1
\$\begingroup\$

Here's a picture of the R+RLC circuit:

enter image description here

I'm trying to find the frequency response of this low pass filter, my solution is below and please let me know if there is anything wrong with it.

My solution: $$ \\ x=V_{in} \\ y=V_{out} \\ x-y = (i_1+i_2+i_3)R_1 \\ y=i_1R_2 \\ y=Li_2'\\ i_3 = Cy' \\ x'-y' = (i_1'+i_2'+i_3')R_1 \\ x'-y' = (\frac{y'}{R_2} +\frac{y}{L} + Cy'')R_1 \\ LR_2x' = R_1R_2y + L(R_1+R_2)y' + LR_1R_2C y'' \\ x(t) = e^{st} , y(t) = H(s)e^{st} \\ H(s) = \frac{LR_2s}{R_1R_2+L(R_1+R_2)s + LR_1R_2Cs^2} $$

Thanks for your help!

\$\endgroup\$
2
  • \$\begingroup\$ Yes, your solution is correct! However, this is a bandpass filter, not a lowpass. \$\endgroup\$
    – hryghr
    Feb 13 '15 at 21:01
  • \$\begingroup\$ It is correct. You are good yo go. \$\endgroup\$
    – nidhin
    Feb 13 '15 at 21:05
1
\$\begingroup\$

I did voltage division using Z-terms and then converted using the laplace transforms and I ended up with the same result you got. Looks good to me!

$$ H(s)=\frac{Z||}{Z_1+Z||} $$ $$ Z|| = \left(\frac{1}{R_2}+\frac{1}{Ls} +\frac{1}{(1/Cs)}\right)^{-1} = \frac{R_2Ls}{R_2LCs^2+Ls+R_2} $$ $$ H(s)=\frac{R_2Ls}{R_1R_2CLs^2 + (R_1+R_2)Ls+R_1R_2}$$

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

There is a fast and clean way to get there by using the fast analytical circuits techniques or FACTs. If you consider natural time constants of the circuits obtained with the input source is zeroed (replace \$V_{in}\$ by a short circuit), then you can determine the transfer function swiftly without writing a single line of algebra. You can rearrange the final expression to make it fit a low-entropy format, a term forged by Dr. Middlebrook. The below pictures show the steps for the time constants:

enter image description here

You can see that for \$s=0\$ the gain is 0 indicating the presence of a zero at the origin. Then you temporarily disconnect the energy-storing elements and "look" through their connecting terminals to determine the resistance \$R\$ driving the considered element. For \$C_2\$, \$R=0\$ and the time constant is 0. For \$L_1\$, the time constant involves the parallel combination of \$R_1\$ and \$R_2\$. Then you calculate the gain \$H^1\$ when the inductor is set in its high-frequency state (an open circuit). Finally, you combine all these time constants as in the below Mathcad sheet. Follow the guidelines from the book and rearrange the equation to unveil the peaking gain, a quality factor and a resonant frequency. This is the ultimate goal when deriving a transfer function.

enter image description here

\$\endgroup\$

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.