This time we used Sagemath to model an LC bandpass filter, centered on 100MHz, and calculate the insertion loss in dB at various frequencies. It’s based on what the filter calculator defaults to for a 1st order shunt-first Chebyshev bandpass filter.
# Complex impedance for capacitor and inductor C(c,f) = -I / (2 * pi * f * c) L(l,f) = I * 2 * pi * f * l # Composition rules for impedance recip(x) = 1 / x ipar2(a,b) = recip( recip(a) + recip(b)) ipar3(a,b,c) = recip( recip(a) + recip(b) + recip(c)) iser(a,b) = a + b # A simple RC lowpass filter V = 10 Rs = 50 # source impedance Rl = 50 # load impedance # These are just the default values from https://rf-tools.com/lc-filter/ # They give a bandpass filter centered at 100MHz, with 20MHz bandwidth. C1 = 48.58e-12 L1 = 52.67e-9 # This is the overall LC bandpass filter, with source impedance and load. impedance = iser( Rs, ipar3(C(C1,f),L(L1,f),Rl) ) # The filter website shows insertion loss, which looks at how the power in the load # is affected by putting in the LC filter. power(v,r) = (v^2 / r).abs() # Without our filter ("device under test"), the power is simple - half the voltage # is dropped over the source impedance, leaving half for the load. power_without_dut = power(V/2,Rl) # To calculate power in load with our filter, we need to know the current current = V / impedance V_over_load = V - (current * Rs) power_in_load = power(V_over_load, Rl) as_dB(frac) = log(frac,10) * 10 insertion_loss_dB = as_dB(power_in_load / power_without_dut) insertion_loss_dB(51750194).n() # ans: -3.29183565814456 insertion_loss_dB(41896869).n() # ans: -5.05202056128256
These values for insertion loss match what the calculator website shows in the “S parameters” tab.
How to think about this? If you take away the LC bit, you’re just left with a “maximum power theorem” situation, ie. source impedance and load impedance are both constant and equal.
Now, let’s introduce the LC again but imagine that there’s some magic circumstance in which it’s impedance ends up zero – meaning we’ll still be in “maximum power theorem” situation. Since the L and C are of “opposite nature”, there’ll be some frequency where the L has reactance-blah reactance at the same time as the C has negative-blah impedance. With the L and C in parallel we sum their impedances (actually, reciprocal of sum of reciprocals) so they cancel out.
In all other circumstances, either the L will “win” with a larger reactance or the C will “win” and we’ll net out to the L-par-C having some non-zero impedance. This will pull us away from our perfect 50ohm “maximum power” point and the power in the load will be less.
At low frequencies, the inductor (L) will present a low-impedance path, meaning that the overall impedance will be much lower than 50. This results in a higher current, a greater voltage drop over the source impedance leaving our parallel RLC bundle seeing a lower voltage.
At high frequencies, it’s the capacitor which presents a low-impedance path, and by the same argument, we get a lower voltage over our load.
But when we’re dealing with RF we’re dealing with waves, and when there’s a change in impedance you get reflections. If our bandpass filter only had 50ohms impedance at it’s centre frequency, then it has a non-50ohm impedance everywhere else – which causes reflections. Actually, this is kinda obvious from conservation of energy: if there’s energy coming in, but it’s not ending up in the load then either it’s absorbed in the filter itself or it must be reflected. If our filter was absorbing energy, it would be heating up (unlikely, since we build filters with low-resistance components), or it would have to be radiating (unlikely unless the filter was physically very big).
If we’re using a bandpass filter at the input of a receiver, all the signals outside the passband are getting bounced back towards the antenna (the amount of energy here is tiny so it’s not a problem).
If we use a lowpass filter to “get rid of” harmonics at the output of a transmitter, the energy of those harmonics is heading straight back towards the power amp.
If we use an IF bandpass filter in a superhet receiver, then those reflected signals will be heading back towards a mixer where they can re-mix and cause problems.
If we’re worried by those reflections, what can we do? The answer is to use a diplexer – essentially a fork in the road leading to a lowpass and highpass filters respectively. Whichever part (low or high frequencies) we don’t want can be sent to a 50ohm terminator to be absorbed and turned into heat – and overall the whole thing looks like 50ohms across all frequencies This is something I’ve seen used in the “High Performance Direct Conversion Receivers” ARRL article. Also in this “Popcorn DC Receiver”, this one and this one.
When learning about amateur radio, there’s loads of discussion about reflections that happen when the impedance of your antenna isn’t 50ohm at the frequency you’re sending on. This leads to reflections back up your 50ohm coax, and the sum of the outgoing and reflected wave leads to standing waves. The standing waves can be characterised by the “standing wave ratio” (SWR) which is a measurable quantity, and can therefore be used to indicate that your antenna impedance isn’t right. However, if your feeder cable is attenuating the waves, the reflected wave might be weak by the time it gets back to the home end of the cable – which could wrongly lead you to infer that you have the impedance spot on.
Despite all this chatter about reflections from the antenna, I have seen almost no mention of reflections from filtering steps. I guess it’s just a question of degree. Reflections from antennas are bad because its potentially the full transmit power at your transmit frequency that’s bouncing back. Your lowpass filter is also after the power amp, but in normal operation the reflected power is much small since it’s only the power from the harmonics that bounces back. If your power amp was very very nonlinear you might have more power in your harmonics. Or if you had been sending at 7MHz with a 8MHz lowpass filter, and then switched up to sending at 14MHz but didn’t change to an appropriate lowpass filter you’d end up getting all your transmit power bouncing back.