0
\$\begingroup\$

I'm having difficulties finding complex impedance of the following circuit:

enter image description here

What I did first is find the impedance of \$C\$ and \$R\$ in parallel, which is: $$\underline{Z}_{CR}=\frac{R}{1+j\omega CR}$$ Then, I added two inductor impedances: $$\underline{Z}_{CRL}=2j\omega L+ \frac{R}{1+j\omega CR}$$ and finally calculated the parallel of that with capacitor impedance \$\frac{1}{j\omega C}\$ which yields: $$\underline{Z}_{in}=\frac{\omega CR+2 \omega L(1+(\omega CR)^2)-j(1+2\omega CR^2)}{\omega CR + j(2\omega ^2 LC(1+(\omega CR)^2) -2(\omega CR)^2 -1)}$$ However, I'm not getting the correct result: $$\underline{Z}_{in_{correct}}=\frac{R(1-2LC \omega ^2) + j2L\omega}{1-2LC\omega ^2 +j2RC\omega (1-LC\omega ^2)}$$

\$\endgroup\$
4
  • 1
    \$\begingroup\$ I would suggest that it's much easier to use Laplace i.e. sL and 1/sc, and do s->jw at the very end. This reduces the probability of algebraic errors. \$\endgroup\$
    – Chu
    Dec 7 '18 at 0:15
  • \$\begingroup\$ I get the correct result and I used @Chu's approach (as I always do, for things like this.) I agree with your work up to the point where you get your Zin result that is NOT the correct result. So, perhaps you could show more steps from the Zcrl to the Zin? \$\endgroup\$
    – jonk
    Dec 7 '18 at 1:02
  • \$\begingroup\$ I think you have two excellent answers now and you should be able to find your own error easily now. \$\endgroup\$
    – jonk
    Dec 7 '18 at 1:49
  • \$\begingroup\$ In words: I think, R and C are in parallel. And this parallel block is in series with 2L, is it not? And the whole stuff is, again, in paralel to C. Am I wrong? \$\endgroup\$
    – LvW
    Dec 7 '18 at 8:44
3
\$\begingroup\$

\$\small R//C\$: $$\small \frac{R/sC}{R+1/sC}=\frac{R}{1+sCR}$$

Add series L's: $$\small \frac{R}{1+sCR}+2sL=\frac{R+2sL+2s^2RLC}{1+sCR} $$

Parallel C: $$\large \frac{\frac{R+2sL+2s^2RLC}{sC(1+SCR)}}{\frac{1}{sC}+\frac{R+2sL+2s^2RLC}{1+sCR}}$$

Clear denominators:

$$\small \frac{{R+2sL+2s^2RLC}}{1+sRC+sRC+2s^2LC+2s^3RLC^2} $$

Simplify: $$\small \frac{{R+2sL+2s^2RLC}}{1+2sRC+2s^2LC+2s^3RLC^2} $$

\$\small s \rightarrow j\omega\$:

$$\small \frac{{R(1-2\omega^2LC)+j2\omega L}}{(1-2\omega^2LC)+j2\omega RC(1-\omega^2LC)} $$

\$\endgroup\$
3
\$\begingroup\$

I think Chu missed a "2" factor in there somewhere. Here's my approach:

$$\begin{align*} Z_\text{IN}\left(s\right)&=\left(2sL + \frac{R}{1+sRC}\right)\,\biggl|\biggr|\:\frac{1}{sC}=\frac{\left(2sL + \frac{R}{1+sRC}\right)\cdot\frac{1}{sC}}{\left(2sL + \frac{R}{1+sRC}\right)+\frac{1}{sC}}\\\\ &=\frac{\left(2sL + \frac{R}{1+sRC}\right)\cdot\frac{1}{sC}}{\left(2sL + \frac{R}{1+sRC}\right)+\frac{1}{sC}}\cdot\left[\frac{sC}{sC}\right]=\frac{2sL + \frac{R}{1+sRC}}{\left(2sL + \frac{R}{1+sRC}\right)sC+1}\\\\ &=\frac{2sL + \frac{R}{1+sRC}}{\left(2sL + \frac{R}{1+sRC}\right)sC+1}\cdot \frac{1+sRC}{1+sRC}=\frac{R+s2L+s^22RLC}{1+sRC+sC\left(s^22RLC+2sL + R\right)}\\\\ &=\frac{R+2L\,s+2RLC\,s^2}{2RLC^2\,s^3+2LC\,s^2+2RC\,s+1}\\\\ \therefore\\\\ Z_\text{IN}\left(j\omega\right)&=\frac{R\left(1-2LC\,\omega^2\right)+j\,2L\,\omega}{\left(1-2LC\,\omega^2\right)+j\,2RC\,\omega\left(1-LC\,\omega^2\right)} \end{align*}$$

\$\endgroup\$
2
  • \$\begingroup\$ Yes I did miss a 2. Corrected now. Thank you, Jonk. \$\endgroup\$
    – Chu
    Dec 7 '18 at 1:39
  • \$\begingroup\$ @Chu Ah. Good. It's so easy to miss something like that. ;) \$\endgroup\$
    – jonk
    Dec 7 '18 at 1:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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