Question

The CTFT of a continuous time signal x(t) = e^-A|t|, A>0 is _________

(a) \(\frac{2A}{ω^2} \)

(b) \(\frac{A}{A^2+ω^2} \)

(c) \(\frac{2A}{A^2+ω^2} \)

(d) \(\frac{A}{ω^2} \)

This question was addressed to me in quiz.

I need to ask this question from Fourier Analysis in division Fourier Analysis on Complex Spaces of Signals and Systems

Answer 1

Correct option is (c) \(\frac{2A}{A^2+ω^2} \)

The explanation is: CTFT {x (t)} = X (jω) = \(\int_{-∞}^∞ x(t) e^{-jωt} \,dt\)

= \(\int_{-∞}^∞ e^{-A|t|} e^{-jωt} \,dt\)

= \(\int_{-∞}^0 e^{-A(-t)}  e^{-jωt} \,dt + \int_0^∞ e^{-At} e^{-jωt} \,dt\)

= \(\int_{-∞}^0 e^{At}  e^{-jωt} \,dt + \int_0^∞ e^{-At} e^{-jωt} \,dt\)

= \(\int_{-∞}^0 e^{(A-jω)t} \,dt + ∫_0^∞ e^{-(A+jω)t} \,dt\)

= \([\frac{1}{A-jω} e^{(A-jω)t}]_{-∞}^0 + [\frac{1}{-(A+jω)} e^{-(A+jω)t}]_0^∞\)

= \([\frac{1}{A-jω} +  \frac{1}{A+jω} = \frac{2A}{A^2+ω^2}]\)

∴ x(jω) = \(\frac{2A}{A^2+ω^2} \).