0
\$\begingroup\$

enter image description heremaybe a stupid question. But let us say we have a ideal LPF filter like this: $$H\left(j\omega \right)=\begin{cases}1&-\omega _m<\omega <\omega _m\\ 0&else\end{cases}$$ If I shift it , which means like that: $$H\left(j\omega \right)=\begin{cases}1&\omega _m<\omega <3\omega _m\\ 0&else\end{cases}$$ Is it still ideal LPF filter? Because we know the "shape" of LPF fitler is like a step function of $$u(t+1)-u(t-1)$$

I know my question is a little stupid, but it is important to me.

EDIT: I was said in comments it will be Band Pass Filter, but how? band pass filter looks like double shifted LPF, not one. enter image description here

EDIT:

\$\endgroup\$
7
  • 3
    \$\begingroup\$ The shifted LPF will be a band pass filter. \$\endgroup\$ Jun 27, 2023 at 17:27
  • \$\begingroup\$ but band pass filter has two squares, I mean, two places like LPF How is it possible? \$\endgroup\$
    – user323806
    Jun 27, 2023 at 17:28
  • \$\begingroup\$ @StefanWyss Hi, edited in post the BPF, that is not logical to me how it will be BPF, please explain to me \$\endgroup\$
    – user323806
    Jun 27, 2023 at 17:30
  • 1
    \$\begingroup\$ Forget about negative frequencies. They have no relevance other than math. \$\endgroup\$ Jun 27, 2023 at 17:32
  • 1
    \$\begingroup\$ This filter passes a band from 100 - 200 Hz. \$\endgroup\$
    – tobalt
    Jun 27, 2023 at 17:58

1 Answer 1

0
\$\begingroup\$

The filter response cannot be shifted this way.

The negative values of angular velocity are a mirror image of the positive values.

The definition for a low pass filter is based on the positive values of \$\omega\$. The 0 Hz boundary cannot be shifted.

When a corner frequency of a filter is shifted to the right, the image corner frequency is shifted to the left, and vice versa.

Perhaps frequency should be calculated as \$f=\left|\frac{\omega}{2\pi}\right|\$.

\$\endgroup\$
2
  • \$\begingroup\$ Hmm, if I am understanding correct. When you said that if I shift it, it cannot be shifted. So when I shift it, I basically make it BPF ( as before it was LPF ). Am I right? \$\endgroup\$
    – user323806
    Jun 27, 2023 at 19:54
  • \$\begingroup\$ You can draw whatever you like. In a real filter, there is only one corner frequency. The negative sign indicates 180 degree phase shift. \$\endgroup\$
    – RussellH
    Jun 27, 2023 at 20:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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