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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
)
=
∫
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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