**It is common to use dB scale in EMC analyses.** Let alone EMC, almost everybody who owned an audio amplifier - back in the good old days, when people listened to music from CDs - recognize "decibels" as the units in which a change in volume is shown after adjusting volume potentiometer. To understand decibels, we first need to appreciate the fact, that they allow us to see the data in a slightly different manner. It's all about transformation, really. But why this transformation is useful? Is it at all? Sometimes one needs to review the basics. Let's review them.

**The key fact to uderstanding decibels is to realize that they are used for comparison. **Imagine that you are carrying a bag weighing 1kg. You remember how much strength you need to carry such a bag. Now, somebody has added additional items weighing 1kg to this bag. You immediately recognize that to carry this bag you need twice as much strength as in previous case. Next, you pick additional bag weighing 2kg and you feel that the amount of strength you need to carry those bags (now weighing 4kg) is again doubled comparing to the previous case. That is to say, the strength that you needed has risen proportionally to the weight of the bags and this relationship was linear.

However, not all our senses (maybe even less of them) perceive the world in a linear manner. Think of your hearing. Imagine that you find yourself on a motorbike showcase. Let's assume that it is very boring showcase, where all motorbikes are the same. Now, you hear one of them turning on its engine. You perceive the noise level and you keep this level in mind. Next, another engine is turned on. You immediately identify that the noise level has not increased twofold. Instead, it has increased by a smaller amount, say \(\Delta l\). After that another two motorbikes start their engines and you notice that the noise level has again increased by \(\Delta l\). This process repeats with another four engines turned on and the perceived noise level has again risen by \(\Delta l\). So, each time the number of engines running doubles, the perceived difference between successive noise levels remains the same.

This is what we call a logarithmic process. Let's plot it on a 2-D plot with x-axis representing the number of vehicles and y axis the perceived noise level.

Okay, what if the process of turning the engines on continues?

Notice how first observations clash at the beginning of the plot. If we wanted to see them clearly, we could set the difference between units on the x-axis to be 5mm, being a standard square width in a checkered notebook. Then, in order to see the noise level when 1000 engines are running our x axis would have to be 5 meters long!

Okay, it would be nice to change our x axis so that the data for both small and large number of engines is visible. For example we could set a constant difference between 1 and 2, 2 and 4, 4 and 8 and so on. Why not?

Now it seems to be quite clear. We can see data for both small and large number of engines. What is more we observe an interesting linear-like relationship between the number of engines and the noise level. To see what we have actually done here, let us introduce a logarithm \(\log\): \[\log_b(a) = c \iff b^c = a,\]

for \(a > 0, b > 0\) and \(b\neq 1.\) For now let's not dwelve into meaning of all this. Let's instead remember the logarithm symbol \(\log\) and maybe also notice the fact, that we can treat it as a function, i.e. when you provide two numbers \(a\) and \(b\), logarithm spits out another number \(c\). For now we are also not much interested in the number \(b\). We may even omit it for some time and instead of writing \(\log_b(a),\) write simply \(\log(a)\).

Having introduced logarithm, we can say that in our case, the noise level \(l\) is a logarithm of number of engines \(n\), i.e: \[l = \log(n).\]

Let us see some properties of \(\log\), based on our example. First, remember that our reference value was when number of engines running was equal to \(1\). Since the increase in noise level in any subsequent cases will be referred to this reference value, we can safely assume it is equal to \(0\), i.e \(\log(1) = 0.\) Now, when the second engine was turned on, the noise level increased by \(\Delta l\). Therefore, \(\log(2) = \Delta l\). As you may remember, with \(4\) engines running, the noise level has again increased by \(\Delta l\), thus \(\log(4) = 2\Delta l\). The same goes for 8 engines, where we have to add \(\Delta l\) to the previous value \(\log(4)\) and obtain \(\log(8) = 3\Delta l\), etc. Now, you may notice that we have picked our engines turned in such a way that their number doubles. Therefore, each number \(2\), \(4\), \(8\)... is a power o \(2\). We have \(2 = 2^1\), \(4 = 2^2\), \(8 = 2^3\), \(16 = 2^4\) and so on. Combining these two observations, we get:

\begin{align}\log(2) &= \Delta l\\ \log(4) &= \log(2^2) = 2\log(2) = 2\Delta l\\ \log(8)&= \log(2^3) = 3\log(2) = 3\Delta l\\ \log(16) &= \log(2^4) = 4\log(2) = 4\Delta l \\ &\vdots \\ \log(2^N) &= N\log(2) = N\Delta l. \end{align}

