Cutoff frequency derivation

By using our site, you acknowledge that you have read and understand our Cookie PolicyPrivacy Policyand our Terms of Service. Electrical Engineering Stack Exchange is a question and answer site for electronics and electrical engineering professionals, students, and enthusiasts. It only takes a minute to sign up. I've been having some problems with how to find the cutoff frequency and the phase shift for a simple RC low pass filter circuit.

I've read this post, but I need a better understanding of what frequency response to fully be able to understand it. A simple RC circuit:. Sign up to join this community. The best answers are voted up and rise to the top. Home Questions Tags Users Unanswered. Derivation of cutoff frequency and phase shift for RC low pass filter Ask Question. Asked 2 years, 5 months ago. Active 2 years, 5 months ago. Viewed 10k times. Bassusour Bassusour 11 1 1 silver badge 4 4 bronze badges.

Look at my answer to the question the 2nd one and does what I say in the first line mean anything to you? And I just realized I'm stupid. Always best to be honest with yourself LOL. Active Oldest Votes. Mitu Raj Mitu Raj 6, 2 2 gold badges 10 10 silver badges 30 30 bronze badges. So phase is tan inverse of -infinity. I don't how to write " infinity ". So I put alpha symbol It is there actually. Check wiki link : en.Below this cut-off frequency the waveguide is unable to support power transfer along its length.

When choosing a waveguide it is important to bear this frequency in mind, especially if any changes to the system may be likely. In view of the critical nature of the cut-off frequency, it is one of the major specifications associated with any waveguide product. Waveguides will only carry or propagate signals above a certain frequency, known as the cut-off frequency. In order to carry signals a waveguide needs to be able to propagate the signals and this is dependent upon the wavelength of the signal.

If the wavelength is too long, then the waveguide will not operate in a mode whereby it can carry the signal. As might be imagined, the cut-off frequency depends upon its dimensions. In view of the mechanical constraints this means that waveguides are only used for microwave frequencies. Although it is theoretically possible to build waveguides for lower frequencies the size would not make them viable to contain within normal dimensions and their cost would be prohibitive.

As a very rough guide to the dimensions required for a waveguide, the width of a waveguide needs to be of the same order of magnitude as the wavelength of the signal being carried. As a result, there is a number of standard sizes used for waveguides as detailed in another page of this tutorial.

How to Select the Cutoff Frequency of Your Low-Pass Filter

Also other forms of waveguide may be specifically designed to operate on a given band of frequencies. Although the exact mechanics for the cut-off frequency of a waveguide vary according to whether it is rectangular, circular, etc, a good visualisation can be gained from the example of a rectangular waveguide. This is also the most widely used form. Signals can progress along a waveguide using a number of modes. However the dominant mode is the one that has the lowest cut-off frequency.

For a rectangular waveguide, this is the TE10 mode. The TE means transverse electric and indicates that the electric field is transverse to the direction of propagation. The diagram shows the electric field across the cross section of the waveguide. The lowest frequency that can be propagated by a mode equates to that were the wave can "fit into" the waveguide.

cutoff frequency derivation

As seen by the diagram, it is possible for a number of modes to be active and this can cause significant problems and issues. All the modes propagate in slightly different ways and therefore if a number of modes are active, signal issues occur. It is therefore best to select the waveguide dimensions so that, for a given input signal, only the energy of the dominant mode can be transmitted by the waveguide.

For example: for a given frequency, the width of a rectangular guide may be too large: this would cause the TE20 mode to propagate. As a result, for low aspect ratio rectangular waveguides the TE20 mode is the next higher order mode and it is harmonically related to the cut-off frequency of the TE10 mode. This relationship and attenuation and propagation characteristics that determine the normal operating frequency range of rectangular waveguide.

Although waveguides can support many modes of transmission, the one that is used, virtually exclusively is the TE10 mode. If this assumption is made, then the calculation for the lower cut-off point becomes very simple. The cut-off frequency for a rectangular waveguide can be calculated using the formula given below:. It is worth noting that the cut-off frequency is independent of the other dimension of the waveguide.

