Simple RC and RL Transfer Functions

[tastypotato]

[Solved] Simple RC and RL Transfer Functions

Homework Statement

Given the transfer function:

TF= (200)/(1 + j0.001w)

a. Determine the cuttoff frequency in radians per second and in hertz

Homework Equations

There are a couple other equations like this in the book, but I just don't grasp the concept or the formulas that the book offers which looks like TF(w)= Vout/Vin = A < Theta degrees.

The Attempt at a Solution

TF= Vout/Vin = 200 / (1 + j0.001w) --> converted to rectangular form

(200 < 0)/(1 < .057) = 200 < -.057

Which I know is horribly wrong, but these formulas just don't make sense to me. :/

 

[tastypotato]

Figured it out and did some deeper reading online, came up with:

Wc= 1/t

Where in the equation TF=200/(1+j.001w), t=.001

So, 1/.001 = 1000

Then you take Wc/(2*pi)

Which then gives you 159.15 Hz

 

[MATLABdude]

Looks like you've already answered your own question, but I thought I'd elaborate...

To find the corner frequencies of a transfer function (incidentally, the transfer function you give is that of a low-pass filter--Wikipedia for it), you just have to find its poles ('w' which make the denominator go to zero).

In this case, you just set 1+0.001j*w to zero, and determine that w_c is 1000j, which you interpret as 1000 rad/s, or with the formula that w_c = 2*pi*f_c, 159 Hz. (Technically, it should be when the magnitude of the transfer function goes to zero, which gives you the two poles at +/-1000j, and hence, a real frequency of 1000 rad/s).

What is the physical significance of this divide by zero? (Assuming you haven't taken any complex analysis courses)--the system's 'quirks' emerge and have a significant impact (in the case of a low-pass filter, that frequencies higher than the cut-off get significantly attenuated or, cut-off, so to speak). Yes, yes, this is an idealized view of things, but it does give you significant useful information.