# Signal Strength and Power

The measurement of the signal strength at various points in a communication system will give us an idea of how well the system is performing. We can compare the power at two different points to see how much the signal is attenuated (reduced) as it passes through the system, or if it is passing through an amplifier, we can see how much the amplifier is boosting the signal. Another critical measurement of signals compares the strength of the signal to the strength of any noise in the system. For all of these comparisons, we are looking at the strength of one signal with respect to the strength of another (i.e. a ratio of one signal strength divided by another). A convenient way of looking at these ratios is to use the decibel.

The decibel (dB) scale measures ratios of things logarithmically. Some common examples of decibel scales are measuring magnitudes of earthquakes and measuring the intensity of sound. The decibel scale is also used extensively for measuring amplitudes and power in electronic communication systems. The formulae are as follows:

$Signal Power (dB) =10log_{10}(\frac{Signal Power}{Reference Power})$

$Signal Amplitude (dB) =20log_{10}(\frac{Signal Amplitude}{Reference Amplitude})$

Note that in electronic communications, amplitude is usually in VRMS.

The most important thing to notice here is that the dB scale is not an absolute scale. The value is always in reference to something, and therefore it is used to express relative gains and losses, and not absolute gains and losses.

For example, if you have an amplifier of some kind, you can measure the power of the signal into the amplifier (in Watts), and the power of the signal coming out of the amplifier (in Watts).  The gain of the amplifier (Ap) can then be expressed in dB as

$A_p(dB) = 10 log_{10}(\frac{SignalPower_{out}}{SignalPower_{in}})$

Going back to the power and amplitude formulae; to understand why the power formula multiplies by 10, and the amplitude formula multiplies by 20, we have to look at (1) what exactly a decibel is, (2) the relationship between amplitude and power, and (3) the properties of logarithms.

1) The decibel originated as the Bel which is named after Alexander Graham Bell.  The Bel is defined as:

$Bel = log_{10}(\frac{Signal Power}{Reference Power})$

It was soon found that the Bel unit gave a very low range of values.  For example, when the gain is 0.000000001, the value is only -8 Bel and when the gain is 100,000,000, the measurement is only 8 Bel. This gives a range of 16 Bel when the absolute range is in the order of 1016. In order to create a larger range, the Bel is multiplied by 10, and the decibel is used instead (deci means one tenth).

It’s kind of like if you have a big picture of Alexander Graham Bell, but you don’t need one so big, so you have to cut it into 10 smaller pieces where each piece is an Alexander Graham deciBell

2) Power is proportional to the square of the amplitude. In electronics:

$Power = \frac{{Voltage}^2}{Resistance} = \frac{V^2}{R}$

3) One rule of logarithms is that the log of a number squared is equal to 2 times the log of the number, or in equation form:

$log(x^2) = 2log(x)$

So, point (1) tells us why we multiply by 10 for calculating dB for power, and points (2) and (3) tell us why we multiply by 20 for calculating amplitude.

The decibel is used to express the ratio of two different power levels, but sometimes, it is useful when the reference power level is a standard, fixed level. There are several such references in common use in communications systems.

dBm: A dBm is a decibel relative to 1 milliWatt (1mW = 0dBm). It is commonly used to identify power levels for most electromagnetic frequency bands, from ultralow frequencies to light-wave frequencies. Mathematically, a dBm is represented by:

$dBm = 10 log_{10}(\frac{Signal Power in mW}{1mW})$

dBW: a dBW uses 1 Watt as its reference (i.e., 1 W = 0dBW) Mathematically, a dBW is represented by:

$dBm = 10 log_{10}(\frac{Signal Power in W}{1W})$

dBV: a dBV uses 1Volt as its reference. Note that the equation is multiplied by 20 instead of 10

$dBV = 20 log_{10}(\frac{Signal Voltage(RMS)}{1V_{RMS}})$

dBc: This is a decibel relative to the power of a carrier signal. Mathematically:

$dBc = 10 log_{10}(\frac{Signal Power}{Carrier Power})$

dBi: This is a measure of antenna gain. It is a measure of the power of the signal from an antenna in the direction the signal is strongest relative to the power that would be transmitted by an isotropic antenna emitting the same total power (an isotropic antenna is an ideal antenna that radiates power uniformly in all directions).

$dBi = 10 log_{10}(\frac{Antenna Power}{Isotropic Antenna Power})$

dBd:  Another measure of antenna gain. It is similar to dBi, except that instead of measuring relative to a theoretical perfect antenna, it measures relative to a theoretical half-wave dipole antenna. It is easy to convert dBd can be converted to dBi, simply add 2.14 (i.e., 0dBd = 2.14 dBi)

$dBd = 10 log_{10}(\frac{Antenna Power}{Dipole Antenna Power})$

## Examples

1. A base station has an input sensitivity of -112dBm. This means that the lowest signal strength that it can receive and use is -112dBm. How many Watts of power is this minimum?

$-112dBm = 10log_{10}(\frac{Power(mW)}{1mW})$

$-11.2 = log_{10}(\frac{Power(mW)}{1mW})$

$10^{-11.2} = 10^{log_{10}(\frac{Power(mW)}{1mW})}$

$5\times3=15$

$10^{-11.2} = \frac{Power(mW)}{1mW}$

$Power(mW) = (10^{-11.2})(1mW)=6.3\times10^{-12}mW$

$Power(W) = 6.3\times10^{-15}W$

2. What is the ratio of Pout to Pin for an amplifier that has a gain of 25dB?

$25dB = 10log \frac{P_{out}}{P_{in}}$

$2.5 = log \frac{P_{out}}{P_{in}}$

$10^{2.5} = 10^{log \frac{P_{out}}{P_{in}}}$

$\frac{P_{out}}{P_{in}} = 10^{2.5} = 316.2$

3. For the amplifier in the above question, what is the output power if the input power is -80dBm. (2 ways to calculate)

a) First method convert -80dBm to Watts and then multiply by the amplifier gain

$P_{in(watts)} = 10^{-8} \times 0.001 = 10^{-11} W$

$P_{out(watts)} = P_{in(watts)} \times (Amplifier Gain) = 10^{-11} W \times 316.2 = 3.162\times 10^{-9}W$

b) Second method: Using rules of logarithms:

$P_{out(dBm)} = P_{in(dBm)} + A_{p(dBm)} = -80dBm + 25dB = -55dBm$

$P_{out(watts)} = 10^{-5} \times 0.001 = 3.162\times10^{-9} W$