1
$\begingroup$

Actually I got confused about definition of frequency response of Ideal low-pass filters because in some books they mention that phase of H(f) should be linear in pass band and it's value is $ -2\pi t_{0}$ enter image description here

However in other books they mention that phase should be zero in pass band.

enter image description here

So I want to know their differences and where does $ -2\pi t_{0}$ comes from?

$\endgroup$
3
$\begingroup$

An ideal low-pass filter has frequency response $H(f)$ equal to the brickwall function: $$H(f)=\mathbb I[|f|<f_c]$$ where $\mathbb I[s]$ is the indicator function, which is equal to 1 if the statement $s$ is true and 0 otherwise, and $f_c$ is the cutoff frequency. Since this frequency response is real, its phase is zero.

This ideal filter is not stable and non-causal. It also has zero delay; it corresponds to $t_0=0$ in the first definition you list above. Now consider the case where we want an ideal LPF, but with some delay $t_0>0$. If the filter input is a sinusoid of frequency $f_0<f_c$, then the filter output is delayed a time $$-\frac{-2\pi f_0t_0}{2\pi f_0}=t_0.$$

In other words, the first filter in your question is ideal with a delay $t_0$ that you can measure or specify; the second filter is ideal with delay equal to 0. As pointed out in another answer, this helps to make the filter "more causal" -- it allows you to implement a filter that approximates the ideal.

See here for another explanation of phase and delay.

$\endgroup$
  • $\begingroup$ Thanks Mbaz for your good explanation and your contribution $\endgroup$ – Ehsan Zakeri Nov 24 '16 at 5:49
2
$\begingroup$

This difference stems from the (approximated) fact of causality:

Looking at the time-domain response of an ideal lowpass with linear phase, it is a sinc function that has its peak at time $t=t_0$. The LP without the phase has the peak at $t=0$. Now, if you assume that you can neglect the relatively low values of the sinc at times $|t|>t_0$ the first filter (with time-shift) becomes causal. HOwever, this is just an approximation (because sinc is infinitely long in time).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.