What are the applications of decibel Representation in Wireless Communications?

Digital ElectronicsElectronics & ElectricalElectron

What are the applications of decibel (dB) representations?

For convenient representation of data for analysis, the decibel representation is used for various parameters such as amplifier gain, return loss, path loss, signal-to-noise ratio (SNR) and so on. Some of such representations are presented here.

Quantity in decibel (dB) = log10(Compared Quantity)/(Reference Quantity)

Amplifier Gain

RF (Radio Frequency) power amplifier gains are also usually expressed in ‘dB’. It is a practice to arrange amplifiers in series to achieve high gain at the output. This form of arrangement is called cascading. In a cascaded amplifier system, the output of one amplifier is fed as input to its succeeding amplifier. The overall gain is calculated as the product of the gains of the individual amplifiers.

The gains of individual amplifiers, when expressed in ‘dB’, are added to compute the overall gain.

Gain’ is a performance metric of an amplifier. Let us consider that an amplifier is designed to provide 10 dB gain.

What Does a Power Gain Of 10 dBW Mean?

Let X = 1 W and Y = 10 W. Thus, we can say that Y is ten times more than X.

Interestingly; this ten-fold increase in the linear scale corresponds to a 10 dB gain in the logarithmic scale. Here, X can be thought of as a reference. If X is 1W Y turned out to be 10 W. Since 1 W is the reference, we express power gain in dBW (decibel with respect to 1 Watt). The power gain is 10 dBW

$$10dB=10log_{10}(\frac{Y}{X})=10log_{10}(\frac{10X}{X})=10log_{10}(\frac{10}{1})$$

This is what the ‘dB’ representation essentially conveys to us. It compares two quantities and tells us how much a quantity relatively higher or lower than the other is.

In our ‘X-Y’ example, if X was to be static and Y was to change from time to time, at each point of time, the ‘dB’ gain tells how high or low (or how strong or weak) Y is when compared to X. If Y is less than X, the gain will have a negative sign! (It can be thought of as a loss). Even attenuation can be expressed in ‘dB’.

Let there be a two amplifier cascaded system with gain of the first amplifier being 10 dB and the second amplifier being 15 dB. Then, the overall gain of the amplifier system comes out as 25 dB (15 dB+10 dB). Let us understand more about this now!

$$Power\:gain(dB)=10log_{10}(\frac{P_{output}}{P_{input}});\frac{P_{output}}{P_{input}}=10^{\frac{power gain(dB}{10}}$$

This is the standard formula for calculating power gain in dB.

We include the multiplication factor of 10 in the front of log operator. For finding voltage gains, we use a multiplication factor of 20 instead of 10.

A power gain of 25 dB corresponds to an output to input power ratio of 316.22 and this means that the amplified output power is 316.22 times greater than the input power.

  • A 3-dB power gain translates to ‘output to input power’ ratio of 2.

  • A 10-dB power gain translates to ‘output to input power’ ratio of 10.

Since the ‘dB’ is based on logarithms, it is important to note two important properties of logarithms −

$$Multiplication\:\longleftrightarrow\:Addition;\:log(A_{1}A_{2})=log(A_{1})+log(A_{2})=(A_{1})_{dB}+(A_{2})_{dB}$$

$$Division\:\longleftrightarrow\:Subtraction;\:log(\frac{A_{1}}{A_{2}})=log(A_{1})-log(A_{2})=(A_{1})_{dB}-(A_{2})_{dB}$$

Thus, logarithms have proven to be a convenient way of expressing values that vary over dynamic ranges.

Signal-to-Noise Ratio (SNR)

This is a performance metric that describes the signal power in the presence of wireless channel noise (interference). In the linear scale, the SNR is the ratio of the signal power to the noise power. The wireless channel is never noise-free. There always exist noise floors that the signal must combat to reach the receiver successfully.

Whenever the signal power drops below a certain threshold, it cannot be detected and decoded by the receiver. This is what SNR indicates us. It tells us how stronger the signal is when compared to the channel noise. A positive SNR indicates that the signal power is greater than the noise power while a negative SNR indicates the opposite. Let us see the expression of SNR below −

$$SNR(dB)=10log_{10}(\frac{S_{p}}{N_{p}})$$

Where SP denotes signal power and NP denotes noise power.

SNR- Numeric Example

At a communication receiver, the SNR of the signal is measured as 15 dB. What does this specify?

Soln. We know that signal-to-noise ratio represents the ratio of the signal power to the noise power. SNR is given as 15 dB. Hence,

$$15dB=10log_{10}(\frac{S_{p}}{N_{p}})$$

$$log_{10}(\frac{S_{p}}{N_{p}})=1.5$$

$$(\frac{S_{p}}{N_{p}})=10^{1.5}\approx\:31.622$$

So, we can observe that the ratio of signal to noise power is 31.622 and this indicates that the signal power is nearly 32 times stronger than the noise power and hence the signal doesn’t get buried under the noise.

Ideal Value Of SNR

There is no fixed value but it varies depending on channel environments. If the channel is too noisy, a high SNR is desired. What happens when the signal power and noise power equal each other? This is a worse case. Unfortunately, the receiver can’t detect and decode the signal. Let us see look at this scenario below −

$$SNR(dB)=10log_{10}(\frac{S_{p}}{N_{p}})=10log_{10}(\frac{N_{p}}{N_{p}})=0$$

The SNR is 0 dB when the signal power equals the noise power. Any SNR value including 0 dB and below this is considered worse.

Path Loss (PL)

An electromagnetic wave undergoes attenuation during its journey from the transmitter to the receiver. The signal takes multiple paths to reach the receiver, and this is called multi-path propagation. During its journey, the signal undergoes phenomena such as reflection, diffraction, scattering and so on. All these lead to attenuation of the signal strength.

Consequently, the received power may never be the same as the transmitted power. Put in other words, the received signal can never be of the same strength as it was at the transmitter. In practice, the received power level is often multiple orders lower in magnitude than the transmitted signal. Let us look at the formula of path loss.

$$path\:loss(PL)(dB)=20\:log_{10}(\frac{4\pi\:d}{\lambda})$$

Where, λ is the wavelength of the transmitted signal and d is the distance between the transmitter and receiver. As ‘d’ increases, the path loss increases. The received power level thus decreases. However, as long as the received power level is above the noise floor, the signal can be detected and decoded by the receiver assuming that the receiver has high sensitivity.

raja
Published on 23-Jun-2021 14:11:56
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