Finding the Uncertainties in Frequency with Given Capacitor and Inductor Values

[Aristotle]

Homework Statement

I am given a frequency value of 95 GHz (9.5x10^10 Hz), C= 25 F, L=1.12x10^(-25) H.
The question is to find the uncertainties in frequency by taking account of inductor being 5% accurate & capacitor being 8% accurate.

Homework Equations

I believe this is the correct formula to use--since f= 1/ 2pi * sqrt(LC) (frequency formula) is a division/fraction.

(σf)^2= [ (df / L)^2 * σL^2 + (df / C)^2 * σC^2 ]

The Attempt at a Solution

Well taking the partial derivatives of f respect to L, I get: C sqrt(LC) / 2pi.
For f respect to C I get: L sqrt(LC) / 2pi.
So taking the partial derivatives that I had found, I plugged into the equation above Relevant equations & got:

(σf)^2= [ (C sqrt(LC) / 2pi)^2 * σL^2 + ( L sqrt(LC) / 2pi)^2 * σC^2 ]

I know that the percentage of accuracy should be substituted in σL^2 and σC^2 with respect to the given capacitor and inductor values, however I need a little guidance whether I am on the right track. Thank you!

 

[ehild]

Homework Statement

I am given a frequency value of 95 GHz (9.5x10^10 Hz), C= 25 F, L=1.12x10^(-25) H.
The question is to find the uncertainties in frequency by taking account of inductor being 5% accurate & capacitor being 8% accurate.

Homework Equations

I believe this is the correct formula to use--since f= 1/ 2pi * sqrt(LC) ...

check the formula, it is not correct! The partial derivatives are also wrong.

You get the absolute accuracy of a quantity X if if you multiply X by the percent accuracy divided by 100.

 

[Aristotle]

check the formula, it is not correct! The partial derivatives are also wrong.

You get the absolute accuracy of a quantity X if if you multiply X by the percent accuracy divided by 100.

Sorry Ehild, but what would be a correct formula to use? My teacher hasn't really been that thorough with the subject & had us with that equation to work with, so I wasn't really sure.

What exactly is "absolute accuracy" and after that what would I have to do?

 

[STEMucator]

Is this an $LC$ circuit? If so, you may want to be careful about how you are taking your partial derivatives.

For example, here is the partial with respect to $L$:

 

[Aristotle]

I mean isn't

Is this an $LC$ circuit? If so, you may want to be careful about how you are taking your partial derivatives.

For example, here is the partial with respect to $L$:

View attachment 83008

Thank you for your response, Zondrina.
Yes it is indeed an LC circuit.

Ah, yeah I see where I went wrong with the derivative...but just a curiosity, would the formula that I used be incorrect?

 

[STEMucator]

I mean isn'tThank you for your response, Zondrina.
Yes it is indeed an LC circuit.

Ah, yeah I see where I went wrong with the derivative...but just a curiosity, would the formula that I used be incorrect?

You have the correct formula I believe, just not written properly.

When you write:

$$f = 1/2pi * sqrt(LC)$$

People interpret it as:

$$f = \frac{1}{2} \pi \sqrt{LC}$$

When you really meant:

$$f = \frac{1}{2 \pi \sqrt{LC}}$$

You could have wrote it as:

$$f = 1/(2pi * sqrt(LC))$$

To signify everything is included in the denominator.

You need to find the partial with respect to $C$ now, as well as the errors $\sigma_L$ and $\sigma_C$. Then plug and chug into:

$$\sigma_f = \sqrt{ \left(\frac{\partial f}{\partial L} \right)^2 \sigma_L^2 + \left(\frac{\partial f}{\partial C} \right)^2 \sigma_C^2 }$$

 

[Aristotle]

Is this an $LC$ circuit? If so, you may want to be careful about how you are taking your partial derivatives.

For example, here is the partial with respect to $L$:

View attachment 83008

Are you sure that you can just pull the sqrt C out like that? I mean if we hypothetically multipled C^(1/2) + L^(1/2), we would get LC now, not sqrt LC, right?