So, with exponentially changing number of engines the noise level changes linearly! It is, in fact, true in general, for any \(a, b\) and \(N\): \[\log_b(a^N) = N\log_b(a).\]

Based on this simple observation we can also see that:

\begin{align}\log(4)&=\log(2\cdot 2) - \log(2) + \log(2) = 2\Delta l\\ \log(8)&=\log(2\cdot 4) = \log(2) + \log(4) = \Delta l + 2\Delta l = 3\Delta l\\ \log(16)&=\log(4 \cdot 4) = \log(4) + \log(4) = 2\Delta l + 2\Delta l = 4\Delta l.\end{align}

Which shows us how the multiplication of two number turns into sum of their logarithms. This is as well true in a more general case, for any \(a_1\) and \(a_2\):

\[\log_b(a_1\cdot a_2) = \log_b(a_1) + \log_b(a_2).\]

Note, that the above is by no means a proof, but rather a simple observation. Using those general relations we can conlude the following:

\begin{align}\log_b\left(\frac{a_1}{a_2}\right)&=\log_b(a_1\cdot a_2^{-1})\\ &= \log_b(a_1) + \log_b(a_2^{-1})\\ &=\log_b(a_1)-\log_b(a_2)\end{align}

for \(a_2 \neq 0.\) The latter fact is really the clue for introducing decibels. Before going further into that, let us explain what is the mysterious number \(b\) in \(\log_b(a)\). To this end, let us revise the definition \[\log_b(a)= c \iff b^c = a\]

The number \(b\) is called the base of the logarithm. As an example we could change the way we represent our number of engines as the exponents of 2. In such a case our base \(b\) would be equal to \(2\). We have:

\begin{align}\log_2(2) &=1 &\ \text{because}&\ \ 2^1 = 2\\ \log_2(4) &= 2 &\ \text{because}&\ \ 2^2 = 4\\ \log_2(8) &= 3&\ \text{because}&\ \ 2^3 = 8\\ \log_2(16)&= 4 &\ \text{because} &\ \ 2^4 = 16\\ \vdots \end{align}

We can therefore change the labeling in our plot. Can you see why?

Of course, base does not have to be equal to \(2\). Maybe you want to check the noise level after running 1, 5, 25, 125, ... \(5^N\) engines? Or maybe you analyse the data from bank withdrawals and want to check how many people are withdrawing sums with one decimal place (from \(10\) to \(100\)), two decimal places (from \(100\) to \(1000\)), three decimal places (from \(1000\) to \(10000\)) etc.? With small sums it is feasible, but the more decimal places you have the harder it becomes to count them. It is much simpler to represent them as logarithms. Compare the x axes between two plots below. Which one is simpler to read?

Maybe you haven't noticed, but we have just introduced bels! Let's look at the definition below

\[x~\text{[B]} = \log_{10}\left(\frac{a_2}{a_1}\right)\]

for any \(a_2\) and \(a_1 \neq 0\). Using the properties of a logarithm we can rewrite it and obtain

\[x~\text{[B]} = \log_{10}(a_2) - \log_{10}(a_1)\]

What does it say? The multiplicative difference between \(a_1\) and \(a_2\) translates to an additive difference \(x\) expressed in bels. Notice that according to the above definition bels don't represent any unit in particular, what makes them perfect for comparison. They are rather corresponding to what you could call a factor.

However, it is possible to use bels to express particular quantities with units. To this end, we would need to introduce a quantity which we can refer to. In the cash withdrawal example this quantity could be \(a_2 = 1\$.\) Then people withdrawing 10$ from the cash machine (which is 10 times more than 1$) could be said to withdraw

\[\log_{10}\left(\frac{10~\text{\$}}{1~\text{\$}}\right) = 1~\text{[B\$]}\]

Those withdrawing 100\$ could be said to have withdrawn 2B$ and those withdrawing 1000$ could be said to withdraw 3B$, etc.

