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I have the circuit below. How can I see that it is a band-pass filter?

I tried calculating the equivalent impedance for the parallel part of the circuit:

$$\bar{Z}_{parallel} = \frac{\bar{Z}_C\bar{Z}_L}{\bar{Z}_L + \bar{Z}_C} = \frac{j\omega L(\frac{1}{j\omega C})}{j\omega L + \frac{1}{j\omega C}} = \frac{j\omega L}{1 - \omega^2LC}$$

Now applying it to the gain I have:

$$A = \frac{\bar{V_o}}{\bar{V_i}} = \frac{\bar{Z}_{parallel}}{R + \bar{Z}_{parallel}} = \frac{j\omega L}{R - \omega^2RLC + j\omega C}$$

All I can do now is see that \$A \rightarrow 0\$ when \$\omega \rightarrow \infty\$ and also \$A \rightarrow 0\$ when \$\omega \rightarrow 0\$, which would mean at least that it doesn't pass very high frequencies and very low frequencies - but I'm not able to tell if it passes mid-range frequencies or not.

I'd like to know if this is the right approach, and how I can tell if this is a selective band-pass filter.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ @ludicrous Do you think there might be an error in the last term of the right-side denominator of your 2nd equation? Also, are you familiar with putting these 2nd order transfer functions into one of the two standard forms (one using \$Q\$ and the other using \$\zeta\$?) Standard form makes identification easy. \$\endgroup\$
    – jonk
    Commented Aug 28, 2022 at 18:12
  • \$\begingroup\$ Your intuition is correct although the form is not. But this example must have some resistive load that also dampens and attenuates, so more often the BPF has some resistor also shown on the output or a series LC version is used. sim.okawa-denshi.jp/en/Fkeisan.htm \$\endgroup\$ Commented Aug 28, 2022 at 19:06
  • \$\begingroup\$ also cnx.org/contents/Cy752CuW@4/Second-Order-Filters \$\endgroup\$ Commented Aug 28, 2022 at 19:17
  • \$\begingroup\$ "How can I see that it is a band-pass filter?" - I wonder why there is no experimental answer yet. I think we would connect the RLC circuit to a VNA and have a look at the response curve. \$\endgroup\$ Commented Aug 30, 2022 at 13:47

6 Answers 6

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Long way around the barn, but I'll get there....

Sallen & Key: "A Practical Method of Designing RC Active Filters"

I want to start out by reflecting on the TR-50 paper by R. P. Sallen & E. L. Key, dated 6 May 1954. The authors' focus is on active networks, with active gain stages using vacuum tubes, and therefore only give a small nod towards passive networks (as a basis upon which to build these active network filters.) But they do provide some useful thoughts about parsing 2nd order transforms of the form (where \$a_i\$ and \$b_i\$ are all real, positive constants such that \$a_i\ge 0\$ and \$b_i\gt 0\$:

$$G_s = \frac{N_s}{D_s}=\frac{a_2s^2+a_1s+a_0}{b_2s^2+b_1s+b_0}$$

Zeroing in on \$D_s\$ provides that if \$\omega_{_0}=\sqrt{\frac{b_0}{b_2}}\$ and \$d=\frac{b_1}{\sqrt{b_2 \,b_0}}\$, then \$D_s\$ can be factored out as:

$$D_s=b_0\cdot\left[\left(\frac{s}{\omega_{_0}}\right)^2+d\cdot\left(\frac{s}{\omega_{_0}}\right)+1\right]$$

where the zeros of \$D_s\$ (poles, when placed in the denominator) lay on a circle with radius \$\omega_{_0}\$, with the real part at \$-\frac12 d\,\omega_{_0}=\zeta\,\omega_{_0}\$ in the under-damped case when \$\zeta\le 1\$. (The over-damped case has all the zeroes directly located on the negative real axis.)

\$\omega_0\$ determines the positions of the zeroes in the frequency domain and \$b_0\$ is merely a relative amplitude value.

\$d\$ vs \$\zeta\$ and \$Q\$

The transfer function shape (given a log-log plot with [angular] frequency on the x-axis and magnitude on the y-axis) is determined solely by \$d\$. Back then, Sallen & Key used \$d\$. Today, we use \$\zeta=\frac{d}2\$ or \$Q=\frac1{d}\$ and their term \$d\$ has fallen out of use.

Which, \$\zeta\$ or \$Q\$, is preferred, I think, depends mostly on where your brain is currently at. (I don't think of it as an always \$Q\$ or always \$\zeta\$ kind of thing.) When dealing with under-damped situations, I tend to think more in terms of \$Q\$. When dealing with wide bandpass situations (over-damped, for sure) then I tend to think more in terms of \$\zeta\$.

Your Transfer Function

You made a mistake in the expression found on the right side of your 2nd equation. Before I get there, I'd just like to say that I prefer writing one letter, \$s\$, over two, \$j\,\omega\$. So I'll continue using \$s\$ (as did Sallen & Key.)

Your expression should have been written out as:

$$\frac{s\,L}{s^2\,L\,C\,R_1+s\,L+R_1}$$

You got a small part of yours wrong. This one is correct.

Putting the denominator into standard form, you can use Sallen & Key's approach mentioned at the outset, above, and find that \$\omega=\frac1{\sqrt{L\,C}}\$ and \$d=\frac{\sqrt{L}}{R_1\,\sqrt{C}}\$.

2nd Order Transfer Function Development

Now, you can also do the same thing for the numerator (if it were in 2nd order form, which yours isn't.) But let's assume it was in 2nd order form for a moment and do some transformation steps that combines both the numerator and denominator into a new structure that you can always achieve.

$$N_s=a_0\cdot\left[\left(\frac{s}{\omega_{_0}^{\:'}}\right)^2+d^{\,'}\cdot\left(\frac{s}{\omega_{_0}^{\:'}}\right)+1\right]$$

where \$\omega_{_0}^{\:'}=\sqrt{\frac{a_0}{a_2}}\$ and \$d^{\,'}=\frac{a_1}{\sqrt{a_2 \,a_0}}\$.

Watch what now happens:

$$\begin{align*} G_s &= \frac{N_s}{D_s}=\frac{a_2s^2+a_1s+a_0}{b_2s^2+b_1s+b_0} \\\\ &=\frac{a_0}{b_0}\cdot\left[\frac{\left(\frac{s}{\omega_{_0}^{\:'}}\right)^2+d^{\,'}\left(\frac{s}{\omega_{_0}^{\:'}}\right)+1}{\left(\frac{s}{\omega_{_0}}\right)^2+d\left(\frac{s}{\omega_{_0}}\right)+1}\right] \\\\ &=\frac{a_0}{b_0}\cdot\left[\frac{\left(\frac{s}{\omega_{_0}^{\:'}}\right)^2}{\left(\frac{s}{\omega_{_0}}\right)^2+d\left(\frac{s}{\omega_{_0}}\right)+1}+\frac{d^{\,'}\left(\frac{s}{\omega_{_0}^{\:'}}\right)}{\left(\frac{s}{\omega_{_0}}\right)^2+d\left(\frac{s}{\omega_{_0}}\right)+1}+\frac{1}{\left(\frac{s}{\omega_{_0}}\right)^2+d\left(\frac{s}{\omega_{_0}}\right)+1}\right] \\\\ &=\frac{a_0}{b_0}\cdot\left[\frac{\left(\frac{\omega_{_0}}{\omega_{_0}^{\:'}}\right)^2\left(\frac{s}{\omega_{_0}}\right)^2}{\left(\frac{s}{\omega_{_0}}\right)^2+d\left(\frac{s}{\omega_{_0}}\right)+1}+\frac{\left(\frac{\omega_{_0}}{\omega_{_0}^{\:'}}\right)\left(\frac{d^{\,'}}{d}\right)d\left(\frac{s}{\omega_{_0}}\right)}{\left(\frac{s}{\omega_{_0}}\right)^2+d\left(\frac{s}{\omega_{_0}}\right)+1}+\frac{1}{\left(\frac{s}{\omega_{_0}}\right)^2+d\left(\frac{s}{\omega_{_0}}\right)+1}\right] \\\\ &=\underbrace{\overbrace{\frac{a_2}{b_2}}^{\text{gain}}\frac{\left(\frac{s}{\omega_{_0}}\right)^2}{\left(\frac{s}{\omega_{_0}}\right)^2+d\left(\frac{s}{\omega_{_0}}\right)+1}}_{\text{high-pass}} + \underbrace{\overbrace{\frac{a_1}{b_1}}^{\text{gain}}\frac{d\left(\frac{s}{\omega_{_0}}\right)}{\left(\frac{s}{\omega_{_0}}\right)^2+d\left(\frac{s}{\omega_{_0}}\right)+1}}_{\text{band-pass}} + \underbrace{\overbrace{\frac{a_0}{b_0}}^{\text{gain}}\frac{1}{\left(\frac{s}{\omega_{_0}}\right)^2+d\left(\frac{s}{\omega_{_0}}\right)+1}}_{\text{low-pass}} \end{align*}$$