 

[Aristotle]

When I did ∂F/∂L, I get:

-[2π (LC)^1/2 ]^(-2) * ∂/∂L[(2π (LC)^(1/2)]
= -[(2π(LC)^1/2 ]^(-2) * (π (LC)^(-1/2) * C ]

**the 2s cancel out on last step.

 

[STEMucator]

Are you sure that you can just pull the sqrt C out like that? I mean if we hypothetically multipled C^(1/2) + L^(1/2), we would get LC now, not sqrt LC, right?View attachment 83010

Recall from math:

$$\sqrt{xy} = \sqrt{x} \sqrt{y}$$

So we can write:

$$\sqrt{LC} = \sqrt{L} \sqrt{C}$$

Now, because the frequency $f(L, C)$ is a function of $L$ and $C$, when we take the partial derivatives with respect to a variable, we treat all the other variables as constants. That's why I was able to pull out the $\sqrt{C}$ when taking the derivative with respect to $L$; the variable $C$ is treated as a constant.

Cleaning up that other partial derivative would give you:

Notice the similarities to the other partial derivative.

 

[Aristotle]

xy−−√=x√y√​

Ah I understand that its one of those properties to know. However I thought if you multiplied two integers/variables, you add their exponents?
Like √(xy) = √x * √ y = xy <----

 

[Aristotle]

Oh wait no silly me, they're two different variables so you can't add the exponents together.

 

[STEMucator]

Ah I understand that its one of those properties to know. However I thought if you multiplied two integers/variables, you add their exponents?
Like √(xy) = √x * √ y = xy <----

You can only add the exponents if the bases are the same.

You can do:

$$\sqrt{x} \sqrt{x} = x^{\frac{1}{2}} x^{\frac{1}{2}} = x^{\frac{1}{2} + \frac{1}{2}} = x^1 = x$$

You can't do:

$$\sqrt{x} \sqrt{y} = x^{\frac{1}{2}} y^{\frac{1}{2}} = xy^{\frac{1}{2} + \frac{1}{2}} = xy^1 = xy$$

 

[Aristotle]

You can only add the exponents if the bases are the same.

You can do:

$$\sqrt{x} \sqrt{x} = x^{\frac{1}{2}} x^{\frac{1}{2}} = x^{\frac{1}{2} + \frac{1}{2}} = x^1 = x$$

You can't do:

$$\sqrt{x} \sqrt{y} = x^{\frac{1}{2}} y^{\frac{1}{2}} = xy^{\frac{1}{2} + \frac{1}{2}} = xy^1 = xy$$

Oh yeah, that's right! I really appreciate your help Zondrina! I don't know where my head is today.

Anyways...
Okay so I did get the same answer you got for partial f with respect to L----> -1/(4piLC) * sqrt(C/L)

For partial f with respect to C I got answer of ----> -1/(4piLC) * sqrt(L/C)In terms of this equation, partial derivatives are determined.

So the absolute errors we know is that inductor is 5 percent and capacitor is 8 percent...& know that C= 25 F and L=L=1.12x10^(-25) H.

Then taking C=25 F...I would take that number and multiply it with 8 percent and divide by 100 correct?

 

[STEMucator]

$\sigma_L$ and $\sigma_C$ represent the errors associated with measuring the inductance and capacitance of each component.

For example, we know if we measure the inductance, we may write it as:

$$L = (1.12 \times 10^{-25} \pm \sigma_L) \space H$$

If we measure the capacitance we may write it as:

$$C = (25 \pm \sigma_C) \space F$$

You are told the inductor is $5 \%$ accurate and the capacitor is $8 \%$ accurate. By accurate, they mean how much the inductance/capacitance may vary according to the nominal value.

So if the capacitor is $8 \%$ accurate, then $\sigma_C = 25 \times 0.08 = 2$ means we can write the capacitance as:

$$C = (25 \pm 2) \space F$$

 

[Aristotle]

$\sigma_L$ and $\sigma_C$ represent the errors associated with measuring the inductance and capacitance of each component.

For example, we know if we measure the inductance, we may write it as:

$$L = (1.12 \times 10^{-25} \pm \sigma_L) \space H$$

If we measure the capacitance we may write it as:

$$C = (25 \pm \sigma_C) \space F$$

You are told the inductor is $5 \%$ accurate and the capacitor is $8 \%$ accurate. By accurate, they mean how much the inductance/capacitance may vary according to the nominal value.

