--- title: "Fourier transform of a trapezoidal pulse" subject: "Signals and Systems" chapter: 3 key: "semester_4/signals_and_systems:ft-trapezoidal-signal" source: https://www.bctengineer.com/q/semester_4/signals_and_systems/ft-trapezoidal-signal site: BCT Engineering --- # Fourier transform of a trapezoidal pulse Signals and Systems · Chapter 3: Fourier Transform - Asked in: 2070 Magh, 8 marks - Video: [Fourier transform - trapezoidal signal - signals and systems](https://www.youtube.com/watch?v=Dxq25tw36fQ) — faculty 4u - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/ft-trapezoidal-signal - Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-3 ## Question Find the Fourier transform of a trapezoidal signal shown below: [Figure: trapezoid x(t) rising linearly from 0 at t = 0 to 1 at t = 1, constant at 1 from t = 1 to t = 2, falling linearly to 0 at t = 3; zero elsewhere] ## Short answer Differentiate twice: x''(t)=δ(t)−δ(t−1)−δ(t−2)+δ(t−3). Then (jω)^2 X(jω)=1−e^{-jω}−e^{-j2ω}+e^{-j3ω}, giving X(jω) = 4 sin(ω/2) sin ω e^{-j1.5ω} / ω^2. ## Answer **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): $$ x''(t) = \delta(t) - \delta(t-1) - \delta(t-2) + \delta(t-3) $$ ```text 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 $\sin(\omega/2)$). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the main idea of the method? A. Integrate twice B. Use frequency shifting C. Use Parseval's theorem D. Differentiate twice to get impulses, then use (jω)^2 X(jω) Answer: D. Piecewise-linear signals give only impulses after two derivatives. (Key point: Method) 2. What is x''(t)? A. δ(t) - δ(t-1) - δ(t-2) + δ(t-3) B. δ(t) + δ(t-1) + δ(t-2) + δ(t-3) C. δ(t) - δ(t-3) D. u(t) - u(t-1) - u(t-2) + u(t-3) Answer: A. Slope changes are +1 at 0, -1 at 1, -1 at 2, +1 at 3. (Key point: Second derivative) 3. What is X(jω)? A. 4 sin(ω/2) sin(ω)/ω^2 · e^(-j3ω) B. 4 sin(ω/2) sin(ω)/ω^2 · e^(-j1.5ω) C. 2 sin(ω/2) sin(ω)/ω · e^(-j1.5ω) D. 4 sin(ω/2) sin(ω)/ω^2 Answer: B. The delta sum factors as (1-e^(-jω))(1-e^(-j2ω)); dividing by -ω^2 gives the result with centre delay 1.5. (Key point: Final answer) 4. What is X(0) and why? A. 1 B. 3 C. 2, the area of the trapezoid D. 0 Answer: C. Area = (3+1)/2 × 1 = 2, matching the limit of the formula. 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