Resonance issue in capacitive inductive low-pass filter

Thread Starter

Kurious

Joined Sep 15, 2017
10
I'm trying to understand what's going on in the filter in figure 8.29 of the AC textbook by Kuphardt.

On page 211 of the PDF version, he says, "A little more reflection reveals that if L1 and C2 are at resonance, they will impose a very heavy (very low impedance) load on the AC source, which might not be good either."

But there is no C2 in the diagram. Probably a scrivener's error. Anyway, what was meant?

It looks to me like C1 is in parallel with the series branch of L2 and Rload. And all that is added in series to L1. If I am mistaken, or any insights you can offer, please do. I'm putting the math into a spreadsheet to try to grasp it better.
 

WBahn

Joined Mar 31, 2012
30,298
It might help if you posted an image of the figure and of any relevant text.

That is, unless you only want assistance from someone who just happens to have the specific PDF file of a book with an unspecified title (assuming "the AC textbook" isn't the actual title of the book) from an author about whom we only have a last name.

When I look on Amazon I find that "Kuphardt" appears to refer Tony R. Kuphardt, but I only see a three-volume set on Lessons in Industrial Instrumentation.

I do also see a Tony R. Kuphaldt, who has a six volume set on Lessons in Electric Circuits. One of those, Vol 2, Alternating Current. Is that "the AC textbook by Kuphardt" to which you refer?

Despite Amazon listing the first author as "Kuphardt", the book cover itself says that the author is "Kuphaldt". So I can't find any texts by a "Kuphardt".
 

Thread Starter

Kurious

Joined Sep 15, 2017
10

WBahn

Joined Mar 31, 2012
30,298
Much better. Thanks.

People ask questions about a lot of different texts, not just the e-book here. Providing a link to the page is always much appreciated.

I assume you are talking about the sentence, "A little more reflection reveals that if L1 and C2 are at resonance,..."

I agree, this would seem to be a typo and should be C2.

I'll see if I can bring it to the appropriate person's attention for correction in the next round of updates.
 

Thread Starter

Kurious

Joined Sep 15, 2017
10
I think I'm beginning to understand what's happening, but please correct me if I'm mistaken.

If we take L1 and C1 as though they're in series by themselves, we get resonance at 503.3 Hz. And with impedances 180° out of phase with each other, it's as though there is no resistance at all. But it doesn't work exactly that way because between them is the tap for the L2 and R load branch. I don't have the math figured out for it yet, but the author's SPICE plots show 526 Hz is what's being passed through.

If I'm correctly understanding this part, my next question is how the 526 Hz signal is passing to load. It looks like it should not be, because it's either coming from or going to the same node. Or does this also have to do with C1 and L2 being 180° out of phase--the current goes toward node 2 from C1 when L2 is drawing from node 2 (and vice versa)?



I think if I can understand what is happening in this circuit, I'll be better able to understand the author's instruction later for getting the desired effect.
Thank you for any insights you are able to provide.
 

MrAl

Joined Jun 17, 2014
11,717
I think I'm beginning to understand what's happening, but please correct me if I'm mistaken.

If we take L1 and C1 as though they're in series by themselves, we get resonance at 503.3 Hz. And with impedances 180° out of phase with each other, it's as though there is no resistance at all. But it doesn't work exactly that way because between them is the tap for the L2 and R load branch. I don't have the math figured out for it yet, but the author's SPICE plots show 526 Hz is what's being passed through.

If I'm correctly understanding this part, my next question is how the 526 Hz signal is passing to load. It looks like it should not be, because it's either coming from or going to the same node. Or does this also have to do with C1 and L2 being 180° out of phase--the current goes toward node 2 from C1 when L2 is drawing from node 2 (and vice versa)?



I think if I can understand what is happening in this circuit, I'll be better able to understand the author's instruction later for getting the desired effect.
Thank you for any insights you are able to provide.
Hi,

What is it exactly you want to understand about this circuit?
 

Thread Starter

Kurious

Joined Sep 15, 2017
10
What is it exactly you want to understand about this circuit?
The part above

https://www.allaboutcircuits.com/textbook/alternating-current/chpt-8/resonant-filters/#02128.png

that begins with "A word of caution" and continues down to "The problem is that an L-C filter has an input impedance and an output impedance which must be matched."

Essentially, my question is why what's supposed to be a low pass filter has become a band pass filter as a result of overdesign. The author is teaching about avoiding a pitfall in filter design. But I need help understanding what's going on so I know how to avoid it.
 

MrAl

Joined Jun 17, 2014
11,717
The part above

https://www.allaboutcircuits.com/textbook/alternating-current/chpt-8/resonant-filters/#02128.png

that begins with "A word of caution" and continues down to "The problem is that an L-C filter has an input impedance and an output impedance which must be matched."

Essentially, my question is why what's supposed to be a low pass filter has become a band pass filter as a result of overdesign. The author is teaching about avoiding a pitfall in filter design. But I need help understanding what's going on so I know how to avoid it.
Hi again,

What kind of solutions are you looking for such as:
1. Equation(s) that describe the situation.
2. Simulations.

What it sounds like is that it takes looking at different solutions to see what happens when we change some elements, such as the input and output resistors.

