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An actual BPSK signal can be expressed as $$\begin{align} s(t) &= \operatorname{Re}\left[\sum_{n=-\infty}^\infty (-1)^{b_n}p(t-nT)e^{j(2\pi f_c t + \theta)}\right]\\ &= \left[\sum_{n=-\infty}^\infty (- … and say that $$\sum_{n=-\infty}^\infty (-1)^{b_n}p(t-nT)e^{j(2\pi f_c t + \theta)}$$ is a BPSK signal containing positive frequencies only. …

output to leave only one BPSK signal. … If one desires to have only the BPSK signal at $f_1+f_0 = f_c$ Hz, say, then the BPSK signal at the image frequency $f_1-f_0$ must be filtered out by passing the mixer output through a bandstop filter …

Turning to the question of why the trick of using the coherent BPSK error formula as a guide and simply replacing the SNR for BPSK with the SNR for DBPSK in the error probability formula doesn't give the … Note, for example, that $P_e = Q(\sqrt{2E_b/N_0})$ for coherent BPSK but only $Q(\sqrt{E_b/N_0})$ for coherent FSK. …

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 … $f_c$ 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. …

The mixer/LPF filter that you describe as the receiver is typically one that is designed (possibly for optimal performance e.g. matched filter) using a mathematical model that assumes its input is the …

OBPSK in the way that you describe it is also called $\pi/2$-BPSK meaning that the signal constellation is different from bit to bit, rotating by $\pi/2$ after it is used. …

You are taking 64 samples per second, consisting of 8 samples per period of a sinusoid of frequency 8 Hz. The information that you need to determine is which of the 8 possible phases of the sinusoid y …

The OP's model of white noise as a process with finite power is where the problem is. The standard model of continuous-time white noise that is used in communications and signal processing has infinit …

Two successive symbols in the demodulator are $Z_1 = (X_1,Y_1)$ and $Z_2 =(X_2,Y_2)$ where $X$ is the output of the I branch and $Y$ the output of the Q branch of the receiver. The hard-decision DBP …