Calculating pi network impedance

Question

I am practising for my radio license exam. One of the questions from an old exam:

This filter belongs to a 3.7MHz transmitter. Approximate the impedance Z for a load of 50Ω.

Answers: 1kΩ; 50Ω; 10kΩ; or 10Ω.

I thought I could calculate the reactances of the inductors and capacitors (left is C₁; assuming full values for the caps and 2.5uH for the inductor):

• \$X_L = 2\cdot\pi\cdot3700000\cdot0.0000025 = 58.12\Omega\$
• \$X_{C_1} = \frac1{2\cdot\pi\cdot3700000\cdot0.000000000463} = 92.90\Omega\$
• \$X_{C_2} = \frac1{2\cdot\pi\cdot3700000\cdot0.000000001852} = 23.23\Omega\$

And then proceed to compute the substitution resistance of C₂ with R_load (parallel), then combine with L (series) and finally with C₁ (parallel). I do that with Pythagoras and get:

• With C₂ and R_load: \$Z_1 = \sqrt{50^2 + 23.23^2} = 55.13\Omega\$
• With L: \$Z_2 = \sqrt{58.12^2 + 55.13^2} = 80.11\Omega\$
• With C₁: \$Z = \sqrt{92.90^2 + 80.11^2} = 122.67\Omega\$

This is not one of the answers. The correction model says it should be 1kΩ, but how should I calculate that?

Answer 1

The Fast Analytical Circuits Techniques or FACTs will get you there without writing a single line of algebra, just draw little sketches that you can revisit in case you identify a mistake. First, we start with the network observed at \$s=0\$: open the caps and short the inductor. What remains seen from the input is the the 50-\$\Omega\$ resistance. We have \$R_0=50\;\Omega\$. Then, temporarily remove the energy-storing elements and determine the resistance "seen" from their connecting terminals. Here, for the natural time constants of the denominator, you reduce the excitation to 0 A or open-circuit the test generator used to determine the input impedance. You repeat the exercise as shown in the below figure:

You then assemble the time constants the following way:

\$D(s)=1+s(\tau_1+\tau_2+\tau_3)+s^2(\tau_1\tau_{12}+\tau_1\tau_{13}+\tau_2\tau_{23})+s^3\tau_1\tau_{12}\tau_{123}\$

The zeros are found by nulling the response. Nulling means that the voltage across the current test generator to determine the input impedance is 0 V. 0 V across a current source is a degenerate case and you can replace the source by a short circuit. Again, "look" at the resistance offered in this mode by the energy-storing elements:

Once this is done, you can assemble the time constants to form the numerator \$N(s)\$:

\$N(s)=1+s(\tau_{2N}+\tau_{3N})+s^2(\tau_{3N}\tau_{32N})\$

The final expression for this input impedance appears in a low-entropy form and follows:

\$Z(s)=R_0\frac{N(s)}{D(s)}=R_0\frac{1+s\frac{L_3}{R_1}+s^2L_3C_2}{1+sR_1(C_1+C_2)+s^2L_3C_1+s^3(C_1C_2L_3R_1)}\$

You can further rearrange this expression in a nice canonical form as

\$Z(s)=R_0\frac{1+\frac{s}{\omega_{0N}Q_N}+(\frac{s}{\omega_{0N}})^2}{(1+\frac{s}{\omega_p})(1+\frac{s}{\omega_{0D}Q_D}+(\frac{s}{\omega_{0D}})^2)}\$

The below Mathcad sheet gathers all these equations for you and define all the terms needed in the above equation.

Finally, I plotted the raw expression (the simple paralleling of all the elements) and the two other expressions, the complete one and the canonical form. Not too bad! : )

You may be surprised by the approach here, with these FACTs. You have seen that I did not write a single line of algebra, just small sketches that I individually solved. The neat thing is that if I see a deviation between the raw transfer function and the one I derived, I can go back to the small sketches and solve the one which is wong. Try to do that with the brute-force expression : ) If you want to know more about the FACTs, you can check out this tutorial from APEC 2016 and also the list of transfer functions I derived in the book:

http://cbasso.pagesperso-orange.fr/Downloads/PPTs/Chris%20Basso%20APEC%20seminar%202016.pdf

http://cbasso.pagesperso-orange.fr/Downloads/Book/List%20of%20FACTs%20examples.pdf

Answer 2

One of the comments to your question mentioned performing the analysis using complex mathematics. Here is an outline:

The impedance of a resistor is R $$$$ The complex impedance of a capacitor is $$\frac{-j}{\omega \cdot C}$$

