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I ask this question because it seems to me that if pulse shaping has been applied in the transmitter, the signal that is actually transmitted is not phase modulated at all, rather, it is amplitude modulated. I have made some pictures to illustrate my thinking, which goes as follows: Most literature about BPSK illustrates the modulated wave as follows, with sudden changes in amplitude that result in a large bandwidth. BPSK signal with no shaping applied to baseband signal

Pulse shaping attempts to address the problem by modifying the baseband signal. We start with a polar, non return to zero, digital signal that looks like this: Polar NRZ Digital data signal

The signal is sampled: sampled digital data

A sinc function is calculated for each sample. This is done carefully to ensure that if the amplitude of a particular sample is measured, the amplitudes of all other samples are zero at that point in time. sinc functions from samples

These sinc shapes are added together to create a new baseband signal (shown in green): smoothed baseband signal

Here is the new smoothed signal without the clutter. smoothed signal without clutter

The new baseband signal is then multiplied by a continuous wave carrier to generate the modulated signal which is converted into radio waves and transmitted. The modulated signal will look something like this. Its envelope is the pulse shaped baseband signal.
This is what I would expect to see as a result of regular amplitude modulation. modulated signal

Therefore, am I correct in saying that if pulse shaping is used with BPSK, it is not BPSK at all? Am I also correct in assuming that a signal such as that in my fist image (with the sudden changes in amplitude) would never be transmitted in this day and age?

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You are mostly correct. The main problem is that the introductory textbook definition of BPSK as a train of square pulses multiplied by a carrier is so simplistic that it hides what is really going on.

Say you have some binary sequence $\lbrace b_0, b_1, b_2, \ldots \rbrace$ that you want to transmit. First you map each bit $b_k$ to an amplitude $a_k$; for example, one could say that

$\begin{equation}a_k = \begin{cases} 0.5, \quad\text{if $b_k = 1$} \\ -0.5, \quad\text{if $b_k = 0$.}\end{cases}\end{equation}$

Then, one chooses an appropriate pulse shape $p(t)$ -- you used a sinc in your example, but one might as well choose a square-root raised cosine or something else. Each pulse is delayed a time $kT$ and scaled by $a_k$ (here, $R=1/T$ is the bit rate). The baseband signal is then $$s_\text{BB}(t) = \sum_k a_k p(t-kT).$$

This is known as a 2-PAM signal. The "2" indicates that each amplitude $a_k$ is restricted to one of two values, meaning that it corresponds to one single bit. (4-PAM would let $a_k$ take one of four values, conveying two bits, etc.)

Now, the technically correct definition of BPSK is "an upconverted 2-PAM signal", which is simply

$$s_\text{BPSK}(t) = s_\text{BB}(t)\cos(2\pi f_c t) = \left(\sum_k a_k p(t-kT)\right) \cos(2\pi f_c t).$$

You mention that this is simply an amplitude-modulated signal. You are quite correct: BPSK is an AM-SC (suppressed-carrier) signal. What makes BPSK different is the type of message that is being conveyed: in BPSK, the baseband signal has a very specific shape, and it carries digital information (in the amplitudes $a_k$), whereas "normal" AM is assumed to carry an arbitrary message signal $m(t)$.

You also mention phase modulation. BPSK is actually phase modulated, but the phase is limited to two values, $0$ and $\pi$. When $a_k > 0$, the phase is $0$, and when $a_k < 0$, the phase is $\pi$.

Finally, you ask about the possibility of using, in this day and age, BPSK with square waves (and the resulting sudden phase transitions): the answer is that all modern narrowband (BPSK, QPSK, QAM, etc) wireless systems do pulse shaping, mainly to save bandwidth.

There may be some legacy systems still using square-wave BPSK, but I'm not aware of any. When the available technology was much less powerful than it is today, using square waves had the advantage of allowing for a very simple, optimal receiver technique called "integrate and dump" (see here).

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    $\begingroup$ Nice answer. To add, GPS is one example of square wave BPSK— because the signal as it reaches the earth is below the noise floor, pulse shaping is not relevant— and the wider bandwidth of unfiltered (unshaped) BPSK leads to sharper correlation peaks (tighter resolution for position and time). For GPS the transmit bandwidth if I remember correctly is close to 50 MHz… that’s enough harmonics of the 1 MHz switching signal to essentially be a square wave) $\endgroup$ Nov 2, 2023 at 19:39
  • $\begingroup$ Thank you Dan! GPS is indeed a good example of a square-wave BPSK signal. $\endgroup$
    – MBaz
    Nov 2, 2023 at 20:36
  • $\begingroup$ Thank you for this, you've cleared up so much for me. You probably gathered from my question that I like to visualise concepts, so there is still something I’m not sure of. You said "BPSK is actually phase modulated…" but BPSK still seems like something of a misnomer when it is used in conjunction with pulse shaping. Are you saying that the 2-PAM signal has two phases? $\endgroup$
    – Drummy
    Nov 3, 2023 at 12:33
  • $\begingroup$ The concept of phase, as used here, refers to the pulse amplitudes $a_k$ and not the phase of the Fourier transform of $s_\text{PB}(t)$. Furthermore, for a given pulse index $k$, the transmitted signal is $|a_k| p(t)\cos(2\pi f_c t)$ if $a_k$ is positive, and $|a_k| p(t) \cos(2\pi f_c t + \pi)$ if $a_k$ is negative. It is in this sense that BPSK is said to have two phases. $\endgroup$
    – MBaz
    Nov 3, 2023 at 13:15

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