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I'm having trouble determining the value of the parameters of a LC Ladder Filter

Consider the available tables for designing second-order lowpass ladder filters (series first) with a Bessel response. The filter is terminated with resistors on both end with RS = RL/2. Find the values of the components parameters in order to obtain a filter with \$\tau_0\$ = 0.8 ms and adapted for a load resistor of 250 ohm.

So first I consulted some Bessel filter design table to obtain the normalized parameters $$R_S=1/2$$ $$L_1=0.2601$$ $$C_2=3.5649$$

Then I performed the impedance scaling $$R_S=1/2 \times R_L=125 \Omega$$ $$L_1=0.2601 \times R_L=65.025$$ $$C_2=3.5649 \times 1/R_L=0.0142596$$

Everything ok. Now the problem comes when I need to apply the frequency denormalization. I thought I just needed to multiply by \$\tau_0\$ leading to:

$$L_1=52.02 \space m H$$ $$C_2=11.49768 \space \mu F$$

Which is incorrect, as the values should be:

$$L_1=38.2 \space m H$$ $$C_2=8.378 \space \mu F$$

What am I doing wrong?

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You used the tables for I/O loads correctly, but you forgot to apply the frequency scaling, table shown below:

N  | fs
---------
2  | 1.36
3  | 1.75
4  | 2.13
5  | 2.42
6  | 2.7
7  | 2.95
8  | 3.17
9  | 3.39
10 | 3.58

If you divide your values to 1.36 you'll see that the results match.


Passive filter synthesis is a large domain but, in general, the filter is treated like a two-port network, having a voltage source with a series resistor as an input, and a resistor at the output (doubly-terminated passive port). Part of the design procedure is determining what elements the port will have based on a transfer function.

For this case, the requirements are \$R_L=1, R_S=\frac12\$, and a 2nd order Bessel with a group delay of \$\tau=8\text{ms}\$. That makes the whole circuit look like this:

basic

I'll take a different approach in finding the elements. The transfer functions for a 2nd order, normalized, prototype lowpass Bessel, and the circuit above are:

$$\begin{align} H(s)&=\frac{3}{s^2+3s+3}=\frac{1}{\frac{s^2}{3}+s+1} \\ G(s)&=\frac{R_L}{R_LLCs^2+(L+R_SR_LC)s+R_S+R_L}=\frac{1}{LCs^2+(L+0.5C)s+1.5} \end{align}$$

It's apparent that \$\small G(0)=\frac23\$, so \$\small G(s)=\frac23H(s)\$. I've chosen the non-monic representation because now it's easy to make a system of equations from the denominators, only, from which the values of the elements can be readily found, for two values of \$s\$ (which don't need to be complex). Since for \$s=0\$ both \$L\$ and \$C\$ disappear, let's choose \$s=1\$ and \$s=2\$:

$$\begin{align}\left\{ \begin{array}{x} G(1)=\frac23H(1)\quad\Rightarrow\quad LC+L+C+1.5&=1.5(\frac13+1+1) \\ G(2)=\frac23H(2)\quad\Rightarrow\quad 4LC+2(L+C)+1.5&=1.5(\frac43+2+1) \end{array}\right. \end{align} \\ \Rightarrow \\ \begin{array}{x} L=1.31\,&,\quad 0.191 \\ C=0.382\,&,\quad 2.618 \end{array}$$

The solution shows two sets of values, any of them can be chosen (shown below). As mentioned in the comments, the frequency scaling which you had to apply was simply the frequency of the lowpass prototype when:

$$ |H(j\omega)|=\left|\frac{3}{-\omega^2+j3\omega+3}\right|=\frac{3}{\sqrt{\left(3-\omega^2\right)^2+9\omega^2}}=\frac{1}{\sqrt{2}} \\ \Rightarrow \\ \omega_1=\pm\sqrt{\frac32}\sqrt{\sqrt5-1}\quad\omega_2=\pm j\sqrt{\frac32}\sqrt{\sqrt5+1}$$

Since we're dealing with positive, real numbers, \$\omega\approx 1.36\$ remains, which is what is given in the table. The same goes for all the orders.

Now, if you take the value of the normalized elements from the table, \$L=\frac{0.2601}{1.36}=0.19125\$ and \$C=\frac{3.5649}{1.36}=2.62125\$, values which are very close to the ones calculated above, save roundings & co. This can be applied to higher orders; the system of equations will get fluffy, but it can be done. A quick check shows that the results are valid:

test

V(a) is the magnitude-scaled lowpass prototype, V(b) is your corrected result, V(c) and V(d) are the two solutions. The traces have been slightly displaced to avoid overlapping, but they are the same, as shown by the group delay (dotted trace). Whatever minor differences exist, they're due to rounding, nothing more. If they had been calculated with float precision, or more, they would have overlapped perfectly.


For the sake of proof, here's how a 3rd order Bessel could be deduced in the same way (\$R_S=2, R_L=3, \tau=1\text{s}\$):

$$\begin{align} H(s)&=\frac{15}{s^3+6*s^2+15s+15} \\ G(s)&=\frac{R_L}{L_1L_2Cs^3+(R_SL_2C+R_LL_1C)s^2+(R_LR_SC+L_1+L_2)s+R_S+R_L} \end{align}$$

The system of equations can be built in a similar manner and there will be a lot of solutions (32), but many of them will have negative or complex values. Sorting them out leaves these two:

$$\left\{ \begin{array}{x} L_1=0.404\,&,\quad 3.378 \\ L_2=2.87\,&,\quad 0.555 \\ C=0.288\,&,\quad 0.178 \end{array}\right.$$

Both solutions work (this time I let all traces overlap):

test3rd

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  • \$\begingroup\$ Thank you! Can you show me any reference for that table? My professor doesn't mention it in any of the slides :/ \$\endgroup\$ Sep 5 '20 at 12:31
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    \$\begingroup\$ @GrangerObliviate I find that very strange. Point your teacher to this book. It also has the normalized element tables you were mentioning in your question. In Google Books, at book page 48, you'll find the table (if they let you view it). \$\endgroup\$ Sep 5 '20 at 13:57
  • \$\begingroup\$ Thank you so much for the book reference, I was looking for a good Filters book that touched both analog and digital and explained things in a more concise way (I'm taking an Analog and Digital Filter Design course). \$\endgroup\$ Sep 5 '20 at 16:04
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    \$\begingroup\$ It's not a bad book, I only mentioned it for the table. A better one is Analog Electronic Filters: Theory, Design and Synthesis. For passive filters, the best way to calculate them is by the method described in this book. It's more involved, yes, but that's how those tables got to be. The frequency scaling can be calculated as the point where the normalized transfer function, \$\frac{3}{s^2+3s+3}\$, has \$\frac{1}{\sqrt{2}}\$ magnitude. Maybe I'll add a bit on this, later on. \$\endgroup\$ Sep 5 '20 at 16:29
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    \$\begingroup\$ @aconcernedcitizen yes, of course. When I wrote "It brings up some good points" I was referring to discussion points, not forum points. \$\endgroup\$
    – P2000
    Sep 6 '20 at 6:54
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I can help, as I have designed these filters, but I am not sure about how the question is formulated.

RS = RL=2? Or typo, RS = RL/2

Is the input supposed to be matched to RS?

Is this homework, and is the expected result from the answer guide?

This is my attempt. Check the diagram against how your textbook sets up the RS and RL in previous questions. And is \$\tau_0\$ the same as \$1/f_0\$, the cut-off?

enter image description here

https://rf-tools.com/lc-filter/

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  • \$\begingroup\$ Hi! Yes, it was a typo I meant RS=RL/2. This is from a problem set my professor gave us and the answer is from an answer guide. The diagram is correct, I also used that tool to simulate my filter! My question is exactly if \$\tau_0\$ the same as \$1/f_0\$. Apparently it is not, otherwise my answer would have been correct (just like yours) \$\endgroup\$ Sep 5 '20 at 12:34
  • \$\begingroup\$ @GrangerObliviate The group delay for a Bessel filter is \$\tau=\frac{1}{2\pi f_p}\$. Just tried the link and the values are off. I don't know how they calculate them. \$\endgroup\$ Sep 5 '20 at 13:59
  • \$\begingroup\$ That's what I thought, apparently there is an additional frequency denormalization that needs to be performed (as @aconcernedcitizen pointed out) \$\endgroup\$ Sep 5 '20 at 16:02
  • \$\begingroup\$ That page calculates for -3 dB, so they're applying the frequency scaling, but if you want to calculate for a group delay, you need to apply the inverse of the frequency scaling to those values. So for this case, you need to calculate for \$\small f_p=1.36/(2\pi 8\text{ms})\approx 270\text{Hz}\,\Rightarrow\,L=262.7\text{m},\,C=1.226\mu\$, which are like my 1st set of solutions: \$L=1.31\cdot 250\cdot 8\text{m}=262\text{m},\,C=0.382\cdot 8\text{m}/250=1.222\mu\$. The magnitude graph is also misleading because it displays a 0 dB DC (i.e. no attenuation). \$\endgroup\$ Sep 5 '20 at 22:13

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