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For those curious about how impedance matching works,
here's the algebra for what is going on when putting two SAW filters in series.
I believe this is correct, but would be interesting to verify it through the measurement of real parts.

Eric is using the B3550 SAW filter, which recommends using a 33nH coil at each end,
plus a 5.6pf cap from the far end of each coil to ground.?
This matching network converts the filter impedance to 50 ohms resistive.

Impedances in series and parallel work exactly the same as resistance, except we must use "complex arithmetic"
when doing the calculations:??
? ??

The freely available python interpreter knows about complex numbers, using "j" to represent the sqrt(-1):
For a specific frequency of operation, we can find the impedance of a coil or cap:??
? ??

Our frequency is 434 Mhz:? ?freq = 434e6? ? ? ? ? ? ? # Using scientific notation, the e6 means add 6 zeros
The impedance of our 33 nH coil is given by? ?Zcoil = 1j * 2 * 3.1416 * freq * 33e-9
And the impedance of our 5.6 pF cap is:? ? ? ? Zcap = 1/(1j * 2 * 3.1416 * freq * 5.6e-12)

Using the same rules for impedance as we do for resistance,
the series combination of the filter plus the coil is:? Zfilter + Zcoil
That impedance in parallel with Zcap is:? ? (Zfilter+Zcoil)*Zcap / ((Zfilter + Zcoil) + Zcap)? =? 50 ohms
(The B3550 datasheet says that a matching network of a 33 uH coil and 5.6 pF cap will?
make the B3550 SAW filter look like a pure resistance of 50 ohms.)

We can solve that equation for Zfilter using some very basic algebra:
? (Zfilter+Zcoil)*Zcap / ((Zfilter + Zcoil) + Zcap)? =? 50
? Zfilter*Zcap + Zcoil*Zcap = 50*Zfilter + 50*Zcoil + 50*Zcap
? Zfilter * (Zcap - 50) = 50*Zcoil + 50 *Zcap - Zcoil*Zcap
? Zfilter =? (50*Zcoil + 50 *Zcap - Zcoil*Zcap) / (Zcap - 50)

Putting it all together as a python script:

freq = 434e6? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? # Our operating frequency in Hz
Zcoil = 1j * 2 * 3.1416 * freq * 33e-9? ? ? ? ? ? ? ? ? ? # The complex impedance of the coil in ohms
Zcap = 1/(1j * 2 * 3.1416 * freq * 5.6e-12)? ? ? ? ? ?# The complex impedance of the cap in ohms
Zfilter =? (50*Zcoil + 50 *Zcap - Zcoil*Zcap) / (Zcap - 50)? ? ? ? ? ? ? ?# Compute the SAW filter's impedance
print(Zfilter)
??
That script gives me this result:

(31.59 - 65.87j)

So the bare B3550 SAW filter without a matching network
should have a resistive impedance of 31.59 ohms.
In series with that resistance is 65.87 ohms of capacitive reactance.

We could cancel out the 65.87 ohms of capacitve reactance by?
by adding 65.87 ohms of inductive reactance.
(Note that the capacitive reactance is -65.87j ohms, our inductance will be +65.97j ohms,
when added together the sum is zero ohms).

The impedance of a coil is? Zcoil = 1j * 2 * 3.1416 * freq * Henrys
so to get an impedance of +65.97j ohms:
65.97j = 1j * 2 * 3.1416 * 434e6 * Henrys
Henrys = 65.97 / (2 * 3.1416 * 434e6) = 24.2e-9

So if we add a 24.2 nH coil in series with one end of a B3550 SAW filter,
it should then present an impedance of around 31.59 ohms, purely resistive.


In the case of two B3550 SAW filters in series, we can have a matching network
of a single 48.4 nH coil in between the two SAW filters, and no capacitors.
Conceptually, this is two 24.2 nH coils in series, and in between those two coils
we see a pure resistance of 31.59 ohms when looking in either direction.??
The capacitive reactance of both SAW filters is canceled out by the added inductive reactance.
?
Jerry, KE7ER