The main point here is that the reactive elements always provide a somewhat wild response, and would always produce some wild response if it where not for the damping action of the energy absorbers in the circuit which here are the resistor(s). The circuit will start to damp with almost any resistance, but what we might see here is a point where we reach critical damping or perhaps over damping or even somewhat under damping. The tradeoff is where we might see a decrease in slope toward the end of the response as we add more damping, and that is not usually desirable, so there may be some optimum point where we see just a little peaking (the part just before the response starts to die down) and decent slope at the end. The end slope is usually the part that is used to quote the response of the system, such as -20db/decade or something like that.
A secondary issue which may be actually a primary issue in some applications is the energy transfer. The maximum power transfer theorem says that the impedances must match in order to get the most energy to transfer from the input to the output.

We can work up an equation(s) if you like and possibly do a couple simulations to see what is going on and what changes as the resistances vary.

So i guess this is not really homework then right?
 

Thread Starter

Kurious

Joined Sep 15, 2017
10
We can work up an equation(s) if you like and possibly do a couple simulations to see what is going on and what changes as the resistances vary.

So i guess this is not really homework then right?
I'm already working on the equations. I like the challenge of figuring them out myself. I know SPICE can do wonders if someone needs just the results, but I'm looking to go deeper. I have been putting equations in spreadsheets as I work through the book. Key to putting in the right equation is understanding what's happening. I saw a video lecture in which the teacher explaining the math of series and parallel circuits briefly describes complex circuits and says that'll come up much later. This would seem to count as a complex circuit. So maybe it's too much for right now?

This is self-guided study. Since I'm working on a lesson in a textbook, this seemed to be the right forum. But if there's a more appropriate forum for my questions, please point me in that direction. Thank you, in any case.
 

MrAl

Joined Jun 17, 2014
11,717
I'm already working on the equations. I like the challenge of figuring them out myself. I know SPICE can do wonders if someone needs just the results, but I'm looking to go deeper. I have been putting equations in spreadsheets as I work through the book. Key to putting in the right equation is understanding what's happening. I saw a video lecture in which the teacher explaining the math of series and parallel circuits briefly describes complex circuits and says that'll come up much later. This would seem to count as a complex circuit. So maybe it's too much for right now?

This is self-guided study. Since I'm working on a lesson in a textbook, this seemed to be the right forum. But if there's a more appropriate forum for my questions, please point me in that direction. Thank you, in any case.
Here are a few equations that might help but you can try to come up with them yourself too.

Vout/Vin=R/((R+s*L)*(s*C*R+s^2*C*L+2))
Vc/Vin=1/(s*C*R+s^2*C*L+2)
V1/Vin=(s^2*C*L*R+R+s^3*C*L^2+2*s*L)/((R+s*L)*(s*C*R+s^2*C*L+2))

In this set we assume the input resistor equals the output resistor so they are both R.
Vin is the input voltage with no series resistance.
Vout is the output voltage.
Vc is the cap voltage.
V1 is the voltage between the input source resistance R and the first inductor L.
Both inductors have value L.
The capacitor has value C.

From these equations you can do almost anything. For example, you can calculate the input impedance looking into the first inductor (which is considered the input impedance for this circuit). You can then set that equal to R, and then solve for R, and that gives you the 316 Ohms used in the text, assuming the frequency is w=1/sqrt(LC) and w=2*pi*f.

One interesting point is that after we add the resistors (as the above formulas apply) we get a -3db cutoff frequency that is higher than the assumed LC resonant frequency.
 

RBR1317

Joined Nov 13, 2010
715
What is s in the equations?
The "s" in the equations is the complex variable (s=σ+jω) as used in the Laplace transform. If transient events are not of interest, such as when finding the Bode plot of the filter transfer function, then it is customary to set σ=0 and use "ω" as the frequency variable.
 

MrAl

Joined Jun 17, 2014
11,717
Thank you. I'm sure I wouldn't have figured these out by myself. What is s in the equations?
Hi,

As RBR pointed out, 's' is just a complex number a+b*j and is usually referred to as complex frequency. For a frequency domain analysis we can set 's' to the imaginary part of the complex frequency, so s=j*w where w=2*pi*f with 'f' the frequency in Hertz.
The magnitude of the response is then the amplitude which is the square root of the sum of the squares of the imaginary part and real part: sqrt(real^2+imag^2) and the phase shift is atan2(imag,real).
The analysis proceeds just as you would with all resistors in place of the L's and C's, but with the L changing to s*L and the C changing to 1/(s*C). That leads to all three of those equations. The input source just stays at whatever amplitude you use such as 1v and is in units of what you choose such as peak volts.
We can do an example if you like.
 

MrAl

Joined Jun 17, 2014
11,717
I might like. If it's something I can't handle at the moment, I will save the notes for later.
Hi again,

Well you think you can calculate the output at your node labeled "3" if the components were all resistors? If you can, then you can come up with those three equations. You can also use automated math software which many people use these days. Everything works the same except you substitute:
L goes to s*L
C goes to 1/(s*C)

and sometimes we call these the 'impedances' so we might declare them as:
zL=s*L
zC=1/(s*C)
zR=R

and note R does not change.

So if you know how to put two resistors in parallel:
Rp=R1*R2/(R1+R2)

then you know how to put two impedances in parallel:
Zp=Z1*Z2/(Z1+Z2)

and so everything works the same except you get some 's' variables in there, which at this point are just algebraic variables just like any other like a, b, c, x, y, etc.

For a quick example, if you wanted to put an inductor L in parallel with a resistor R you would just do:
Zp=zL*zR/(zL+zR)

which comes out to:
Zp=s*L*R/(s*L+R)

and then simplify that either by hand or using automated math software.

When it comes time to evaluate these expressions, we have different methods depending on what it is you want to know about the circuit such as response to a sine wave or step wave.
 
Top