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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]
     
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