Graduate How Can I Linearize \( f(t) = \sin(\Phi(t)) \) Using Laplace Transform?

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The discussion revolves around linearizing the function \( f(t) = \sin(\Phi(t)) \) using the Laplace transform to fit it into a larger transfer function. Participants suggest using a Taylor expansion for linear approximation, but emphasize that this approach is effective only if \( \Phi(t) \) changes minimally over time. There is confusion regarding the order of the polynomial generated by the Taylor expansion, with some asserting that a linear approximation should yield a first-order function. Clarification is sought on the specific problem being addressed and the desired form of the linearization. The conversation highlights the need for precise definitions and context to effectively assist in the conversion process.
Jarfi
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By using the laplace transform:

$f(t)=sin(Φ(t))$

I want it in the form:

F(S)/Φ(S)

The purpose is to linearize it in order to put it into a larger transfer function, so far my only solution is to simplify it using taylor expansion.
 
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Yes, you can linearize - just Taylor expand about the relavent point. However, since you really care about the Laplace transform of this, the linear approximation will only help you if ##\phi(t) ## changes very little for ##0<t<\infty##.
 
jasonRF said:
Yes, you can linearize - just Taylor expand about the relavent point. However, since you really care about the Laplace transform of this, the linear approximation will only help you if ##\phi(t) ## changes very little for ##0<t<\infty##.
Taylor expansion the sin function would yield a polynomial of several orders, say I had two 3 order approximated sinus functions in series(with an output between them), then I have a 6th order polynomial, increasing complexity. So i was hoping for another solution.

Φ is from 0-45°
 
I don't understand what you mean. A linear approximation is, by definition, linear. How would it be 6th order?

If you want a linear approximation, it obviously must be of the form ##\sin\phi(t) \approx a \, \phi(t) + b##. How you select ##a## and ##b## depends upon what criterion makes the most sense for the problem you are trying to solve. Unless you tell us what problem you are actually trying to solve I doubt I can help much more.

jason
 
jasonRF said:
I don't understand what you mean. A linear approximation is, by definition, linear. How would it be 6th order?

If you want a linear approximation, it obviously must be of the form ##\sin\phi(t) \approx a \, \phi(t) + b##. How you select ##a## and ##b## depends upon what criterion makes the most sense for the problem you are trying to solve. Unless you tell us what problem you are actually trying to solve I doubt I can help much more.

jason

It's linear in the frequency domain, NOT in the time domain: My preferrence is exactly not to approximate it, but to convert it. I already know how to approximate-linearize a sinus function
 
Jarfi said:
It's linear in the frequency domain, NOT in the time domain: My preferrence is exactly not to approximate it, but to convert it. I already know how to approximate-linearize a sinus function

Jarfi said:
By using the laplace transform:

$f(t)=sin(Φ(t))$

I want it in the form:

F(S)/Φ(S)

The purpose is to linearize it in order to put it into a larger transfer function, so far my only solution is to simplify it using taylor expansion.

I still don't know what you are asking - you use the word "it" a lot, and now it is clear that the word "it" may refer to different things in different places. So … what exactly are you trying to approximate and in what form? What do you mean by "convert it"? When you say linear, do you mean a linear function of ##\Phi(s)##, or a linear function of ##s##?

I would be happy to help if you answer these questions, or (probably better) just state a more explicit question that helps us understand what you are doing.

jason
 

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