Electronic – lead high pass filter and lag low pass filter
filterhigh pass filterlow pass
I have a question regarding lead high pass filter and lag low pass filter.
I do not quite understand the reasoning behind these two special case (when z=0) ?
Thanks!
Best Answer
Lag low-pass and lead high-pass are in fact the "standard" low-pass and high-pass filters, in the sense that an ideal low-pass filter should have a gain of zero for \$\omega\rightarrow\infty\$, and an ideal high-pass filter should have a gain of zero at \$\omega=0\$. These conditions are satisfied by the lag low-pass filter (with a zero at \$s\rightarrow\infty\$), and by the lead high-pass filter (with a zero at \$s=0\$).
The phase of the lead low-pass filter is greater (i.e., less negative) than the phase of the lag low-pass filter (\$\Rightarrow\$ "lead"), but the magnitude is worse because its gain only decays from \$z_1/p_1>1\$ to \$1\$ for \$\omega\rightarrow \infty\$. A similar thing is true for the lag high-pass filter. Its gain is not zero at \$\omega=0\$ but it equals \$z_1/p_1<1\$. Its phase is smaller (i.e., less positive) than the phase of the lead high-pass filter (\$\Rightarrow\$ "lag").
I'm afraid you're up against a fundamental limitation of real, causal physical systems. You're asking for a filter that can distinguish between signals that have periods of 12 to 16 seconds, but only allowing it to "look at" a 2-second segment (1/6 to 1/8 of the period) of the waveform in question. It simply isn't possible to get no phase shift and low delay under these circumstances.
Can you provide some more context for this problem? What is the nature of the signals, and why is such low processing latency required?
A second order bandpass filter has a first order roll off rate on each side of the pass band. A second order low pass filter in series with a second order high pass filter has second order roll off rates on both sides of the pass band. The two are not equivalent.
Best Answer
Lag low-pass and lead high-pass are in fact the "standard" low-pass and high-pass filters, in the sense that an ideal low-pass filter should have a gain of zero for \$\omega\rightarrow\infty\$, and an ideal high-pass filter should have a gain of zero at \$\omega=0\$. These conditions are satisfied by the lag low-pass filter (with a zero at \$s\rightarrow\infty\$), and by the lead high-pass filter (with a zero at \$s=0\$).
The phase of the lead low-pass filter is greater (i.e., less negative) than the phase of the lag low-pass filter (\$\Rightarrow\$ "lead"), but the magnitude is worse because its gain only decays from \$z_1/p_1>1\$ to \$1\$ for \$\omega\rightarrow \infty\$. A similar thing is true for the lag high-pass filter. Its gain is not zero at \$\omega=0\$ but it equals \$z_1/p_1<1\$. Its phase is smaller (i.e., less positive) than the phase of the lead high-pass filter (\$\Rightarrow\$ "lag").