How to invert a Transfer function in Simulink?

Hi everyone!
I am trying to invert my transfer function: 1/(7.5*s+1) in simulink to 1/G(s) = 7.5*s+1. The goal here is to use the output of the system, pass it through the inverse of the transfer function and get the original input of the system.
However, I have some difficulties establishing the transfer runction in Simulink. Any help is greatly appreciated. Thank you!

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Bill Cobb
Bill Cobb 2024년 8월 11일
What I've done is produce time signals from x/y from the transfer function x by y. Then chirp this x/y function and recompute a y/x transfer function and drive iy by y to get x.
1/(7.5*s+1) in simulink to 1/G(s) = 7.5*s+1.
syms G s
eqn = G == 1/(7.5*s+1)
eqn = 
inveqn = solve(eqn, s)
inveqn = 
So the inverse of G is 2(1/15G - 1) not 7.5*G+1
Paul
Paul 2024년 8월 11일
편집: Paul 2024년 8월 11일
The OP is referring to the multiplicative inverse of G(s), which is, in fact, 1/G(s) = 7.5*s + 1.

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Les Beckham
Les Beckham 2021년 2월 4일

0 개 추천

Transfer functions are not allowed to have a higher order in the numerator than in the denominator (more zeros than poles). This is because that makes them 'non-causal' which means that they depend on the future value of the input (which doesn't make sense).
Why would you want to "get the original input ot the system" when you already know what it is?

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Thanks! yeah that makes sense, we originally wanted to reverse the output as the input to re-calibrate the system, but it seems like inverting the transfer function isn't doable. Thanks again!

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