# 8-PSK baseband signal with real and imaginary part

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.

• The wikipedia image out of context is confusing; this is not the value of the symbols in baseband. As the text above the image on the wikipedia page says: "The modulated signal is shown below for a short segment of a random binary data-stream. The two carrier waves are a cosine wave and a sine wave"; carrier waves, not baseband. If you're not sure about the difference, please read up on complex baseband first. – Marcus Müller Dec 15 '16 at 17:58

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.

• thank you, where does your sine and cosine curve for the value $011 \equiv -0.7071 + 0.7071j$ start? I assume that sine starts at $-0.7071$ at the beginning of the interval T and cosine starts at $0.7071$ at the the beginning of the interval T. – autoship Dec 15 '16 at 13:48
• There's no cosine curve in a constellation mapping – that happens due to signal shaping through limited channel bandwidth or pulse shaping – Marcus Müller Dec 15 '16 at 14:26
• you're confusing 8-PSK (a constellation) with a 8-PSK-modulated carrier – Marcus Müller Dec 15 '16 at 14:26
• In baseband, there's no cosines/sines here at the symbol periods – it's just the values. – Marcus Müller Dec 15 '16 at 14:27
• my exercise asks me to draw on two diagrams: 1st diagram: vertical axis: $i(t)$, horizontal axis: $t[\mu s]$ 2nd diagram: vertical axis: $q(t)$, horizontal axis: $t[\mu s]$ It looks like I have to draw something like this mathworks.com/help/comm/ug/qpsk_transition_timing.png – autoship Dec 15 '16 at 15:07

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