How can I generate offset QPSK symbols from QPSK symbols

Question

I am trying to generate Offset-QPSK modulated symbols. But I am facing some difficulties. I know how to make QPSK symbols but I am unable to figure out at what stage I have to bring in changes to make it OFFSET QPSK.

The following are the steps I did:

1. Generated a stream of 1s and 0 s
2. I have paired at 2 consecutive bits and mapped it to one among the integers 0,1,2,3.
3. Now I map these integers to $(1+j), \ (-1+j), \ (-1-j), \ (1-j) $ following grey code based ordering.

I know that these are the steps for QPSK. But I am not sure If I need to generate OQPSK symbols at which step should I make changes and what are the changes?

Any help is appreciated.

Answer 1

The idea of OQPSK is very simple:

1. Consider your QPSK as two BPSK, one orthogonal to the other. It helps remembering that you're free to rotate the constellation, so that the constellation points are on the I and Q axis. Instead of choosing one point of a four-points constellation based on two bits, you now choose two separate constellations points from two two-point constellations.
2. pulse shape I and Q independently with a pulse shaper that has more than 2 Samples per Symbol. Typically, use 5 or more.
3. offset (in time) either component so that it lags half a symbol time behind the other

If you already have QPSK-modulated baseband, you can still do step 3.

Answer 2

You already have a serial stream of bits, say one bit every $T$ seconds. Think of the bits as being numbered consecutively $b_0, b_1, b_2, \ldots$ and modulate the even-numbered bits onto the inphase carrier and the odd-numbered bits onto the quadrature carrier with each modulation lasting for $2T$ seconds. That is, bit $b_{2n}$ creates a BPSK signal $(-1)^{b_{2n}}\cos(2\pi f_c t)$ that lasts from $t=2nT$ to $t=(2n+2)T$ while bit $b_{2n+1}$ creates a BPSK signal $(-1)^{b_{2n+1}}(-\sin(2\pi f_c t))$ that lasts from $t=(2n+1)T$ to $t=(2n+3)T$. The OQPSK signal is then the sum of these two BSPK signals, that is, at any time, we have that $$s(t) = (-1)^{b_I}\cos(2\pi f_c t) - (-1)^{b_Q}\sin(2\pi f_c t)$$ as in this answer of mine but what $b_I$ and $b_Q$ are is different between standard QPSK and OQPSK.

In OQPSK, during the $T$-second time interval $[2kT, (2k+1)T)$, the signal is

$$s(t) = (-1)^{b_{2k}}\cos(2\pi f_c t) - (-1)^{b_{2k-1}}\sin(2\pi f_c t), ~2kT \leq t < (2k+1)T,$$ while during the $T$-second interval $[(2k+1)T, (2k+2)T)$, the QPSK signal is

$$s(t) = (-1)^{b_{2k}}\cos(2\pi f_c t) - (-1)^{b_{2k+1}}\sin(2\pi f_c t), ~(2k+1)T \leq t < (2k+2)T.$$

In contrast, in standard QPSK, during the entire $2T$-second interval from $2kT$ till $(2k+2)T$, we have that

$$s(t) = (-1)^{b_{2k}}\cos(2\pi f_c t) - (-1)^{b_{2k+1}}\sin(2\pi f_c t), ~2kT \leq t < (2k+2)T.$$

That is, in OQPSK, at all times the signal is a QPSK signal but the modulating bit in the I branch changes at even multiples of $T$ while the modulating bit in the Q branch changes at odd multiples of $T$. Thus, the QPSK bit pair changes from $0*$ to $1*$ or from $1*$ to $0*$ at even multiples of $T$ and from $*1$ to $*0$ or from $*0$ to $*1$ at odd multiples of $T$ (in contrast to standard QPSK in which both bits can change at even multiples of $T$ and there are no changes at odd multiples of $T$). Hence, in OQPSK, at no time can the transition complement both the I branch bit and the Q branch bit, and there is never a phase change of $180^\circ$.

Answer 3

Okay, as an addendum to my other answer (that no one liked), here is a little bit more formal answer:

Let your serial bit stream be $a[n] \ \in \{$0, 1$\}$ and the discrete-time bipolar binary signal be

$$\begin{align} x[n] &\triangleq -1 + 2 a[n] \quad \in \{-1, 1 \} \\ &= -(-1)^{a[n]} \\ \end{align}$$

So it's the negative of the other bipolar binary convention of $x[n] \triangleq (-1)^{a[n]} $ which some folks like to use.

The modulated IQ signal is:

$$\begin{align} s(t) &= \Re\Big\{ (i[n] + j q[n]) \, e^{j 2 \pi f_\text{c} t} \Big\} \\ & = i[n] \cos(2 \pi f_\text{c} t) - q[n] \sin(2 \pi f_\text{c} t) \\ \end{align}$$

where $f_\text{c}$ is the carrier frequency and

$$ n = \Big\lfloor \tfrac{t}{T} \Big\rfloor = \operatorname{floor}\Big(\tfrac{t}{T} \Big) \ .$$

$T$ is the bit period and $\frac{1}{T}$ is the baud rate or bps (bits per second).