Okay, but we rarely use bels, what you find mostly is the usage of decibels [dB]. What is a decibel? It is one tenth of a bel \(1~\text{dB}=0.1~\text{B}\). That is to say that \(1~\text{B} = 10~\text{dB}\), which means that if something changes by any \(x\) amount of bels, it changes by \(10x\) decibels. Hence, we have

\[x~\text{[dB]} = 10\log_{10}\left(\frac{a_2}{a_1}\right).\]

Now, how many decibels is 10$? Answer: 10dB$. How many decibels is 100$? Answer: 20dB$. Seems a bit of a stretch? At least, I hope you understand that decibel is not a unit in itself and it can be used for any purposes (even if it is something strange, such as usage of dB$), especially when you find that the quantities of interest can be succesfully shown on a logarithmic scale.

Let's come back to engineering. Decibels can be used to compare power ratios. Say we have a power amplifier, which aim is to amplify the input power ten times. That is, the ratio \(P_2/P_1 = 10.\) How many decibels does this relate to?

Let's use symbols \(L_1\) and \(L_2\) to represent respectively \(P_1\) and \(P_2\) in decibels. We have \[L_2 = 10\log_{10}(10\cdot P_1) = 10\log_{10}(10) + 10\log_{10}(P_1) = 10+L_1~\text{[dB]}.\]

Therefore adding 10 decibels to \(L_1\) gives us \(L_2\). You could also see it by taking the ratio \(P_2/P_1\) instead:

\[10\log_{10}\left(\frac{P_2}{P_1}\right) = 10\log_{10}\left(\frac{10\cdot P_1}{P_1}\right) = 10\log_{10}(10) = 10~\text{[dB]}.\]

Notice that in both cases, we are not very interested in the value of \(P_1\), it can be virtually any value. However, what if we wanted to express our nominal values using decibels, just as in previous strange example with cash withdrawals? Then we have to set a specific value to which we compare. In case of power, it can be 1 watt. Let's say that our input value \(P_1\) is 2W. How many dBW does this relate to?

\[10\log_{10}\left(\frac{2~\text{W}}{1~\text{W}}\right) = 3~\text{[dBW]}.\]

How to find \(L_2\), a decibel value for \(P_2\)? We have just calculated the difference in decibels between \(P_1\) and \(P_2\) (or rather \(L_1\) and \(L_2\)) and we know it is equal to 10 dB. Therefore, the value of \(L_2\) is simply \(3 + 10 = 13~\text{[dB]}.\) Let's calculate \(P_2\) based on this to see, whether it makes sense:

\[10\log_{10}\left(\frac{P_2}{1~\text{W}}\right) = 13~\text{[dB]} \iff P_2 = 10^\frac{13}{10}\cdot 1~\text{W} \approx 20~\text{W}. \]

And we know that \(P_2 = 10 P_1 = 10 \cdot 2~\text{W} = 20~\text{W}.\) It makes sense.

Take a look at \(P_1\) again. Its value was 2W, which means twice as 1W. Having said that we can conlude the following:

\[10\log_{10}\left(\frac{2 P_1}{P_1}\right) = 10\log_{10}(2) = 3~\text{[dB]}\]

Hence, doubling any value corresponds to increasing it by 3 decibels. What if, instead of doubling, we want to half a value?

\[10\log_{10}\left(\frac{1/2 P_1}{P_1}\right) = 10\log_{10}(1/2) = 10\log_{10}(2)^{-1} = -10\log_{10}(2) = -3\text{[dB]}\]

As we see it corresponds to decreasing the value by 3 decibels. Those two cases are important for the practical purposes. Now you know that if you wanted to amplify your input signal by 4, then you have to add 3 decibels twice, which means you end up adding 6 decibels to the input value.

