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If a practical filter can't remove all unwanted frequency components like ideal filter, does that mean unwanted frequency component are still present after filtering?

how can we use such distorted output in practical use ?

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A real filter wont perfectly remove all unwanted frequencies. But then, a real $X_\mathrm{anything,whatever}$ won't do its job perfectly. The real world isn't perfect, you only get perfection in mathematics.

What's important though is that a filter can approximately remove the unwanted frequencies. For example if you have some interference at 100 kHz, then a simple 2nd-order lowpass filter at 20 kHz will not be able to remove that completely, but it will reduce its amplitude by more than 24 dB. If the signal you're interested in is strong enough, this smaller inteference level may then not bother you anymore. Or maybe it will still bother you and you therefore need to switch to a higher-order filter. Judging something like this and deciding which filter is most appropriate for the application is the bread and butter of linear signal processing.

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You can increase the order of the practical filter to obtain an ideal filter like behavior. Example: an ideal rectangular filter's magnitude response can be accurately approximated by a high-order (6th or 7th) Butterworth or Chebyshev Filter.

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  • $\begingroup$ I have same question brother still not finding answer $\endgroup$ – user48391 Oct 7 '18 at 3:23
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Filtering has 2 sides. Decreasing what you don’t want and increasing what you do want. It’s a game of gain. It’s not a situation where anything less than infinite SNR gain, which is perfect elimination, isn’t useful. Perfect filters aren’t required.

If the achievable gain of a filter is insufficient, you look someplace else like increasing the power of a transmitter or finding a code that has gain.

At some point no more gain is possible and the limit of a system is reached. Signal Processing isn’t magic.

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