# transmitting the real/imaginary parts of pass-band signal

As I know, the transmitted pass-band signal whose carrier frequency is $$f_c$$ can be written as follow:

$$s_1(t) = \text{Re}(a(t)e^{j2πf_ct})$$ , where $$a(t)$$ is the complex modulated base-band signal. It's clear to notice that at the receiving end, we can recover the complex modulated symbol $$a(t)$$ based on the real transmitted part $$s_1(t)$$.

But the question which I couldn't get a sure reply for it, what's about if we transmitted the imaginary part of our signal? .. it means that $$s_2(t) = \text{Imag}(a(t)e^{j2πf_ct})$$. Will we be able to recover the complex modulated symbol based on $$s_2(t)$$ too? How can we prove that mathematically?

Will we be able to recover the complex modulated symbol based on $$s_2(t)$$ too?

Yes, no information is lost between the two.

How can we prove that mathematically?

One way to think about it is to write out each individually and see what you have. Let $$a(t)=a_I(t)+ja_Q(t)$$, expanding $$s_1(t)$$ and $$s_2(t)$$ we have:

\begin{align} s_1(t) &= a_I(t)\text{cos}(2\pi f_ct)-a_Q(t)\text{sin}(2\pi f_ct) \\ s_2(t) &= a_I(t)\text{sin}(2\pi f_ct)+a_Q(t)\text{cos}(2\pi f_ct) \end{align}

The in-phase and quadrature parts are present in each so we just need to bring them back to baseband to get $$a(t)$$ back. Further, whatever workings you have to do this for $$s_1(t)$$ you can reuse for the processing of $$s_2(t)$$ by first applying a simple $$+\frac{\pi}{2}$$ to it ($$\text{cos}(\theta + \frac{\pi}{2})=-\text{sin}(\theta)$$).

Clearly it doesn't make a difference if you use the real part or the imaginary part (as long as you know what you're doing), because

$$\textrm{Im}\left\{a(t)e^{j2\pi f_ct}\right\}=\textrm{Re}\left\{b(t)e^{j2\pi f_ct}\right\}\tag{1}$$

with

$$b(t)=-ja(t)=a_Q(t)-ja_I(t)\tag{2}$$

So you basically just exchange the in-phase and quadrature components (apart from a sign).