types of filters and properties

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

  1. templeavenue

    Thread Starter New Member

    Dec 2, 2004
    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

    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


    Oct 31, 2004
    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
    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
    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.


    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.