Given the Fourier transform (t) → sinc(f), the inverse Fourier transform of X(f) = 400sinc(10f)e-jf is: 01. x(t) = 40n(-0.5) 10 x(t) = 4π(10(t - 0.5)) O 2. O 3. None of the answers O 4. t x(t) = 40m(- 1 10 5. x(t) = 4(10t - 0.5)

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Given the Fourier transform (t) → sinc(f), the inverse Fourier transform of
X(f) = 400sinc(10f)e-jf is:
○ 1.
O 2.
O 3. None of the answers
x(t) = 40m(-0.5)
x(t) = 4(10(t - 0.5))
O 4.
1
x(t) = 40n(-
t-5
10
○ 5. x(t) = 4π(10t - 0.5)
Transcribed Image Text:Given the Fourier transform (t) → sinc(f), the inverse Fourier transform of X(f) = 400sinc(10f)e-jf is: ○ 1. O 2. O 3. None of the answers x(t) = 40m(-0.5) x(t) = 4(10(t - 0.5)) O 4. 1 x(t) = 40n(- t-5 10 ○ 5. x(t) = 4π(10t - 0.5)
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