design an active filter 2nd order

[funjoke]

• No ripple can exist in the output, but non-linear phase shift is allowed.
• The frequency response must be as close as possible to the ideal filter response.
• Signals of frequency 0 ~ 5 KHz are only allowed to pass through the filter with a desired gain of 4dB . Other frequencies should reduce at least 100 times of gain for every 10 times of frequency increase.
• capacitor of 0.01uF must use in the active filter design due to its high availability in the stock.
• Minimize offset error of your design


My answer:
no ripple means butterworth, 4 db means this is HPF

fc=1/2pie RC=7957OHM
Apb =4
20logApb=4
Apb-1.585

Rf=Rapb=7957 x 1.585 =12.6kohm

Apb=1+Rf/Rs
1.585=1/(12.6k/Rs)
Rs=21.54kohm


i just know this few steps .if there any mistake

 

[LvW]

Very very confusing !
How can you calculate resistors without knowing the circuitry ?

Just two short comments:
1.) Gain of 4 dB in the passband of the lowpass means: Ao=+1.58
2.) Factor 100 per decade means: 40 dB/Dec >>> low pass of second order.

 

[funjoke]

How you know the Ao is 1.58 and what are the following steps ?any guidance for me in this question ?

Added after 1 minutes:

and low pass and high pass is see the Apb ,if positive value it is high pass,if negative value it is low pass.am i correct ?

 

[LvW]

How you know the Ao is 1.58 and what are the following steps ?any guidance for me in this question ?
Added after 1 minutes:
and low pass and high pass is see the Apb ,if positive value it is high pass,if negative value it is low pass.am i correct ?

to 1.) 4 dB is equivalent to 1.58. Do you know the meaning of dB ?
to 2.) What is the source of your assumption ? It is completely nonsense. Sorry.

Recommendation: If you like to design a filter read some introductory papers resp. books before.

 

[Audioguru]

A Sallen-Key second-order filter has a Butterworth response if its two resistors are the same and its two capacitors are the same and its gain is 1.58 (+4db).

 

[FvM]

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