Low pass double pole filter module calculation

I shoud calculate module of this transfer fucntion....
http://img508.imageshack.us/img508/5108/fsdgsdow2.jpg
is it correct or not?
thanks in advance

 

[Bob Penoyer]

On Sat, 3 Jan 2009 00:30:04 -0800 (PST), silusilusilu@gmail.com wrote:

I shoud calculate module of this transfer fucntion....
http://img508.imageshack.us/img508/5108/fsdgsdow2.jpg
is it correct or not?
thanks in advance

First, the function that you represent is called a "synchronous
filter." That is, the function you've used is a simple first-order
filter cascaded with an identical first-order filter. Using your
example, one filter is represented by
1 / (1 + Ts)
and the second filter is also represented by
1 / (1 + Ts)
so that cascading them (one circuit feeding another, with no loading
of the first filter by the second filter) is represented by
1 / ((1 + Ts)^2), your original function.

This is a poor filter because the -3dB frequency of one filter section
corresponds to the -3dB frequency of the other filter section so that
the attenuation at the corner frequency is -6dB. A good filter will be
down no more than about -3dB at the corner frequency.

Second, your math is wrong. The best thing you can do is convert s
into jw BEFORE you begin your calculations. You're more likely to keep
the real and imaginary parts of the calculation intact when you do
that. If you do, here's what you'll get (use a fixed-point font such
as Courier to view the following):

Taking one section,
1 1 - jwT 1 - jwT
------- * ------- = ----------
1 + jwT 1 - jwT 1 + (wT)^2

Squaring this to cascade the two functions,
1 - jwT 1 - 2jwT - (wT)^2
(----------)^2 = -----------------------
1 + (wT)^2 (wT)^4 + 2 * (wT)^2 + 1

Now separate the real and imaginary parts and calculate the magnitude.

Good luck.