Computer Science - Networking - Discussion
Discussion Forum : 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?
Discussion:
3 comments Page 1 of 1.
Abhishek said:
1 decade ago
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
Using Nyquist's Theorem,
C = 2B log2M
C = 2 x 6 x 106 x log24
C = 24 Mbps
Hence, C = 24 Mbps
SANTOSH said:
1 decade ago
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.
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:
8 years ago
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.
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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