Solving for Angular Frequency in Series R to Parallel LC

[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.

Homework Statement

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.

Homework 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

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

 

[STEMucator]

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.

Homework Statement

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.

Homework 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

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

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

 

[ChasW.]

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

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.

X_T = jωL / (1 - ω²LC) I don't think can be used because in order to set a value for X_T, 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 V_O.

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

Charles

 

[STEMucator]

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.

X_T = jωL / (1 - ω²LC) I don't think can be used because in order to set a value for X_T, 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 V_O.

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

Charles

That's fine.

You should get something along the lines of:

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

 

[ehild]

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.

Homework Statement

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.

Homework 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

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

 

[ChasW.]

That's fine.

You should get something along the lines of:

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

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 into 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 X_T 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

 

[ChasW.]

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

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 X_C and X_L 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

 

[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

 

[STEMucator]

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

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

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

 

[ChasW.]

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

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.

 

[ChasW.]

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

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.

 

[ehild]

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.


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

.

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

 

[ChasW.]

It was a mistake, solve for ω.

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

Any ideas where I might be going wrong?

 

[ehild]

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

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

 

[ehild]

What is the solution of Ω^2+2Ω-1=0?

ehild

 

[ehild]

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,

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

ehild

 

[STEMucator]

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

ehild

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

 

[ehild]

@Zondrina: Read my post #5

ehild

 

[STEMucator]

@Zondrina: Read my post #5

ehild

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

 

[ChasW.]

What is the solution of Ω^2+2Ω-1=0?

ehild

-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.

 

[ehild]

-1000000(√2-1)

The minus sign in front of 10⁶ is wrong.
So ω=0.4142*10⁵ and f= ω/(2pi)=6.5924*10⁴ Hz. You messed up the magnitudes.

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

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

when I am expecting ≈384235Hz and ≈65924Hz

The results are just those you expected.

ehild

 

[ChasW.]

The minus sign in front of 10⁶ is wrong.
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

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.

 

[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!

 

[ehild]

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?

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 ω_1,2 = 10⁶(±1+√2) at the end.

I suggested you to use a new variable, replacing ω: It was Ω=10^-6ω.
With that, your equations become

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

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