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?**

## Best Answer

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

really goodlibrary.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:

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!)