# Make Sense of Attenuation

For RF and EMC engineers, terms such as dB, attenuation, insertion loss, impedance characteristics are their mother tongue (although some speak better than others). But if you are an electronics engineer, a power electronics engineer, or a system engineer, you might be confused by these EMC jargons. In this article, we use simple math equations to demystify the term ‘attenuation’. You will see a few practical day-to-day examples which might confuse you in the past, but hopefully after reading this article, they will make perfect sense.

Case 1 : The 50 Ohm Puzzle

If you have a spectrum analyser with TG (Tracking Generator) function, or even better, a network/impedance analyser, you can measure the attenuation of a simple 50 Ω feedthrough terminator. Figure 1 shows the test set-up, frequency sweep range is between 10 kHz and 1 GHz. Cable was normalised before the 50 Ω feedthrough terminator is connected. One can notice two things from this simple test set up, first is the frequency response (which correlates to the characteristics impedance of the terminator) is not flat (with oscillation rather than a straight flat line). Second is that the 50 Ω feedthrough terminator has about 3.5 dB attenuation.

If a 3 dB attenuator is tested, the result is very similar, but with much flat response than the curve shown in Figure 1. So why does 50 Ω feedthrough terminators give 3.5 dB attenuation?

Notice, both the TG output and the RF input has 50 Ω impedance. If the coaxial cable is 50 Ω impedance, and assuming it is a lossless transmission line, the ‘resistive’ part of the coaxial cable is ignored, we have a potential divider circuit shown in Figure 2 (a). The attenuation is defined by VRF/(Vsource/2), not VRF/Vsource! (You will see why) So in this case, attenuation is V/(2V/2)=1, which correlates to 0 dB attenuation. In Figure 2(b), the 50 Ω feedthrough effectively is paralleled with the 50 Ω receiving impedance, this changes the ratio of the potential divider. As a result, we have VRF=2V×[25/(25+50)]=2/3V. So the attenuation is (2/3V)/(2V/2)=2/3, which correlates to 3.5 dB attenuation.

In comparison, a 3 dB attenuator often has a π (sometimes T) impedance network shown in Figure 3. We can work out the attenuation by simple calculation. 292.4 Ω in parallel with 50 Ω gives 42.7 Ω. VRF is then calculated as V×42.7/60=0.708V. The attenuation is 0.854V/(2V/2)=0.708, which correlates to -3dB.