fourier transform of sin


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β(t) = { 1 if t < 1 0 if t > 1 2 sinc(ω)=2 sin(ω) ω Boxcar in time (6) 1 π sinc(t) β(ω) Boxcar 

  • Answer and Explanation:
    a) The Fourier sine series expansion of a function on the interval is: f ( x ) = ∑ n = 1 ∞ b n sin ⁡ n π x L , where b n = 2 L ∫ 0 L f ( x ) sin ⁡ Here.

  • What is the equation for the Fourier sine transform?

    (4) (a) f(x) = [ B(w) sin wx dw, where (b) B(w) = 00 2 [ f(v) sin wv dv. f(x) sin wx dx.
    This is called the Fourier sine transform of f(x).

  • Is the Fourier transform of sine imaginary?

    This makes more sense when you remember that sin(-θ) = -sin(θ), and cos(-θ) = cos(θ).
    The second piece that should jump out is that the Fourier transform of the sine function is completely imaginary, while the cosine function only has real parts.

  • the answer is: πi[δ(ω−2πA)−δ(ω+2πA)].
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