Fourier Transform of x(t) = ae^(bt)*u(-t)

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SUMMARY

The discussion focuses on calculating the Fourier Transform of the function x(t) = ae^(bt)*u(-t). The Fourier Transform is derived as F[x(t)] = a/(b-jw), where the integral is evaluated from -infinity to 0. The unit step function u(-t) is clarified as a time-reversed function, which does not directly affect the Fourier Transform's result in this context. The calculation is confirmed to be correct, emphasizing the importance of understanding the role of the unit step function in signal processing.

PREREQUISITES
  • Understanding of Fourier Transform principles
  • Familiarity with the unit step function u(t)
  • Knowledge of complex exponentials in signal processing
  • Basic calculus for evaluating integrals
NEXT STEPS
  • Study the properties of the Fourier Transform in relation to unit step functions
  • Learn about the implications of time-reversal in signal processing
  • Explore applications of Fourier Transform in analyzing signals
  • Investigate advanced topics such as Laplace Transform and its relationship with Fourier Transform
USEFUL FOR

This discussion is beneficial for students and professionals in electrical engineering, signal processing, and applied mathematics, particularly those working with Fourier analysis and time-domain signals.

Larrytsai
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Hey I am trying to figure out this easy problem, and I don't know if I am doing this properly or not here's the questions & the work.

x(t) = ae^(bt)*u(-t)

F[x(t)] = a*integral[(e^bt)*e^(-jwt)*dt] upper bound = 0 lower bound = -infinity
= [a*e^(t(b-jw))] / (b-jw)
= a/(b-jw)
 
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What's u(-t), and why is it not part of the Fourier transform? Otherwise, your work is correct.
 
u(-t) is the unit step function with a time reversal.
 

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