- #1

- 38

- 7

- Homework Statement
- Cut off frequency

- Relevant Equations
- 1/2*pi*L*C

How would i find the cut of frequency of this RLC circuit .I have used LTspice to find it but what is the equation for finding it

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter IronaSona
- Start date

In summary, the cutoff frequency of an RLC circuit can be found using the equation ##f_{\rm c.o.} = \frac{1}{2\pi}\left[\frac{1}{LC} - \frac{R^2}{2L^2}+\sqrt{\frac{2}{L^2C^2}-\frac{R^2}{L^3C}+\frac{R^4}{4L^4}} \right]^{1/2}## which is derived from the voltage divider magnitude being set equal to ##\frac {1}{\sqrt{2}}##. This equation may give a different result than the more commonly known 1/(2*pi*sqrt(L*C

- #1

- 38

- 7

- Homework Statement
- Cut off frequency

- Relevant Equations
- 1/2*pi*L*C

Physics news on Phys.org

- #2

Homework Helper

Gold Member

- 14,226

- 4,326

You should look for at least one more source on the web in order to check the formulas.

- #3

- 5,695

- 2,477

- #4

- 38

- 7

when i search it i get 1/2*pi*L*C formula , and when i use it i get answer or 8kHz but the simulation is showing 11.1kHzTSny said:

You should look for at least one more source on the web in order to check the formulas.

- #5

- 5,695

- 2,477

##\frac{1}{2\pi\sqrt{LC}}## is the frequency of resonance, not the cutoff frequency. I suggest you read carefully the link of post #2, in post #3 I give a summary of the most important points.IronaSona said:when i search it i get 1/2*pi*L*C formula , and when i use it i get answer or 8kHz but the simulation is showing 11.1kHz

- #6

- 38

- 7

what is the w is that the 2*piDelta2 said:

- #7

- 5,695

- 2,477

##\omega## is the angular frequency is is equal to ##\omega=2\pi f## where ##f## the frequencyIronaSona said:what is the w is that the 2*pi

- #8

- 38

- 7

o ok thank youDelta2 said:##\omega## is the angular frequency is is equal to ##\omega=2\pi f## where ##f## the frequency

- #9

- 5,695

- 2,477

- #10

Science Advisor

Gold Member

- 3,419

- 3,053

It will be the solution(s)* to this equation: ## L^2C^2\omega^4+(R^2C^2-2LC)\omega^2-1=0 ##

I suspect that they are really asking for the resonant frequency, or they expect you to make an approximation based on the extension of the high and low frequency asymptotes. Both of which are much easier to derive.

The other approach would be to put in the values immediately, which makes the algebra easier for that equation. But it isn't very instructive; not much different than asking LTSpice, really.

Ask me if you want to see how I derived that equation, it is really just a voltage divider magnitude (with complex impedances) set equal to ##\frac {1}{\sqrt{2}}##

*there should be only one solution that is physically realizable.

- #11

Homework Helper

Gold Member

- 14,226

- 4,326

As @DaveE noted in the previous post, finding the -3dB cutoff frequency involves solving the quadratic equation that he wrote down. The solution is a little messy. (I let Mathematica find it!)

$$f_{\rm c.o.} = \frac{1}{2\pi}\left[\frac{1}{LC} - \frac{R^2}{2L^2}+\sqrt{\frac{2}{L^2C^2}-\frac{R^2}{L^3C}+\frac{R^4}{4L^4}} \right]^{1/2}$$ You can plug in your values for ##R, L## and ##C## to see if you get a cutoff frequency of about 11 kHz.

- #12

- 38

- 7

thank you , ill try it .TSny said:

As @DaveE noted in the previous post, finding the -3dB cutoff frequency involves solving the quadratic equation that he wrote down. The solution is a little messy. (I let Mathematica find it!)

$$f_{\rm c.o.} = \frac{1}{2\pi}\left[\frac{1}{LC} - \frac{R^2}{2L^2}+\sqrt{\frac{2}{L^2C^2}-\frac{R^2}{L^3C}+\frac{R^4}{4L^4}} \right]^{1/2}$$ You can plug in your values for ##R, L## and ##C## to see if you get a cutoff frequency of about 11 kHz.

- #13

Homework Helper

Gold Member

- 14,226

- 4,326

If ##\large \frac R L## is *small* compared to ##\large \frac{1}{\sqrt{LC}}##, the expression for ##f_{\rm c.o.}## can be approximated as $$f_{\rm c.o} \approx \frac{1}{2 \pi}\left[\frac{1+\sqrt{2}}{LC}\right]^{1/2}= \left(1+\sqrt{2}\right)^{1/2}f_{\rm resonance}$$

This approximation works well for your setup.

This approximation works well for your setup.

Last edited:

Share:

- Replies
- 43

- Views
- 1K

- Replies
- 10

- Views
- 341

- Replies
- 20

- Views
- 852

- Replies
- 3

- Views
- 861

- Replies
- 2

- Views
- 617

- Replies
- 2

- Views
- 627

- Replies
- 3

- Views
- 825

- Replies
- 4

- Views
- 617

- Replies
- 10

- Views
- 1K

- Replies
- 6

- Views
- 848