Butterworth HW help

Discussion in 'General Electronics Chat' started by jeni4vijay, Aug 6, 2011.

  1. jeni4vijay

    Thread Starter New Member

    Jun 24, 2011
    13
    0
    I need a help in Butterworth filter design. The question is as follows.

    A high frequency signal is susceptible to 50Hz mains pick-up. A second order Butterworth active filter with a corner frequency 800Hz is used to reduce the pick up signal. Sketch the magnitude characteristic of the filter (dB versus log f) and estimate the filter output if a 50Hz, 1 volt input is applied to it.

    I had started doing it as follows:

    800Hz = 1/2∏RC
    RC = 1/1600∏

    I cannot quit understand what is mentioned by "A high frequency signal is susceptible to 50Hz mains pick-up"

    Thank you in advance
     
  2. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    The first thing I would suggest is that you look in your textbook or on the internet for the general expression for the second-order low-pass Butterworth active filter. These are typically available designed around 1 rad/sec corner frequency. You can then use filter scaling method to translate it to 800 Hertz.

    hgmjr
     
  3. t06afre

    AAC Fanatic!

    May 11, 2009
    5,939
    1,222
    That means that some of the mains 50 Hz is picked up by the signal cable. If you need to picture how this happened. Think about how a transformer work. In this case we will have what is named inductive coupling.
     
  4. jeni4vijay

    Thread Starter New Member

    Jun 24, 2011
    13
    0
    Thank you for the reply
     
  5. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    A second order Butterworth filter is down -12dB per octave. Since the cutoff frequency is 800Hz then 400Hz is -12dB, 200Hz is -24db, 100Hz is -36dB and 50Hz is -48dB.
    Simply make a graph of it.

    You can lookup or calculate how much 48dB attenuates the 50Hz interference. It is a little less than 0.005 of the original.
     
Loading...