# Homework Help: Solving for Angular Frequency in Series R to Parallel LC

1. Aug 27, 2014

### ChasW.

This is my first post here. I hope I have found the correct place to do it. This is not actual coursework, but as it is this type of work, I am posting here per the guidelines.

1. The problem statement, all variables and given/known data

I am trying to solve for angular frequency for a given output voltage, so I am seeking to get ω on its own.

This problem applies to a series R to parallel LC circuit.

The formula in bold at the end of this post is what I am trying to solve for ω.

2. Relevant equations

1) Vo = Xtotal / ((Rs + Xtotal)(Vi)) see below
2) Xtotal = jωL / (1 - ω2LC) see below
3) Vo/Vi = jωL / ((Rs - ω2RsLC) + jωL)

where,
Vo = voltage out
Vi = voltage in
ω = angular frequency
Rs = series resistance
L = inductance
C = capacitance

Formula #3 in its above form can be used for plotting band pass filter response for various frequencies of interest.

So what I am seeking to do is identify frequencies for voltage output levels of interest.

For example if the voltage input was 5V and I wanted the frequencies for the -3dB point, I would input for Vo 5V x 1/(√2) or 3.5355. Proof: 20log10(3.5355/5) ≈ -3.0103

3. The attempt at a solution

The above formula #3 is derived from #1:

Vo = Xtotal / ((Rs + Xtotal)(Vi))

where,
Xtotal is the total inductor capacitor reactance
Vo is output voltage
Vi is input voltage

where,
Xtotal = XcXL / Xc + XL

where,
XC = 1/jωC
XL = jωL

Xtotal becomes formula #2

= jωL / (1 - ω2LC)

Assuming Vo < Vi, I believe there are going to be 2 solutions for ω which would make sense given that for example, there are 2 -3dB points for a given center frequency of this circuit type.

How do I solve for or begin to solve for ω when
Xtotal = jωL / (1 - ω2LC) ?

A strong nudge in the right direction is most welcome.

Charles

2. Aug 27, 2014

### Zondrina

Hmm, I'm not certain about this, but couldn't you solve: $(X_T LC) \omega^2 + (jL) \omega - X_T = 0$

3. Aug 27, 2014

### ChasW.

You gave me exactly what I asked for, so Thank you. I now see the quadratic form of the problem, although regrettably I should have been more careful in what I was asking.

XT = jωL / (1 - ω2LC) I don't think can be used because in order to set a value for XT, a frequency must first be known since the complex total reactance is frequency dependent. I really should have been more careful and I do apologize.

I think this leaves me with formula #3.
Vo/Vi = jωL / ((Rs - ω2RsLC) + jωL)

Here, the total reactance as a variable has been substituted away so that perhaps I can still solve for ω for a given VO.

I will try and see if this formula has a quadratic form as well.

Charles

4. Aug 27, 2014

### Zondrina

That's fine.

You should get something along the lines of:

$$(V_0 R_s LC) \omega^2 - (jL(V_0 - V_i)) \omega - V_0 R_s = 0$$

5. Aug 27, 2014

### ehild

The output voltage is usually not in phase with the input voltage, so Vi/Vo is a complex number.
You can plot the ratio of the magnitudes, |Uo|/|Ui|, in terms of frequency.

$$\frac{|lU_o|}{|U_i|}=|\frac {jX}{Rs+jX}|=\frac{|X|}{\sqrt{R_s^2+X^2}}$$ where X is your Xtotal divided by j.
$$X=\frac {ωL}{1-ω^2(LC)}$$

If you want to know the angular frequency ω where the output voltage is a given ratio to the input voltage, |Uo|/|Ui|=A, take the square of the equation above

$$A^2=\frac{X^2}{R_s^2+X^2}$$

Isolate X2, and then solve the equation $$|X|=\frac {ωL}{|1-ω^2(LC)|}$$
for ω2.

ehild

Last edited: Aug 27, 2014
6. Aug 27, 2014

### ChasW.

I did. Thank you.

