Cutoff frequency of RLC band pass and band stop filter

RLC Passband and Stopband

Your question doesn't explicitly take note of the fact that there are four distinct RLC filters that are either band-pass or band-stop. But I'd like to list them for those interested in a slower pace:

^{simulate this circuit – Schematic created using CircuitLab}

The diagrams with the capacitor and inductor in parallel are called in-parallel bandpass or in-parallel bandstop. The diagrams with the capacitor and inductor in series are called in-series bandpass or in-series bandstop.

The in-parallel arrangement has infinite (in theory) impedance at its resonant frequency. The in-series arrangement has zero (in theory) impedance also at its resonant frequency. Looking at #1 above, this means that all of the input gets to the output, so this is a bandpass. From #2 above, this means that the input impedance is zero, so again this is a passband. From #3 above, this means the input impedance is infinite, so none of the input gets to the output and this must be a stopband. Finally from #4 above, the series impedance is zero and effectively grounds-out the input so none of the input gets to the output and this also must be a stopband.

Returning to your website

Your website concludes the following two equations for case #2, above: the in-series bandpass case.

$$\begin{align*} \omega_{c_1}&=-\frac12\frac{R}{L}+\sqrt{\left(\frac12\frac{R}{L}\right)^2+\frac1{L\,C}} \\\\ \omega_{c_2}&=\frac12\frac{R}{L}+\sqrt{\left(\frac12\frac{R}{L}\right)^2+\frac1{L\,C}} \end{align*}$$

Let's pick a specific case, using reasonable part values, in order to test (or demolish) the above website equations for \$\omega_{c_1}\$ and \$\omega_{c_2}\$. I'll use \$R=1\:\text{k}\Omega\$, \$L=2.2\:\text{mH}\$, and \$C=220\:\text{nF}\$.

Let's put this into a spice program (LTspice) and see what it shows us:

(The above image can be clicked in order to see it in more detail.)

From the above, using LTspice, I find numerical values of \$f_{c_1}=716.37682 \:\text{Hz}\$ and \$f_{c_2}=73.066561 \:\text{kHz}\$.

Let's now compute the results using the website's equations:

$$\begin{align*} \omega_{c_1}&=-\frac12\cdot\frac{1\:\text{k}\Omega}{2.2\:\text{mH}}+\sqrt{\left(\frac12\cdot\frac{1\:\text{k}\Omega}{2.2\:\text{mH}}\right)^2+\frac1{2.2\:\text{mH}\cdot 220\:\text{nF}}} \\\\ &\approx 4.501 \:\text{k}\frac{\text{rad}}{\text{s}} & (f_{_0}\approx 716.3384\:\text{Hz}) \\\\ \omega_{c_2}&=\frac12\cdot\frac{1\:\text{k}\Omega}{2.2\:\text{mH}}+\sqrt{\left(\frac12\cdot\frac{1\:\text{k}\Omega}{2.2\:\text{mH}}\right)^2+\frac1{2.2\:\text{mH}\cdot 220\:\text{nF}}} \\\\ &\approx 459.05 \:\text{k}\frac{\text{rad}}{\text{s}} & (f_{_0}\approx 73.0595\:\text{kHz}) \end{align*}$$

Remarkably close results. I think we can say that your website has the right calculations.

NOTE:

I want anyone reading this to read and think for yourself, but trust nothing you read. Verify everything, relentlessly. Don't trust anything you read, even when you've read it from 100 different sources and every one of them says the same thing.

Can you imagine what would have happened in science if Ted Maiman (1960) had believed all of the experimental results by many different scientists, including the results of his own team when he asked them to replicate earlier results, regarding the viability of the ruby laser? Everyone was saying it simply could not reach the critical point of lasing. When he finally examined the experimental setups used to reach those conclusions he realized some flaws in thinking and was then able to perform improved experimental designs and demonstrate that a ruby could, in fact, lase.

So trust nothing. Least of all yourself. (It's probably easiest for you to fool yourself!)

There's another feature of the cutoff frequencies. It must be the case that \$\omega_{_0}=\sqrt{\omega_{c_1}\cdot \omega_{c_2}}\$ and also that \$f_{_0}=\sqrt{f_{c_1}\cdot f_{c_2}}\$. We know that \$\omega_{_0}=\frac1{\sqrt{L\,C}}=45.\overline{45}\:\frac{\text{rad}}{\text{s}}\$ (\$f_{_0}\approx 7.2343\:\text{kHz}\$.) I'll leave it to you to test out these ideas.

I think we have taken the time to verify the results of your website.

