Fourier Transform of Cosine Function with Step Function Constraints

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SUMMARY

The discussion focuses on the Fourier Transform of a cosine function constrained by a step function. The function g(t) is defined as g(t) = cos(t) for 0 ≤ t < π and g(t) = 0 for t ≥ π. The transformation is expressed as F(s)e^{-as}, leading to the conclusion that g(t) can be represented as g(t) = cos(t)u(t) - cos(t)u(t-π), resulting in the Fourier Transform F(s) = \frac{s}{s^2-1} + \frac{s}{s^2-1}e^{-π}. The application of the trigonometric identity cos(t-π) = -cos(t) is also discussed, confirming the correctness of the transformation.

PREREQUISITES
  • Understanding of Fourier Transform principles
  • Familiarity with step functions and unit step functions (u(t))
  • Knowledge of trigonometric identities, particularly cos(t-π)
  • Basic calculus and integration techniques
NEXT STEPS
  • Study the properties of Fourier Transforms of piecewise functions
  • Explore the application of the Heaviside step function in signal processing
  • Learn about the implications of phase shifts in Fourier Transforms
  • Investigate the use of trigonometric identities in simplifying Fourier Transform calculations
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Mathematicians, engineers, and students in signal processing or applied mathematics who are interested in Fourier analysis and its applications to piecewise functions.

electron2
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[tex] f(t-a)u_a(t)->F(s)e^{-as}[/tex]
[tex] g(t)={\cos (t),0<=t<\pi}[/tex]
[tex] g(t)={0,t>=\pi}[/tex]

[tex] g(t)=\cos(t)[ u(t)-u(t-\pi)]=\cos(t) u(t)-\cos(t) u(t-\pi)=\frac{s}{s^2-1}+??\\[/tex]

i can't apply the formula in the last part
 
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Trig formula [itex]cos(t-\pi)= -cos(t)[/itex].
 
[tex]g(t)=\cos(t)[ u(t)-u(t-\pi)]=\cos(t) u(t)+\cos(t-\pi) u(t-\pi)=\frac{s}{s^2-1}+\frac{s}{s^2-1}e^{-\pi}\\[/tex]

is it correct?
 

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