I've fed into the equation the following values:
Vi = 5
Vo = 3.5356
L = 100*10-6 (100μH)
P = 10*10-9 (10πF)
Rs = 50

In setting up the quadratic form, the values I got were:
For a I got: 0.00000000017678
For b I got: 0.00014644
For c I got: -176.78

For x I got: 668194.96 and −1496569.21

When I feed these values back in to the original formula as frequencies to solve for Vo,
Vo/Vi = jωL / ((Rs - ω2RsLC) + jωL)
I do not get a resulting Vo of 3.5356 for either.

I think one correct frequency should be ≈384235Hz and the other should be ≈65924Hz with a center frequency ≈159154.9Hz (1/2∏√LC). Which for those cases do result in XT of -50j Ω and 50j Ω respectively, which makes sense since the series resistance is 50 and we are seeking the -3dB point (half power point).

I am not certain what I missed here, but I am suspecting perhaps I mishandled the imaginary numbers going into the quadratic calculations.

Any assistance with checking my application of the quadratic formula using the imaginary numbers would be most helpful.

Charles

7. Aug 27, 2014

### ChasW.

Thank you for this reply but I need to clarify something.

I do want to know the angular frequency ω for the given ratio of output voltage to input voltage, but what do you mean by isolate X2 exactly?

If X2 is the square a quantity dependent on XC and XL which are both frequency dependent reactances, how can I hope to use X? Since it is the frequency I am seeking and cannot expect to provide one as input, don't I have to substitute X out? I apologize ahead of time if the premise of my question is not valid. I am clearly confused.

Charles

8. Aug 27, 2014

### ehild

You want |Uo/Ui| = A=1/√2.

$$1/2=\frac{X^2}{R_s^2+X^2}$$

that is, $$R_s^2+X^2=2X^2 \rightarrow X^2=R_s^2$$.

you got the expected value for X: X=± 50 Ω.
Now you know X, find ω.

ehild

9. Aug 27, 2014

### Zondrina

I am unsure as to how you have fed values into the equation.

I'm assuming $P$ is the capacitance $C$. A value for $j$ is not known, so I am unsure as to how you would have obtained those answers.

http://www.wolframalpha.com/input/?i=-%283.5356*50*10^%28-4%29*10^%28-8%29%29x^2+%2B+%28j*10^%28-4%29%283.5356+-+5%29%29x+%2B3.5356*50+%3D+0

10. Aug 27, 2014

### ChasW.

Yes the P was meant to be C. Still not quite right yet. Even with your input, the resulting values for X are probably revealing that the formula I provided was not usable this way. I am still working my way though ehild's response as well. When I can correctly solve for ω2 I should at least know if I am any closer. Thank you again.

11. Aug 27, 2014

### ChasW.

I think I am doing this incorrectly. When trying to solve for ω this is what I am getting:

|X|= ωL / |1-ω2(LC)|
|X - (LCX)ω2| = ωL
|-(LCX)ω2| - (L)ω + |X| = 0

I am coming up with ω = 1000000.

http://www.wolframalpha.com/input/?i=%7C-%2850*%28100*10%5E-6%29*%2810*10%5E-9%29%29w%5E2%7C-%28100*10%5E-6%29w%2B%7C50%7C%3D0

I realize you asked me to solve for ω2. I am not sure how to do that any other way.

When I convert angular degrees to rotational, I notice the value is 159154.9431 which is the resonant or center frequency, not one of the -3dB frequencies. This may mean I am close to getting this, but I am not sure. I may have an error in my frequency translations somewhere else.

Last edited: Aug 27, 2014
12. Aug 27, 2014

### ehild

It was a mistake, solve for ω.

Substitute the numerical data into the equation.
$$50=\frac{10^{-4}ω}{|1-10^{-12}ω^2|}$$

a)$$1-10^{-12}ω^2=2 * 10^{-6}ω$$

Let be Ω=10-6ω

Ω2+2Ω-1=0

b)$$10^{-12}ω^2-1=2* 10^{-6}ω$$

Ω2-2Ω-1=0.

ehild

Last edited: Aug 27, 2014
13. Aug 27, 2014

### ChasW.

Ok then I am coming up with ω = 1000000 which is the resonant frequency where XT is infinite, not the -3dB frequency for when XT is +- 50 Ω.