I also very much encourage you to perform your own calculations of the above. Perhaps I made a mistake in performing the formula, as I interpreted it from the website. Or perhaps I didn't properly transfer the formula here and got that wrong. Verify everything you see. Even when I seem to be verifying something someone else said! Leave no stone unturned.

Q and \$\zeta\$

Bandpass filters, like all 2nd order filters, will have an important variable that is a description of the shape of their behavior. (Their shapes vary and it helps to know which shape you are looking at.) This is either \$Q\$ or \$\zeta\$. Their relationship is \$Q=\frac1{2\,\zeta}\$ or \$\zeta=\frac1{2\,Q}\$.

\$\zeta\$ is the damping factor. It's special because when \$\zeta=1\$, then the system is critically-damped and that's one particular shape. If \$\zeta\gt 1\$, then the system is over-damped and that's another set of particular shapes that gradually change in particular ways away from the critically-damped shape. If \$\zeta\lt 1\$, then the system is under-damped and that's yet another set of particular shapes that gradually change in particular ways, again away from the critically-damped shape. Together, with \$\zeta\$ you can specify all of the possible shapes that exist. Just one parameter tells you all of that.

\$Q\$ is just another way of saying similar things. If you go to this Wikipedia page on the Q factor and bandwidth, you will see a statement that says, "The 2-sided bandwidth relative to a resonant frequency of \$f_{_0}\:\text{Hz}\$ is \$\frac{f_{_0}}{Q}\$." This is one of the uses for \$Q\$: \$\frac{f_{_0}}{Q}=f_{c_2}-f_{c_1}\$.

When I chose the values \$R=1\:\text{k}\Omega\$, \$L=2.2\:\text{mH}\$, and \$C=220\:\text{nF}\$, I chose them in such a way that \$Q=0.1\$ and \$\zeta=5\$.

This would be a good time to now spend a moment and verify what the above Wikipedia page says, too. Double-check their computation suggestion and see if it really works, given the values already calculated above.

2nd order bandpass, general form

All of the above topologies are 2nd order, as you've got two energy-storage devices in each of them.

All 2nd order transfer functions -- and I mean all of them -- fit a single general structure:

$$\mathcal{H}\left(s\right)=\frac{a_2 s^2 + a_1 s + a_0}{b_2 s^2 + b_1 s + b_0}$$

That may look a bit daunting to start, especially when you realize that \$s=\sigma + j\,\omega\$. But there are some simplifications to help you out.

• All of the above co-efficients are real-valued. They are not complex-valued. This isn't mathematics and when taking some complex analysis course. This is electronics. So you need to put your electronics hat on when reading these things.
• You can divide up the above fraction into the following three terms, which can be separately considered:

$$\mathcal{H}\left(s\right)=\frac{a_2 s^2}{b_2 s^2 + b_1 s + b_0}+\frac{a_1 s}{b_2 s^2 + b_1 s + b_0}+\frac{a_0}{b_2 s^2 + b_1 s + b_0}$$

In fact, those three terms are, respectively, a highpass, a bandpass, and a lowpass.

We are discussing bandpass and bandstop filters. So this either means your transfer function will be like the middle term above (bandpass) or else like the sum of the first and last terms above (bandstop.)

And since your question is specifically about the in-series bandbass, I'll ignore the rest.

Derivation of standard form

For a bandpass, we are interested in this:

$$\mathcal{H}\left(s\right)=\frac{a_1 s}{b_2 s^2 + b_1 s + b_0}\tag{1}$$

Before continuing with that, you'll often see people say that \$s=j\,\omega\$. It's really the case that \$s=\sigma+j\,\omega\$. (At least, so far.) Multiplication in the complex domain involves two things (in our feeble human minds, anyway; but in an alien who can think in complex numbers as a single concept it's only one thing): scaling and rotation. The \$\sigma\$ value is actually part of a factor: \$e^{\sigma\,t}\$. When \$\sigma=0\$ then the factor stays \$1\$ for all time and everyone is happy. If \$\sigma\lt 0\$ then the factor shrinks (scales downward) with time and this means the circuit might hick-up or burble a bit, but given enough time it will settle down. And again, everyone can be happy. But if \$\sigma\gt 0\$ then we have trouble. Stuff goes to heck, given enough time. And that's not a happy circumstance. This is one of the reasons that folks like the left-hand complex plane and feel less comfortable when things are playing out over in the right-hand complex plane. So for limiting ourselves to just looking at the frequency behavior, we'll often just set \$\sigma=0\$ because we aren't interested in the scaling aspects of the problem, only the frequency aspects when assuming no scaling is going on.

