Show that the Fourier transform of the truncated sine...

Transcript

00:01 In this question we are asked to do so that the fourier transform of this function is equal to this one.

00:07 So let's start to solve this problem.

00:10 We know fourier transform of fx is equal to 1 over root 2 pi integration negative infinity to positive infinity fx e to the power negative z omega x d x.

00:28 Here foray transform of fx comes equal to fomega.

00:32 Equals to 1 over root 2 pi integration negative infinity to negative pi for negative negative infinity to negative pi f x is equal to 0 to the power negative z omega x d x plus 1 over root 2 pi integration negative pi to pi 4 negative pi to pi f x is equal to sign 3x e to the power negative z omega x d x plus 1 over root 2 pi pi to infinity for pi to infinity f x is equal to 0 e to the power negative j omega x d x now from this we obtain f omega is equal to 1 over root 2 pi integration negative pi to pi sine 3x e to the power negative z j omega x d x d here don't confuse between j and iota.

01:31 Here both j and iota are representing same thing.

01:36 That means for z and iota is equal to square root of negative 1.

01:43 Now we solve this integration by using this formula.

01:48 By comparing both these integrations, we obtain a equals to negative z omega and b equals to 3.

01:57 Now by using this formula, the value of this integration, becomes equal 1 over root 2 pi e to the power negative z omega x over negative j omega square plus 3 square negative j omega sine 3x minus b cos 3x lower limit is negative pi and upper limit is pi now this will become equal to 1 over root 2 pi 9 and negative j omega square becomes equal to negative omega square over upper limit this will become equal to e to the power negative j omega pi negative j omega sine 3 pi minus b cause 3 pi minus e to the power z omega pi negative j omega sine negative three pi negative three five minus three here b equals to three three three cause negative three pi now from this we obtain f omega is equal to one over root two pi two pi nine minus omega square a to the power negative j omega pi, we know, sine 3 pi is equal to 0...

Question

Show that the Fourier transform of the truncated sine function f(x) = { sin 3x, -? ? x ? ? 0, x < ??, ? < x is ?6i sin ?? / ?(2?(9 ? ?^2)). [50 marks]

Question

Show that the Fourier transform of the truncated sine function f(x) = { sin 3x, -? ? x ? ? 0, x < ??, ? < x is ?6i sin ?? / ?(2?(9 ? ?^2)). [50 marks]

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