# How to match two impedances which both have an imaginary component?

I am designing a matching network for an inverted F antenna, the source RF pin impedance is 30 Ω + j10 Ω (source IC is an ESP32-D0WDQ6, datasheet here, RF pin impedance in section 2.5 on page 7). I will be designing the antenna using the openEMS simulator software. The datasheet calls for a pi-network close to the IC and a 50 ohm transmission line (datasheet in section 3.1.5 on page 16) the antenna will be as close to 50 ohms as possible, frequency is 2.4 GHz. The PCB layout is very similar to what is shown in the image below, I have the output capacitor of the pi-filter connected directly to the antenna input with no transmission line, and the transmission line coming from the IC is only 1 mm in length.

I asked a similar previous question to which someone offered a solution on how to match the source impedance to 50 ohms using this method:

If your antenna is 50 Ω and your source is 30 Ω + j10 Ω then, add a series capacitor of -j10 Ω to cancel out the +j10 Ω effect of the inductor.

This now means you are trying to match 30 Ω resistive to an antenna of 50 Ω resistive.

Then use an L-pad calculator like this: -

You can double check the formula derivation on that site.

Series inductance needs to be about 1.6 nH Parallel capacitance needs to be about 1.1 pF Then, if you went back to the start of the problem and analysed what value of inductance is needed to produce the j10 Ω in your driver output impedance, you'd calculate it to be 0.663 nH at 2.4 GHz.

This means that you can actually dispense with the added series capacitor of -j10 Ω (as originally proposed) because you need 1.624 nH from the above calculator. The upshot of this is that 1.624 nH might as well be 1.663 nH so, the external series inductor you need to add is 1 nH.

But an antenna is rarely ever purely 50 ohms resistive once it is on the board, and will involve some imaginary component. My question is, how will I match the source impedance to the antenna impedance if the antenna impedance also has an imaginary component, say it ends up being 55 Ω + j 15 Ω.

HERE IS MY 2 PART QUESTION:

1. How would I match 30 Ω + j 10 Ω to 55 Ω + j 15 Ω using a pi-filter? Or any other filter topology if another would be any advantage?

2. The datasheet says the pi-filter should be placed close to the IC and the output transmission line should be 50 ohms to the 50 ohm antenna, but I essentially do not have a transmission line, only the 1 mm trace leading to the pi-filter. Since the source is not 50 ohms, would there be a more ideal impedance to set the transmission line at which leads to the pi-filter from the IC? Such as 30 or 40 ohms?

Here is an image of the layout they use on the ESP32 module:

• The previous question you have linked does actually tell you how to match a complex impedance antenna to a resistive value. You seem to be linking the picture and words from an earlier L-pad question. The linked answer covers a pi-network and then considers adding capacitance in series with the antenna should it be presenting an inductive impedance. Not sure if you have got things muddled up a bit. Mar 30, 2021 at 5:47
• For part 2 of your question, you just use reactance cancellation by adding the appropriate sized series capacitor (or inductor) to cancel the reactive part. Wasn't this clear from earlier answers? Do it on both sides of the pi-filter when matching a complex source to a complex load. Mar 30, 2021 at 5:50
• @Andyaka For this question I eliminated the t-line, so I was wondering about building everything into only one filter such as on the module. Also see my comment to the person's answer where I mention the T-filter, does this sound like a good solution? Mar 30, 2021 at 6:14
• Your link to a previous question takes you to my pi-filter answer and not to what you have embedded following the link. A T filter works just as well as a pi filter and, if that helps you mop up the extra needed capacitance to cancel the reactive parts then it should be simpler. Mar 30, 2021 at 6:28
• I believe the T filter would be simpler in terms of component count. Mar 30, 2021 at 6:51

For question 1: the previous answer suggested putting a capacitor (negative reactance) in series with the load to make the load entirely real. Then you can match the real source to a real load. If both the source and load are inductive, then place one capacitor in series with the load to cancel out the reactance, then put another capacitor in series with the source to cancel out the source reactance. When you do that, you can match a real source to a real load.

For question 2: your transmission line impedance should match both the load and the source impedances in order to deliver maximum power to the load. So if your source is 30 ohms, and you impedance transformed the load to be 30 ohms, then you should use a 30 ohms transmission line. That said, the impedance mismatch between 30 and 50 ohms isn't that much $$\\Gamma=\frac{50-30}{50+30}=1/4\$$ so the power delivered is $$\1-|\Gamma|^2=94\%\$$. So if you just made you load 50 ohms, used a 50 ohm transmission line, and had a 30 ohm source, you'd still deliver 94% of the power to the load.

• So a T-filter topology would be required? Say I was trying to match 30Ω+j10Ω to 50Ω+j10Ω, would I just use the same values for the L-pad filter in the other answer I received, 1nH series inductor followed by 1.1pF shunt capacitor, then add a series 6.63pF cap for the output of the T-filter which would equal 10 Ω capacitive reactance to cancel out the +j10 from the antenna? Would there be a way to incorporate the value of the output capacitor into the first series inductor for the L-pad filter as the person did with the +j10 coming from the source? Or would the LCC T-filter be required? Mar 30, 2021 at 5:45
• And with question 2, with my exact layout, if the source is 30 ohms and the antenna is 50 ohms, and the matching network is transforming 30 ohms to 50 ohms, are you saying in this situation the line should be 30 ohms or 50 ohms? Mar 30, 2021 at 5:46
• For question 1, you can use any impedance matching. Just "pretend" that they capacitor+load is one lumped object. If you use a T, you can combine the two capacitors on either side if the T into a single one to save components. You just find the equivalent capacitance is the two in series. So, to start, you can try whatever matching you are comfortable with, then reduce the circuit if you have two of the same elements in parallel or series. Mar 30, 2021 at 11:35