Fourier transform of a CT signal

The signal was not printed on the paper, so a common exam signal is assumed here: the two-sided exponential $x(t) = e^{-a|t|}$, $a > 0$. The same method applies to any other signal.

Definition used:

$$X(j\omega) = \int_{-\infty}^{\infty} x(t)\,e^{-j\omega t}\,dt$$

Step 1: split the signal. $e^{-a|t|} = e^{at}$ for $t<0$ and $e^{-at}$ for $t>0$:

$$X(j\omega) = \int_{-\infty}^{0} e^{at}e^{-j\omega t}dt + \int_{0}^{\infty} e^{-at}e^{-j\omega t}dt$$

Step 2: integrate each part.

$$\begin{aligned} \int_{-\infty}^{0} e^{(a-j\omega)t}dt &= \left[\frac{e^{(a-j\omega)t}}{a-j\omega}\right]_{-\infty}^{0} = \frac{1}{a-j\omega} \\ \int_{0}^{\infty} e^{-(a+j\omega)t}dt &= \left[\frac{e^{-(a+j\omega)t}}{-(a+j\omega)}\right]_{0}^{\infty} = \frac{1}{a+j\omega} \end{aligned}$$

Both limits at $\pm\infty$ go to zero because $a>0$.

Step 3: add.

$$X(j\omega) = \frac{1}{a-j\omega} + \frac{1}{a+j\omega} = \frac{2a}{a^2+\omega^2}$$

Result:

• $X(j\omega)$ is real, positive and even, because $x(t)$ is real and even. So the phase is $0$ for all $\omega$.
• Peak value $X(0) = \frac{2}{a}$ (equal to the area under $x(t)$); it falls to half at $\omega = \pm a$.
• A larger $a$ (faster decay in time) gives a wider spectrum: time and frequency widths are inversely related.

  X(jw)
 2/a ^
     |    ..
     |  .    .
 1/a |.--------.   (half value at w = +/- a)
    .|          .
 ..  |            ..
-----+-------------------> w
    -a   0    a

If the question paper shows a different signal, use the same steps: write $x(t)$ piece by piece, apply the definition (or known pairs and properties such as shifting and scaling), and simplify.