Now before I defined the in-phase, $i[n]$, and quadrature, $q[n]$ signals as:

$$\begin{align} i[n] \ &= \ g[n] x[n] \ + \ (1-g[n]) x[n-1] \\ q[n] \ &= \ (1-g[n]) x[n] \ + \ g[n] x[n-1] \\ \end{align}$$

where $g[n]$ is an even/odd gating signal defined as

$$ g[n] \triangleq \tfrac{1}{2}\left( 1 + (-1)^n \right) $$

and

$$ 1-g[n] = \tfrac{1}{2}\left( 1 - (-1)^n \right) $$

(Note that for $n$ even, $g[n]=1$ and only $i[n]$ can change, while for $n$ odd, $g[n]=0$ and only $q[n]$ can change. The bit rates for $i[n]$ and $q[n]$ is half the bit rate for $x[n]$.)

But let's do that a little differently this time and define $i[n]$ and $q[n]$ more generally.

$$\begin{align} i[n] &= \sum\limits_{m=-\infty}^{\infty} x[2m] \, p[n-2m] \\ \\ q[n] &= \sum\limits_{m=-\infty}^{\infty} x[2m+1] \, p[n-(2m+1)] \\ \end{align}$$

Here $p[n]$ is our bandlimited pulse shape. For $i[n]$ only even-indexed samples of $x[n]$ are used and for $q[n]$ only odd-indexed samples are used. This means that the effective baud rate for $i[n]$ and $q[n]$ is half of the bit rate for $a[n]$ or the bipolar $x[n]$. So only half of the bandwidth is needed for $i[n]$ and $q[n]$.

Now above (and previously) I was defining the pulse shape to be:

$$ p[n] = \begin{cases} 0 \qquad & n < 0 \\ 1 \qquad & 0 \le n < 2 \\ 0 \qquad & 2 \le n \\ \end{cases} $$

Note the pulse width is $2T$, two bit widths, wide. So it would occupy half of the bandwidth than a similarly shaped pulse that is one bit width wide.

So try this for a pulse shape:

$$ p[n] = \operatorname{sinc}\left( \tfrac{n}{2} \right) $$

where

$$ \operatorname{sinc}(u) \triangleq \begin{cases} \frac{\sin(\pi u)}{\pi u} \qquad & u \ne 0 \\ 1 & u = 0 \\ \end{cases}$$

This will result in a bandwidth of half of the bandwidth necessary for a baud rate of $\tfrac{1}{T}$. And it's a flat (brickwall) bandwidth, emphasizing or de-emphasizing no frequencies (within the bandwidth) over any others.

Note that for $i[n]$ or $i[2m]$, because none of the other even-indexed samples will contribute to the value of $i[n]$ at the even index of $n=2m$.

For $q[n]$ or $q[2m+1]$, none of the other odd-indexed samples will contribute to the value of $q[n]$ at the odd index of $n=2m+1$. So there is no inter-symbol interference and these are strictly bandlimited pulses.

Now that $\operatorname{sinc}(\cdot)$ function goes on forever, which means that the you'll have to add an infinite number of non-zero terms in the summations for $i[n]$ and $q[n]$ above, so you'll have to window it with a window function, $w[n]$, of non-zero width $M+1$:

$$ p[n] = \operatorname{sinc}\left( \tfrac{n}{2} \right) w[n] $$

This changes the summations for $i[n]$ and $q[n]$ to be finite:

$$\begin{align} i[n] &= \sum\limits_{m=\lfloor n/2-M/4 \rfloor}^{\lfloor n/2+M/4 \rfloor} x[2m] \, \operatorname{sinc}\left( \tfrac{n-2m}{2} \right) w[n-2m] \\ \\ q[n] &= \sum\limits_{m=\lfloor (n-1)/2-M/4 \rfloor}^{\lfloor (n-1)/2+M/4 \rfloor} x[2m+1] \, \operatorname{sinc}\left( \tfrac{n-(2m+1)}{2} \right) w[n-(2m+1)] \\ \end{align}$$

You could define the window function as a Hamming window:

$$ w[n] \triangleq \begin{cases} \tfrac{27}{50} + \tfrac{23}{50} \cos\left(2\pi \tfrac{n}{M}\right) \quad \quad & |n| \le \tfrac{M}{2} \\ 0 & |n| > \tfrac{M}{2} \\ \end{cases} $$

but I would suggest a Kaiser window:

$$ w[n] \triangleq \begin{cases} \frac{1}{J_0(\beta)} J_0\left(\beta \sqrt{1 - \left(\frac{n}{1+M/2}\right)^2} \right) \quad \quad & |n| \le \tfrac{M}{2} \\ 0 & |n| > \tfrac{M}{2} \\ \end{cases} $$

$J_0(\cdot)$ is the 0th-order modified Bessel function of the first kind.

$$ J_0(u) = 1 \ + \ \sum\limits_{k=1}^{\infty} \frac{1}{(k!)^2} \left(-\frac{u^2}{4}\right)^{k} $$

$M+1$ is the number of non-zero samples or FIR taps. $\beta$ is a "shape parameter" and O&S recommend this heuristic:

$$ \beta = \begin{cases} 0.1102 \cdot (A-8.7) & A>50 \\ 0.5842 \cdot (A-21)^{2/5} + 0.07886 \cdot (A-21) \quad & 21 < A \le 50 \\ 0.0 & A \le 21 \\ \end{cases}$$

$$ M = 2 \left\lceil \frac{A-8}{4.57 \cdot \Delta\omega} \right\rceil - 1 $$

$A$ is the desired stopband attention in dB and $\Delta\omega$ is the desired width of the transition band in normalized angular frequency. $\lceil \cdot \rceil$ is the ceiling function (i.e. "round up").