How are they normalizing a frequency of this transfer function ?

[MathWizard]

In this transfer function example, H(s)=110s/(s^2+110s+1000) = 0.11s/((1+s/10)*(1+s/100))

once they have it in standard form, with a pole at 10 and 100, they normalized the 1st pole, to be 1, and the 2nd pole became 10. I don't see how the how they get the 0.011 as the new K

0.011s/((1+s/1)*(1+s/10))

I tried letting new p1=(p1/p1)=1 and new p2=(p2/p1)=10, but then new k would still be 110/(1*10*10*10)=0.11 not 0.011

And just switching to jw and letting new w=p1=10, well that doesn't get the 0.011 either.

from page 11
https://www.mcmsnj.net/cms/lib07/NJ01911694/Centricity/Domain/134/ENZ%20-%20Transfer%20Functions%20and%20Bode%20Plots.pdf

[moffy]

To shift the frequency but maintain the same shape, the 's' terms all need to be multiplied by the same factor, which in this case is 10. So the upper term becomes 1.1 not 0.011 as they stated. Just to check I ran a quick test in LTSpice which I have included.

[thermistor-guy]

...So the upper term becomes 1.1 not 0.011 as they stated. Just to check I ran a quick test in LTSpice which I have included.

Agree that this coefficient equals 1.1. Just substitute 10s for s in the transfer function, top and bottom.

[MathWizard]

Ok thanks, they used the same circuit in most that pdf but always had the 0.011. I didn't know ltspice does Laplace transforms like that, I'll have to try it. I'm just learning how to plot the complex poles/zero's of transfer functions. I have GNU octave, it looks pretty easy to plot them in that too.

I'm making a modified copy of a Heathkit Curve tracer, and in the simulations, there's a lot of nasty over/undershoot mainly from the step generator, that makes it's way into the other opamp circuits. So I want to actually calculate my way to a better looking circuit, and model.

[bson]

0.11 is the sole zero, when it's scaled by 10 it becomes 0.011.
Poles and zeros are in the same plane, so when scaling the plane they all change.