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GATE 2009
December 14, 2023
UGC NET CS 2015 Dec – paper-3
December 14, 2023
GATE 2009
December 14, 2023
UGC NET CS 2015 Dec – paper-3
December 14, 2023

UGC NET CS 2013 Dec-paper-2

Question 3
An analog signal carries 4 bits in each signal unit. If 1000 signal units are sent per second, then baud rate and bit rate of the signal are and
A
4000 bauds \ sec & 1000 bps
B
2000 bauds \ sec & 1000 bps
C
1000 bauds \ sec & 500 bps
D
1000 bauds \ sec & 4000 bps
Question 3 Explanation: 
→ Bit rate is nothing but number of bits transmitted per second
→ Baud rate is nothing but number of signals units transmitted per unit time.
Given data,
— Bit rate of a signal=4000 bps
— Each signal unit carries=4 bits
— Baud rate=?
Step-1: Baud rate= Bit rate of signal / baud rate
= 4000/4 bps
= 1000 bps
Step-2: Bit rate= Number of bits transmitted per second*number of signals units transmitted per unit time
= 4*1000 bps
= 4000 bps
Correct Answer: D
Question 3 Explanation: 
→ Bit rate is nothing but number of bits transmitted per second
→ Baud rate is nothing but number of signals units transmitted per unit time.
Given data,
— Bit rate of a signal=4000 bps
— Each signal unit carries=4 bits
— Baud rate=?
Step-1: Baud rate= Bit rate of signal / baud rate
= 4000/4 bps
= 1000 bps
Step-2: Bit rate= Number of bits transmitted per second*number of signals units transmitted per unit time
= 4*1000 bps
= 4000 bps

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