Now, I want you to go back up above and make absolutely certain that I didn't make any mistakes. I want you to see how it is that I arrived at the last (bottom) right-hand side expression.

It's correct. But please note that we start out treating the numerator completely independently from the denominator, developing a different angular frequency and different shape factor for each, to begin. But the above process shows how to completely remove those 'special' values originally developed solely for the numerator, throwing them away and leaving for you only those that were originally created for the denominator.

The only remaining place for the numerator's polynomial coefficients, now, is in the gain factors before each term of the new expression. Those numerator coefficients are no longer found anywhere else. What does this suggest about the impact of the 2nd order numerator?

You need to see how this happens and why it is that the denominator is the important characteristic equation, determining the frequency domain shape and the key frequency around which the shape presents itself.

The numerator then plays a role in determining the gain for each term. But do take note that we started out having zeros in the numerator, which are the roots of \$N_{\text{s}}\$ at the very start of my writing above. But these roots have been replaced with the prefixed gain fractions for each of the three terms. In short, there are no zeros in the standard form.

Please do the algebra, at least once, yourself and by hand. Get out some paper and just walk through it. This will deepen what I've written out for you. It's worth a moment of your life. I promise.

The last result annotated above carries only a few interesting parameters: \$d\$, \$\omega_{_0}\$, and the three gains needed for each term. This is so much better than seeing six constants, three in the numerator and three in the denominator, none of which do much to help you understand meaning.

So the result is worth the work to get there. It has taken what earlier appeared to be an abstract pair of different 2nd order polynomials, each with what may have initially seemed to be independent behaviors where at first glance their combined behaviors would seem almost impenetrable, and then 'magically' transformed the whole mess into far more simplified key ideas, neatly separated out.

This insight is incredibly important to grasp. Apply some of your time and walk through this. If you need to, use a Spice program (like LTspice) to plot out different transfer functions and see their shape unfold. (You can directly provide Laplace equations in Spice and plot them without needing a circuit.) Change some parameter values. Check again. The effort is worth every moment.

Detection of Filter Type

Now we can finally discuss this issue. I said I'd take the long way around the barn. And I did. But we are here, now.

The first term shown is that for a high-pass, the middle term is a band-pass, and the final term is a low-pass. And the gains for each are separated out, as well.

You can recognize the high-pass because its numerator has an \$s^2\$ factor. You can recognize the band-pass because its numerator has an \$s\$ factor. Finally, you can recognize the low-pass because its numerator doesn't have an \$s\$ factor. (Put yet another way, look at the numerator for \$s^2\$, \$s^1\$ or \$s^0\$ as factors.)

From this, you can always tell what you are looking at.

