0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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 $e^{j2\pi (f_1-f_2)t}$. $\endgroup$ – Dilip Sarwate Jul 25 '15 at 13:04
  • $\begingroup$ @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? $\endgroup$ – Dimitri C Jul 25 '15 at 13:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.