High Transmission Power : BPSK or QPSK?

Question

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.

Answer 1

Since for QPSK each symbol carries two bits, it is not surprising that the average energy per symbol must be greater for QPSK than for BPSK to achieve the same error probability. What counts is the energy per bit that is necessary to achieve a certain error probability, and this energy per bit is the same for BPSK and QPSK.

Answer 2

That's right. When we compare BPSK and QPSK at equal energy per bit, we find that QPSK requires twice the power of BPSK. From wikipedia:

Although QPSK can be viewed as a quaternary modulation, it is easier to see it as two independently modulated quadrature carriers. With this interpretation, the even (or odd) bits are used to modulate the in-phase component of the carrier, while the odd (or even) bits are used to modulate the quadrature-phase component of the carrier. BPSK is used on both carriers and they can be independently demodulated.

As a result, the probability of bit-error for QPSK is the same as for BPSK:

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

However, in order to achieve the same bit-error probability as BPSK, QPSK uses twice the power (since two bits are transmitted simultaneously).