# Why are ideal filters not particularly useful/ hard to do?

Why is a LPF ideal filter:

a) Not useful?

b) Hard to make?

I was asked to make a Low pass filter (simulink block) that will take an input of a cutoff frequency, and for a signal with a frequency higher then that cutoff frequency, the amplitude would be zero & with zero roll-off. Hence ideal. Is this an impossible task?

• Regarding a), one disadvantage of an ideal LPF is that it's not stable (dsp.stackexchange.com/questions/1032/is-ideal-lpf-bibo-unstable). Regarding b), I think you should try and see for yourself how hard it really is. – MBaz Jun 2 '15 at 17:22
• Question (b) is a misconception: it is not possible to construct an ideal filter. Many people assume that they can create one by transforming a discrete-time signal with a DFT, zeroing out some bins, and inverse-transforming back to the time-domain. Doing this does not apply an ideal filter frequency response to the original signal; try it, look at an interpolated version of the output signal's spectrum, and see what you get. – Jason R Jun 2 '15 at 19:17
• The causality is the problem with an ideal filter. However, if you have non-causal access (e.g. you have the sequence you want to filter stored as a file), you can implement it (say, by padding the signal before and after with zeros). – Batman Jun 2 '15 at 20:28
• @Batman, I disagree... you still need to contend with the fact that the impulse response's duration is infinite. – MBaz Jun 3 '15 at 0:42
• @Batman: If the length of the input signal is finite, your output signal is still infinitely long if the system's impulse response is infinitely long. Furthermore, the problem is not only causality but also stability, because an ideal LPF is not stable, i.e. there are bounded input signals for which the output signal is unbounded. – Matt L. Jun 3 '15 at 16:09

With a finite-order filter, the Z-transform will be an analytic function over the complex domain. The frequency response is the Z-transform evaluated over the closed path $z = e ^{j\Omega}$ (i.e. the unit circle), where $\Omega$ is the frequency.