Fourier transform of e^(−2|t−1|)

Method: Find the FT of $e^{-2|t|}$, then use the time-shifting property.

Step 1: FT of $x_1(t) = e^{-2|t|}$.

$$\begin{aligned} X_1(j\omega) &= \int_{-\infty}^{0} e^{2t}e^{-j\omega t}dt + \int_{0}^{\infty} e^{-2t}e^{-j\omega t}dt \\ &= \frac{1}{2 - j\omega} + \frac{1}{2 + j\omega} \\ &= \frac{(2+j\omega) + (2-j\omega)}{4 + \omega^2} = \frac{4}{4 + \omega^2} \end{aligned}$$

Step 2: time shift. $x(t) = x_1(t-1)$, and $x_1(t - t_0) \leftrightarrow e^{-j\omega t_0}X_1(j\omega)$ with $t_0 = 1$:

$$X(j\omega) = \frac{4}{4 + \omega^2}\,e^{-j\omega}$$

Answer:

$$e^{-2|t-1|} \;\longleftrightarrow\; \frac{4\,e^{-j\omega}}{4 + \omega^2}$$

• Magnitude: $|X(j\omega)| = \frac{4}{4+\omega^2}$ (same as the unshifted signal; peak 1 at $\omega=0$, equal to the area under $x(t)$).
• Phase: $\angle X(j\omega) = -\omega$ (linear phase due to a delay of 1 s).

 x(t)                |X(jw)|
 1 ^    .            1 ^   .
   |   . .             |  . .
   | .     .           | .   .
 ..+'       '..     ...|'     '...
---+---+---+---> t  ---+---+---+---> w
   0   1   2          -2   0   2