Post History
Filter design 101 given hypothetical high-pass (HPF) filter -3dB gain HPF at fc attenuation of -8dB at fc/2 Assumptions gain = 0 dB at f>> fc Ripple between -3dB and 0 dB...
Answer
#2: Post edited
- Filter design 101
- -
- 1. given hypothetical high-pass (HPF) filter -3dB gain HPF at fc
- 2. attenuation of -8dB at fc/2
- 3. Assumptions gain = 0 dB at f>> fc
- 4. Ripple between -3dB and 0 dB is unknown but assume 0dB max for simpler case.
- 5. steepness of skirts << fc is unknown but we know 1st order slope is 6 dB/octave maximum
6. the attenuation at fc on a 1st order filter at fc/2= -7 dB which almost satisfies -8dB so is slightly greater than 1st order, which means a 2nd order filter that results in -1dB at fc/2 may be added to solve this problem. That frequency may be computed from impedance ratios to obtain the final transfer function, but I can tell you -1dB is about 2xfc.- 7. There are also an infinite number of other solutions if the assumptions change above.
- The breakpoint is defined as the half-power point where the voltage drop is 0.707 or -3dB approx.
- Using my assumptions in 6. above I declare the 2nd order HPF transfer function is;
- $$H(s)=\dfrac{s^2}{(s+\omega_0)(s+2\omega_0)}$$
- See if that suits your specs.
- Filter design 101
- -
- 1. given hypothetical high-pass (HPF) filter -3dB gain HPF at fc
- 2. attenuation of -8dB at fc/2
- 3. Assumptions gain = 0 dB at f>> fc
- 4. Ripple between -3dB and 0 dB is unknown but assume 0dB max for simpler case.
- 5. steepness of skirts << fc is unknown but we know 1st order slope is 6 dB/octave maximum
- 6. the attenuation at fc on a 1st order filter at fc/2= -7 dB which almost satisfies -8dB so is slightly greater than 1st order, which means another 1st order filter that results in -1dB at fc/2 may be added to solve this problem. That frequency might be computed from impedance ratios to obtain the final transfer function, but I can tell you -1dB is about 2xfc.
- 7. There are also an infinite number of other solutions if the assumptions change above.
- The breakpoint is defined as the half-power point where the voltage drop is 0.707 or -3dB approx.
- Using my assumptions in 6. above I declare the 2nd order HPF transfer function is;
- $$H(s)=\dfrac{s^2}{(s+\omega_0)(s+2\omega_0)}$$
- See if that suits your specs.
#1: Initial revision
Filter design 101 - 1. given hypothetical high-pass (HPF) filter -3dB gain HPF at fc 2. attenuation of -8dB at fc/2 3. Assumptions gain = 0 dB at f>> fc 4. Ripple between -3dB and 0 dB is unknown but assume 0dB max for simpler case. 5. steepness of skirts << fc is unknown but we know 1st order slope is 6 dB/octave maximum 6. the attenuation at fc on a 1st order filter at fc/2= -7 dB which almost satisfies -8dB so is slightly greater than 1st order, which means a 2nd order filter that results in -1dB at fc/2 may be added to solve this problem. That frequency may be computed from impedance ratios to obtain the final transfer function, but I can tell you -1dB is about 2xfc. 7. There are also an infinite number of other solutions if the assumptions change above. The breakpoint is defined as the half-power point where the voltage drop is 0.707 or -3dB approx. Using my assumptions in 6. above I declare the 2nd order HPF transfer function is; $$H(s)=\dfrac{s^2}{(s+\omega_0)(s+2\omega_0)}$$ See if that suits your specs.