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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?

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  • \$\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

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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.

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