Many transfer functions will only have one of these terms -- not all three. But once in a while you will see two of them combined. Rarely, all three. In such cases, you have something not quite just one, or another, but a composite.

Simple Examples can be Deceptive

Just by way of an example, either of the following simpler passive networks will result in transfer functions including all three terms:

schematic

simulate this circuit – Schematic created using CircuitLab

Annoyingly simple-looking.

The transfer function for both sides is the same, where in both cases, \$k_2=1\$, \$k_0=1\$, and \$\omega_{_0}=\frac1{\sqrt{R_1\,R_2\,C_1\,C_2}}\$:

$$\begin{align*} &\underbrace{\overbrace{k_2}^{\text{gain}}\frac{\left(\frac{s}{\omega_{_0}}\right)^2}{\left(\frac{s}{\omega_{_0}}\right)^2+d\left(\frac{s}{\omega_{_0}}\right)+1}}_{\text{high-pass}}+ \underbrace{\overbrace{k_1}^{\text{gain}}\frac{d\left(\frac{s}{\omega_{_0}}\right)}{\left(\frac{s}{\omega_{_0}}\right)^2+d\left(\frac{s}{\omega_{_0}}\right)+1}}_{\text{band-pass}}+\underbrace{\overbrace{k_0}^{\text{gain}}\frac{1}{\left(\frac{s}{\omega_{_0}}\right)^2+d\left(\frac{s}{\omega_{_0}}\right)+1}}_{\text{low-pass}} \end{align*}$$

If we set \$k=C_1\left(R_1+R_2\right)\$ in the left side case and set \$k=R_2\left(C_1+C_2\right)\$ in the right side case, then we can find for both cases that \$k_1=\frac1{1+\frac{R_1\,C_2}{k}}\$ and \$d=k\cdot \omega_{_0}\$.

What does this mean? Well, we'd expect a gain of 1 at very low frequencies relative to \$\omega_{_0}\$ and a gain of 1 at very high frequencies relative to \$\omega_{_0}\$. But in between? We'd expect some kind of attenuation (notched) because in both cases a term in the denominator of the gain has \$\frac{R_1\,C_2}{k}\gt 0\$.

Returning to Your Transfer Function

Your transfer function, as it turns out, only has \$s\$ of the first power in it:

$$G_s=\frac{d\left(\frac{s}{\omega_{_0}}\right)}{\left(\frac{s}{\omega_{_0}}\right)^2+d\left(\frac{s}{\omega_{_0}}\right)+1}$$

with \$\omega=\frac1{\sqrt{L\,C}}\$ and \$d=\frac{\sqrt{L}}{R_1\,\sqrt{C}}\$.

So it is a band-pass transfer function. Simple as that.

Final Note

I stayed with using Sallen & Key's \$d\$. Please feel free to replace it in terms of \$\zeta\$ or \$Q\$, which is the more modern way to see these in standard form. But \$d\$ is fine, as well. It's just that most textbooks don't use it, today. The same arguments I made still apply, regardless, of course.

