2019-02-25T17:21:58-05:00
Determine the Fourier transform of the function
f(t)=1-t. 0 ≤t≤1
1+t. -1≤t≤0
0. Otherwise
1
2019-03-02T11:12:16-0500
f ^ ( k ) = ∫ R f ( t ) e − 2 π i k t d t \hat{f}(k)=\int\limits_{\mathbb{R}}f(t)e^{-2\pi i kt} \, dt f ^ ( k ) = R ∫ f ( t ) e − 2 πik t d t
= ∫ − 1 0 ( 1 + t ) e − 2 π i k t d t + ∫ 0 1 ( 1 − t ) e − 2 π i k t d t =\int\limits_{-1}^0 (1+t) e^{-2\pi i kt} \, dt+\int\limits_0^1 (1-t) e^{-2\pi i kt} \, dt = − 1 ∫ 0 ( 1 + t ) e − 2 πik t d t + 0 ∫ 1 ( 1 − t ) e − 2 πik t d t
= 2 π i k − e 2 π i k + 1 4 π 2 k 2 + 2 π i k + e 2 π i k − 1 4 π 2 k 2 =\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} = 4 π 2 k 2 2 πik − e 2 πik + 1 + 4 π 2 k 2 2 πik + e 2 πik − 1
= sin 2 ( π k ) π 2 k 2 =\dfrac{\sin^2(\pi k)}{\pi^2 k^2} = π 2 k 2 sin 2 ( πk )
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