So if the capacitor is $8 \%$ accurate, then $\sigma_C = 25 \times 0.08 = 2$ means we can write the capacitance as:

$$C = (25 \pm 2) \space F$$

Ah I see.
I understand that the measurement includes the best value and error and also that the +/- indicates a range of the possible correct value...but silly question...when you take C=(25±2)F and plug it into the absolute error of the equation...how do I go on about inputting that in my calculator? Would I just use the "2" ?

 

[STEMucator]

Ah I see.
I understand that the measurement includes the best value and error and also that the +/- indicates a range of the possible correct value...but silly question...when you take C=(25±2)F and plug it into the absolute error of the equation...how do I go on about inputting that in my calculator? Would I just use the "2" ?

I think you are referring to this equation from prior:

$$\sigma_f = \sqrt{ \left(\frac{\partial f}{\partial L} \right)^2 \sigma_L^2 + \left(\frac{\partial f}{\partial C} \right)^2 \sigma_C^2 }$$

Perhaps it would be insightful if I included some extra information:

$$\sigma_f = \sqrt{ \left(\frac{\partial f}{\partial L} (1.12 \times 10^{-25}, 25) \right)^2 \sigma_L^2 + \left(\frac{\partial f}{\partial C} (1.12 \times 10^{-25}, 25) \right)^2 \sigma_C^2 }$$

You need to evaluate the partial derivatives at the point $(L, C) = (1.12 \times 10^{-25}, 25)$. You also need to plug in $\sigma_C = 2$ and $\sigma_L = ?$. The answer that comes out is in $Hz$, so make sure to convert to $GHz$.

 

[Aristotle]

I think you are referring to this equation from prior:

$$\sigma_f = \sqrt{ \left(\frac{\partial f}{\partial L} \right)^2 \sigma_L^2 + \left(\frac{\partial f}{\partial C} \right)^2 \sigma_C^2 }$$

Perhaps it would be insightful if I included some extra information:

$$\sigma_f = \sqrt{ \left(\frac{\partial f}{\partial L} (1.12 \times 10^{-25}, 25) \right)^2 \sigma_L^2 + \left(\frac{\partial f}{\partial C} (1.12 \times 10^{-25}, 25) \right)^2 \sigma_C^2 }$$

You need to evaluate the partial derivatives at the point $(L, C) = (1.12 \times 10^{-25}, 25)$. You also need to plug in $\sigma_C = 2$ and $\sigma_L = ?$. The answer that comes out is in $Hz$, so make sure to convert to $GHz$.

Here is what I did:

Does this seem correct to you?

***EDIT: Excuse my error in typing from the picture I attached. I forgot to multiply (2)^2 at the far end of the square root. And I meant to have the " ]^2 " part at the very end to be INSIDE the square root.

Just to be 100% clear, the σL and σC still contain its units of H and F respectively right?

 

[STEMucator]

Just to be 100% clear, the σL and σC still contain its units of H and F respectively right?

Yes.

I suggest just converting everything to standard S.I. units for calculations, and omitting the units when you do the actual calculation. This makes it a bit easier to look at, and it will give the answer in standard S.I. units.

 

[Aristotle]

Yes.

I suggest just converting everything to standard S.I. units for calculations, and omitting the units when you do the actual calculation. This makes it a bit easier to look at, and it will give the answer in standard S.I. units.

Thank you so much for your help Zondrina! This site needs more of you

 

[NascentOxygen]

I am given a frequency value of 95 GHz (9.5x10^10 Hz), C= 25 F, L=1.12x10^(-25) H.
The question is to find the uncertainties in frequency by taking account of inductor being 5% accurate & capacitor being 8% accurate.

Are your sure you needed to go the partial derivatives route?

Because if you just needed to know the result, you could have calculated the upper and lower extremes of f using the extreme values of the elements, e.g., for the upper

$\dfrac 1{2\pi\sqrt{L*0.95*C*0.92}}\\\\=\ \dfrac 1{2\pi\sqrt{LC}}*\dfrac1{\sqrt{0.95*0.92}}\\\\\\=\ \dfrac 1{2\pi\sqrt{LC}}*1.07\\\\\\$

The other extreme of f is less percent, so I'd keep the error as ± 7%

How does this compare with your calculation using partial derivatives?