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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.

enter image description here

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 enter image description here

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!

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  • $\begingroup$ "I understand for QPSK there are symbols 00, 01, 10, 11 with phase shifts of 45,135,225 and 315." Stop right there! You need to tell us what the I and Q bits are. See this answer for some details of how QPSK really works. $\endgroup$ – Dilip Sarwate Aug 17 at 20:33
  • $\begingroup$ Thanks. I understand it now $\endgroup$ – TheCoolest2 Sep 1 at 2:43
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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.

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  • $\begingroup$ No, this is all wrong. There is no phase shift of 180 degrees in QPSK, and the description of the carrier phase is hopelessly muddled. $\endgroup$ – Dilip Sarwate Aug 17 at 22:56

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