# Is Phase Locked Loop is essential for BPSK signal reception?

I'm noob in DSP studying BPSK communication between speaker and microphone using acoustic signal. I read article of this BPSK communication which says it needs Phase Locked Loop(PLL) sequence to match both ends communicate precisely. But in my textbook, there is no mention of PLL in BPSK demodulation.

I want to communicate Tx/Rx within close range to get channel tap corresponding to its multipath effects in my room. In this case, does PLL is essential to synchronize the signal in both ends? Or I can communicate well without PLL and can get channel tap?

• "I read [an] article" <- Which one? – Marcus Müller Jan 15 '18 at 19:28

There are three different kinds of synchronization in a passband digital communications system:

• Carrier synchronization: the receiver needs to know the exact frequency and phase of the carrier used by the transmitter.

• Symbol synchronization: the receiver needs to know the optimum instants to sample the matched filter's output.

• Frame synchronization: the receiver needs to know where each transmitted "character" (or word, or byte) starts.

So, if you're doing passband BPSK, you'll need to perform all three. Carrier synchronization is typically done with a PLL (or a Costas loop) and you can find the details in several answers on this website.

Many textbooks ignore these subjects. A textbook that does a great job of teaching these three aspects of a receiver is "Telecommunications Breakdown" by Johnson and Sethares, an early version of which is available for free from the authors' website.

Having said that, there is a way to avoid doing carrier synchronization, at the cost of energy efficiency, by upconverting the baseband BPSK signal using AM DSB-LC instead of the more common AM DSB-SC. So, you first design your baseband signal: $$s_{BB}(t) = \sum_k a_k p(t-kT_P),$$ where $a_k \in \lbrace -1, 1 \rbrace$, $T_p$ is the pulse interval, and $p(t)$ is the pulse shape. Then, the DSB-LC passband signal is $$s_{PB}(t) = (A_m + s_{BB}(t)) A_c \cos(2\pi f_c t),$$ where $A_m > -\min(s_{BB}(t))$ (so that $(A_m + s_{BB}(t))>0$) and $A_c$ is the carrier amplitude. Then, the baseband signal can easily be recovered by a simple envelope detector (identical to conventional analog modulation).

• Thank you for answering my question! I'll try what you told me. – JayHeo Jan 16 '18 at 16:24
• PAM would approve of this message – Robert L. Dec 20 '18 at 15:24