# QPSK implementation with 2 BPSK [duplicate]

I have a tiny problem with understanding a thing here.

for the 1st the phase shift is 0 degrees, so on its modulated wave we put 180-degree phase shift.

for the 2nd 0, the oscillator puts a 90-degree phase shift in the carrier signal and as it's a 0-degree we have to put 180-degree shift.

When we sum up the two 0's in the final wave, how are we getting the -135 degree???

I'm unable to understand the summing. https://prnt.sc/gP0OypJ3n_jB

• also, removed the numerous unrelated tags. Jun 4 at 11:34

Your constellation points are the sum of {+1, -1} and {+1j, -1j}, so you get the points {+1+1j, +1-1j, -1+1j, -1-1j}, which have these odd multiples of 45° as angles.

That's one way to look at it.

The other is to realize that if you had only multiples of 90° as angles, only the inphase- or quadrature component can be active at a time, as the amplitude of the other must be 0. But in your schematic, you see that both are active with a non-zero amplitude at the same time.

In the end, it really doesn't matter. The receiver gets the signal with a different, arbitrary phase, anyways, and will rotate it however it sees fit for best decidability. So, maybe it is assumed that the 90° point on transmit side gets simply rotated to be 135° by the receiver (just as any other points are rotated by the same amount).

• Hello Sir, prnt.sc/HrQ6niQhkfYC please see... I'm able to understand how 180 +270 phase shift results in -135 Jun 4 at 12:16
• sum of the two phases (180+270)/2 = 225 Jun 4 at 12:16
• Have you read the first paragraph of my answer? Jun 4 at 12:16
• 225 - 360 = -135 Jun 4 at 12:17
• @DanBoschen finally gotten around to fixing this. That was much too late, sorry, Jency. Jun 23 at 7:56