Question #85505

Determine the Fourier transform of the function
f(t)=1-t. 0 ≤t≤1
1+t. -1≤t≤0
0. Otherwise

Expert's answer

f^(k)=Rf(t)e2πiktdt\hat{f}(k)=\int\limits_{\mathbb{R}}f(t)e^{-2\pi i kt} \, dt

=10(1+t)e2πiktdt+01(1t)e2πiktdt=\int\limits_{-1}^0 (1+t) e^{-2\pi i kt} \, dt+\int\limits_0^1 (1-t) e^{-2\pi i kt} \, dt

=2πike2πik+14π2k2+2πik+e2πik14π2k2=\dfrac{2\pi i k-e^{2\pi i k}+1}{4\pi^2 k^2}+\dfrac{2\pi i k+e^{2\pi i k}-1}{4\pi^2 k^2}

=sin2(πk)π2k2=\dfrac{\sin^2(\pi k)}{\pi^2 k^2}


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