I have gotten some help from the community here and thus I have reached this.

This is part of the circuit that I want to find the impedance


simulate this circuit – Schematic created using CircuitLab

and as a result of the Fourier Transformation looking at the circuit we have the following.

enter image description here

My confusions are how did we come wit those three results for the three cases. Is there a way I can plot into a calculator those three cases of ω.

Just trying to prove the results

  • \$\begingroup\$ Do you not understand what limits are? Just plug the limit into the equation. If you get a finite number, that's your answer. If you get 0/0 then you have to use math tricks to work around the 0/0 since it is indeterminate. \$\endgroup\$
    – DKNguyen
    Feb 8, 2020 at 0:02
  • \$\begingroup\$ @DKNguyen, For case I would replace ω=0 and then 0/1= 0 , For ω =∞ I would get ∞/∞(squared). So not sure for we get zero: For this case ω=1/sqrtLC I am not sure. \$\endgroup\$
    – E199504
    Feb 8, 2020 at 0:11
  • \$\begingroup\$ Think again. What is \$ \frac{x}{x^2}\$? It's \$\frac{1}{x}\$. Now what if \$x = \infty\$? Should it matter whether I wrote it the first way or the second way? For the ω=1/sqrtLC case, literally plug it in and try simplifying the numerator and denominator. If simplifying one seems like its going nowhere...try the other one... \$\endgroup\$
    – DKNguyen
    Feb 8, 2020 at 0:26
  • \$\begingroup\$ @DKNguyen, thank you. If there any online website where I could plug it it the ω=1/sqrtLC case just so I can prove that I get infinite ? \$\endgroup\$
    – E199504
    Feb 8, 2020 at 0:32
  • \$\begingroup\$ Why would you need that? Did you even try plugging it into the denominator to see what happens? It's trivial. Junior high school math would be able to simplify that denominator. \$\endgroup\$
    – DKNguyen
    Feb 8, 2020 at 7:26

1 Answer 1


The capacitor and inductor are in parallel, which means that the equivalent impedance of these two elements is

$$Z_{EQ} = \frac{Z_LZ_C}{Z_L+Z_C}$$ So, if either \$Z_L\$ or \$Z_C\$ approaches zero then \$Z_{EQ}\$ must approach zero. Now, \$Z_L \propto \omega\$ and \$Z_C \propto 1/\omega\$, so as \$\omega\$ approaches 0 or \$\infty\$ we know that \$Z_{EQ}\$ will approach 0.

When \$\omega = \frac{1}{\sqrt{LC}}\$ we find that \$Z_L = -Z_C\$, so the denominator of the \$Z_{EQ}\$ equation approaches 0, and the magnitude of \$Z_{EQ}\$ approaches \$\infty\$.

I'm not sure what you want to do with your calculator, but if you can program the formula for \$Z_{EQ}\$, using \$Z_L = j\omega L\$ and \$Z_C = \frac{1}{j\omega C}\$, then sweep \$\omega\$ you will see this behavior. Note that you must do these calculations using complex arithmetic, and your calculator may have trouble at the point where \$\omega = \frac{1}{\sqrt{LC}}\$.


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