This is because the major dimension governs the lowest frequency at which the waveguide can propagate a signal. Although it is possible to provide more generic waveguide cut-off frequency formulae, these ones are simple, easy to use and accommodate, by far the majority of calculations needed.

The cut-off frequency for a waveguide is one of the most important parameters.Electric filters have many applications and are extensively used in many signal processing circuits. It is used for choosing or eliminating signals of selected frequency in a complete spectrum of a given input.

So the filter is used for allowing signals of chosen frequency pass through it or eliminate signals of chosen frequency passing through it. At present, there are many types of filters available and they are differentiated in many ways. And we have covered many filters in previous tutorialsbut most popular differentiation is based on. We know signals generated by the environment are analog in nature while the signals processed in digital circuits are digital in nature.

We have to use corresponding filters for analog and digital signals for getting the desired result. So we have to use analog filters while processing analog signals and use digital filters while processing digital signals.

The filters are also divided based on the components used while designing the filters. On the other hand, if we use an active component op-amp, voltage source, current source while designing a circuit then the filter is called an active filter.

More popularly though an active filter is preferred over passive one as they hold many advantages. A few of these advantages are mentioned below:. The components used in the design of filter changes depending on the application of filter or where the setup is used. For example, R-C filters are used for audio or low-frequency applications while L-C filters are used for radio or high-frequency applications.

All signals above selected frequencies get attenuated. The frequency response of the low pass filter is shown below. Here, the dotted graph is the ideal low pass filter graph and a clean graph is the actual response of a practical circuit. This happened because a linear network cannot produce a discontinuous signal. As shown in figure after the signals reach cutoff frequency fH they experience attenuation and after a certain higher frequency the signals given at input get completely blocked.

All signals above selected frequencies appear at the output and a signal below that frequency gets blocked. The frequency response of a high pass filter is shown below.In my experience, the most common filter design tasks—i. Electrical engineers often find themselves dealing with signals that have low-frequency information and high-frequency noise or interference. Perhaps we need to suppress unpleasant sounds in an audio signal, or remove spurious variations from a sensor signal, or eliminate the undesired spectrum created during a demodulation process.

Choosing the cutoff frequency of a low-pass filter initially seems quite simple, but when you think about it more carefully—such as when a real-life design forces you to think about it more carefully—you realize that there actually are some subtle details and complexities.

Low-pass filters always transition smoothly from the passband to the stopband. Nevertheless, the more fundamental portion of the design task is thoroughly understanding the signals that will enter the filter and the signals that should come out of the filter.

To optimize your low-pass filter, you have to know as much as possible about the expected frequency content of your input and the desired frequency content of your output. If you plan to use a first-order filter, the frequency response will always have the same basic characteristics, and consequently there are only two generic scenarios that occur to me:.

In this first situation, there are frequencies toward the end of the passband that cannot experience significant attention. For example, you know that all of your signals will be below 10 kHz, but you have an important sensor output that tends to stay around 7. You may not want a cutoff frequency of 10 kHz because this would apply almost 2 dB of attenuation to the 7.

In this case it would be better to increase the cutoff frequency until 7. For example, if you push the cutoff out to 20 kHz, the attenuation at 7. Further increases in cutoff frequency would produce corresponding reductions in the 7. By increasing the cutoff, you include more unnecessary and possibly noisy frequencies in the passband and you reduce the attenuation of frequencies in the stopband.

The second scenario is when the priority is to suppress a particular frequency in the stopband, rather than to preserve a particular frequency in the passband. For example, you might have a clock signal or RF transmitter that is contaminating your fragile analog signal. Within the limitations of a first-order filter, all you can do to increase the attenuation at a particular frequency is move the cutoff closer to 0 Hz.

The important thing to understand in the context of this article is that second-order filters are more malleable. A second-order filter can be adjusted so as to offer a flatter passband a Butterworth filtera steeper roll-off a Chebyshev filteror a more-linear phase response a Bessel filter. The following three plots provide a visual comparison of Butterworth, Chebyshev, and Bessel responses.

