# Spectral Effeciency of BPSK

The null-to-null bandwidth of the main lobe of a BPSK signal is twice the symbol rate.

BPSK transmits one bit per symbol, so the signal bandwidth is twice the bit rate.

That has been shown on a few stack overflow questions. This implies the spectral efficiency is 0.5 bits/second/Hz.

However, a quick Google for the spectral efficiency of BPSK gives an answer of 1 bits/second/Hz. Why?

The spectral efficiency depends on the pulse shape. The basepand BPSK signal can be written as $$s(t)=\sum_k a_k p(t-kT_b),$$ where $a_k$ is equal to either $\sqrt{E_b}$ or $-\sqrt{E_b}$, $E_b$ is the bit energy, $T_b$ is the bit interval (so that the bit rate is $R_b=1/T_b$), and $p(t)$ is a Nyquist pulse. The bandwidth of $s(t)$ is equal to the bandwidth of $p(t)$.

Unfortunately, many textbooks insist on limiting the discussion to square pulses, where $$p(t)=\begin{cases}1,\text{ if 0\leq t <T_b}\\0,\text{ otherwise.}\end{cases}$$ This means that the spectrum of $s(t)$ is infinite, strictly speaking. This would seem to preclude any discussion of spectral efficiency.

However, when bandwidth is understood as an engineering (not mathematical) concept, we see that we can truncate the spectrum to a desired bandwidth $B$ and, if the pulse $p(t)$ is not distorted too much, communication is still possible.

Different authors make different choices for $B$. As you've seen, some say $B=2R_b$ and others say $B=R_b$. Personally, I use $B=5R_b$ (I'm not as optimistic as most book authors). Fortunately, you can make up your mind: using Matlab (or similar), you can create a close-to-ideal BPSK signal $s(t)$ by sampling it much faster than required, and then filter it using different values of $B$. You can see the resulting distortion and intersymbol interference, and decide what value of $B$ you're comfortable with.

If you have the time and inclination, you can do the experiment with an analog signal generator and analog low-pass filters; it's more work but it feels more "real" than a Matlab simulation.

• Nice answer @MBaz. I always saw this intuitively by considering a square wave which is the fastest 1010101 data rate that can be transmitted, with insight that the first harmonic is halfway into the main lobe of the sinc--thus to me that proved that that bandwidth was the boundary of what was needed to transmit the 101010 pattern (1/2T) .Every other patterns involve lower frequencies and will occupy the remainder of the spectrum below 1/(2T)-- everything above 1/(2T) is just higher harmonics of signals we already have, which we reduce with pulse shaping! Apr 5 '17 at 1:11
• @DanBoschen I dislike the approach that pulse shaping is used to reduce the harmonics of a square pulse -- I see pulse shaping as fundamental, along with matched filtering, which is optimal. Also, just filtering out the harmonics of a square pulse will introduce ISI. Finally, to me it's just simpler to take the pulse shaping approach from the start -- you can even use square pulses if you want, since they're Nyquist (when they don't overlap)! :)
– MBaz
Apr 5 '17 at 1:31
• Yes agree- it is not an approach but more a way to view the higher frequencies, they are not necessary to convey the information so can be viewed as higher harmonics (with a Fourier series expansion view if you transmit a repeated pattern they are indeed exactly that) Apr 5 '17 at 1:33
• But not suggesting the purpose of pulse shaping is to filter harmonics but more specific regardless of the pulse shaping was the point to where fundamental frequencies are given the possible patterns that you can transmit to see how the highest possible frequency is 1/(2T) for the max rate pattern of 10101010; just a way to see or explain it quickly why the bandwidth is just 1/2 the main lobe (if we could brick wall filter) Apr 5 '17 at 1:37
• @DanBoschen There is no such thing as no pulse shaping :) You can write any linear modulation (and many nonlinear) schemes with my equation for $s(t)$ above. Even if the pulses are rectangular, you're doing pulse shaping. You choose a pulse shape because it's simple (rectangular, half-sine), or because it's spectrally efficient (sinc, raised cosine). A matched filter at the receiver always optimizes the SNR, so you can't do better than that over the AWGN or fading channels. I'm unaware of how GPS uses the higher frequencies -- is that Doppler-shift related?
– MBaz
Apr 5 '17 at 2:07