channel capacity

•A very important consideration in data communications is how fast we can send data, in bits per second, over a channel.

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DATA RATE LIMITS• The maximum data rate limit over a medium is

decided by following factors:1. Bandwidth of channel.2. Signal levels3. Channel quality (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.1. For noiseless channel- Nyquist bit rate2. For noisy channel- Shannon capacity.

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Nyquist Theorem• Nyquist theorem states that “to produce the

original analog signal ,the sample rate must be at least twice the highest frequency in the original signal”.

Nyquist rate=2 ˟ fmax

Nyquist rate is also called as Nyquist sample rate.

• Sampling rate is inverse of sampling interval. Sampling rate or sampling frequency.

fs = 1/TS . 3

Noiseless Channel : Nyquist bit rate

• Nyquist bit rate defines the theoretical maximum bit rate for a noiseless channel or ideal channel.

• The formula for maximum bit rate in bits per second(bps) is:

Maximum bit rate = 2˟BW ˟log2L

Where, BW =bandwidth at channel L= number of signed levels used to represent

data.

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Noisy Channel : Shannon capacityAn ideal noiseless channel never exists. The maximum data rate for any noisy channel is:

C = BW ˟ log2 (1+S/N)

Where,

C= Channel capacity in bits per secondBW= bandwidth of channelS/N= signal to noise ratio.

The channel capacity is also called as Shannon capacity. The channel capacity do not depend upon the signal levels used to represent the data.

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Decibel• The decibel (dB) measures the relative

strength of two signals or one signal at two different points

• The decibel is positive if signal is strengthened & it is negative when signal attenuates.

it is calculated asdb = 10 log10 (P2/P1)

• Variables P1 & P2 are the power of signal at two points.

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Noiseless Channel: Nyquist RateNoiseless Channel: Nyquist Rate

For a noiseless channel, the Nyquist bit rate formula defines the theoretical maximum bit rate.

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Does the Nyquist theorem bit rate agree with the intuitive bit rate described in baseband transmission?

Example 3.33

SolutionThey match when we have only two levels. We said, in baseband transmission, the bit rate is 2 times the bandwidth if we use only the first harmonic in the worst case. However, the Nyquist formula is more general than what we derived intuitively; it can be applied to baseband transmission and modulation. Also, it can be applied when we have two or more levels of signals.

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Consider a noiseless channel with a bandwidth of 3000 Hz transmitting a signal with two signal levels. The maximum bit rate can be calculated as

Example

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Consider the same noiseless channel transmitting a signal with four signal levels (for each level, we send 2 bits). The maximum bit rate can be calculated as

Example

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We need to send 265 kbps over a noiseless channel with a bandwidth of 20 kHz. How many signal levels do we need?

Example

SolutionWe can use the Nyquist formula as shown:

Since this result is not a power of 2, we need to either increase the number of levels or reduce the bit rate. If we have 128 levels, the bit rate is 280 kbps. If we have 64 levels, the bit rate is 240 kbps.

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Noisy Channel: Shannon CapacityNoisy Channel: Shannon Capacity

In reality, we cannot have a noiseless channel; the channel is always noisy. In 1944, Claude Shannon introduced a formula, called the Shannon capacity, to determine the theoretical highest data rate for a noisy channel:

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Consider an extremely noisy channel in which the value of the signal-to-noise ratio is almost zero. In other words, the noise is so strong that the signal is faint. What is the Capacity C of the channel?

Example

This means that the capacity of this channel is zero regardless of the bandwidth. In otherwords, we cannot receive any data through this channel.

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3.14

We can calculate the theoretical highest bit rate of a regular telephone line. A telephone line normally has a bandwidth of 3000 Hz (300 to 3300 Hz) assigned for data communications. The signal-to-noise ratio is usually 3162. For this channel the capacity is calculated as

Example

This means that the highest bit rate for a telephone line is 34.860 kbps. If we want to send data faster than this, we can either increase the bandwidth of the line or improve the signal-to-noise ratio.

Calculation of SNR in db

• The signal to noise ratio is often given in decibels also.

• SNRdb = 10 log10 SNRor

• SNRdb = 10 log10 S/N

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Assume that SNRdB = 36 and the channel bandwidth is 2 MHz. Calculate the theoretical channel capacity?

Example

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Assume that SNRdB = 45 and the channel bandwidth is 10 MHz. Calculate the theoretical channel capacity?

3.18

Using Both LimitsUsing Both Limits

In practice, we need to use both methods to find the limits and signal levels. Let us show this with an example.

3.19

We have a channel with a 1-MHz bandwidth. The SNR for this channel is 63. What are the appropriate bit rate and signal level?

Example

SolutionFirst, we use the Shannon formula to find the upper limit.

The Shannon formula gives us 6 Mbps, the upper limit. For better performance we choose something lower, 4 Mbps. Then we use the Nyquist formula to find the number of signal levels.

Calculate L for 6 mbps also.

Some More ExamplesExample : Determine the data rate for a noiseless channel having

BW of 3kHz and two signal levels are used for signal transmission.

Solution:For a noiseless channel, the maximum data rate is given by

Nyquist bit rate as

Bit rate= 2˟BW ˟log2L

Bit rate = 2 ˟(3˟103 ) ˟ log2 (2)

Bit rate = 6 ˟ 103 ˟ log10 (2) / log2 (2)

Bit rate = 6000bps 20

Example : Calculate the BW of a noiseless channel having maximum bit rate of 12kbps and four signal levels.

Solution:Nyquist bit rate is given by

Bit rate = 2˟BW ˟log2L

Given, Bit rate = 12 ˟ 103 bps & L= 4

12 ˟ 103 = 2 ˟ BW ˟ [log2 (4)]

BW = 12 ˟ 103 / 2[log10 (4)/log10 (2)]

BW = 3000Hz = 3kHz.21

Example: Compute the channel capacity of a noisy channel having BW = 4Khz, & S/N =0.Solution:

C= BW˟ Log2 (1+S/N)

C=(4˟ 103 )Log2 (1+0)

C=0

i.e Channel capacity is zero hence channel is not able to transmit data.

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Ex4. Calculate the capacity of a telephone channel. The channel BW is 3000HZ & S/N is 3162.Solution:

The telephone channel is a noisy channel

C=BW ˟ log2 (1+S/N)

C= 3000 ˟ log2 (1+3162)

C= = 3000 ˟ (log10 3162/log10 2)

C= 34,860 bps

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Ex5. A system sends a signal that can assume 8 different voltage levels. it sends 400 of these signals per second, what is baud rate?

Solution:

Baud rate is defined as number of signals (symbols)transmitted per second since 400 symbols are transmitted per second.

Baud rate = 400 symbols/sec

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Ex6. A system sends a signals that can assume 2 different voltage levels. It sends 100 of signals per seconds, what is baud rate?

Solution:

Baud rate = number of symbols/secBund rate = 100 symbols /sec.

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Ex7. calculate the maximum bit rate for a channel having bandwidth 3100Hz and SNR ratio 20 Db.Solution:

We have to use formula (SNRdb = 10 log10 SNR)

GivenGiven BW =3100Hz & SNR ratio =20 Dbi.e. 20 Db =10log[SNR], SNR = 100 Maximum bit rate for noisy channel is given by

C=BW ˟ Log2 (1+S/N)

= 3100 ˟ Log2 (1+100)

= 20,640 bits/sec 26

Ex8. Calculate the maximum bit rate for a channel having bandwidth 3100Hz and S/N ratio 10db.Solution:

BW=3100HzS/N =10dB

10dB =10 LOG[S/N]S/N = 10

C=BW ˟ log2 (1+S/N)=3100 ˟ log2 (1+10)

C=3100 ˟ [log10 (11)/log10 2)=10,724bits/sec. 27

Ex9. Calculate the maximum bit rate for a channel having bandwidth =1600Hz if (a)S/N ratio 0 dB (b) S/N ratio 20 dB

Solution:Given BW=1600Hz

a)S/N = 0dB0 = 10 LOG(S/N)S/N =1Maximum bit rate is given by

C=BW ˟ LOG2 (1+S/N)=1600 ˟ log2 (1+1)

C= 1600 ˟ [log10 (2)/log10 (2)]=1600bits/sec

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b) S/N =20 dB

i.e. 20dB = 10log(S/N)

S/N =100

C =1600 ˟ log2 (1+100)

= 1600 ˟ [log10 (101)/log10(2)]

C= 10,654 bits/sec

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3.30

Q. Suppose a signal travels through a transmission medium and its power is reduced to one half. Calculate signal power in db.Ans: Since power is reduced to half this means that P2 = 0.5 P1. Hence, the attenuation (loss of power) can be calculated as

Example

A loss of 3 dB (−3 dB) is equivalent to losing one-half the power.

3.31

A signal travels through an amplifier, and its power is increased 10 times. This means that P2 = 10P1. In this case, the amplification (gain of power) can be calculated as

Example

3.32

Figure: Decibels for given example

3.33

One reason that engineers use the decibel to measure the changes in the strength of a signal is that decibel numbers can be added (or subtracted) when we are measuring several points (cascading) instead of just two. In Figure a signal travels from point 1 to point 4. The signal is attenuated by the time it reaches point 2. Between points 2 and 3, the signal is amplified. Again, between points 3 and 4, the signal is attenuated. We can find the resultant decibel value for the signal just by adding the decibel measurements between each set of points. In this case, the decibel value can be calculated as

Example

Sometimes the decibel is used to measure signal power in milliwatts. In this case, it is referred to as dBm and is calculated as dBm = 10 log10 Pm, where Pm is the power in milliwatts.

Q. Calculate the power of a signal if its dBm = −30.

SolutionWe can calculate the power in the signal as

dBm = 10 log10 Pm

=> -30 = 10 log10 Pm

=> log10 Pm = -3=> Pm = 10-3 mW

Example

3.35

The loss in a cable is usually defined in decibels per kilometer (dB/km). If the signal at the beginning of a cable with −0.3 dB/km has a power of 2 mW, what is the power of the signal at 5 km?

SolutionThe loss in the cable in decibels is 5 × (−0.3) = −1.5 dB. We can calculate the power as

Example