Noise, Data Rate And Frequency Bandwidth

Noise, Data Rate and Frequency Bandwidth

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The Relationship Between Data Rate Capacity, Noise, and Frequency Bandwidth (Morikawa)

The maximum data rate capacity of a digital communications system is a factor of the noise environment, frequency bandwidth, and modulation scheme. Nyquist, Shannon, and Hartley who were pioneers in the science of information technology, discovered equations that took these factors into consideration. Today, their work helps us to better understand how these factors influence one another.

The Nyquist formula below provided a relationship between capacity and bandwidth under idealized conditions where noise is not considered.

C(bps) = 2B * log2M (Nyquist)

C is the capacity in bits per second, B is the frequency bandwidth in Hertz, and M is the number of levels a single symbol can take on. This "idealized" capacity equation shows us that data rate is proportional to twice the bandwidth and logarithmically proportional to M. It is considered "idealized" since environmental influences, particularly noise, is not considered.

Example 1: What is the Nyquist capacity for a signal with a frequency bandwidth of 1kHz, using Binary Phase Shift Keying (BPSK) modulation?

Answer: First, BPSK is the binary form of PSK, where each signaling symbol can take on one of two values (i.e., a phase shift of either 0 radians mapped to a logical "0", or π radians mapped to a logical "1"), therefore M=2. So:

C = (2 * 1kHz) * log2 (2) = 2000 * 1 = 2000 bps

Note: log2M is in binary form. In order to enter this problem into your calculator, you should use the following conversion: log2M = (log10M/log102).

Shannon-Hartley developed a similar equation for capacity; however they included the Signal-to-Noise ratio (SNR) which provides a slightly more realistic answer. Shannon-Hartley's equation provides the "theoretical" maximum capacity for a signal given its frequency bandwidth and SNR. Typically, digital communication systems do not reach this theoretical capacity due to many factors such as modulation scheme and the overall noise environment. However, it remains the goal of many communications engineers to get as close to the theoretical maximum capacity as possible in designing their systems.

C(bps) = B * log2(1+SNR) (Shannon-Hartley)

Note: Read the tech note on SNR (Signal-to-Noise ratio).

SNR is the ratio of the received power in watts, to thermal noise. Noise other than thermal, such as impulse, interference from other sources, etc., is not considered. So it's worth taking a closer look at what thermal noise is.

Thermal noise is caused by the agitation of electrons in a material. As temperature increases, so does the agitation or activity of the electrons, thus resulting in greater thermal noise. This noise impacts our ability to distinguish signal power from noise power. Since electrons exist in all materials, the noise produced cannot be eliminated. Thermal noise spans across all frequencies and is usually called the "noise floor". We use thermal noise when we determine SNR, not only because of its prevalence but also because we can measure it.

N = k*T*B (Noise Power)

N is noise power, k is Boltzmann's constant (1.38E-23 J/K), T is temperature on the Kelvin scale, and B is frequency bandwidth in Hertz. By looking at the units of the equation, we see that N is in watts.

N = (1.38E-23 J/K) * T(K) * B (Hz), where Hz = 1/seconds, watts=J/seconds

Exercise: check the units to see if you come up with watts.

SNR can be re-written as: SNR = P(received)/N = P/kTB

Example 2: What is the Shannon-Hartley theoretical capacity for a signal with a frequency bandwidth of 1kHz, and a SNR = 200?

Answer: C = 1kHz * log2 (1+200) = 1000 * 4.39 = 7651 bps

Note: log2M is in binary form.

If we compare Nyquist's solution in example 1 and Shannon's in example 2, we get two very different answers: C=2000 bps (Nyquist) and C=7651 bps (Shannon). So why the difference even though we used the same bandwidth of 1kHz?

The way to get more insight to this difference is to take the answer we calculated in example 2, and determine M using Nyquist:

Example 3: Using the Nyquist equation, what is M, given C=7651 bps?

Answer:

C = B * log2M, we rearrange the equation to solve for M:

Log2M = 7651/1kHz = 7.651 (number of bits represented by 1 symbol)

M = 27.651 = 200 signaling levels per symbol

Example 3 informs us that the different answers resulted from different values of M. With Hartley, we assumed M=2, however with Shannon, we used a theoretical maximum M of 200. So this example shows us how to use both Nyquist and Shannon-Hartley to determine maximum theoretical throughput based upon M signaling levels per symbol.

One last word on thermal noise: Often times we normalize thermal noise to 1Hz bandwidth. This makes signal calculation possible without requiring the direct knowledge of received frequency bandwidth. Called noise density, the equation follows:

No = kT (Noise Density)

The relationship between N and No is simply the bandwidth:

N = No * B (Noise Power)

Noise density (No) is used to calculate carrier over noise density (C/No), and energy bit per noise density (Eb/No), which are two very important equations in digital communications theory.

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