Question

Find the Fourier transform of $f(t) = \begin{cases} -(1+t), & -1 \le t \le 0; \\ t-1, & 0 < t \le 1; \\ 0, & |t| > 1. \end{cases}$ Solution: $F(\omega) = \frac{2(\cos \omega - 1)}{\omega^2}$

          Find the Fourier transform of
$f(t) = \begin{cases} -(1+t), & -1 \le t \le 0; \\ t-1, & 0 < t \le 1; \\ 0, & |t| > 1. \end{cases}$
Solution: $F(\omega) = \frac{2(\cos \omega - 1)}{\omega^2}$

Find the Fourier transform of
f(t) = -(1+t), -1 ≤ t ≤ 0;
t-1, 0 < t ≤ 1;
0, |t| > 1.
Solution: F(ω) = (2(cosω - 1))/(ω^2)

Added by Travis B.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Ishana K

05/31/2022

Step 1

Step 1: The Fourier transform of a function f(t) is defined as F(\omega) = \int_{-\infty}^{\infty} f(t)e^{-i\omega t} dt. Show more…

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Find the Fourier transform of f(t)={(-(1+t),-1<=t<=0),(t-1,01):} Solution: F(omega )=(2(cosomega -1))/(omega ^(2)). Find the Fourier transform of -(1+t), -1t0: f(t)=3 t-1, 0<t1; 0. It|>1. 2(cos w - 1) Solution: F(w)= w2