# Question about frequency response of a transfer function

Discussion in 'Math' started by hunterage2000, Jun 26, 2013.

1. ### hunterage2000 Thread Starter Active Member

May 2, 2010
438
1
Can someone please tell me the equation or the reason how the transfer function

G(s) = 1/(s+2) is converted to

G(jω) = 2 - jω / ω^2 + 4

2. ### WBahn Moderator

Mar 31, 2012
20,059
5,656
How us the function:

f(x) = 3 + 9x

converted to

f(2y) = 3 + 18y

Having said that, what you have written is wrong.

G(jω) = 2 - (jω / (ω^2)) + 4

Don't forget good ole order of operations!

3. ### hunterage2000 Thread Starter Active Member

May 2, 2010
438
1
This is what the book says. It somehow goes from G(S) = 1/S+2
to G(jω) = 2 - jω / ω^2 + 4. I have attached what it says.

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4. ### LvW Active Member

Jun 13, 2013
674
100
It is correct - from a formal mathematical point of view.

Multiply numerator as well as denominator by (s-2) and replace s=jw.
Then you arrive at the final expression.
As far as the transfer function (and its interpretation) is concerned you now have pole-zero pair that cancels out.

5. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
784
I would think the correct form is

$G(j\omega)=\frac{(2-j\omega)}{(\omega^2+4)}$

What pole-zero cancellation? Isn't this simply a case of rationalization of the denominator?

Last edited: Jun 27, 2013
6. ### LvW Active Member

Jun 13, 2013
674
100
Yes - in post#1 the brackets are suppressed.

The original expression has no zero at all.
However, the final expression has one zero at s=2 and a pole at s=2.

7. ### WBahn Moderator

Mar 31, 2012
20,059
5,656
$G(j\omega)=1s^{-1}\frac{(2s^{-1}-j\omega)}{(\omega^2+4s^{-2})}=1s^{-1}\frac{(2s^{-1}-j\omega)}{(2s^{-1}+j\omega)(2s^{-1}-j\omega)}$

Here, s is 'seconds', which is why I think it was a stupid convention to adopt s as the variable for the Laplace transform. The 1s^-1 out front is to remove time from the transfer function as a whole, which is important. Other than that, I have to assume that the transfer function is a straight gain and not a transgain. Let the mathematicians do what they want, the EE community should have adopted something else, just like they adopted 'j' instead of 'i'. But, they didn't ask me, so....

A cleaner way to express this is to define

$\omega_0=2\frac{r}{s}$

resulting in

$G(j\omega)=0.5\omega_0\frac{(\omega_0-j\omega)}{(\omega^2+\omega_0^2)}=0.5\omega_0\frac{(\omega_0-j\omega)}{(\omega_0+j\omega)(\omega_0-j\omega)}$

or even better

$G(j\omega)=0.5\frac{(1-j\frac{\omega}{\omega_0})}{(1+\left( {\frac{\omega}{\omega_0}} \right) ^2)}=0.5\frac{(1-j\frac{\omega}{\omega_0})}{(1+j\frac{\omega}{\omega_0})(1-j\frac{\omega}{\omega_0})}=0.5\frac{1}{(1+j \frac{ \omega }{ \omega_0 } )}$

The first form makes it appear that there is only a zero at 2r/s, though if you notice the ω^2 in the denominator then you have a big hint that something else is going on. But by expressing things in pole-zero form, it becomes clear that this is a system with a DC gain of 0.5 and a first-order low-pass characteristic with a cutoff frequency of 2r/s.

8. ### LvW Active Member

Jun 13, 2013
674
100
Yes - sometimes it is confusing.
By the way - that is the reason I prefer to write
(The fact that the circuit is a first order lowpass can be derived, of course, also from the first form of the transfer function in the s-domain).

9. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
784
Yes I see your point. I guess one could then arbitrarily state that any transfer function can be regarded as an irreducible function multiplied by an infinite array of unique matched pole-zero pairs.

I tend to reserve the idea of pole-zero cancellation as a design "tool" for making improvements in system stability.

10. ### WBahn Moderator

Mar 31, 2012
20,059
5,656
The final expression is not a function of s, but of jω. It has a zero at jw=2r/s and two poles at jw=±2r/s. If left as a function of s, it would still have two poles, one at +2r/s and one at -2r/s.

11. ### LvW Active Member

Jun 13, 2013
674
100
OK - from the mathematical point of view this is correct.
However, does this makes sense? What means jw=2rad/sec ?
(left side imaginary, right side real).
Can we measure - after setting s=jw - that the transfer function has a zero or a pole, respectively, at w=2rad/sec? No - we cannot.
Thus, determination of poles and zeros for a transfer function makes sense only in the s-domain (complex s-plane).

Present example:
* Original function: Real pole at sp=-2rad/sec
* After manipulation: