# How to Calculate Complex Low Pass Passive RLC Filters?

I have designed a FPGA based DDS and now I want to add a low pass filter to its output. Previously I was working on a project based on AD9850 DDS(0-50MHz) and I had a commercial module for it that had a low pass RLC filter which worked excellent ( schematic below). Now I want to use the same design for my own frequency (fc= 70,100 ,120 MHz with different setups) but I can not analyze its values.

I think R1 and R2 serve as parts of two back-to-back RLC+LCR filters (something like designs in rows 3,5 here ) but I don't know what those small capacitors C5-C7 are doing there? .

I also don't know why L1 inductance differs from L2,L3 (From answer of Andy Aka in this thread I just can guess it is related to the impedance between steps but can't realize how to calculate them? )

I am looking for a formula or method for calculating these values ( R1,2 C1-C7, L1-L3) for other frequencies ( for example for fc=70, 90, 100, 110 ,.. MHz).

For example assume we have fc=100MHz. How do you select part values in this design?

• I can't help with the specifics of this filter but you might like to look at Elsie, the electronic filter design software. – David Nov 11 '13 at 8:42
• @David I looked at it. Is it free? – Aug Nov 11 '13 at 8:45
• Yes, for student/hobbyist use it is free up to 7 stage filters. For professional use with larger filters you will need to purchase a key. I don't know if they specify non-commercial without a key, you'll have to check the license. – David Nov 11 '13 at 8:48
• have you tried doing some s-domain analysis? obtaining the transfer function of the filter may help a great deal determining the component values in relation to filter characteristics. – deadude Nov 11 '13 at 23:10

The classic answer to this question must be "Zverev". But that might be overkill, unless you have access to a really good library.

A simpler and non-mathematical answer to some of your questions is possible, which may help:

R1 and R2 provide impedance matching; the original filter is designed to accept a signal driven from a specific source impedance, and deliver its output to a specific load impedance (R1,R2 are also mentioned later). These impedances are:

• normally the same
• known as the "characteristic impedance" of the circuit
• usually the same as the characteristic impedance of the application's standard cables (e.g. coaxial cable in RF applications)
• commonly (but not always) 50 ohms. (you will see 75 ohms in video applications, and (rarely nowadays) 600 ohms in audio and telephony.

Check the original filter info for its characteristic impedance, but 50 ohms is most likely. So - the impedance of the L-C network was not exactly 50 ohms, and R1,R2 reduced the input and output impedances to match.

C5,C6,C7 ... Consider that C5 and L1 on their own form a parallel L/C resonant circuit. This acts as an inductor (L1) at low frequencies, and as a pure capacitor at high frequencies (VERY high since it is 1 pf!)

But at the resonant frequency, the impedance is infinite. Therefore at this frequency, the filter will have infinite attenuation. (Over-simplification! all the components interact with each other, so the actual frequency is slightly different from this calculation)

There are three such notches in the frequency response; and you can learn a bit about this filter by calculating C5/L1, C6/L2, C7/L3. Usually 2 are quite close together and the third will be significantly higher; without doing the math I can already see that here.

That makes this a 7th order Cauer filter (or Cauer/Chebyshev) and the art of getting good stopband rejection (or the reason for 592 pages of Zverev) is the art of tuning C5-C7 to place those notches (last picture on Wiki page) the right distance apart so the peaks between them are all the same height.

Theory apart, circuit tolerances virtually guarantee tweaking trimmer caps or inductor cores while watching a spectrum analyzer for best results!

C1 to C4 also resonate with L1 to L3; in this case, the main effect is on the passband flatness as well as the actual cutoff frequency (which must be below the first notch!) It can be understood as a cascade of 2nd-order sections with different characteristics and one first-order section. Look at Figure 3 in that article (embedded below, hope that's OK)

It shows underdamped sections (with peaks) and overdamped ones (which just roll off). A skilful combination of these will give an (approximately) flat response up to the cutoff. Again, I cannot cover the details here, but I hope it is clear how different values of inductor forming different 2nd-order filters are part of the puzzle. Getting R1 and R2 wrong will principally affect the passband flatness, by affecting the Q (damping) of the input and output sections (L1 etc and L3 etc).

Here is a more typically mathematical explanation

Now to the most important part of the question:

How do I select part values for 100 MHz?

Given all the above, usually not from scratch... You can take an existing filter, and simply scale it.

Given Xl=jwL and Xc=1/jwC,
assuming the current filter is set for 50MHz,
assuming you want the new filter set for 100 MHz
and assuming the characteristic impedance is to remain the same,

you can simply halve all the inductances and capacitances, so that Xl is the same at twice the frequency, and ditto for Xc. Resistances remain the same, since the characteristic impedance is the same, and a resistor's impedance is not a function of frequency. (Check both versions in simulation!)

• Thank you very very much for such nice explanation! I really learned a lot from this answer but I have a bit confusion with the last part: "scaling the filter". It was the first thing I thought about. All the scalings work nice only on the paper! Practically I received sub-optimal results by this method. I think I need designing a Cauer or M-derived or Constant-K family from the scratch but don't know which one is easiest to design. BTW your answer is a complete one. – Aug Nov 16 '13 at 23:08
• If the scalings also work in simulation but not real wire, then parasitic components are getting in the way. (You may have more than 0.5pf stray capacitance!) The original filter may have been very carefully laid out on PCB, perhaps with tricks to neutralise stray components : there is a LOT of practice involved and as a young engineer I had some dismal failures at it. Now there are specialist RF filter design tools but I am the wrong person to describe them. – Brian Drummond Nov 16 '13 at 23:26

I cannot give you a fixed formula to design your filter, because it all depends on what exactly you want from it. You are essentially looking at an optimization problem that includes much more than just a corner frequency.

If you cannot be bothered with additional details, you could just try this dumb hack: Scale all values to get to your desired frequency. Ignoring the small capacitors C5 through C7, the circuit you have drawn is a multi-pole low-pass filter with a corner frequency around 50 MHz. To go to 100 MHz = 2 * 50 MHz, divide the values of all capacitors and inductors by 2. This approach will shift the corner frequency without changing the impedance (at the corner frequency). So watch out if the impedance that matters to you is defined at a frequency that does not scale the same way as the corner frequency!

If you do want to improve your chances of getting a particularly good result, you will need to understand the requirements you have (or the problems this circuit solves). For example, one effect of slightly spreading the individual resonance frequencies (L1 * C1 != L2 * C2, ...) is smoothing out the frequency response around the corner. Another affected characteristic that may or may not matter for you is signal dispersion and related quantities (phase shifts, delay time), etc. If you choose nominally identical values, some of these end up essentially undefined at the corner frequency because they may depend extremely sensitively on component variations. Spacing resonance frequencies by what corresponds to slightly more than component tolerances hence helps you in reliably getting at least qualitatively the same behavior from nominally identical circuits. But this may not matter in your application.

I think you should at least try to figure out what C5 to C7 do. My best guess is that they shift an internal resonance of L1 to L3 out of a frequency band that matters for the application. If you either change the inductors or the frequency range you wish to use the circuit in, you may have to adjust these accordingly. And in a different way if they serve a different purpose---after all, it just might be that my guess of their purpose is wrong and they instead should flatten the frequency response in some range or compensate some undesired phase shift...

• If you want to go into more depth with filter design, have a look at some tutorial. This one looks promising, but I am sure there must be lots of good ones out there. – pyramids Nov 13 '13 at 12:19

This filter features 9 energy-storage elements but you can see that when the excitation voltage $V_{in}$ reduces to 0 V, $C_5$ comes in // with $C_2$ and you lose an order. Then, if you further observe the other capacitors, like $C_2$, $C_6$ and $C_3$, they form a capacitive mesh whose states variables are not independent. Same for $C_3$, $C_7$ and $C_4$. You lose two more orders. As a result, the denominator $D(s)$ is of degree 6. For the numerator, we can apply the fast analytical techniques or FACTs immediately. If the associations of $C_5||L_1$ or $C_6||L_2$ or $C_7||L_3$ become transformed opens, you have zeros. Therefore, the poles of each of these resonating network become the zeros of the transfer function. We can therefore immediately express the numerator $N(s)$ without writing a line of algebra, just a quick impedance calculation:

$N(s)=(1+s^2C_5L_1)(1+s^2C_6L_2)(1+s^2C_7L_3)$

The dc gain of this circuit is 1, then the transfer function is given by:

$H(s)=\frac{(1+s^2C_5L_1)(1+s^2C_6L_2)(1+s^2C_7L_3)}{1+b_1s+b_2s^2+b_3s^3+b_4s^4+b_5s^5+b_6s^6}$

To determine all these coefficients, you can either go for the FACTs or use Thévenin and rearrange quickly with Mathcad. This is what I did and the result appears in a well-ordered form below:

Then you can check how the Thévenin expression compares with the low-entropy form. Of course, further arrangement are possible in the denominator to form more compact canonical forms but the whole expression is correct.

FACTs are truly the way to go to analyze any type of transfer function. Sometimes you can combine them with a more classical approach as I did here but you always save time. You can discover an introduction to FACTs here http://cbasso.pagesperso-orange.fr/Downloads/PPTs/Chris%20Basso%20APEC%20seminar%202016.pdf

## Basic Stuff

1: Clock Filtering

DDS generators work at a constant clock rate, with a [programmable] phase increment between samples. The main point of the filter is - to remove that clock.

2: Reduce Noise / Distortion

The practical limit of a DDS is less than 1/2 the clock rate. The closer you get to 1/2, the worse the signal gets. A better limit is 2.5 or even 3 (ie. 1/3 of clock rate)

3: Summary- change the clock = change filter parameters If you design a filter for Fc=70MHz, it will still work for Fc=120MHz. You'll just be filtering out a whole load of frequencies that would have been ok.

I'd suggest finding the highest clock frequency that works with you FPGA, then stick with it. This will give you the widest range of output frequencies, and simplify the filter design.

## Filter Part Selection

On to the good stuff! Filters can be analysed in Excel if you know circuit theory. There are whole books on this topic, so I'm going to limit myself to your questions.

R1, R2 = Source & Load impedance If you started out with AD9850, you may be aware there's a limit how much current it can provide. There's a similar limit on the currents out of your FPGA, and we haven't even discussed what DAC (Digital to Analog Converter) you are using.

L1 is different

Plenty of ways of looking at this. Chebyshev filters (for example) almost never have the same values. Or you could just think of L1 doing the "heavy lifting" because it's the first inductor to get the signal.

C5-C7 help block the clock

A capacitor in parallel with an inductor form a "tuned circuit" at a particular frequency (like tuning a radio). Picking the right values here will help block your sample clock. Changing the sample clock rate, as you suggest, makes these less effective.

## In Conclusion

If you really want to play with filter design, it's a heck of a lot easier to download and use a program. In this instance, the program I use is from Iowa Hills Software and is called RF Filters (they do digital filter designs, too)