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http://zone.ni.com/images/reference/en-XX/help/371025N-01/loc_eps_rectangular_modulation.gifenter image description here

I just wanted to make sure that my assumption that in this figure $\phi(t)$ is representing $m(t)$, and $R(t)$ is the amplitude of the carrier frequency is correct.

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Your understanding is not correct. The baseband processing maps the message to a complex-valued signal

$$x(t)=R(t)e^{j\phi(t)}\tag{1}$$

or, equivalently, to an in-phase component $I(t)=\text{Re}\{x(t)\}$, and a quadrature component $Q(t)=\text{Im}\{x(t)\}$. It is not clear from the figure how $R(t)$ and $\phi(t)$ are related to the message signal $m(t)$. The modulated signal is constructed by multiplying $(1)$ by a (complex) carrier, and taking the real part:

$$s(t)=\text{Re}\{x(t)e^{j\omega_ct}\}=R(t)\cos(\omega_ct+\phi(t))\tag{2}$$

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  • $\begingroup$ thank you, is the mapping done by Hilbert transformation $\endgroup$ – Jack Aug 3 '16 at 18:37
  • $\begingroup$ @MaryannEthan: That would be a possibility, resulting in single sideband modulation. $\endgroup$ – Matt L. Aug 3 '16 at 18:42
  • $\begingroup$ So in that case I is the information signal in your case R(t) and Q is its Hilbert transform? Does it mean that after demodulation the signal we are looking for is I? $\endgroup$ – Jack Aug 3 '16 at 19:25
  • $\begingroup$ @MaryannEthan: For SSB, $I(t)$ is indeed the information signal, but it's not the same as $R(t)$. Look at the figure, you have $I(t)=R(t)\cos(\phi(t))$. $\endgroup$ – Matt L. Aug 4 '16 at 8:36
  • $\begingroup$ Downvoter, why don't you just leave a comment to express what you think is wrong with this answer? $\endgroup$ – Matt L. Jan 31 '17 at 7:12

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