# High Transmission Power : BPSK or QPSK?

In BPSK, the constellation consists of 2 points (equiprobable) spaced at distance of $\sqrt{E_b}$ each from the origin.

So the average power from the constellation can be obtained by:

• For BPSK, $P = \frac{E_b+E_b}{2} = E_b.$

• For QPSK, 4 points which are equiprobable are spaced at distance of $\sqrt{2E_b}$ from origin each, that is they are placed on vertices of a square of side $2\sqrt{E_b}$. So, for QPSK, $P = \frac{2\cdot4\cdot E_b}{4}=2E_b.$

So, does this mean that the power required for QPSK is more than BPSK when the probability of error is same for both?

Note: $E_b$ is the energy per bit and $P$ is the power which is assumed as proportional to the distance of the signal point from origin in the constellation.

$P_b = Q\left(\sqrt{\frac{2E_b}{N_0}}\right).$