Lowpass filter

Discussion in 'Homework Help' started by Anthony Quah, Sep 9, 2008.

  1. Anthony Quah

    Thread Starter Active Member

    Dec 1, 2007
    hi guy,

    I am currently study RF electronic, I got this ABDC matrix which i find it hard to understand how do i obtain the graph as show in my attachment..

    Example how do RL = 5 ohms have attenuation of 23dB? how do i get value from the formula. Can show someone show me the example of calculation, i would find it helpful .. Thanks

  2. guitarguy12387

    Active Member

    Apr 10, 2008
    To find the attenuation factor (alpha) we need to find the transfer function. That is defined as 1/A = V2/VG.

    Plug in given values for R, Rg, C into the formula for V2/VG and you'll end up with a function depending only on the frequency (omega) which is your transfer function. Actually, it will also depend on RL. Differing values of RL will give you different transfer functions (and thus different attenuation factors).

    As you can see, the smaller the load resistance, the more the attenuation. I think that is the point of the exercise.