6.13 Binary phase shift keying

A commonly used example of a signal set consists of pulses that are negatives of each other ( [link] ).

This way of representing a bit stream---changing the bit changes the sign of the transmitted signal---is known as binary phase shift keying and abbreviated BPSK. The name comes from concisely expressing this popular way of communicating digital information. The word "binary" is clear enough (one binary-valued quantity is transmitted during a bitinterval). Changing the sign of sinusoid amounts to changing---shifting---the phase by $\pi$ (although we don't have a sinusoid yet). The word "keying" reflects back to the first electrical communication system, which happened to be digital as well: the telegraph.

The datarate $R$ of a digital communication system is how frequently an informationbit is transmitted. In this example it equals the reciprocal of the bit interval: $R=\frac{1}{T}$ . Thus, for a 1 Mbps (megabit per second) transmission, we must have $T=1\mathrm{\mu s}$ .

The choice of signals to represent bit values is arbitrary to some degree. Clearly, we do not want to choosesignal set members to be the same; we couldn't distinguish bits if we did so. We could also have made the negative-amplitude pulserepresent a 0 and the positive one a 1. This choice is indeed arbitrary and will have no effect on performance assuming the receiver knows which signal represents which bit. As in all communication systems, we designtransmitter and receiver together.

A simple signal set for both wireless and wireline channels amounts to amplitude modulating a baseband signal set (moreappropriate for a wireline channel) by a carrier having a frequency harmonic with the bit interval.

$$s_1(t)=-(A{p}_T(t)\sin \left(\frac{2\pi kt}{T}\right))$$

This signal set is also known as a BPSK signal set. We'll show later that indeed both signal sets provide identical performancelevels when the signal-to-noise ratios are equal.

What is the transmission bandwidth of these signal sets? We need only consider the baseband version as the second is anamplitude-modulated version of the first. The bandwidth is determined by the bit sequence. If the bit sequence isconstant—always 0 or always 1—the transmitted signal is a constant, which has zero bandwidth. Theworst-case—bandwidth consuming—bit sequence is the alternating one shown in [link] . In this case, the transmitted signal is a square wave having a period of $2T$ .

From our work in Fourier series, we know that this signal's spectrum contains odd-harmonics of the fundamental, which hereequals $\frac{1}{2T}$ . Thus, strictly speaking, the signal's bandwidth is infinite. In practical terms, we use the 90%-power bandwidth toassess the effective range of frequencies consumed by the signal. The first and third harmonics contain that fraction ofthe total power, meaning that the effective bandwidth of our baseband signal is $\frac{3}{2T}$ or, expressing this quantity in terms of the datarate, $\frac{3R}{2}$ . Thus, a digital communications signal requires more bandwidth than the datarate: a 1 Mbps baseband systemrequires a bandwidth of at least 1.5 MHz. Listen carefully when someone describes the transmission bandwidth of digitalcommunication systems: Did they say "megabits" or "megahertz"?