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:

    [​IMG]

    Solving for the transfer function :

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

    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.
     
    Last edited: Apr 19, 2015
  2. MikeML

    AAC Fanatic!

    Oct 2, 2009
    5,450
    1,066
    Which transfer function V(out)/V(in) are you trying to describe?

    193.gif

    193a.gif

    193b.gif
     
    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
    10,138
    1,786
    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
    10,138
    1,786
    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
    17,722
    4,788
    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:

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

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

    :(
     
    Last edited: Apr 20, 2015
  9. WBahn

    Moderator

    Mar 31, 2012
    17,722
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    Note that your 0.707 comes from 1/sqrt(2). Leverage that knowledge.

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

    Now apply the quadratic formula and see what happens.
     
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  10. WBahn

    Moderator

    Mar 31, 2012
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    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 !

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

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