# Transfer Function S and j

Discussion in 'General Electronics Chat' started by Mechatronical, Feb 19, 2014.

1. ### Mechatronical Thread Starter Member

Feb 15, 2014
30
0
In the link below I've added the transfer function of an (active, inverted) band pass filter in the form H(s) and H(jω)

http://s28.postimg.org/kw5cnrxi5/Transfer_Function.png

Inverted Band Pass Filter:
http://www.electronics-tutorials.ws/filter/fil51.gif

Does this mean ω0/Q s = jωR2C1 = R2C1s?
Replacing jω with s and dissolving (sR2C2+1)(sR1C1+1) =
R2C2R1C1s^2 + (R2C2+R1C1)s + 1
So (R2C2+R1C1)s = ω0/Q s. But this can't be true.
Because R2C2+R1C1 = R2C1 can not be correct.

So what do I need to do to get ω0/Q s on one side of the equal sign and resistors and capacitors on the other side?

Last edited: Feb 19, 2014
2. ### LvW Active Member

Jun 13, 2013
674
100
Mechatronical - the answer is simple:
The form of the left expression applies to a 2nd order bandpass functions only that has a conjugate-complex pole pair. However, the expression to the right (and the given circuit) has to two REAL poles only.
Thus, you must not compare both equations.

3. ### Mechatronical Thread Starter Member

Feb 15, 2014
30
0
In that case, how should the transfer function on the left look like?

4. ### LvW Active Member

Jun 13, 2013
674
100
In this case (two real poles) such a form is not possible because only a 2nd order function has one single pole frequency wo.
In contrast, the function on the right side has two (real) poles.

5. ### Mechatronical Thread Starter Member

Feb 15, 2014
30
0
Since the step response is overdamped it has roots: -ζω +/- sqrt(ζ^2 - 1)
and the transfer function can be described as having poles:
(R2C2s+1)(R1C1s+1) I have poles at 1E-4 and 1E-5.
If I know ω, then I should be able to find the damping ratio ζ:

-ζω + sqrt(ζ^2 - 1) = 1E-4
-ζω + sqrt(ζ^2 - 1) = 1E-5

Do you agree?
ω = 2∏f, but would f be the incoming frequency passing through the bandpass filter?

6. ### LvW Active Member

Jun 13, 2013
674
100
What is the purpose of your investigations?
What do you intend to show?
The expression in the first line must be wrong (look at the dimensions of both parts).
You have two real poles with a negative sign (yours are positive).
You can treat the transfer function as a series connection of a 1st-order highpass and a 1st-order lowpass.
What is the damping ratio of a first-order circuit?

7. ### Mechatronical Thread Starter Member

Feb 15, 2014
30
0
My goal is to be able to calculate the Quality Factor of the band pass filter by acquiring the damping ratio of the filter. But it is not a first order circuit.
Because it has a second order polynomial expression. This is expressed throughout the scientific media. Now leave it by that. The first thing I take into consideration is plotting the transfer function in hopes to find either the pike time, settling time or overshoot, either is fine. By using either of the two equations:

H(jω) = R2C1jω / (R2C2jω+1)(R1C1jω+1)
H(s) = ω0/Q s / (s^2 + (ω0/Q)s + ω0^2)
But you might argue the H(s) above is wrong.
The H(s) above is for band pass filters with high Q factor.
Otherwise H(s) = lowpass*highpass = [ω0 / (s+ω0)] * [s / (s+ω0)]= ω0 s / (ω0 + s)^2

The values are R2=R1=10K ohm and C1=10nF and C2= 1 nF.
I've tried to plot either of these transfer functions, but they don't appear well.

Bode plot of H(s) = ω0/Q s / (s^2 + (ω0/Q)s + ω0^2)
http://www.wolframalpha.com/input/?i=bode+plot+10033s%2F%28s^2%2B10033s%2B374964496%29

Bode plot of H(s) = ω0 s / (ω0 + s)^2
http://www.wolframalpha.com/input/?i=bode+plot+19364s%2F%28s^2%2B%2819364*2%29s%2B374964496%29

I also want to know the relationship between the H(s) and H(jω),
so I can set ω0 or ω0/Q for example equal to RC values.

I've also tried to plot the response of the transfer function in Scilab with this setup:
http://s27.postimg.org/ihw6d12ab/Scilab.png
I get a flat line with these values:
H(jω) = R2C1jω / (R2C2jω+1)(R1C1jω+1)
Replacing with s
H(s) = 0.0001s / (0.00001s+1)(0.0001s+1)

Last edited: Feb 19, 2014
8. ### LvW Active Member

Jun 13, 2013
674
100
I think, we shouldn`t discuss in two different threads.