Consider the following circuit:

enter image description here

By inspection, it can be said that the complex impedance of the circuit will be \$R_2\$ for very low frequencies and \$R_1\$ for very high ones, as the inductor will behave as an open circuit for very high frequencies and the capacitor will do the same for very low ones.

I have tried to demonstrate the second statement (the impedance will be \$R_1\$ for very high frequencies) mathematically but I can't. Below are my calculations (let \$s\$ be \$j\omega\$):

$$Z_t = \frac{1}{\frac{1}{R_1 + \frac{1}{sC}} + \frac{1}{R_2 + sL}} = $$

$$\frac{1}{\frac{sC}{sCR_1 + 1} + \frac{1}{R_2 + sL}} = $$

$$\frac{(sCR_1 + 1)(R_2 + sL)}{sC(R_2+sL) + sCR_1 + 1} = $$

$$\frac{sCR_1 R_2 + s^2CR_1 L + R_2 + sL }{s^2L + 1 + s(CR_1 + CR_2)} = $$

$$\frac{-w^2CR_1 L + R_2 + j\omega(L + CR_1 R_2)}{-w^2L + 1 + j\omega(CR_1 + CR_2)}$$

Then, I calculate the module of \$Z_t\$ as:

$$|{Z_t}| = \frac{\sqrt{(R_2 - \omega^2CR_1L)^2 + (\omega(L+CR_1R_2))^2}}{\sqrt{(1-\omega^2L)^2 + (\omega(CR_1+CR_2))^2}}$$.

For very high frequencies, I do:

$$\lim_{\omega \to \infty}\frac{\sqrt{(R_2 - \omega^2CR_1L)^2 + (\omega(L+CR_1R_2))^2}}{\sqrt{(1-\omega^2L)^2 + (\omega(CR_1+CR_2))^2}}$$

As \$(\omega^2)^2 = \omega^4\$, I cancel out the squared terms to the right, leaving the following:

$$\lim_{\omega \to \infty}\frac{\sqrt{(R_2 - \omega^2CR_1L)^2}}{\sqrt{(1-\omega^2L)^2}}$$

which yields

$$\frac{\sqrt{C^2R_1^2L^2}}{\sqrt{L^2}} = CR_1$$

But it should be \$R_1\$. What am I missing?

  • \$\begingroup\$ Try and concentrate on being an EE (rather than trying to find a mathematical solution for something that basic-EE-common-sense tells us is true). By the way, it's only a filter if it has an input node and a different output node (and a common point). \$\endgroup\$
    – Andy aka
    Jan 13 at 16:08
  • \$\begingroup\$ Where is the ouput node for the filter? \$\endgroup\$
    – LvW
    Jan 13 at 16:34
  • \$\begingroup\$ Probably @Verbal Kint could give a good answer with FACTs. \$\endgroup\$
    – internet
    Jan 13 at 16:45

1 Answer 1


In the last formula with s ( in denominator),

enter image description here

the term \$s^ 2 L\$ should be \$ s^2 L C \$ -> \$-w^2LC\$.

NB: you don't need passing to \$w\$.
Just search the "limit" when s-> infinity = \$(s^2 C R1 L)/(s^2 L C) = R1\$.

And it is R2 when s-> zero.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.