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Find the Fourier Cosine transform of the function;๐‘“๐‘ฅ=๐‘˜,ย ย 0<๐‘ฅ<๐‘Ž0,ย ย ๐‘ฅ>๐‘Ž

Question

Find the Fourier Cosine transform of the function;๐‘“๐‘ฅ=๐‘˜,ย ย 0<๐‘ฅ<๐‘Ž0,ย ย ๐‘ฅ>๐‘Ž

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Solution

The Fourier Cosine transform of a function f(x) is given by the formula:

F(ฯ‰) = โˆš(2/ฯ€) โˆซ from 0 to โˆž [f(x) cos(ฯ‰x) dx]

Given the function f(x) = k for 0 < x < a and f(x) = 0 for x > a, we can split the integral into two parts:

F(ฯ‰) = โˆš(2/ฯ€) [ โˆซ from 0 to a [k cos(ฯ‰x) dx] + โˆซ from a to โˆž [0 cos(ฯ‰x) dx] ]

The second integral is zero because the integrand is zero. So we only need to compute the first integral:

F(ฯ‰) = โˆš(2/ฯ€) โˆซ from 0 to a [k cos(ฯ‰x) dx]

This is a standard integral, and its solution is:

F(ฯ‰) = โˆš(2/ฯ€) [ (k/ฯ‰) sin(ฯ‰x) ] from 0 to a

Evaluating this at the limits gives:

F(ฯ‰) = โˆš(2/ฯ€) [ (k/ฯ‰) sin(ฯ‰a) - (k/ฯ‰) sin(0) ]

Since sin(0) = 0, this simplifies to:

F(ฯ‰) = โˆš(2/ฯ€) (k/ฯ‰) sin(ฯ‰a)

So the Fourier Cosine transform of the given function is F(ฯ‰) = โˆš(2/ฯ€) (k/ฯ‰) sin(ฯ‰a).

This problem has been solved

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Find the Fourier sine and cosine transform of f (t) = eโˆ’at, a > 0

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