What happens to the Low Pass Filter as frequency approaches 0 and infinity?

[houlahound]

Low pass filter function

1/x(sqrt(R^2+1/(x^2)))

Just to show the limits as x approaches 0 and infinity, where x is frequency.

 

[BvU]

Concise question ! Looks like homework, so PF culture requires some attempt at solution on your part. What do you propose ?

 

[houlahound]

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.

 

[BvU]

Self studier here

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.

 

[houlahound]

thanks, I cast the problem of x approaching 0 as

(1/x)*1/(stuff)

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.