Fourier transform of a trapezoidal pulse

Signal (from the figure):

$$x(t) = \begin{cases} t, & 0 \le t \le 1 \\ 1, & 1 \le t \le 2 \\ 3 - t, & 2 \le t \le 3 \\ 0, & \text{otherwise} \end{cases}$$

Method: differentiation property. If $x(t) \leftrightarrow X(j\omega)$, then $\frac{d^2x}{dt^2} \leftrightarrow (j\omega)^2X(j\omega)$. Differentiating a piecewise-linear signal twice gives only impulses, whose transforms are easy.

Step 1: first derivative.

$$x'(t) = \begin{cases} 1, & 0 < t < 1 \\ 0, & 1 < t < 2 \\ -1, & 2 < t < 3 \end{cases}$$

Step 2: second derivative (jumps of $x'(t)$ become impulses):

$$x''(t) = \delta(t) - \delta(t-1) - \delta(t-2) + \delta(t-3)$$

 x(t)               x'(t)              x''(t)
 1    ______        1 +--+             ^        ^
     /      \         |  |             |  |  |  |
    /        \     ---+--+--+--+---    +--+--+--+--> t
---+--+--+--+--      0  1  2| |3       0  1  2  3
   0  1  2  3      -1       +-+           v  v

Step 3: transform. Using $\delta(t - t_0) \leftrightarrow e^{-j\omega t_0}$:

$$\begin{aligned} (j\omega)^2X(j\omega) &= 1 - e^{-j\omega} - e^{-j2\omega} + e^{-j3\omega} \\ &= (1 - e^{-j\omega})(1 - e^{-j2\omega}) \end{aligned}$$

Step 4: simplify. $1 - e^{-j\omega} = 2j\sin(\omega/2)\,e^{-j\omega/2}$ and $1 - e^{-j2\omega} = 2j\sin\omega\,e^{-j\omega}$, so

$$\begin{aligned} -\omega^2X(j\omega) &= (2j)^2\sin(\omega/2)\sin\omega\;e^{-j3\omega/2} \\ X(j\omega) &= \frac{4\sin(\omega/2)\,\sin\omega}{\omega^2}\,e^{-j3\omega/2} \end{aligned}$$

Answer:

$$X(j\omega) = \frac{4\sin(\omega/2)\sin(\omega)}{\omega^2}\,e^{-j1.5\omega}$$

Checks:

• At $\omega \to 0$: $\frac{4(\omega/2)(\omega)}{\omega^2} = 2$, which equals the area of the trapezoid $\frac{(3+1)}{2}\times1 = 2$.
• The trapezoid is symmetric about $t = 1.5$, so the phase is linear, $-1.5\omega$ (plus $\pi$ jumps where the real factor changes sign).
• Alternative: the trapezoid is the convolution of a unit pulse on $(0,1)$ with a unit pulse on $(0,2)$. Their transforms $\frac{2\sin(\omega/2)}{\omega}e^{-j\omega/2}$ and $\frac{2\sin\omega}{\omega}e^{-j\omega}$ multiply to the same answer.
• Zeros of $|X|$: $\omega = \pm\pi, \pm2\pi, \dots$ (from $\sin\omega$) and $\omega = \pm2\pi, \pm4\pi,\dots$ (from $\sin(\omega/2)$).