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.
ANUSHKA said:
8 years ago
DATA RATE=2*BANDWIDTH*LOG(L BASE 2),
ACCORDING TO THE PROBLEM BANDWIDTH=3KHZ.
AND L=2(BINARY LEVEL).
SO, DATA RATE=2*3*LOG(2 BASE 2).
DATA RATE=6KBPS.
ACCORDING TO THE PROBLEM BANDWIDTH=3KHZ.
AND L=2(BINARY LEVEL).
SO, DATA RATE=2*3*LOG(2 BASE 2).
DATA RATE=6KBPS.
(1)
Messi said:
8 years ago
Wifi adapter : 150 kbps.
What is kbps? Discuss the meaning of kbps.
What is kbps? Discuss the meaning of kbps.
(1)
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.
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