I have a question regarding BPSK. If we want to modulate the bit stream $[0, 0]$ with $T_\text{ bit}=0.25 \text{ sec}$ and the carrier frequency of the cosine that carries the BPSK signal is $2\ \rm Hz$ then the resulting waveform would be the one with the discontinuity or would the signal continue as it is until it reaches $0.5\ \text{sec}$? See pics below, it makes sense to me that we would have a discontinuity at $0.25\ \text{sec}$ because otherwise the receiver which samples the transmitted signal at every $T_\text{bit}$ would confuse the second zero as an 1:


  • $\begingroup$ Are you sure it makes much sense to have a symbol rate greater (4 Hz) than the carrier (2 Hz) for BPSK? $\endgroup$ Jan 16 at 17:31
  • $\begingroup$ That is what is given to us $\endgroup$
    – Jmk
    Jan 16 at 17:40
  • $\begingroup$ then what is given to you makes no sense. The general assumption is that the carrier frequency is much larger than the signal bandwidth, and that's not the case here. $\endgroup$ Jan 16 at 17:52
  • $\begingroup$ I understand that but what i need to know is whether the graph would have a discontinuity, as i understand from the receiver block diagram it samples the signal for every Ts so if there is no "break" it would guess the bit to be 1 not 0. $\endgroup$
    – Jmk
    Jan 16 at 17:57
  • $\begingroup$ We can assume for the sake of the question just that carrier frequency is not a multiple of the bit rate and we would still have the same question. $\endgroup$
    – Jmk
    Jan 16 at 18:01

1 Answer 1


The canonical baseband version of BPSK is that we wish to transmit a sequence $\{a_k\}$ of bits ($0$ or $1$) at a rate of $1$ bit every $T$ seconds using a pulse $g(t)$ of duration $T$ and so the transmitted baseband signal is $$s_{\text{baseband}}(t) = \sum_{k=-\infty}^\infty (-1)^{a_k}g(t-kT).$$ This is a sequence of nonoverlapping pulses of the form $\pm g(\cdot)$ spaced $T$ seconds apart. If, for example, $g(t)$ has value $1$ for $t \in [0,T)$, then $s(t)$ is a sequence of rectangular pulses of duration $T$ with the $k$-th pulse lasting from $t=kT$ to $t=(k+1)T$. Let's use this specific $g(t)$ for the rest of this answer. Note that $s_{\text{baseband}}(t)$ has value $0$ or $1$ for all $t$, and that it changes from one value to the other at times tenant are multiples of the bit duration $T$. Some elementary books on digital communication extend these notions to the passband signal too, saying that the passband BPSK signal during $[0,T)$ is of the form (say) $\pm g(t)\cos(2\pi f_ct)$ (where $f_c$ is the carrier frequency) and so during $[T,2T)$, the passband BPSK signal is just this passband RF burst delayed in time by $T$ seconds. That is, such texts take the passband BPSK signal to be $$s_{\text{passband}}(t) = \sum_{k=-\infty}^\infty (-1)^{a_k}g(t-kT)\cos(2\pi f_c(t-kT)).$$ Such a re-setting of the carrier phase to $0$ every $T$ seconds leads to discontinuities such as seen in the OP's first figure which correspond to $a_0=a_1 =1$.

In practice, and even in more advanced textbooks on digital communications, the BPSK passband signal is of the form $$s_{\text{passband}}(t) = s_{\text{baseband}}(t)\cos(2\pi f_ct) = \left(\sum_{k=-\infty}^\infty (-1)^{a_k}g(t-kT)\right)\cos(2\pi f_ct)$$ which makes the modulation aspect very clear: there is an oscillator that is producing the carrier signal $\cos(2\pi f_ct)$, and this is being modulated by the baseband BPSK signal. Sure looks like amplitude modulation to the untrained eye, but it is indeed phase modulation, because the baseband signal has value $\pm 1$ and since $\cos(2\pi f_ct-\pi)=-\cos(2\pi f_ct)$, we can write
$$s_{\text{passband}}(t) = \cos\left(2\pi f_ct - \pi\sum_{k=-\infty}^\infty a_kg(t-kT)\right)$$ where that inner sum has value $1$ or $0$ for all $t$ and so $s_{\text{passband}}(t)$ is a sinusoidal signal whose phase at all times is either $0$ or $\pi$, and the phase can flip from $0$ to $\pi$ or vice versa only at $T$ second intervals. With $a_0=a_1=1$, the phase remains at $\pi$ during $[0,2T)$ leading to the second plot provided by the OP.

  • $\begingroup$ Is the last equation missing a $pi$ in front of the sum? Also - one thing I think people find confusing (at least I have before) - I know you noted at the top that you're using a pulse g(t) =1 over [0,T) for your answer, but the final equations showing how it is actually phase modulation does not hold when a different g(t) is used. This is often done, for example when doing RRC filtering. I would argue this is true of most real-world implementations of BPSK (or at least what we tend to call BPSK) and its one thing that's not really well spelled out in textbooks. Interested in your thoughts. $\endgroup$
    – user67081
    Jan 18 at 20:09
  • 1
    $\begingroup$ @user67081 I agree with regard to RRC signals which technically do both phase and amplitude modulation, but at the level of what the OP Is asking about (and he didn't define what he understands BPSK signaling to be doing), it is best to keep things simple and not clutter up the explanation. $\endgroup$ Jan 18 at 20:16

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