Q&A

# Design high -pass filter with 2 points of the bode plot

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Im designing a high-pass filter which has a gain of -8dB at half of the roll-off frequency and I am stuck ,I dont know how to continue the design.

In the bode plot of that filter we have 2 points:1 is at (fc,-3dB) and the other is at (fc/2,-8dB).What information must I extract to find the transfer function of the filter?

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## 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.

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You still haven't said why you need, or even what order, type, etc. Assuming it's a 2nd order, an exact solution involves creating a generic transfer function and then solving a system of equations with imposed conditions (use squared to get rid of radical):

\begin{align} H(s)&=\dfrac{s^2}{s^2+as+b} \tag{1} \\ &\begin{cases} |H(j)|^2&=\dfrac12 \\ |H(j/2)|^2&=\left(10^{-8/20}\right)^2 \end{cases} \end{align}

You wil get four solutions (4 combinations): $$\begin{cases} a_{1,2,3,4}&=[+,-,-,+]0.33035 \\ b_{1,2,3,4}&=[+,+,-,-]0.47976 \end{cases} \tag{2}$$

Since the denominator needs to be a Hurwitz polynomial only the positive values are chosen (the 1st pair), which results in a perfect match:

\begin{align} |H(j)|&=0.70711\space(0.70597) \\ |H(j0.5)|&=0.39811\space(0.39165) \end{align}

In parenthesis are the results of @TonyStewart's solution, tweaked to have $f=0.42\space(2f=0.84)$. And these are the plots (Tony's is dashed):

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