] \ = \ (-1)^{a[n]} \ = \ 1 - 2 a[n]\ \in \ \{+1,-1\} $$

then define these two gating functions:

even samples:

$$ g[n] \ = \ \tfrac{1}{2} \left(1 + (-1)^n \right) \ = \ \tfrac{1}{2} \left(1 + e^{j \pi n} \right) \ = \ \begin{cases} 1, & n\text{ even} \\ 0, & n\text{ odd} \end{cases} $$

odd samples:

$$ 1 - g[n] \ = \ g[n-1] \ = \ \tfrac{1}{2} \left(1 - (-1)^n \right) \ = \ \tfrac{1}{2} \left(1 - e^{j \pi n} \right) \ = \ \begin{cases} 0, & n\text{ even} \\ 1, & n\text{ odd} \end{cases} $$

the I/Q quadrature pair is:

$$ i[n] \ = \ g[n] \ x[n] \ + \ (1-g[n]) \ x[n-1] $$

$$ q[n] \ = \ (1-g[n]) \ x[n] \ + \ g[n] \ x[n-1] $$

note that $i[n+1]=i[n]=x[n]$ for even $n$ and $q[n+1]=q[n]=x[n]$ for odd $n$. so for either $i[n]$ or $q[n]$, the bit rate is half the bit rate is for $a[n]$ or $x[n]$. so the bandwidth needed for the analog reconstructed signals $i(t)$ and $q(t)$ need only be half of the bandwidth needed for the reconstruction of $x(t)$ from $x[n]$ is.

then the discrete-time OQPSK modulating signal is

$$ s[n] \ = \ i[n] \ + \ j \ q[n] $$

returning to the real $i[n]$ and $q[n]$:

$$\begin{align} i[n] \ & = \ g[n] \ x[n] \ + \ (1-g[n]) \ x[n-1] \\ & = \ \tfrac{1}{2} \left(1 + e^{j \pi n} \right) \ x[n] \ + \ \tfrac{1}{2} \left(1 - e^{j \pi n} \right) \ x[n-1] \\ & = \ \tfrac{1}{2} (x[n] + x[n-1]) \ + \ \tfrac{1}{2} e^{j \pi n} (x[n] - x[n-1]) \\ \end{align}$$

$$\begin{align} q[n] \ & = \ (1-g[n]) \ x[n] \ + \ g[n] \ x[n-1] \\ & = \ \tfrac{1}{2} \left(1 - e^{j \pi n} \right) \ x[n] \ + \ \tfrac{1}{2} \left(1 + e^{j \pi n} \right) \ x[n-1] \\ & = \ \tfrac{1}{2} (x[n] + x[n-1]) \ - \ \tfrac{1}{2} e^{j \pi n} (x[n] - x[n-1]) \\ \end{align}$$

the OQPSK modulating signal is

$$\begin{align} s[n] \ & = \ i[n] \ + \ j \ q[n] \\ & = \ \tfrac{1}{2} (x[n] + x[n-1]) \ + \ \tfrac{1}{2} e^{j \pi n} (x[n] - x[n-1]) \ + \ j \ \left( \tfrac{1}{2} (x[n] + x[n-1]) \ - \ \tfrac{1}{2} e^{j \pi n} (x[n] - x[n-1]) \right) \\ & = \ \frac{1+j}{2} (x[n] + x[n-1]) \ + \ \frac{1-j}{2} e^{j \pi n} (x[n] - x[n-1]) \\ \end{align}$$

now compute the Discrete-time Fourier Transform (DTFT)

$$\begin{align} S(\omega) \ & = \ \frac{1+j}{2} \left(X(\omega) + e^{-j\omega}X(\omega) \right) \ + \ \frac{1-j}{2} \left(X(\omega-\pi) - e^{-j(\omega-\pi)}X(\omega-\pi) \right) \\ & = \ \frac{1+j}{2} X(\omega) \left(1 + e^{-j\omega} \right) \ + \ \frac{1-j}{2} X(\omega-\pi) \left(1 - e^{-j(\omega-\pi)} \right) \\ & = \ \frac{1+j}{2} X(\omega) \left(1 + e^{-j\omega} \right) \ + \ \frac{1-j}{2} X(\omega-\pi) \left(1 + e^{-j\omega} \right) \\ & = \ \frac{1 + e^{-j\omega}}{2} \Big( (1+j)X(\omega) \ + \ (1-j)X(\omega-\pi) \Big) \\ & = \ e^{-j\omega/2}\frac{e^{j\omega/2} + e^{-j\omega/2}}{2} \Big( (1+j)X(\omega) \ + \ (1-j)X(\omega-\pi) \Big) \\ & = \ e^{-j\omega/2} \ \cos(\omega/2) \ \Big( (1+j)X(\omega) \ + \ (1-j)X(\omega-\pi) \Big) \\ \end{align}$$

so the next question to ask is what does the spectrum of $X(\omega)$ look like for a bit stream $a[n]$ that looks like 00000000...? or 11111111...? or 01010101...? or 10101010...? or 00110011...? or 01100110...? and then what is the spectrum of the quadrature modulating signal $S(\omega)$ with those same bit sequences for $a[n]$?

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  • $\begingroup$ Life is so much easier if one does not insist that $0$ map into $-1$ and $1$ into $+1$. Try using the transformation $x \to (-1)^x$ for $x \in \{0,1\}$ to get a much simpler system. $\endgroup$ – Dilip Sarwate Jan 23 '14 at 23:20
  • $\begingroup$ Dilip, i agree for the most part. like with maximum-length sequences, when you compute autocorrelation of the bipolar binary signal, multiplication becomes XOR of the exponent. we can do that here, too, but eventually i want to show something about spectrum and it's easier with the expression shown (like how do you compute the spectrum of $(-1)^{a[n]}$ when you know the spectrum of $a[n]$?), however it is the negative of $(-1)^{a[n]}$, so maybe i should flip the sign on $x[n]$. i ain't done with this yet. $\endgroup$ – robert bristow-johnson Jan 24 '14 at 1:33

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