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    \$\begingroup\$ Thank you very much for your answer! It makes sense and I'll keep coming back to this answer to reassure and understand the meaning of what you just taught me. I have only one question: what exactly is the meaning of d? w0 is the resonant frequency, but does d have a physical meaning? \$\endgroup\$
    – ludicrous
    Commented Aug 29, 2022 at 17:47
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    \$\begingroup\$ @ludicrous \$d\$ is the shape of the response. But today it is better to remember that \$\zeta=\frac{d}2\$ and \$Q=\frac1{d}\$, because those are in use today. If \$\zeta=1\$ the shape of the response is critically damped. Larger? Overdamped. Smaller? Underdamped. Look up Q/zeta for 2nd order equations on the web to see. Or watch a video. But they specify the frequency domain shape and the time domain response behavior. And they are the only factor that does this for 2nd order. The concept doesn't exist in 1st order and only has the possibility of an approximation concept in higher order. \$\endgroup\$
    – jonk
    Commented Aug 29, 2022 at 18:28
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    \$\begingroup\$ @ludicrous Sometimes in higher order you can decompose them into a bunch of nice 2nd order factors (plus, for odd order, a 1st order factor.) These 2nd order factors will each have damping. But how they combine is a study of its own. Sometimes, for example, a 3rd order case will have a cutoff frequency of its 1st order factor that is so low or so high that it doesn't matter and the 2nd order factor is what determines things (mostly.) In such cases you may have a damping you can reasonably consider as "dominant." \$\endgroup\$
    – jonk
    Commented Aug 29, 2022 at 18:35
  • \$\begingroup\$ @ludicrous regarding your question about "d": I think, it was not yet mentioned that the pole quality factor Qp (identical to 1/d) is identical to the classical "bandwidth-Q" with Q=fo/B (with centerfrequency fo and B=3dB-bandwidth). This is true, of course, for a second-order bandpass only. \$\endgroup\$
    – LvW
    Commented Sep 1, 2022 at 10:47
  • \$\begingroup\$ @jonk why did you choose 'that' \$\omega_0\$ and \$d\$ substitution? I know they do work for the purpose of seeing that only the denominator affects the poles of the transfer function, but are those substitutions meaningful. In other words, is \$\sqrt{\frac{a_0}{a_2}}\$ the resonant frequency for that generic second degree signal, and is \$\frac{b_1}{\sqrt{b_2b_0}}\$ the inverse of its quality factor? (I told I would come back here :) ) \$\endgroup\$
    – ludicrous
    Commented Oct 4, 2022 at 8:46
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For this particular circuit, you know that there must be some mid-band frequency that does pass, because there is some frequency where

$$j\omega L = -\frac{1}{j\omega C}$$

and therefore the impedance of the parallel combination of L and C goes to infinity, leading to unity gain (assuming the load impedance is much greater than R1).

More concisely, we know there is a parallel resonance between the L and C elements in this circuit, and therefore there must be some frequency with no attenuation.

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  • \$\begingroup\$ Is this resonance relation true for every RCL circuit or just for this parallel (and I presume series also) type of circuit? \$\endgroup\$
    – ludicrous
    Commented Aug 28, 2022 at 17:38
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    \$\begingroup\$ I believe it's safe to say there's a resonance in every RCL circuit. However if there was an R directly in parallel with the L and C in this case then the gain wouldn't reach 1 at resonance. And depending on the R value and position the resonance can be very strongly damped. \$\endgroup\$
    – The Photon
    Commented Aug 28, 2022 at 17:40
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I'm not able to tell if it passes mid-range frequencies or not.

Well, you were nearly there with your first line of algebra that produced this: -

$$Z_{parallel} = \dfrac{jωL}{1−ω^2LC}$$

The next step is to realize that if the denominator became zero the impedance becomes infinite and, that happens when \$ω^2LC = 1\$. Rearranged: -

$$ω_{resonant} = \dfrac{1}{\sqrt{LC}}$$

So, at that frequency the parallel LC network becomes invisible and output = input.

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I will start with an intuitive explanation: At DC the capacitance is an open-circuit, while the inductor is a short, hence \$v_o=0V\$. At very high frequencies the inductor will act as an open-circuit, while the capacitor will behave like a short-circuit, hence \$v_0=0V\$. It is clear that both inductor and capacitor will have some finite impedance at different frequencies, so there will be a non-zero output voltage. If one also knows that a parallel LC circuit will resonate and have an infinite effective impedance at the resonance frequency, it is clear that there will be nothing loading the output after the series resistor. Therefore \$v_o=v_i\$ at this frequency \$\rightarrow\$ this is the centre of the passband.

