## Example 5.6

#### A DPSK receiver has an error in the delay element equivalent to 10% of the carrier period. Given that the expression for bit error probability in a perfect (no delay error) DPSK receiver is of the form:
Pe = 0.5exp(Eb/N0)
where Eb is the energy per bit, and N0 is the noise power density, what is the percentage increase in transmitter power required to counteract the degradation in receiver performance caused by the delay error?

### Solution

#### The time delay error = 10% of carrier period = 36o of phase error in the notional carrier reference.
The voltage output of the mixer used to compare the present symbol cos wct with the previous symbol cos(wct + Tb + 36o) will thus be reduced by a factor cos(36o) from its maximum value as a result of this timing error.
This in turn equates to a reduction in symbol energy at the input to the receiver of cos2(36o).
The noise components passing through the mixer will also be affected by the phase error in the carrier reference, but since the noise vectors are assumed to be randomly distributed through 360o, the carrier phase error will reduce the effect of some noise vectors and enhance others, with the net effect that the average noise voltage at the mixer output will remain unchanged. Hence it is only the symbol energy that is truly affected by the timing error and not the noise power. The bit error probability can thus be written as:
Pe = 0.5exp(Eb cos2(36o) / N0)
This means that the transmitted symbol energy must be increased by a factor 1 / cos2(36o) = 1.52, or 1.85 dB, to maintain the same performance as a system with no timing error.