The relationship between cutoff frequency and the characteristics of second-order filters is the following: Your choice of cutoff frequency might be influenced by the type of filter that you use. If you use the Chebyshev filter, which has a rapid transition from passband to stopband, you may not have to decrease your cutoff frequency as much as you initially thought.

cutoff frequency derivation

Choosing a cutoff frequency begins with having a vague idea of which frequencies should pass through the filter and which should be blocked by the filter. In Partnership with Vicor. Don't have an AAC account? Create one now. Forgot your password? Click here. Latest Projects Education. This article provides some tips on how to fine-tune the characteristics of a low-pass filter. Inputs and Outputs To optimize your low-pass filter, you have to know as much as possible about the expected frequency content of your input and the desired frequency content of your output.

Focusing on the Stopband The second scenario is when the priority is to suppress a particular frequency in the stopband, rather than to preserve a particular frequency in the passband.By using our site, you acknowledge that you have read and understand our Cookie PolicyPrivacy Policyand our Terms of Service. Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics.

It only takes a minute to sign up. I am studying low and high pass RL filters for Navy schooling. From my oscilloscope lab results I got the following. Conceptually, I understand that the root mean square method is used to determine AC's efficiency in comparison to DC. AC are about Plots of frequency responses are call Bode plots try the wikipedia page though its not so good on this topic and it is much simpler to use linear approximations than worry about the exact response.

If you plot these linear approximation curves on a log-log graph or you plot the dB gain versus log of the frequency they will intercept at the cut off frequency point. In fact as long as you divide the input and output then it doesn't matter if you use peak values, RMS value, or peak-peak values.

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Electrical Engineering Stack Exchange is a question and answer site for electronics and electrical engineering professionals, students, and enthusiasts. It only takes a minute to sign up. I am trying to derive an equation describing the cutoff frequency of a high pass filter op amp, as seen below:.

I have tried deriving the cut off frequency from it's transfer function:. But I can't get the intended result as indicated at the bottom of working?

Waveguide Cutoff Frequency

Is anyone able to spot where I am going wrong in my working or method of attack please? All this means that for low frequency the circuit behaves like an ordinary op-amp based differentiator. From inspection, you can notice that the inverting input is at a virtual ground because of R2. This effectively isolates R1 from R2, so only R1 and C affect the frequency response.

Note: Cutoff depends on signal source impedance!

What is the 'Cutoff' on a filter? (Quick Tutorial)

Your calculation is only valid for a zero-ohm impedance voltage source. Keep this in mind when using this circuit in the real world Sign up to join this community. The best answers are voted up and rise to the top.

Home Questions Tags Users Unanswered. Asked 3 years, 4 months ago. Active 3 years, 4 months ago. Viewed 4k times. ConfusedCheese ConfusedCheese 4 4 silver badges 13 13 bronze badges.

Active Oldest Votes. G36 G36 8, 1 1 gold badge 9 9 silver badges 21 21 bronze badges. Were you expecting R2 to figure in there? Opamp uses R2 to convert this current back to voltage at the output. R1, C1 control cutoff. Sign up or log in Sign up using Google. Sign up using Facebook. Sign up using Email and Password. Post as a guest Name. Email Required, but never shown.

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cutoff frequency derivation

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Thread starter billybud Start date Nov 2, Status Not open for further replies. LvW Advanced Member level 5. Yes, it's not easy - however, rather straightforward.

Use the complex transfer function 2nd order and calculate the magnitude. But I do not recommend this procedure cause you will learn not too much.

Low Pass Filter Calculator

Therefore, the normal and classical approach is to use tables in textbooks computer derived based on pole data or filter simulation programs. Last edited: Nov 3, Have you assumed that the gain is 1? Part and Inventory Search. Welcome to EDABoard. Design Fast. This site uses cookies to help personalise content, tailor your experience and to keep you logged in if you register. By continuing to use this site, you are consenting to our use of cookies.

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