FT of periodic signals; e^(−2|t|)sin t

Fourier transform of a CT periodic signal

A periodic signal is not absolutely integrable, so its Fourier transform is built from impulses.

Step 1: consider the spectrum $X(\omega) = 2\pi\delta(\omega - k\omega_0)$. Its inverse transform is

$$x(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty}2\pi\delta(\omega - k\omega_0)e^{j\omega t}d\omega = e^{jk\omega_0 t}$$

So

$$e^{jk\omega_0 t} \xleftrightarrow{F} 2\pi\delta(\omega - k\omega_0)$$

Step 2: a periodic signal with period $T$ ($\omega_0 = 2\pi/T$) has the Fourier series

$$x(t) = \sum_{k=-\infty}^{\infty}a_k e^{jk\omega_0 t}, \qquad a_k = \frac{1}{T}\int_T x(t)e^{-jk\omega_0 t}dt$$

Step 3: apply linearity term by term:

$$\boxed{X(\omega) = \sum_{k=-\infty}^{\infty}2\pi\,a_k\,\delta(\omega - k\omega_0)}$$

The FT of a periodic signal is a train of impulses at the harmonic frequencies $k\omega_0$. The area of each impulse is $2\pi$ times the Fourier series coefficient.

 X(w)
          2pi a0
   2pi a-1  ^   2pi a1
      ^     |     ^
  ^   |     |     |   ^
 -+---+-----+-----+---+-- w
 -2w0 -w0   0     w0  2w0

Examples:

• $\cos\omega_0 t \leftrightarrow \pi[\delta(\omega - \omega_0) + \delta(\omega + \omega_0)]$
• $\sum_n\delta(t - nT) \leftrightarrow \frac{2\pi}{T}\sum_k\delta\!\left(\omega - \frac{2\pi k}{T}\right)$

FT of x(t) = e^{−2|t|} sin(t)

This signal is aperiodic (a decaying envelope times a sine). Use a known pair and the frequency-shift property.

Known pair: with $a = 2$,

$$e^{-a|t|} \leftrightarrow \frac{2a}{a^2 + \omega^2} \quad\Rightarrow\quad e^{-2|t|} \leftrightarrow G(\omega) = \frac{4}{4 + \omega^2}$$

Euler's formula: $\sin t = \frac{e^{jt} - e^{-jt}}{2j}$, so

$$x(t) = \frac{1}{2j}\left[e^{-2|t|}e^{jt} - e^{-2|t|}e^{-jt}\right]$$

Frequency shift ($e^{j\omega_0 t}g(t) \leftrightarrow G(\omega - \omega_0)$, here $\omega_0 = 1$):

$$X(\omega) = \frac{1}{2j}\left[G(\omega - 1) - G(\omega + 1)\right] = \frac{1}{2j}\left[\frac{4}{4 + (\omega - 1)^2} - \frac{4}{4 + (\omega + 1)^2}\right]$$

Simplify: over a common denominator, the numerator is

$$4\left[(4 + (\omega + 1)^2) - (4 + (\omega - 1)^2)\right] = 4\cdot 4\omega = 16\omega$$

So

$$X(\omega) = \frac{16\omega}{2j\,[4 + (\omega - 1)^2][4 + (\omega + 1)^2]} = \frac{-j\,8\omega}{[4 + (\omega - 1)^2][4 + (\omega + 1)^2]}$$

Expanding the denominator, $(\omega^2 - 2\omega + 5)(\omega^2 + 2\omega + 5) = \omega^4 + 6\omega^2 + 25$:

$$\boxed{X(\omega) = \frac{-j\,8\omega}{\omega^4 + 6\omega^2 + 25}}$$

Checks:

• $x(t)$ is real and odd (an even envelope times an odd sine), so $X(\omega)$ must be purely imaginary and odd. It is.
• $X(0) = 0$, which agrees with $\int x(t)dt = 0$ for an odd signal.
• A numerical integration at $\omega = 0.7$ gives $X = -j0.1987$, matching the formula: $-j\frac{5.6}{0.2401 + 2.94 + 25} = -j0.1987$.

Spectrum: $|X(\omega)| = \frac{8|\omega|}{\omega^4 + 6\omega^2 + 25}$, which is zero at $\omega = 0$, peaks where $3\omega^4 + 6\omega^2 - 25 = 0$, i.e. at $\omega = \pm1.434$ rad/s (where $|X| \approx 0.276$), and decays as $8/|\omega|^3$.