types of filters and properties

Discussion in 'Homework Help' started by templeavenue, Dec 2, 2004.

  1. templeavenue

    Thread Starter New Member

    Dec 2, 2004
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    I’m just taking an electrical (ppb introductory) course and cracking for test. I’m really confused by frequency response sections(with filters). I don’t know how to approach such problems, where it asked me to find out the frequency response of the circuit (given in a regular circuit or op-amp) and identify the type of filter. There are different types of frequency (low pass, high-pass, band-pass. Etc.), but I don’t know how to determine it (neither graphing or some other method). In the text book, it mentions that when the load is in series with a resistor and the load is in parallel with capacitor, it gives low-pass. Also, when use capacitor and inductor, it gives bandpass filter). I can’t see what filter it is just looking at a circuit. Also, I want to know how to compute “omega-not” wo, the cut-off frequency. In the book, it used (1/RC) for cut-off frequency for H(jw) = 1 / ( 1+jwCR). Is the the same for any type of circuit(which I think not). To my understanding, it has to do with the denominator. But how about when w is in both numerator and denominator? Also, I want to know more about dB when graphing. Is the slope always (+20 or -20 dB), no matter what? Any help will be appreciate. (Or any link that explains very well). Thanks in advance
     
  2. mozikluv

    AAC Fanatic!

    Jan 22, 2004
    1,437
    1
    hi

    i would like to refer you to my post about filters. just click on my user name and then click on "find members topic" :)
     
  3. xmen205

    Member

    Oct 31, 2004
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    hey dude,filters is one of THE most simplest topic which i have come across.
    first about low pass filter(LPF) .we connect the capacitance to slot no 3 of opamp.the capacitance is connectd in parallel with the source resistance and the ac voltage source.
    secondly for a high pass filter,we connect the capacitor in series with the voltacge source and connect the resistance parallel to them.
     
  4. phibes

    New Member

    Dec 20, 2004
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    Dude, sounds like you know filters!
    Can you tell me how you get from jwRC/(1+jwRC) to 1/(1-jwRC)?
    Both expressions are equivalent in defining a single-pole series RC highpass-filter.
     
  5. Brandon

    Senior Member

    Dec 14, 2004
    306
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    I don't think that is correct. I believe this is the right route. Someone correct me if I am wrong. Haven't done analog filters in a while. Only digital ones.

    =jwRC/(1+jwRC)
    =(jwRC/jwRC)/(1/jwRC+jwRC/jwRC)
    =1/(1/jwRC+1)
    =1/(1-j/wRC)
    =1/(1-jw0/w)

    where w0=1/RC (The cutoff frequency)
    w is the frequency of the signal

    From the above you can plot a frequency curve fairly easily. From this transfer function we have a high pass filter. At w->infinity, w0/w ->0 giving a gain of 1.

    When w=w0, we have 0.707 as the gain (1/sqrt(2)), i.e. the -3db point. Also called the half power point since 0.707^2 = .5 the power.

    when w<w0, the gain tapers off at -20db/decade.
     
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