# Computer Science - Networking - Discussion

### Discussion :: Networking - Section 7 (Q.No.3)

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

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.