Examveda

The Fourier transform F{e-t u(t)} is equal to $${1 \over {1 + j2\pi f}}.$$   Therefore, $$F\left\{ {{1 \over {1 + j2\pi f}}} \right\}$$   is

A. ef u(f)

B. e-f u(f)

C. ef u(-f)

D. e-f u(-f)

Answer: Option C


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