1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: On deriving response function in simple low pass filter

  1. Oct 11, 2014 #1
    1. The problem statement, all variables and given/known data
    I have a problem where the circuit is as follows: (pic attached I hope) but if you can't see it it's just a power source (AC), resistor and inductor with 2 terminals across the inductor (from were you measure the voltage).

    I want to derive the response function, and I am trying to see if I did something off. I am doing it in terms of frequency, rather than [itex]\omega[/itex].

    2. Relevant equations
    So I know that the resistance in an inductor is [itex]Z_L = iL\omega[/itex]
    Resistance from a resistor is just R

    Response function [itex]H ( \omega) = \frac{V_{out}}{V_{in}}[/itex]
    3. The attempt at a solution

    So I take the whole circuit and see these are in series. So the total resistance ([itex]Z_{total} = R + iL\omega [/itex].

    That means the current in the circuit is [itex]\frac{V_{in}}{R + iL\omega }[/itex]

    and the [itex]V_{out} = \frac{V_{in}R}{R + iL\omega } [/itex] because we are measuring the voltage across the inductor.

    Substitute omega with f/2pi and we get
    [tex]V_{out} = \frac{V_{in}R}{R + iL\frac{f}{2\pi} } \rightarrow \frac{V_{out}}{V_{in}} = \frac{2\pi R}{2\pi R + iLf} = H ( f)[/tex]

    Is there anything wrong with this? I ask because I'm doing a lab and even accounting for experimental error my Bode plot diverges a lot from the measured numbers. The shape of the curves is all good; just the one I plotted above seems moved to the right and up a bit from the values I got. Same shape exactly, tho. Not a big deal I guess, but I wanted first and foremost to make sure I did this right.

    Attached Files:

  2. jcsd
  3. Oct 11, 2014 #2
    But this is not the voltage across inductor what you wrote
  4. Oct 12, 2014 #3
    Let me guess: it should have been [itex]V_{out} = \frac{V_{in} \omega L}{R+ \omega L}[/itex], right?
  5. Oct 12, 2014 #4
    What happend to i ?
  6. Oct 12, 2014 #5
    Typo. [itex]V_{out} = \frac{V_{in} i \omega L}{R+i \omega L}[/itex]
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted