Simplifying Laplace Transform of Cosine with Angular Frequency and Phase Shift

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SUMMARY

The discussion focuses on the Laplace Transform of the function \( f(t) = \cos(\omega t + \phi) \). The initial approach simplifies the function using trigonometric identities, resulting in the transform \( Lf(s) = \frac{s \cos(\phi) - \omega \sin(\phi)}{s^2 + \omega^2} \). An alternative method using the time-shifting property is also explored, leading to the expression \( Lf(s) = e^{\frac{\phi}{\omega} s} \frac{s}{s^2 + \omega^2} \). However, the application of the time-shifting property is critiqued, emphasizing the importance of correctly handling the limits of integration in the Laplace Transform.

PREREQUISITES
  • Understanding of Laplace Transforms
  • Familiarity with trigonometric identities
  • Knowledge of the time-shifting property in Laplace Transforms
  • Basic calculus, specifically integration techniques
NEXT STEPS
  • Study the properties of Laplace Transforms, focusing on the time-shifting property
  • Learn about the Heaviside step function and its applications in Laplace Transforms
  • Practice solving Laplace Transforms of trigonometric functions with phase shifts
  • Explore advanced integration techniques for improper integrals in Laplace Transform calculations
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Students studying differential equations, engineers working with control systems, and mathematicians interested in transform methods will benefit from this discussion.

dRic2
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Homework Statement


I have to find the L-transform of ##f(x) = cos(\omega t + \phi)##

Homework Equations


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The Attempt at a Solution


The straightforward approach is to write ##cos(\omega t + \phi)## as ##cos(\omega t)cos(\phi) - sin(\omega t)sin(\phi)## and it becomes: $$Lf(s) = \frac {s cos(\phi) - \omega sin(\phi)} {s^2 + \omega ^2}$$.

But can I try this other way ?
##cos ( \omega t + \phi ) = cos \left[ \omega \left( t - \left( - \frac { \phi} { \omega} \right) \right) \right]## and now I can use the t-shift relation to get: $$ Lf(s) = e^{- \left( - \frac {\phi} {\omega} \right) s} L(cos(\omega t)) = e^{ \frac {\phi} {\omega} s} \frac {s} {s^2 + \omega ^2}$$

I don't know if there is a way to simplify my last result or if it is wrong...
 
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You are applying the time-shifting property incorrectly.

Going back to the definition, <br /> F(s) = \int_0^\infty e^{-st} \cos(\omega t + \phi)\,dt. If I set u = t + \frac{\phi}{\omega} I get <br /> F(s) = e^{s\phi/\omega}\int_{\phi/\omega}^\infty e^{-su} \cos(\omega u)\,du. The expression on the right hand side is not e^{s\phi/\omega} times the laplace transform of \cos(\omega t) because the lower limit of the integral is no longer zero.

If \phi/\omega &gt; 0 I can make the right hand side into e^{s\phi/\omega} times the laplace transform of something by adding 0 = e^{s\phi/\omega}\int_0^{\phi/\omega} 0e^{-su}\,du to both sides, which leaves me with the laplace transform of g(t) = \begin{cases} 0 &amp; 0 \leq t &lt; \frac{\phi}{\omega} \\<br /> \cos(\omega t) &amp; t \geq \frac{\phi}{\omega}\end{cases}
 
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Oh, I see, Thank you! In fact now that I pay attention to it, in the book I'm reading the author takes care of the lower limit of the integral not being zero by writing the t-shift like this:
$$L(f(t-t_0)u(t-t_0)) (s) = e^{-t_0s} L(f)(s)$$
where ##u(t)## is the Heaviside step function. Of course here I can not use this property because I can not shift the "starting point" of my function with such a "trick".
 

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