The complex impedance of an inductor is $$j \cdot \omega \cdot L$$

The definition of j is: $$j=\sqrt{-1}$$

Therefore: $$j\cdot j=-1$$

and: $$-j\cdot j=1$$

What you need, is to work out the impedance using your understanding of series and parallel connected components, substituting these impedance expressions. If you call the cap on the left C1 and the cap on the right C2, you will end up with the j terms which are complex numbers. It'll end up looking like:

$$\frac{\left( \frac{R \cdot \frac{-j}{\omega \cdot C_2}}{R+\frac{-j}{\omega \cdot C_2}} +( j \cdot \omega \cdot L)\right) \cdot \frac{-j}{\omega \cdot C_1}}{\frac{R \cdot \frac{-j}{\omega \cdot C_2}}{R+\frac{-j}{\omega \cdot C_2}} +( j \cdot \omega \cdot L)+\frac{-j}{\omega \cdot C_1}}$$

Expand the mathematic equation, multiply by the complex conjugate and take the square root. The complex expression z $$\frac{a+b\cdot j}{c - d\cdot j }$$

Has the conjugate z^* $$\frac{a-b\cdot j}{c + d\cdot j }$$

You flip the polarity of the j expressions from positive to negative or vice versa

Multiplying the two $$ Z \cdot Z^*=\frac{a+b\cdot j}{c - d\cdot j }\cdot \frac{a-b\cdot j}{c + d\cdot j }=\frac{a^2+b^2}{c^2+d^2}$$ Yeilds a real expression (not complex), but it is the square of the real magnitude, so you square root the answer

$$Z_j=\frac{\left( \frac{R \cdot \frac{-j}{\omega \cdot C_2}+( j \cdot \omega \cdot L)(R+\frac{-j}{\omega \cdot C_2})}{R+\frac{-j}{\omega \cdot C_2}}\right) \cdot \frac{-j}{\omega \cdot C_1}} {\left( \frac{R \cdot \frac{-j}{\omega \cdot C_2}+( j \cdot \omega \cdot L)(R+\frac{-j}{\omega \cdot C_2})}{R+\frac{-j}{\omega \cdot C_2}}\right) +\frac{-j}{\omega \cdot C_1}}$$

Multiply top and bottom of numerators and denominators by \$j\cdot \omega \cdot C_2\$

$$Z_j=\frac{\left( \frac{R +( j \cdot \omega \cdot L)(j \cdot R\cdot \omega \cdot C_2+1)}{j\cdot R \omega \cdot C_2 +1}\right) \cdot \frac{-j}{\omega \cdot C_1}} {\left( \frac{R +( j \cdot \omega \cdot L)(j \cdot R\cdot \omega \cdot C_2+1)}{j\cdot R \cdot \omega \cdot C_2 +1}\right) +\frac{-j}{\omega \cdot C_1}} =\frac{\left( \frac{R +( j \cdot \omega \cdot L)(j \cdot R\cdot \omega \cdot C_2+1)}{j\cdot R \omega \cdot C_2 +1}\right) \cdot \frac{-j}{\omega \cdot C_1}} {\left( \frac{R +( j \cdot \omega \cdot L)(j \cdot R\cdot \omega \cdot C_2+1)}{j\cdot R \cdot \omega \cdot C_2 +1}\right) +\frac{-j}{\omega \cdot C_1}}$$

$$Z_j=\frac{ \left( R \cdot (1 - \omega^2 \cdot C_2 \cdot L) + ( j \cdot \omega \cdot L) \right) \cdot -j } { \left( R\cdot (1 - \omega^2 \cdot C_2 \cdot L) + ( j \cdot \omega \cdot L) \right)(\omega \cdot C_1) + -j\cdot(j\cdot R \omega \cdot C_2 + 1)}$$

$$Z_j=\frac{ ( \omega \cdot L) -j \cdot R \cdot (1 - \omega^2 \cdot C_2 \cdot L) } { R\cdot (\omega \cdot C_1) \cdot (1 - \omega^2 \cdot C_2 \cdot L) + ( j \cdot \omega \cdot L) (\omega \cdot C_1) + \ R \cdot \omega \cdot C_2 + -j}$$

$$Z_j=\frac{ ( \omega \cdot L) -j \cdot R \cdot (1 - \omega^2 \cdot C_2 \cdot L) } { \left(R\cdot (\omega \cdot C_1) \cdot (1 - \omega^2 \cdot C_2 \cdot L) + R \cdot \omega \cdot C_2 \right)+ j\cdot \left((\omega \cdot L) (\omega \cdot C_1) -1\right)}$$

If my algebra works out... $$\vert Z \vert =\sqrt{\frac{ ( \omega \cdot L)^2 +\left(R \cdot (1 - \omega^2 \cdot C_2 \cdot L) \right)^2} { \left(R \cdot (\omega \cdot C_1) \cdot (1 - \omega^2 \cdot C_2 \cdot L) + R \cdot \omega \cdot C_2 \right)^2+ \left((\omega^2 \cdot L \cdot C_1) -1\right)^2}}$$