Computer Science - Networking - Discussion
Discussion Forum : Networking - Section 4 (Q.No. 49)
49.
A 4 KHz noiseless channel with one sample every 125 per sec, is used to transmit digital signals. Find the bit rate (bits per second) that are sent, if CCITT 2.048 Mbps encoding is used.
Discussion:
6 comments Page 1 of 1.
Shailesh Saxena said:
1 decade ago
Here, we are only referring to one channel.
1 Channel 8 bits in 125 micro sec.
(Each CCITT channel carries 8-bits/channel/frame).
Then 8/125 * 10^6 = 64 KBPS.
1 Channel 8 bits in 125 micro sec.
(Each CCITT channel carries 8-bits/channel/frame).
Then 8/125 * 10^6 = 64 KBPS.
Shewangizaw Bogale said:
8 years ago
@ALL.
Is 4 KHz noiseless channel, given input have no use?
Can anyone clearly put the Formula and describe it so that we all can understand it.
Is 4 KHz noiseless channel, given input have no use?
Can anyone clearly put the Formula and describe it so that we all can understand it.
Manoj said:
6 years ago
Here ,sampling frequency=1/125 = 0.008/sec.
Now, no of sample in 4khz channel= 4khz*.008= 32 sample/sec.
Bit rate= 2.048Mbps/32 = 64kbps.
Now, no of sample in 4khz channel= 4khz*.008= 32 sample/sec.
Bit rate= 2.048Mbps/32 = 64kbps.
Garima said:
1 decade ago
Please expalain how is it 64kbps?
Is it so because we use 64 bit samples in CCITT 2.048 Mbps ?
Is it so because we use 64 bit samples in CCITT 2.048 Mbps ?
Sandeep said:
8 years ago
In this formula 8/125 * 10^6 =64,
10^6 denotes what? Please explain.
10^6 denotes what? Please explain.
User said:
9 years ago
Does 10^6 denote what? Please clarify @Sailesh.
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