The non-calculating math way: Look at the end of your first line of calculation. Both the absolute value of the numerator grows with frequency, while the absolute denominator first decreases than grows with frequency. It reaches zero at the resonance frequency (\$ 2 \pi f_0 = 1/\sqrt{LC} \$), at which frequency the inductance and the capacitance cancel out each other, i.e. it will show an infinite impedance. At frequencies higher than that the speed of growth in the denominator is faster (increases with the 2nd power of the frequency), i.e. the denominator will increase and the impedance of the parallel LC circuit will head towards zero, i.e. to a short circuit. You should plot \$Z_{parallel}\$ for further insight.

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Well, let's analyze this circuit. The transfer function is given by:

$$\mathscr{H}\left(\text{s}\right):=\frac{\text{V}_\text{o}\left(\text{s}\right)}{\text{V}_\text{i}\left(\text{s}\right)}=\frac{\text{sL}\space\text{||}\space\frac{1}{\text{sC}}}{\text{R}+\left(\text{sL}\space\text{||}\space\frac{1}{\text{sC}}\right)}=\frac{\text{Ls}}{\text{CLRs}^2+\text{Ls}+\text{R}}\tag1$$

Where \$\alpha\space\text{||}\space\beta:=\frac{\alpha\beta}{\alpha+\beta}\$.

When working with sinusoidal inputs we know that we can write \$\text{s}:=\text{j}\omega\$ where \$\text{j}^2=-1\$. So we get:

$$\left|\space\underline{\mathscr{H}}\left(\text{j}\omega\right)\right|=\frac{\text{L}\omega}{\sqrt{\left(\text{R}\left(1-\text{CL}\omega^2\right)\right)^2+\left(\text{L}\omega\right)^2}}\tag2$$

Now, let's analyze a few cases:

  1. \$\omega\space\to\space0\$: $$\lim_{\omega\space\to\space0}\left|\space\underline{\mathscr{H}}\left(\text{j}\omega\right)\right|=0\tag3$$
  2. \$\omega\space\to\space\infty\$: $$\lim_{\omega\space\to\space\infty}\left|\space\underline{\mathscr{H}}\left(\text{j}\omega\right)\right|=0\tag4$$
  3. When is \$\left|\space\underline{\mathscr{H}}\left(\text{j}\omega\right)\right|\$ at a maximum: $$\frac{\partial\left|\space\underline{\mathscr{H}}\left(\text{j}\hat{\omega}\right)\right|}{\partial\hat{\omega}}=0\space\Longrightarrow\space\hat{\omega}=\frac{1}{\sqrt{\text{CL}}}\tag5$$

So, we get:

$$\left|\space\underline{\mathscr{H}}\left(\text{j}\hat{\omega}\right)\right|=1\tag6$$

So, the cut-off frequencies are:

$$\left|\space\underline{\mathscr{H}}\left(\text{j}\omega\right)\right|=\frac{1}{\sqrt{2}}\space\Longrightarrow\space\omega_\pm=\frac{\sqrt{1+\frac{4\text{CR}^2}{\text{L}}}\pm1}{2\text{CR}}\tag7$$

So, the bandwidth is equal to:

$$\mathscr{B}:=\left|\omega_+-\omega_-\right|=\frac{1}{\text{CR}}\tag8$$

And the quality factor of this filter is given by:

$$\mathscr{Q}:=\frac{\hat{\omega}}{\mathscr{B}}=\text{R}\cdot\sqrt{\frac{\text{C}}{\text{L}}}\tag9$$

So, this is a band-pass filter.


EDIT, the argument of the transfer function can be found using:

\begin{equation} \begin{split} \arg\left(\underline{\mathscr{H}}\left(\text{j}\omega\right)\right)&=\arg\left(\frac{\text{L}\omega\text{j}}{\text{R}\left(1-\text{CL}\omega^2\right)+\text{L}\omega\text{j}}\right)\\ \\ &=\arg\left(\text{L}\omega\text{j}\right)-\arg\left(\text{R}\left(1-\text{CL}\omega^2\right)+\text{L}\omega\text{j}\right)\\ \\ &=\frac{\pi}{2}-\begin{cases} \frac{\pi}{2}&\text{if}\space\text{R}\left(1-\text{CL}\omega^2\right)=0\\ \\ \arctan\left(\frac{\text{L}\omega}{\text{R}\left(1-\text{CL}\omega^2\right)}\right)&\text{if}\space\text{R}\left(1-\text{CL}\omega^2\right)>0\\ \\ \frac{\pi}{2}+\arctan\left(\frac{\left|\text{R}\left(1-\text{CL}\omega^2\right)\right|}{\text{L}\omega}\right)&\text{if}\space\text{R}\left(1-\text{CL}\omega^2\right)<0 \end{cases}\\ \\ &=\begin{cases} \frac{\pi}{2}-\frac{\pi}{2}&\text{if}\space\text{R}\left(1-\text{CL}\omega^2\right)=0\\ \\ \frac{\pi}{2}-\arctan\left(\frac{\text{L}\omega}{\text{R}\left(1-\text{CL}\omega^2\right)}\right)&\text{if}\space\text{R}\left(1-\text{CL}\omega^2\right)>0\\ \\ \frac{\pi}{2}-\frac{\pi}{2}+\arctan\left(\frac{\text{R}\left(\text{CL}\omega^2-1\right)}{\text{L}\omega}\right)&\text{if}\space\text{R}\left(1-\text{CL}\omega^2\right)<0 \end{cases}\\ \\ &=\begin{cases} 0&\text{if}\space\text{R}\left(1-\text{CL}\omega^2\right)=0\\ \\ \frac{\pi}{2}-\arctan\left(\frac{\text{L}\omega}{\text{R}\left(1-\text{CL}\omega^2\right)}\right)&\text{if}\space\text{R}\left(1-\text{CL}\omega^2\right)>0\\ \\ \arctan\left(\frac{\text{R}\left(\text{CL}\omega^2-1\right)}{\text{L}\omega}\right)&\text{if}\space\text{R}\left(1-\text{CL}\omega^2\right)<0 \end{cases}\\ \\ &=\begin{cases} 0&\text{if}\space\omega=\frac{1}{\sqrt{\text{CL}}}\\ \\ \frac{\pi}{2}-\arctan\left(\frac{\text{L}\omega}{\text{R}\left(1-\text{CL}\omega^2\right)}\right)&\text{if}\space0\le\omega<\frac{1}{\sqrt{\text{CL}}}\\ \\ \arctan\left(\frac{\text{R}\left(\text{CL}\omega^2-1\right)}{\text{L}\omega}\right)&\text{if}\space\omega>\frac{1}{\sqrt{\text{CL}}} \end{cases} \end{split}\tag{10} \end{equation}

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  • \$\begingroup\$ Thank you very much for your response! I understand better now not only what I was looking for but also resonant frequency and bandwidth. I have 2 questions: what would the phase angle be in this case? atan(wL) - atan(wL/(R(1-wLC))) ? How would the transfer function be in polar form? \$\endgroup\$
    – ludicrous
    Commented Aug 29, 2022 at 16:57
  • \$\begingroup\$ How have you calculated the cutoff frequency? It have me R/sqrt(LCR^2 + L^2), not the same as you \$\endgroup\$
    – ludicrous
    Commented Aug 29, 2022 at 17:33
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    \$\begingroup\$ @ludicrous see my edit. And the cut-off frequencies are found by solving equation \$(7)\$, try to work it out step-by-step (I checked mine with Mathematica and mine are correct). \$\endgroup\$ Commented Aug 29, 2022 at 18:56
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The fastest way to answer your question, ``how can I see that it is a band-pass filter'' is to simulate it. For example in LTspice, choosing some random values for the passives:

enter image description here

Comparing to the figure in https://en.wikipedia.org/wiki/Band-pass_filter, the circuit looks like a band-pass filter.

Now that you know that it's a band-pass filter, you can apply techniques from your mathematical toolkit to the circuit, as others have so well described.

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