Data rate governs the speed of data transmission. A very important consideration in data communication is how fast we can send data, in bits per second, over a channel. Data rate depends upon 3 factors:
- The bandwidth available
- Number of levels in digital signal
- The quality of the channel – level of noise
Two theoretical formulas were developed to calculate the data rate: one by Nyquist for a noiseless channel, another by Shannon for a noisy channel.
- Noiseless Channel : Nyquist Bit Rate –
For a noiseless channel, the Nyquist bit rate formula defines the theoretical maximum bit rate
BitRate = 2 * Bandwidth * log2(L)
In the above equation, bandwidth is the bandwidth of the channel, L is the number of signal levels used to represent data, and BitRate is the bit rate in bits per second.
Bandwidth is a fixed quantity, so it cannot be changed. Hence, the data rate is directly proportional to the number of signal levels.
Note –Increasing the levels of a signal may reduce the reliability of the system.
Input1 : Consider a noiseless channel with a bandwidth of 3000 Hz transmitting a signal with two signal levels. What can be the maximum bit rate?
Output1 : BitRate = 2 * 3000 * log2(2) = 6000bps
Input2 : We need to send 265 kbps over a noiseless channel with a bandwidth of 20 kHz. How many signal levels do we need?
Output2 : 265000 = 2 * 20000 * log2(L)
log2(L) = 6.625
L = 26.625 = 98.7 levels
- Noisy Channel : Shannon Capacity –
In reality, we cannot have a noiseless channel; the channel is always noisy. Shannon capacity is
used, to determine the theoretical highest data rate for a noisy channel:
Capacity = bandwidth * log2(1 + SNR)
In the above equation, bandwidth is the bandwidth of the channel, SNR is the signal-to-noise ratio, and capacity is the capacity of the channel in bits per second.
Bandwidth is a fixed quantity, so it cannot be changed. Hence, the channel capacity is directly proportional to the power of the signal, as SNR = (Power of signal) / (power of noise).
The signal-to-noise ratio (S/N) is usually expressed in decibels (dB) given by the formula:
10 * log10(S/N)
so for example a signal-to-noise ratio of 1000 is commonly expressed as:
10 * log10(1000) = 30 dB.
Input1 : A telephone line normally has a bandwidth of 3000 Hz (300 to 3300 Hz) assigned for data communication. The SNR is usually 3162. What will be the capacity for this channel?
Output1 : C = 3000 * log2(1 + SNR) = 3000 * 11.62 = 34860 bps
Input2 : The SNR is often given in decibels. Assume that SNR(dB) is 36 and the channel bandwidth is 2 MHz. Calculate the theoretical channel capacity.
Output2 : SNR(dB) = 10 * log10(SNR)
SNR = 10(SNR(dB)/10)
SNR = 103.6 = 3981
Hence, C = 2 * 106 * log2(3982) = 24 MHz
Data Communications and Networking – Book
- Difference between Bit Rate and Baud Rate
- Channel Allocation Problem in Computer Network
- Difference between Fixed and Dynamic Channel Allocations
- Channel Allocation Strategies in Computer Network
- Multiplexing (Channel Sharing) in Computer Network
- Data Abstraction and Data Independence
- Introduction of ALU and Data Path
- Data Concealment Methods
- Data Structures and Algorithms | Set 30
- Data Structures and Algorithms | Set 31
- Data Structures and Algorithms | Set 38
- Data Structures and Algorithms | Set 32
- Differences between Data paths
- Electronic Data Interchange
- Reliable Data Transfer (RDT) 1.0
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