Derive the CT Fourier transform pair

The Fourier transform of an aperiodic signal is obtained by treating the signal as a periodic signal whose period tends to infinity, and then taking the limit of its Fourier series.

Step 1: Periodic extension

Let $x(t)$ be aperiodic and of finite duration, with $x(t) = 0$ for $|t| > T_1$. Build $\tilde{x}(t)$ by repeating $x(t)$ every $T$ seconds, with $T > 2T_1$ so the copies do not overlap. Over $-T/2 < t < T/2$, $\tilde{x}(t) = x(t)$, and as $T \to \infty$, $\tilde{x}(t) \to x(t)$ for every $t$.

 x(t):          ___/\___
 x~(t):  _/\_______/\_______/\_
           |<--T-->|

Step 2: Fourier series of the periodic extension

With $\omega_0 = 2\pi/T$:

$$\tilde{x}(t) = \sum_{k=-\infty}^{\infty}a_k e^{jk\omega_0 t}, \qquad a_k = \frac{1}{T}\int_{-T/2}^{T/2}\tilde{x}(t)e^{-jk\omega_0 t}dt$$

Inside $|t| < T/2$, $\tilde{x}(t) = x(t)$, and $x(t) = 0$ outside this range. So the limits can be extended to infinity:

$$a_k = \frac{1}{T}\int_{-\infty}^{\infty}x(t)e^{-jk\omega_0 t}dt$$

Step 3: Define the envelope X(ω)

$$X(\omega) = \int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt \quad\Rightarrow\quad a_k = \frac{1}{T}X(k\omega_0)$$

The Fourier series coefficients are equally spaced samples of $X(\omega)$, scaled by $1/T$.

Step 4: Substitute in the synthesis equation

Using $\frac{1}{T} = \frac{\omega_0}{2\pi}$:

$$\tilde{x}(t) = \sum_{k=-\infty}^{\infty}\frac{1}{T}X(k\omega_0)e^{jk\omega_0 t} = \frac{1}{2\pi}\sum_{k=-\infty}^{\infty}X(k\omega_0)e^{jk\omega_0 t}\,\omega_0$$

Step 5: Limit T → ∞

• $\omega_0 = 2\pi/T \to d\omega$ (the spectral lines merge).
• $k\omega_0 \to \omega$, a continuous variable.
• The sum $\sum(\cdot)\omega_0$ becomes the integral $\int(\cdot)d\omega$. This is the area under $X(\omega)e^{j\omega t}$, approximated by rectangles of width $\omega_0$.
• $\tilde{x}(t) \to x(t)$.

 X(w) e^{jwt}
      _|_|_
    _|     |_          area of each strip
  _|  w0    |_   =   X(k w0) e^{jk w0 t} w0
 ---+-+-+-+-+-+--- w

Therefore

$$\boxed{x(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty}X(\omega)e^{j\omega t}d\omega \quad \text{(inverse FT / synthesis)}}$$ $$\boxed{X(\omega) = \int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt \quad \text{(FT / analysis)}}$$

In terms of $f$ (with $\omega = 2\pi f$): $X(f) = \int x(t)e^{-j2\pi ft}dt$ and $x(t) = \int X(f)e^{j2\pi ft}df$.

Remarks

• An aperiodic signal has a continuous spectrum, while a periodic signal has a line spectrum.
• Existence (Dirichlet conditions): $x(t)$ is absolutely integrable ($\int|x(t)|dt < \infty$), and it has a finite number of maxima, minima and finite discontinuities in any finite interval. Signals with finite energy also have a transform, converging in the mean-square sense.
• Example: for $x(t) = 1$ when $|t| < T_1$ (0 otherwise), $X(\omega) = \frac{2\sin\omega T_1}{\omega}$. Its samples $\frac{1}{T}X(k\omega_0) = \frac{\sin k\omega_0 T_1}{k\pi}$ are exactly the coefficients of the periodic square wave.