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!

B How can I expand this out

  1. May 24, 2016 #1
    Low pass filter function


    Just to show the limits as x approaches 0 and infinity, where x is frequency.
  2. jcsd
  3. May 24, 2016 #2


    User Avatar
    Science Advisor
    Homework Helper
    2017 Award

    Concise question ! Looks like homework, so PF culture requires some attempt at solution on your part. What do you propose ?
  4. May 24, 2016 #3
    Self studier here, not enrolled in a course so HW is kind of a misnomer.

    It should equal 1 at x = 0 ie DC, open circuit for capacitor and zero for frequency approaches infinity ie capacitor is short circuit.

    It can't tho cos of the first term in x goes to 1/0 ie infinite at DC.
  5. May 24, 2016 #4


    User Avatar
    Science Advisor
    Homework Helper
    2017 Award

    Kudos !

    Simple recipe: for ##\ \ x\downarrow 0\ \ ## you want to be looking at something like ##\ \ A + x\ \ ## and for ##\ \ x\rightarrow\infty\ \ ## you want something like ##\ \ B + {1\over x}\ \ ##.

    To wit for ##\ \ x\rightarrow\infty\ \ ##:$$ { 1\over x \sqrt{R^2+{1\over x^2}}}$$the last term in the denominator disappears besides R2 so you are left with ##1\over x R##.

    For ##\ \ x\downarrow 0\ \ ## you bring the x inside the root: $${ 1\over \sqrt{x^2 R^2+1}}$$ and now the first term disappears besides the 1.
  6. May 24, 2016 #5
    thanks, I cast the problem of x approaching 0 as


    and wrongly insisted that due to the first term (1/x) that the behaviour had to blow up to infinity.

    I wanted the equation to fit my expectation even tho I knew the behaviour of the low pass in advance.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted