Design high pass filter with 2 points of the bode plot
Im designing a highpass filter which has a gain of 8dB at half of the rolloff 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?
3 answers
Miss Mulan  I am a bit late with my answer. Nevertheless  there is formula which gives you the necessary filter order "n"  as a function of two frequencies and both associated damping figures. If applied to the most simple case (Butterworth approximation) the result is
n>1.2
Hence, a 2ndorder filter topology is required. This sounds plausible because a 1storder filter has a damping characteristic of 6dB/okt.  however, you require 8 dB.
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Filter design 101

given hypothetical highpass (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 is unknown but assume 0dB max for simpler case.

steepness of skirts << fc is unknown but we know 1st order slope is 6 dB/octave maximum

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.

There are also an infinite number of other solutions if the assumptions change above.
The breakpoint is defined as the halfpower 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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