Can the Laplace Inverse be Applied to Divided Transfer Functions?

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SUMMARY

The discussion confirms that the Laplace Inverse can indeed be applied to divided transfer functions. Specifically, when a transfer function is divided into two parts, \(\frac{Y(s)}{Z(s)} = N(s)\) and \(\frac{Z(s)}{U(s)} = \frac{1}{D(s)}\), the Laplace Inverse can be utilized separately for each equation. The condition for this application is that \(Z(s) \neq 0\). This allows for the substitution of \(z(t)\) back into the equations to derive the overall transfer function.

PREREQUISITES
  • Understanding of transfer functions in control systems
  • Familiarity with the Laplace Transform and its properties
  • Knowledge of algebraic manipulation of functions
  • Basic concepts of system stability and response analysis
NEXT STEPS
  • Study the properties of the Laplace Transform in detail
  • Learn about stability criteria in control systems
  • Explore the application of the Inverse Laplace Transform in engineering
  • Investigate the implications of \(Z(s) \neq 0\) in system design
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Control system engineers, electrical engineers, and students studying system dynamics who are looking to deepen their understanding of transfer functions and the application of the Laplace Transform.

Chacabucogod
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I was wondering whether this can be done:

Let's say you have transfer function that goes like this:

\frac{Y(s)}{U(s)}= \frac{N(s)}{D(s)}

Now let's say I divide my transfer into two:

\frac{Y(s)}{Z(s)}= N(s)

\frac{Z(s)}{U(s)}= \frac{1}{D(s)}

Can I apply the Laplace Inverse to these two equation separately and then substitute the value of z(t) on one?

D(s)Z(s)=U(s)

N(s)Z(s)=Y(s)

Thank you!
 
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Of course you can:

You then have:

Y(s)/U(s) = (N(s)*Z(s) ) / (D(s)*Z(s) ) => (shorten right fraction)

Y(s)/U(s) = N(s)/D(s)

No problem, assuming that Z(s) ≠ 0.
 

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