If you substitute \$s=j\,\omega\$ into the above equation, then you have:

$$\mathcal{H}\left(j\,\omega\right)=\frac{j\,a_1 \,\omega}{\left(b_0-b_2 \omega^2\right) + j\,b_1 \,\omega}\tag{2}$$

Just think of the denominator now as having two sides of a right triangle (the imaginary part is orthogonal to the real part) and that the hypotenuse is the magnitude. (The magnitude of the numerator is all in one direction and is just the factor after \$j\$.)

As \$\omega\to 0\$, you just have \$\frac{0}{b_0}=0\$. And as \$\omega\to\infty\$, \$b_0\$ isn't going to matter and you just have \$\frac{j\,a_1}{-b_2 \omega + j\,b_1 }\$ and the hypotenuse of the denominator will be entirely determined by \$\omega\to\infty\$ and again the result is \$\frac{j\,a_1}{-b_2 \omega + j\,b_1 }\to 0\$. So this is certainly a bandpass.

In between, there is a place where the real part in the denominator goes to zero. As that web page mentioned, this is when \$b_0=b_2 \omega^2\$. That's when \$\omega=\sqrt{\frac{b_0}{b_2}}\$. It doesn't matter what \$b_0\$ is. It doesn't matter what \$b_2\$ is. Whatever they are, that's going to be a special value of \$\omega\$. We call this special value, \$\omega_{_0}\$. It is super-important.

What happens when \$\omega=\omega_{_0}\$? Well, we've already said that \$\omega_{_0}\$ causes the real part of the denominator of \$\mathcal{H}\left(j\,\omega\right)\$ to go to zero. So then the equation becomes: \$\frac{j\,a_1 \,\omega_{_0}}{j\,b_1 \,\omega_{_0}}=\frac{a_1}{b_1}\$. This is now also a special value. We call it the gain and label it \$A\$ (or \$K\$ or \$h\$ [in Sallen & Key's paper] or pretty much anything anyone is feeling like using, that day.)

So we've uncovered two interesting values related to bandpass filters: \$A=\frac{a_1}{b_1}\$ and \$\omega_{_0}=\sqrt{\frac{b_0}{b_2}}\$. (Do keep in mind that we set \$\sigma=0\$ to find them. But they are still very special.)

Now, look back at the denominator in equation (1) above.

We could just extract \$b_2\$ out in order to get \$b_2\left[s^2+\frac{b_1}{b_2}\,s+\frac{b_0}{b_2}\right]\$. But, hmm, there's that interesting fraction now: \$\frac{b_0}{b_2}\$. We know that is the same thing as \$\omega_{_0}^2\$. So let's plug that back in and get \$b_2\left[s^2+\frac{b_1}{b_2}\,s+\omega_{_0}^2\right]\$. That seems like it could go somewhere. But we aren't there, yet.

At this point, it's time to consider factoring that quadratic equation. That might help us. There is a very special set of values that would be really nice when solving a quadratic equation and this is when \$b_1^2=4\,b_2\,b_0\$. (Obviously, because of \$\frac{-b_1\pm\sqrt{b_1^2-4\,b_2\,b_0}}{2\,b_2}\$.) This suggests \$b_1=2\sqrt{b_2\,b_0}\$, in this special circumstance. But what about other circumstances?

Well, let's create a new constant, \$\zeta\$, and use it like this: \$b_1=2\zeta\,\sqrt{b_2\,b_0}\$. This means that even if the two sides aren't equal, then there's some kind of factor, \$\zeta\$, that we can plug in to make them equal. So, solving we find \$\zeta=\frac{b_1}{2\sqrt{b_2\,b_0}}\$. And we know for a fact that when \$\zeta=1\$ then it is also the case that \$b_1^2=4\,b_2\,b_0\$ and therefore that the solution is very simple because the quadratic equation's square-root term now goes to zero and there's only one resulting value from it. So, in some fashion, we know that \$\zeta=1\$ is also quite special.

But let's think just a little bit more about when \$\zeta\ne 1\$. When \$\zeta\lt 1\$ then this means we had to reduce the term \$2\sqrt{b_2\,b_0}\$ in order to make things equal. Clearly, this means that \$b_1\lt 2\sqrt{b_2\,b_0}\$. And that means that the square-root will be imaginary and non-zero. Okay. That's interesting. And when \$\zeta\gt 1\$ then this means we had to increase the term \$2\sqrt{b_2\,b_0}\$ in order to make things equal. Clearly, this then means that \$b_1\gt 2\sqrt{b_2\,b_0}\$. And that means that the square-root will be real and non-zero. So, \$\zeta\$ is a very special value that informs us immediately something quite interesting about the quadratic solution, too!