Any ideas where I might be going wrong?

14. Aug 27, 2014

### ehild

You operated with the absolute values incorrectly.
|x|=50, but 1-ω2LC can be either positive or negative.
And |a+b| is not |a|+|b|

ehild

15. Aug 27, 2014

### ehild

What is the solution of Ω^2+2Ω-1=0?

ehild

16. Aug 27, 2014

### ehild

j is the imaginary unit when you calculate with complex impedances.

ehild

17. Aug 27, 2014

### Zondrina

Thank you, does the quadratic approach have any validity then? I wouldn't know, never seen any of this.

18. Aug 27, 2014

### ehild

ehild

19. Aug 27, 2014

### Zondrina

Ahh the voltages are not in phase. I see how to solve for $\omega$ now.

20. Aug 27, 2014

### ChasW.

-1000000(√2-1) which is f=650645.1423Hz and
1000000(1+√2) which is f=3792237.796Hz

when I am expecting ≈384235Hz and ≈65924Hz

oddly enough, the above answers seem to be one order of magnitude off with some small additional error, but I am not certain.

I am using this formula for checking ≈384235Hz and ≈65924Hz so I think they are right, but I am at a loss to explain the difference from your results.

Vo/Vi = jωL / ((Rs - ω2RsLC) + jωL)

and I am multiplying the Hertz by 2∏ to get to angular frequency before using this above formula.

21. Aug 27, 2014

### ehild

So ω=0.4142*105 and f= ω/(2pi)=6.5924*104 Hz. You messed up the magnitudes.

Wrong again. ω=2.4142*106 and f=ω/(2pi)=3.8423*105Hz

The results are just those you expected.

ehild

22. Aug 27, 2014

### ChasW.

Thank you for your patience and assistance with this. I used your equation, or at least attempted to, when I got those answers.

After agreeing with your equation, which I do, I entered it at the location below.

10−12ω2−1=2∗10−6ω

It looks right to me which is why I am still confused, but perhaps you did not mean for it to be used that way.

http://www.wolframalpha.com/input/?i=10^−12x^2−1=2∗10^−6x

Can you explain where you are getting ω=0.4142*105 and ω=2.4142*106?

Also, I am not so sure ω=2.4142*106 is much better than 1000000(1+√2). Both appear to be sufficiently wrong.

Last edited: Aug 27, 2014
23. Aug 27, 2014

### ChasW.

Error was on my part. On the sheet I was using for converting from angular freq. to Hz I neglected to encapsulate /(2*PI()) in one of the locations causing improper order of operations. Problem solved. Thank you both again!

24. Aug 28, 2014

### ehild

Wolframalpha does not know Physics. One of the roots it gave, ω=-1000000(√2-1), is negative. But the frequency is a positive quantity, so you must ignore it.

Because of the absolute value, there is an other equation, $1-10^{-12}ω^2=2 * 10^{-6}ω$ with positive solution 1000000(-1+√2) (the other root is negative.)

So the angular frequencies are ω1,2 = 106(±1+√2) at the end.

I suggested you to use a new variable, replacing ω: It was Ω=10-6ω.

$1-10^{-12}ω^2=2 * 10^{-6}ω \rightarrow 1-Ω^2=2Ω$

and

$-1+10^{-12}ω^2=2 * 10^{-6}ω \rightarrow -1+Ω^2=2Ω$

I hope you can bring a quadratic equation into the standard from and apply the quadratic formula.

Rearranging the equations, you get

$Ω^2+2Ω-1=0$, solution Ω = $\frac{-2\pm \sqrt {4+4}}{2}=-1+\sqrt{2}$

and

$Ω^2-2Ω-1=0$, solution $Ω=1+\sqrt{2}$.

You get the angular frequencies by multiplying the Ω values by 106.

Wolframalpha is a great help, but it can mislead you. You can not avoid to learn how to solve a quadratic equation, how to handle absolute values and also using the normal form of numbers instead of writing out 6 or hundred zeroes...

ehild