# Solving for the frequency

Discussion in 'Math' started by simo_x, Apr 19, 2015.

1. ### simo_x Thread Starter Member

Dec 23, 2010
200
6
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.

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Last edited: Apr 19, 2015
2. ### MikeML AAC Fanatic!

Oct 2, 2009
5,451
1,070
Which transfer function V(out)/V(in) are you trying to describe?

Last edited: Apr 19, 2015
3. ### simo_x Thread Starter Member

Dec 23, 2010
200
6
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 !

Last edited: Apr 19, 2015
4. ### Papabravo Expert

Feb 24, 2006
10,685
1,983
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

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5. ### simo_x Thread Starter Member

Dec 23, 2010
200
6
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...

Last edited: Apr 19, 2015
6. ### Papabravo Expert

Feb 24, 2006
10,685
1,983
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.

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7. ### WBahn Moderator

Mar 31, 2012
19,121
5,170
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?

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8. ### simo_x Thread Starter Member

Dec 23, 2010
200
6
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..

Last edited: Apr 20, 2015
9. ### WBahn Moderator

Mar 31, 2012
19,121
5,170
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.

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10. ### WBahn Moderator

Mar 31, 2012
19,121
5,170
I just cranked it and got 15.9 kHz.

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11. ### simo_x Thread Starter Member

Dec 23, 2010
200
6
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 !