So we've uncovered three interesting values related to bandpass filters: \$A=\frac{a_1}{b_1}\$ and \$\omega_{_0}=\sqrt{\frac{b_0}{b_2}}\$ and \$\zeta=\frac{b_1}{2\sqrt{b_2\,b_0}}\$.

Recall that \$b_1=2\zeta\,\sqrt{b_2\,b_0}\$? Follow along:

$$\begin{align*} \mathcal{H}\left(s\right)&=\frac{a_1 s}{b_2 s^2 + b_1 s + b_0} \\\\ &=\frac{a_1 s}{b_2\left( s^2 + \frac{b_1}{b_2} s + \frac{b_0}{b_2}\right)} \\\\ &=A\frac{\frac{b_1}{b_2} s}{ s^2 + \frac{b_1}{b_2} s + \frac{b_0}{b_2}} \\\\ &=A\frac{ \frac{2\zeta\, \sqrt{b_2\,b_0} }{ b_2 } s} { s^2 + \frac{2\zeta\,\sqrt{b_2\,b_0}}{b_2} s + \frac{b_0}{b_2}} \\\\ &=A\frac{ 2\zeta\,\sqrt{\frac{b_0}{b_2}} s} { s^2 + 2\zeta\,\sqrt{\frac{b_0}{b_2}} s + \frac{b_0}{b_2}} \\\\ \mathcal{H}\left(s\right)&= A\frac{ 2\zeta\,\omega_{_0} s} { s^2 + 2\zeta\,\omega_{_0} s + \omega_{_0}^2}\tag{3} \end{align*}$$

This is an important result. We've created some special values that have meaning to us and now have found a way of combining both the numerator's and denominator's coefficients in a way that completely replaces them with these new, special values. Those old constants are totally gone. And instead of staring at seemingly random, meaningless coefficients that may mysteriously relate to each other in obscure ways, we've replaced all that with new, very meaningful values which help us interpret the equation. The work was worth every moment!

Keep in mind that all three of these values are built entirely from the earlier coefficients. The expressions may seem a bit odd or arbitrary. But as you can see from the above work, they are anything but arbitrary. They are quite purposeful and meaningful and they show you how these earlier coefficients were actually related to each other. This is quite important to recognize and appreciate!

So: \$A\$ is the gain, \$\omega_{_0}\$ is when the real part of the denominator goes to zero, leaving only an imaginary part, and \$\zeta\$ is exactly \$1\$ when the quadratic solution only has one resulting magnitude and is critically damped.

The reason I go through this is to explain why equation (3) above is so important. It's called a standard form for the 2nd order bandpass transfer function for a reason.

But there is another standard form, which I may as well now show:

$$\mathcal{H}\left(s\right)=A\frac{ 2\zeta \left(\frac{s}{\omega_{_0}}\right)} { \left(\frac{s}{\omega_{_0}}\right)^2 + 2\zeta \left(\frac{s}{\omega_{_0}}\right) + 1}\tag{4}$$

You will encounter both of these.

Cutoff

The meaning of a cutoff frequency for a 2nd order bandpass, like this, is when it is at half-power. So long as the filter has \$\zeta>1\$, it's over-damped and will have two distinct shoulders as shown in the LTspice picture, earlier. These are the two cutoff frequncies.

Half-power means \$\frac{v_\text{out}}{v_\text{in}}=\frac1{\sqrt{2}}\$. (I assume you already know why.) So in this case to find the cutoff frequencies we need a transfer function and we need to set its magnitude equal to \$\frac1{\sqrt{2}}\$ and then solve it for \$\omega\$. (The magnitude of a complex-valued transfer function is found by multiplying it by its complex conjugate and then taking the square-root.)

While I want you to try to put things into standard form, as above, it's more important for this question that you focus on creating the transfer function for schematic #2 and then develop an expression for its magnitude value. Then set this equal to \$\frac1{\sqrt{2}}\$ and see if you can solve it for \$\omega\$.

The transfer function will be \$\frac{R}{R+Z_L+Z_C}\$, of course. And your web page does correctly help you out there. So I won't duplicate it. But you need to multiply it by its complex conjugate. You don't need to take the square root of that, as you can just square \$\frac1{\sqrt{2}}\$ to get \$\frac12\$ and use that as the assigned value. Then solve the resulting equation for \$\omega\$. See if you can arrive at the same expression on that web site.

