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Can one consider that a BPSK signal that has not gone through any pulse shaping filter to be equivalent to a signal that has gone through a rectangular filter of which the pass BW is equal to the symbol ( bit in that case) rate?

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  • $\begingroup$ Both answers below are excellent. If you accept one of them, it will show your question in the list as answered and consequently will attract more viewers that benefits the community. $\endgroup$ – Qasim Chaudhari Aug 21 '18 at 1:01
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To answer your question, NO! A BPSK signal with a random data bit pattern that has not gone through any (additional) pulse shaping, meaning the pulse shape is rectangular in time equal to the bit duration, has a $Sinc^2$ power spectral density centered on the carrier frequency (as the Fourier Transform of a rectangular pulse is the Sinc function).

This signal has infinite bandwidth, but the power spectral density is dropping within an envelope of $1/f^2$ where f is the offset from the carrier (as expected from the $Sinc^2$ function). Any reduction of this bandwidth by filtering in the frequency domain, results in a finite transition time between symbols, so therefore induces a pulse shape in the time domain that is non-rectangular.

Another way to see this simply, a symbol that can transition from one state to the next in an infinitely short time (which is what the rectangular pulse shape is required to do) would require infinite bandwidth in order to transition that quickly. We slow down the transition rate on purpose in order to limit the bandwidth occupied since spectrum is such a valuable (and regulated) resource. We just do that very strategically to minimize inter-symbol interference, and optimize the SNR we can receive within the bandwidth used.

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A BPSK signal has the following form:

$$s(t)=\sum_kA_kp(t-kT)\cos(2\pi f_ct)\tag{1}$$

where $A_k$ assumes the values $+1$ and $-1$, $p(t)$ is the pulse, in your case rectangular, $T$ is the symbol/bit duration, and $f_c$ is the carrier frequency.

The equivalent baseband signal is

$$s_B(t)=\sum_kA_kp(t-kT)\tag{2}$$

and this signal can be interpreted as a weighted impulse train filtered with an impulse response $p(t)$:

$$s_B(t)=\left[\sum_kA_k\delta(t-kT)\right]\star p(t)\tag{3}$$

where $\delta(t)$ is the Dirac delta impulse, and $\star$ denotes convolution.

Note that $(3)$ is valid for any pulse $p(t)$, not just for a rectangular pulse.

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    $\begingroup$ I'd add this: as the equations make clear, there is no such thing as a non-pulse shaped digital signal (the closest you can get is the expression inside the square brackets in eq. (3)). Having said that, it's very common to find textbooks that either ignore pulse shaping or say they don't do pulse shaping. In both cases, it's safe to assume that the pulse shape is rectangular. $\endgroup$ – MBaz Aug 20 '18 at 22:41

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