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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


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    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


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    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.
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