Phase shifts of QPSK I'm honestly lost on creating timing diagrams for bit sequences. I understand for QPSK there are symbols 00, 01, 10, 11 with phase shifts of 45,135,225 and 315. 



I'm just not understanding how you get these diagrams. 
I graphed an example on paper but it doesn't seem consistent (using the bit sequence below). 

Maybe I'm a bit lost on the whole even and odd bit sequences. 

For example, I'm looking at a bit stream of 01100101 0 (also I assume that lone bit at the end can be written as 00)

So b(t) = 011001010

be(t)=X0100

bo(t)=X1011

E denoting Even, O denoting Odd

Thanks 

Edit: Also again, for example, I grabbed this from Wikipedia 


https://en.wikipedia.org/wiki/Phase-shift_keying#Bit_error_rate_2

I feel I'm just missing something that will make it all click. Thanks!
In QPSK, the bit information is embedded in the phase of the carrier signal. For example, for bits 00, the phase shift is 45degree. This means the carrier signal $x(t) = cos(2\pi f_c t + \phi)  0 \lt t \lt T$, the phase shift $\phi = \pi / 4$. So $x(t) = cos(2\pi f_c t + \pi / 4) = cos(2\pi f_c t) cos(pi/4) - sin(2\pi f_ct)sin(pi/4)$. The Inphase component corresponding to $cos(2\pi f_c t)$ has factor of $cos(pi/4)$ which is what you see in the I component. Similar logic for Q component corresponding to $sin(2\pi f_c t)$. If the bits are 10, then I component will have factor of $cos(3 \pi/ 4)$ which will be $-\frac{1}{\sqrt{2}}$ so you will see phase shift of 180 degree due to the -1 factor.