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Exchange of information between devices. Communication can be wired or wireless.

4
votes
I'd get a managed switch and configure the port for 802.1x authentication, then install the necessary credentials on the kiosk computer. This uses a standardized and fairly secure authentication meth …
answered Jul 7 '14 by Simon Richter
3
votes
Yes, it is practical. For example, if you have an ASK signal with modulation depth 60%: >> am = [ (ones(1,100) * 0.2) (ones(1,200) * 0.8) (ones(1,200) * 0.2) (ones(1,100) * 0.8)]; Using a low-pa …
answered Jul 29 '14 by Simon Richter
1
vote
The only thing I can offer is that LTE's channel allocation scheme allows new modulation technologies to be used inside existing networks, so it is significantly more futureproof than GSM and UMTS. I …
answered Aug 6 '18 by Simon Richter
2
votes
If it doesn't need to be a high-speed link, RS485 or CAN would be your best bet. For RS485, transceivers are cheap and easily connected electrically, but need a bit of work on the protocol side to mak …
answered Mar 1 '18 by Simon Richter
0
votes
If the transfer system is time invariant, the time between two events in the output is the same as the time between the events that caused them. A wireless transmission channel is time invariant, so t …
answered Apr 10 '15 by Simon Richter
3
votes
The guard bands are required because filters aren't perfect. A filter that can pass 8,000 Hz but block 8,001 Hz must have a group delay of at least 1s. Building a circuit that a signal takes 1s to pa …
answered Jan 17 '15 by Simon Richter
0
votes
A pure delay is not that relevant in most systems. It will affect timeouts, because the receiving system needs to be able to process and verify the incoming data before producing an acknowledgement, …
answered Oct 23 '14 by Simon Richter
1
vote
(a) looks good to me. for (c), keep in mind that \$cos x = \frac{e^{ix} + e^{-ix}}2\$, so multiplying with a real valued sine wave will give you mirror images.
answered Jul 24 '14 by Simon Richter
4
votes
Transmission begins by encoding the discrete-time symbol stream \$s[n]\$ to a continuous-time baseband signal \$y(t) = b(s[\lfloor\frac{t}{T_0}\rfloor])\$. In a simple baseband system, this is transm …
answered Nov 12 '16 by Simon Richter