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In QPSK, the bit information is embedded in the phase of the carrier signal. For example, for bits 00, the phase shift is 45degree. This means the carrier signal $x(t) = cos(2\pi f_c t + \phi) 0 \lt t \lt T$, the phase shift $\phi = \pi / 4$. So $x(t) = cos(2\pi f_c t + \pi / 4) = cos(2\pi f_c t) cos(pi/4) - sin(2\pi f_ct)sin(pi/4)$. The Inphase component ...


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No, this doesn't make much sense. What you can do in the time domain is compute the analytic signal and derive the signal's instantaneous amplitude (envelope) and its instantaneous phase from it. Take as an example $$x[n]=A\sin(\omega_0n+\theta)\tag{1}$$ The corresponding analytic signal is $$x_a[n]=-jAe^{j(\omega_0n+\theta)}\tag{1}$$ Its instantaneous ...


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