Hi all,

I'm in doubt on how to solve for the \(\omega\) for any transfer function \(T(\omega)\), and I would like to solve it.

My question is not related to 1st order functions, but from 2nd order and so on.

For example, suppose I have a simple RLC low pass filter:

Solving for the transfer function :

\(
\begin{align}
T(s) &= {1 \over s^2LC + sRC + 1}\\
s &= \text j \omega \\
T(\text j \omega) &= {1 \over 1-\omega^2LC + \text {j} \omega R C}\\
|T(\text j \omega)| &= {1 \over \sqrt{(1-\omega^2 L C)^2 - (\omega RC)^2}}
\end{align}
\)

for example, I tried to solve for the -3dB frequency

\(0.707 = |T(\text j \omega)|\)

But I cannot find the correct result, and I know by simulation and from the first pole that \(f_c \approx 16\) kHz

I opened this post here because I think my doubt is more related with math concepts.
Thank you in advance.

 

Which transfer function V(out)/V(in) are you trying to describe?

 

Hi Mike,

It's the second one, but you swapped the values of inductance and capacitance between them.

L = 10 uH
C = 10 nF

From the denominator of the transfer function I know that in this case the frequency of -3dB is 15931 Hz, and for the circuits you posted the -3 dB is 15,915 Hz for the LP and Hp.

But I would like to know how to solve the equation analytically, if possible.

Thanks !

 

To solve for ω you would:

1. Substitute the values for R, L, and C.
2. Square both sides of the equation to remove the radical
3. Then solve for ω
4. Check your work by using the value of ω to get a magnitude of 0.707

 

That's what I have done solving a 4th order polynomial:

\(0.707^2 \left ( \left ( 1 - \omega^2\cdot10^{-6}\cdot10^{-9}\right )^2- \left (\omega \cdot 1000\cdot 10^{-9}\right )^2 \right ) -1 = 0\)

but I cannot get the correct result, or I cannot recognize it...

 

If you expand things out you have terms in the fourth, second, and 0th power of ω. That suggests a quadratic polynomial in ω^2 which can be solved with an appropriate substitution.

 

It really helps if you show your work from beginning to end. Think about it -- you say that you believe the problem you are having is related to math concepts -- so wouldn't the most effective way for people to spot where you are having problems with the math for you to show the math you are doing?

 

Hi, sorry for my late response. I had to study other things too..

This is the calculation I have done:

\(
\begin{align}
& 0.707 = {1 \over \sqrt{(1 - \omega^2LC)^2 + (\text j \omega RC)^2}}\\
& 0.707^2 = {1 \over [(1 - \omega^2LC)^2 - (\omega RC)^2}]\\
& 0.707^2((1 - \omega^2LC)^2 - (\omega RC)^2) = 1\\
& 0.707^2(1 - 2\omega^2LC + \omega^4L^2C^2 - \omega^2R^2C^2) -1 = 0\\
& 0.49985 - 0.99970\omega^2LC + 0.49985\omega^4L^2C^2 - 0.49985\omega^2R^2C^2 - 1 = 0\\
& 0.49985\omega^4L^2C^2 - \omega^2(0.99970LC + 0.49985R^2C^2) -0.50015 = 0\\
& -1.0010\cdot 10^8\\
& 1.0010\cdot 10^8 \\
& -7.2858\cdot 10^{-17} + 9.9930^4\text j\\
& -7.2858\cdot 10^{-17} - 9.9930^4\text j\\
\end{align}
\)

But no one brings me to the result of 100100 rad/s..

 

Hi, sorry for my late response. I had to study other things too..

This is the calculation I have done:

Note that your 0.707 comes from 1/sqrt(2). Leverage that knowledge.

\(
\begin{align}
& \frac{1}{\sqrt{2}} = {1 \over \sqrt{(1 - \omega^2LC)^2 + (\text j \omega RC)^2}}\\
& \frac{1}{\sqrt{2}}^2 = {1 \over [(1 - \omega^2LC)^2 - (\omega RC)^2]}\\
& \frac{1}{2}[(1 - \omega^2LC)^2 - (\omega RC)^2] = 1\\
& (1 - \omega^2LC)^2 - (\omega RC)^2 = 2\\
& (1 - \omega^2LC)^2 - (\omega RC)^2 - 2 = 0\\
& 1 - (2LC)\omega^2 + (LC)^2\omega^4 - (RC)^2\omega^2 - 2 = 0\\
& [(LC)^2]\omega^4 + [(2LC) - (RC)^2]\omega^2 + [-1] = 0\\
\end{align}
\)

Now apply the quadratic formula and see what happens.

 

I just cranked it and got 15.9 kHz.

 

Yes ! I got it !

\(
\begin{align}
\begin{cases}
-1.0374\cdot 10^{-15} + \text i1.0010\cdot 10^{5}\\
-1.0374\cdot 10^{-15} - \text i1.0010\cdot 10^{5}\\
\end{cases}& \text{1st pole}\Rightarrow \approx 15.9\,\text{kHz}\\
\begin{cases}
-9.9900\cdot 10^{7}\\
9.9900\cdot 10^{7}\\
\end{cases}& \text{2nd pole}\Rightarrow \approx 15.9\,\text{MHz}
\end{align}\)

Thank you for your help.
I will be more careful with my calculus next time !