You know the procedure. Follow it through.

Summary about standard form

The Laplace transform was invented by a mathematician, Pierre-Simon Laplace, for mathematicians and then systematized by Oliver Heaviside to simplify the solutions for differential equations describing physical processes. It's a fantastically general and powerful tool and it is used in myriad specializations. There are few natural processes for which it doesn't apply, since almost everything in nature is about locality where the change in things depends in some way upon the amount of things. A subset of this powerful tool applies well in electronics, too.

Here's a summary of the genius behind the standard form for 2nd order transfer functions (at least as applied to bandpass.) These are uncovered in the following order, as we proceed:

1. When reducing \$\mathcal{H}\left(s\right)\$ the to \$\mathcal{H}\left(j\,\omega\right)\$, so as to exclude scaling behaviors and focus upon frequency behaviors, we discovered that there was a special value of \$\omega\$ that caused the real part of the denominator to go to zero, leaving only the imaginary part. We call this special value, \$\omega_{_0}\$ (though of course it is also called many other things.)

2. When reducing \$\mathcal{H}\left(s\right)\$ the to \$\mathcal{H}\left(j\,\omega\right)\$ and when setting \$\omega=\omega_{_0}\$, we then also discovered that there is a special value, \$A\$, which is the gain of the transfer function. (There is a more general way of approaching this result, staying in terms of \$s\$. But the bandpass case allows a simpler approach, so I took it.)

3. In returning to the general form, \$\mathcal{H}\left(s\right)\$, and keeping in mind that our coefficients are real and not complex and then applying the standard quadratic solution to it, we uncovered another very useful idea where the quadratic solution's square-root term goes to zero. This is kind of like a dividing line that separates diverging behaviors as things move away from the critically-damped case in either of two directions (over-damped vs under-damped.) We call the measure of the diversion away from the critically damped case, \$\zeta\$. \$\zeta=1\$ for the exact critically-damped point, \$\zeta<1\$ for under-damped cases where the quadratic solution involves complex-valued roots (there will be a damped frequency, too, in this case), and \$\zeta>1\$ for over-damped cases where the quadratic solution involves real-valued roots and where the existence of low and high cutoffs now emerge (for bandpass.)

We were able to take the general form of:

$$\mathcal{H}\left(s\right)=\frac{a_1 s}{b_2 s^2 + b_1 s + b_0}$$

Where the coefficients were arbitrary and difficult to interpret and replace four coefficients with just three uniquely interesting and useful parameters, \$A=\frac{a_1}{b_1}\$, \$\omega_{_0}=\sqrt{\frac{b_0}{b_2}}\$, and \$\zeta=\frac{b_1}{2\sqrt{b_2\,b_0}}\$, to produce a far more meaningful result:

$$\begin{align*} \mathcal{H}\left(s\right)&= A\cdot\frac{ 2\zeta\,\omega_{_0} s} { s^2 + 2\zeta\,\omega_{_0} s + \omega_{_0}^2}\\\\ &=A\cdot\frac{ 2\zeta \left(\frac{s}{\omega_{_0}}\right)} { \left(\frac{s}{\omega_{_0}}\right)^2 + 2\zeta \left(\frac{s}{\omega_{_0}}\right) + 1} \end{align*}$$

This is no mean feat. It's brilliance.

---

There's a final, more philosophical note I want to leave with you. In paraphrase, "I hear and I forget, I see and I remember, I do and I Understand".

When someone hands you an answer on a platter, for example the concepts of mass, volume, and density (the ratio of mass to volume), you don't have any ideas about what led to it. You are just told, "do this and you will get useful answers." So you do that and you do get useful answers. But you've no concept of the struggles that humans went through to uncover the concept of mass as distinct from gravitational attraction or the large number of project failures when people didn't understand the concept of density. You don't know that for some time people thought sharp things sink because they cut water and that blunt things float because they don't. Etc. You've no clue, no deep understanding. All that has happened is that you were handed a tool. And you know how to use it. But it goes no deeper than that (because you were never asked to discover that tool for yourself.) At least, at first.

When you are handed the idea of a resonant frequency, \$\omega_{_0}\$, or a damping factor, \$\zeta\$, or a quality factor, \$Q\$, all of these more as on a silver platter than by doing the hard things to get there, then you learn but you do not understand that something.

How did we arrive at this wonderful place we are being taught?

I hope just a little of that how has leaked out in the above. We do stand on the shoulders of giants. And we should give some conscious nod of appreciation for what we've been given in helping us understand the world around us just a little better.