Q:

How to use input/output termination resistors in filters?

NOTICE: I have added another question here that a solution for each one may help for solving the other one.
Please look at  Fig-1  below. It is a common usage of termination resistors. R1 , R2  work for impedance matching and R2 also reduces the return signals ( specially we have them when Q>> 1/√2).
Firstly, input resistors (R1) are based on line driving. What about the PCB ? in PCB we have not coaxial lines . Should we need them ?
Second question is my specific problem: my filter receives the signal from a source with 75Ω impedance and sends it to a FPGA with a very large input impedance (Fig-1 ,2 ). When I connect R1 as Fig-1, filtering performance is fair ( Q=0.6) with slow slope but if I connect it as Fig-2, it shows satisfactory performance ( No Return Signals ) and if I remove it, many return signals will appear. 
I am confused about the theory: If the device impedance is very high ( say in MΩ region), why putting a relatively small resistor (20KΩ /1MΩ  )in series with that should stop return signals ?
And Finally, how I can relate these resistors to other steps impedances (for best Quality factor across the whole Cauer design)?  
Additional Information:   The main purpose is designing a very high quality LPF for a DDS (that I have designed with FPGA ) for removing aliases that follow Nyquist diagram. The device at the output are the Differential pins of FPGA Xilinx XC3S400. Any help that lead to best design is highly appreciated.
Pictures:
I just put these designs as example , I am using a 3 pole Cauer LC filter but the same result with the circuits in these pictures
Fig-1  Original Design:

Fig-2 My test circuit:

A:

I am confused about the theory: If the device impedance is very high ( say in MΩ region), why putting a relatively small resistor (20KΩ /1MΩ )in series with that should stop return signals ?

Signals approaching a high-impedance connection are akin to waves crashing into a hard surface.  They will bounce, and return in inverted form (ringing).  In electronics the returned wave can be in negative voltage and large enough undershoot to blow the clamping diodes or even the drivers.
A properly calculated source serial terminating resistor placed close the pin prevents many sorts of problems with 'ringing' and EMI.  If your source is 75Ω already, and your termination is Megs'n'Puffs, try this:

make sure your pcb trace is also 75Ω.  You'll likely need change the width and length, and keep in mind it changes with substrate thickness and the layer you are on.   
Place a 66Ω resistor at R2 closest to the pin as possible.
leave everything else out.

This way, all three pieces are the same impedance, nothing should reflect.  Anytimes the line resistance changes, you'll get some reflection.  If you do it right, you can from the scope below where there is 1V undershoot:

To this, by adding just one properly sized resistor:

To get the best result, you'll need to load in a simulator like lineSim (Hyperlynx by mentor) and verify.
My feeling would be that in the above examples you've given, you can do without R1 altogether and make R2 66Ω, place your filter closest to one end, and match your trace impedance 75Ω.

A:

The bits that confuse you are all because of what will probably be a surprising thing to learn about how electrical signals travel. The technical term is transmission line theory, and the link takes you to wikipedia. But stay with me for a moment, because it is not difficult. Indeed I think it is even easier than the prior answers may make it seem.
All of our electronics, even simple cables, have parasitic components. Any wire (when seen together with ground or whatever your return path is) will also have inductance and capacitance. Well, and resistance, but that is often negligible whilst surprisingly the rest often isn't. A signal will see the combination of these parasitics as some characteristic impedance: It is just as if there is a resistance. But it is not an ohmic one, and to emphasize that the more general term impedance is used.
If you want to get more involved, you can characterize an impedance precisely by combining it with the phase difference between current and voltage. But that's not really needed to understand the idea; all we need is that it is going to be either +90° or -90° if everything (well, your transmission line, to sound academic) does not dissipate energy, that is, if there is negligible actual ohmic resistance.
To avoid this system to resonate, which would appear as a signal first travelling forward and then backwards (as a "reflection"), we need to dampen it. Perfect damping requires that you dissipate all of the energy received at the end. This occurs if you provide an ohmic resistance at the end of the transmission line, and only if this resistance matches its impedance.
Does this happen in your example circuits? I will assume the filter (R1, L1, L2, C1) is designed so as to not change (transform) the line's impedance, or else to adjust it to 60 Ohm, the following is the case. But note that this is a gross oversimplification, as you can instantly see from the fact that there is an ohmic resistance (R1) in there.
In the upper diagram, the ohmic resistance is 60 Ohm parallel to the input resistance of your IC, or just a tiny bit below 60 Ohm. That's practically a perfect impedance matching, and should eliminate or at least greatly reduce any reflected "return" signal you might observe.
But now let's have a look at your lower circuit. The terminating resistance is 20 kOhm in series with the input resistance of your IC. That's probably more than a Megaohm in total. I suppose after this explanation you are no longer terribly surprised that there will be some resonance happening? The only place for the energy to travel is back, such that you observe them as what you called "return signals."
So the general answer for a filter like yours that seems to simultaneously be supposed to be a termination is to critically dampen it.

