Computer Science - Networking - Discussion

3. 

A 6-MHz channel is used by a digital signaling system utilizing 4-level signals. What is the maximum possible transmission rate?

[A]. 1.5 Mbaud/sec
[B]. 6 Mband/sec
[C]. 12 Mbaud/sec
[D]. 24 Mbaud/sec
[E]. None of the above

Answer: Option C

Explanation:

No answer description available for this question.

Abhishek said: (May 10, 2012)  
Bandwidth = 6 MHz (given) = 6 x 106
Using Nyquist's Theorem,
C = 2B log2M
C = 2 x 6 x 106 x log24
C = 24 Mbps
Hence, C = 24 Mbps

Santosh said: (Jan 24, 2013)  
Def-Transmission Rate-No. of bits per second.As the question asked is maximum possible transmission rate;

Hence,
Noiseless Channel-Nyquist Bit rate;
C=2BLog2L.

C - Capacity
B - Bandwidth
l - No. of levels used

Therefore;

C=2x6x10 to the power 6xlog2(4)
C=24x10 to the power 6
C=24 mbps.

Taiyaba said: (Aug 28, 2017)  
Def-Transmission Rate-No. of bits per second.As the question asked is maximum possible transmission rate;

Hence,
Noiseless Channel-Nyquist Bit rate;
C=2BLog2L.

C - Capacity
B - Bandwidth
l - No. of levels used

Therefore;
C=2x6x10 to the power 6xlog2(4)
C=24x10 to the power 6
C=24 mbps.

But here the answer is in term of baud and we are using 4 level i. e., 2-bit for each level. Therefore baud rate =(1/2)bit rate. Hence answer is 24Mbps/2=12Mbaud/sec.

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