We might need to memorize decibel values and their correspondence to multiplication factors for at least few cases. A good way of doing that is filling up the following table:

Decibels | Multiplication factor |
---|---|

0 | 1 |

1 | |

2 | |

3 | |

4 | |

5 | |

6 | |

7 | 5 |

8 | |

9 | |

10 | 10 |

11 | |

12 | |

13 |

In order to solve it, you need to remember is that adding 10 decibels corresponds to multiplying the input by 10 and subtracting 3 decibels corresponds to dividing the value by 2. Start by adding 10 decibels to the reference 0 dB (so multiplication factor is 10), then keep on subtracting 3 decibels until you reach a point where you can't subtract anymore - add 10 decibels again and repeat this process. You will end up with something similar to this one:

Decibels | Multiplication factor |
---|---|

0 | 1 |

1 | 1.2 |

2 | 1.5 |

3 | 2.0 |

4 | 2.5 |

5 | 3 |

6 | 3.8 |

7 | 5 |

8 | 6 |

9 | 7.5 |

10 | 10 |

11 | 12 |

12 | 15 |

13 | 20 |

Okay, we have logarithms and their properties, we have decibels, we know that decibel is not a unit in itself and you need to refer it to some value. So far we've been comparing power, but in practice what you measure is voltage and current. You need to compare their amplitudes. How does this translate to decibels? What we know about power \(P\) and its relation to voltage \(V\) and current \(I\)? In most basic setting we have \[P = VI.\]

Assuming an impedance of nominal value \(R\) and applying Ohm's law we get \[P = V^2/R = I^2 R\] and thus, assuming that both the input and output signals have the same impedance, we have:

\begin{align}10\log_{10}\left(\frac{P_1}{P_2}\right)&= 10\log_{10}\left(\frac{V_{1}^2/R}{V^2_{2}}\right)\\ &= 10\log_{10}\left(\frac{V^2_{1}}{V^2_2}\right)\\ &= 10\log_{10}\left(\frac{V_1}{V_2}\right) ^2\\ &=20\log_{10}\left(\frac{V_1}{V_2}\right).\end{align}

A similar calculation follows for current \(I.\) Now, what about the units? Again, the units are dependent on what we would like to compare. If you want to see changes referenced to volts, then you would use dBV calculated as \(20\log_{10}\left(\frac{V_{1}}{1~\text{V}}\right).\) Below is the table showing the amplitude ratios and corresponding decibel values for some important cases.

Amplitude | Decibels |
---|---|

1 | 0 |

\(\sqrt{2}\) | 3 |

2 | 6 |

3 | 9.5 |

4 | 12 |

5 | 14 |

6 | 15.6 |

9 | 19.1 |

10 | 20 |

20 | 26 |

However, if you want to see the values with regard to dBmV or dB\(\mu\)V, then the relations are as follows:

\begin{align}x~\text{[dBmV]}&=20\log_{10}\left(\frac{V_1}{1~\text{mV}}\right)\\ &= 20\log_{10}\left(\frac{V_1}{10^{-3}~\text{V}}\right)\\ &= 20\log_{10}\left(\frac{V_1}{10^{-3}~\text{V}}\right)\\ &= 20\log_{10}{V_1} - 20\log_{10}(10)^{-3}\\ &= 20\log_{10}(V_1) - (-3)\cdot 20\log_{10}(10)\\ &= 20\log_{10}(V_1) + 60\end{align}

and

\begin{align}x~\text{[dB$\mu$V]}&=20\log_{10}\left(\frac{V_1}{1~\text{$\mu$V}}\right)\\ &= 20\log_{10}\left(\frac{V_1}{10^{-6}~\text{V}}\right)\\ &= 20\log_{10}\left(\frac{V_1}{10^{-6}~\text{V}}\right)\\ &= 20\log_{10}{V_1} - 20\log_{10}(10)^{-6}\\ &= 20\log_{10}(V_1) - (-6)\cdot 20\log_{10}(10)\\ &= 20\log_{10}(V_1) + 120\end{align}

As you may easily notice 1 dBV = 60 dBmV = 120dB \(\mu\)V. Now, compare it with: 1 V = 1000 mV = 1000000 \(\mu\)V and choose, which one is easier to read...

This was a brief introduction into decibels. In future posts we will explore this topic a bit further, showing some more interesting examples from electrical engineering.