8-PSK baseband signal with real and imaginary part

Question

I want to draw the complex-valued baseband signal with real $S_{BB,I}(t)$ and imaginary part $S_{BB,Q}(t)$ for the following bit sequence: 011111001000

I have looked around and found on wikipedia such a timing diagram but for QPSK:

I understand that the bit sequence of a QPSK is grouped in pairs of two, the first bit of the pair is assigned to I, the second part of the pair is assigned to Q. If I=1, the sine starts at the top. If Q=1, the cosine rises.

How would I need to draw this diagram for an 8-PSK? The bits are grouped in pairs of three. So my bit sequence is 011,111,001,000. I am puzzled how to assign the pairs of three to only two signals I and Q.

Answer 1

You're making your life actively harder by considering I and Q separate instead of considering them to be the real and imaginary part of a complex signal $S = I + jQ$.

If you then do that, you'd simply get a table with 8 rows, one for every constellation point. Notice that for 8-PSK, you simply divide the full unit circle ($2\pi$) into 8 equal angles – hence, the complex value of all these points is simply $e^{j\angle(\text{constellation point})}= e^{j\frac{2\pi}{8}n}$; I arbitrarily chose that 000 should be mapped to $n=0$, 001 to $n=1$ and so on. There's no reason that this is the right mapping – all 8-PSK says is "there's 8 constellation points on equally distant points on the unit circle of the complex plane".

• 000$_b\equiv 0$:  $e^{j\frac{2\pi}8 \cdot 0}=1+0j$
• 001$_b\equiv 1$:  $e^{j\frac{2\pi}8 \cdot 1}=\frac{\sqrt2}2+\frac{\sqrt2}2j$
• 010$_b\equiv 2$:  $e^{j\frac{2\pi}8 \cdot 2}=0+1j$
• 011$_b\equiv 3$:  $e^{j\frac{2\pi}8 \cdot 3}=-\frac{\sqrt2}2+\frac{\sqrt2}2j$
• 100$_b\equiv 4$:  $e^{j\frac{2\pi}8 \cdot 4}=-1+0j$
• 101$_b\equiv 5$:  $e^{j\frac{2\pi}8 \cdot 5}=-\frac{\sqrt2}2-\frac{\sqrt2}2j$
• 110$_b\equiv 6$:  $e^{j\frac{2\pi}8 \cdot 6}=0-1j$
• 111$_b\equiv 7$:  $e^{j\frac{2\pi}8 \cdot 7}=\frac{\sqrt2}2-\frac{\sqrt2}2j$

You'd then just look up the real and imaginary part from that table. The simple "bit maps to I (or Q)" simply doesn't work for anything but BPSK and QPSK.

Answer 2

You have to map the grouped binary values to the constellation of 8psk from the constellation you will get the value of I and Q.

000 - -1/sqrt(2) - j(1/sqrt(2)) 001 - -1 010 - j 011 - -1/sqrt(2) + j(1/sqrt(2)) 100 - -j 101 - 1/sqrt(2) - j(1/sqrt(2)) 110 - 1/sqrt(2) + j(1/sqrt(2)) 111 - 1