Computer Science - Networking - Discussion
Discussion Forum : Networking - Section 1 (Q.No. 30)
30.
A noiseless 3 KHz Channel transmits bits with binary level signals. What is the maximum data rate?
Discussion:
13 comments Page 1 of 2.
Mint said:
1 decade ago
How can it be 6Kbps?
Channel freq = 3000Hz.
That means that It can transmit 3000bits in 1sec.
Hence channel data rate is 3000bits/sec or 3Kbps.
I think it should be 3Kbps.
Channel freq = 3000Hz.
That means that It can transmit 3000bits in 1sec.
Hence channel data rate is 3000bits/sec or 3Kbps.
I think it should be 3Kbps.
Ansari said:
1 decade ago
Since sampling frequency is Fs=2Fm;.
Fs=2*3000Hz=6000Hz;.
Therefore it is 6000 bits per sec=6kbps.
Fs=2*3000Hz=6000Hz;.
Therefore it is 6000 bits per sec=6kbps.
Sandeep sanmani said:
1 decade ago
W = 2*B*log2
= 2*3000*1
W = 6000 bps = 6 kbps
= 2*3000*1
W = 6000 bps = 6 kbps
Amit said:
1 decade ago
Since sampling frequency is Fs=2Fm;.
Fs=2*3000Hz=6000Hz;.
so, it is 6000 bits per sec=6kbps.
Fs=2*3000Hz=6000Hz;.
so, it is 6000 bits per sec=6kbps.
Pawanrao said:
1 decade ago
W = 2*B*log2
= 2*3000*1
W = 6000 bps = 6 kbps
= 2*3000*1
W = 6000 bps = 6 kbps
Satya said:
1 decade ago
Actually Data rate That is N.
Max data rates Nmax.
We have Nmax=2*B*logL.
Here B=2khz.
L=No.of level=2(as it is binary).
So Nmax=2*3*log2=6(Answer).
Max data rates Nmax.
We have Nmax=2*B*logL.
Here B=2khz.
L=No.of level=2(as it is binary).
So Nmax=2*3*log2=6(Answer).
Bala said:
1 decade ago
Maximum data rate is.
Maximum Capacity in Bits/Second = Bandwidth * Log2 (1 + Signal/Noise).
Where signal = noise = 1. Then apply values.
Maximum Capacity in Bits/Second = Bandwidth * Log2 (1 + Signal/Noise).
Where signal = noise = 1. Then apply values.
Abrar said:
1 decade ago
What is the correct answer? Please tell.
Vivek Maurya said:
1 decade ago
For Noiseless Channel:
Maximum bit rate (Maximum capacity of the channel) is given by Nyquist Bit Rate.
Channel capacity, C = 2*B*Log L {Nyquist Bit Rate formula}.
Where C is channel capacity.
B is bandwidth of channel.
L is number of signal level.
-->> In above formula Log L (is on base 2).
Maximum bit rate (Maximum capacity of the channel) is given by Nyquist Bit Rate.
Channel capacity, C = 2*B*Log L {Nyquist Bit Rate formula}.
Where C is channel capacity.
B is bandwidth of channel.
L is number of signal level.
-->> In above formula Log L (is on base 2).
Vivek Maurya said:
1 decade ago
Correct answer is 6kHz (6 kilo bit per second = 6kbps).
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