RE: final exam (dec 7, 98)

Kevin Oldknow (koldknow@netcom.ca) Thu, 22 Apr 1999 08:14:12 -0400 

From: Kevin Oldknow <koldknow@netcom.ca>
Subject: RE: final exam (dec 7, 98)
Date: Thu, 22 Apr 1999 08:14:12 -0400



Hi there,

I calculated it as follows:

8 data bits, no parity and minimum transmission time suggests 10 bits per character as follows:

1 start bit, 8 data bits, 1 stop bit.  (RS-232 standard from lecture notes & lab 4)

Therefore, 960 characters x 10 bits/character = 9600 bits.

As I understand it, bps does mean bits per second.  i.e. the transmission would take a total of 1 second.

If this is incorrect, can somebody let me know?

Kevin Oldknow

-----Original Message-----
From:	Alfred Lui [SMTP:tlui@ece.ubc.ca]
Sent:	Wednesday, April 21, 1999 10:10 PM
To:	elec379@casas.ece.ubc.ca
Subject:	final exam (dec 7, 98)

Hello!

I'm looking at the last question, part e. I don't know how to find the
time to transmit 960 characters. Is a character consist of 2 words? and
bps means bit per second? Please help me out.

Thanks