Tutorial1_solution - -sin πt(1 We know as a Fourier...

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Solutions to EE3008 Tutorial 1 Problems Problem 1: Using the time scaling property, we have s ( - t ) S ( - f ) . Then, using the time shift property, we have s [ - ( t - 1)] e - j 2 πf S ( - f ) and s [ - ( t + 1)] e j 2 πf S ( - f ) . Therefore, using the linearity property, we obtain s (1 - t ) + s ( - 1 - t ) e - j 2 πf S ( - f ) + e j 2 πf S ( - f ) = 2 cos(2 πf ) S ( - f ) . Problem 2: (a) Using the formula of the Fourier transform, we have S ( f ) = Z + -∞ s ( t ) e - j 2 πft dt = Z + -∞ + X k =0 α k δ ( t - kT ) e - j 2 πft dt = + X k =0 α k Z + -∞ δ ( t - kT ) e - j 2 πft dt = + X k =0 α k e - j 2 πfkT = + X k =0 αe - j 2 πfT k = 1 1 - αe - j 2 πfT . (b) Using the trigonometric identities sin 2 θ = 1 - cos(2 θ ) 2 and sin α cos β = 1 2 [sin( α - β ) + sin( α + β )] 1
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Unformatted text preview: )-sin( πt )] . (1) We know as a Fourier transform pair that cos(2 πf t ) ↔ 1 2 [ δ ( f-f ) + δ ( f + f )] and sin(2 πf t ) ↔ 1 2 j [ δ ( f-f )-δ ( f + f )] . We also know as a Fourier transform pair that A ↔ Aδ ( f ) . Thus, the Fourier transform of s ( t ) in (1) can be obtained as S ( f ) = 1 2 δ ( f )-1 4 [ δ ( f-2) + δ ( f + 2)] + 1 4 j ± δ ² f-3 2 ³-δ ² f + 3 2 ³´-1 4 j ± δ ² f-1 2 ³-δ ² f + 1 2 ³´ . 2...
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