BFSK modulation what signal the transmitter actually sends?

I am trying to understand how modulation with carriers works(transmission part). I will express my question in terms of BFSK for simplicity.

Let us condiser BFSK modulation with symbols {1,2}.

Then for a single symbol period:

the transmitted waveform for symbol 1 is

$$s_1(t)=\sqrt{P\over T}e^{j2\pi f_1 t}$$

the transmitted waveform for symbol 2 is

$$s_2(t)=\sqrt{P\over T}e^{j2\pi f_2 t}$$

where $P , T$ denote signal strength and period respectively and $f_1 , f_2$ denote carrier frequencies for symbols $1 , 2$ respectively .

We also know that $s_1(t)$ has an equivalent $$s_1'(t)=\Re{(s_1(t))}=\Re{(\sqrt{P\over T}e^{j2\pi f_1 t}})=\sqrt{P\over T}cos(2 \pi f_1 t)$$

Same thing for $s_2(t)$

$$s_2'(t)=\Re{(s_2(t))}=\Re{(\sqrt{P\over T}e^{j2\pi f_2 t}})=\sqrt{P\over T}cos(2 \pi f_2 t)$$

Question 1: When I send symbol $1$ which signal is actually transmitted from the transmitter, $s_1(t)$ or $s_1'(t)$?

Question 2: For signals $s_1(t)$ and $s_1'(t)$ which one of two is the low-pass equivalent and which one is the highpass equivalent?

• Question 1: $s_1^\prime (t)$. Transmitting $s_1(t)$ requires a complex transmitter that adds to the cost of the system without providing any advantage in performance. Question 2: Both are band-pass signals. The low-pass equivalent signals are $e^{j2\pi (f_2-f_1)t}$ and $e^{j2\pi (f_1-f_2)t}$. – Dilip Sarwate Jul 25 '15 at 13:04
• @DilipSarwate If transmitting $s_1(t)$ provides no advantage in the performance what is its role and why it is often mentioned? And can you please elaborate on question 2 so I can understand why the signals you wrote are low pass equivalents? – Dimitri C Jul 25 '15 at 13:09