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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 ?
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.
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 $a[n]$?
