Why does the capacitor takes the high-frequency current at the output of a dc-dc converter?
1) Ideally, C1 has the largest value possible yet has a very low ESR rating. This can be from 220uF to 4700uF. If space is an issue then it is easier to install multiple capacitors in parallel (usually tall with a small diameter)to achieve the same goals.2) If you look at a PC motherboard you may see 10 or 20 small capacitors next to each other. These are often filters for 3.3 volt or 1.35 volt power feeds and rated 560uF 4 volts each. One giant capacitor would have taken up too much 3D space and had poor ESR characteristics.3) In the end the design engineer makes compromises about space available vs. the minimum amount of value C1 needs to be, including low ESR values. In compact AC-DC converters C1 maybe several compact low ESR capacitors in parallel to achieve the goal of very low ripple in the DC outputs.4) C1 and L1 must work together to provide very low ac ripple, so certain values of the 2 parts that might resonate with the ripple frequency have to be avoided. This is done with existing performance charts, by basic LC equations and by product testing before the best 'compromise' is found.5) A 'bad' filter  would be one that just approximates the width and/or rate of the pulses, thus it would tend to resonate and cause pulse distortion and excessive loading of the power supply mosfet.6) One could call a good filter 'extreme' pulse shaping as the waveform is flattened down to an almost pure DC voltage-but no more, as it just adds cost to the product.7) Then the supply can go to the next stage which is load testing, but that is to test the servo-loop and min/max currents and voltage stability, vibration test, thermal shock, etc.The following articles may help in understand resonate frequencies:WWW.Wikipedia.org/wiki/LC_circuitWWW.Wikipedia.org/wiki/Pulse_shaping