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EE1003_ClassroomExercise2

# EE1003_ClassroomExercise2 - Classroom Exercise#2 Heres the...

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Classroom Exercise #2 Here’s the exercise that we did in class today. I’ve added an extension to this exercise below as well. Problem: Determine the Fourier transform of v ( t ) = A sin(2 πf 0 t ) . Solution #1: The simplest approach is to express the sin function as a cos function. Recall, sin( θ ) = cos( θ π/ 2) . Thus, v ( t ) = A sin(2 πf 0 t ) = A cos(2 πf 0 t π/ 2) . Applying cos( θ ) = 1 2 ( e + e - ) we have v ( t ) = A 2 e - jπ/ 2 e j 2 πf 0 t + A 2 e jπ/ 2 e j 2 π ( - f 0 ) t . (1) Apply equation (13) in Chap. 1 yields the desired Fourier transform of v ( t ) : V ( f ) = A 2 e - jπ/ 2 δ ( f f 0 ) + A 2 e jπ/ 2 δ ( f + f 0 ) . An alternative approach is to express the given sin function as a sum of complex expo- nentials directly, i.e., we apply the identity sin( θ ) = 1 2 j ( e e - ) . This is described in the second solution below. Solution #2: First, we write v ( t ) as v ( t ) = A 2 j e j 2 πf 0 t A 2 j e j 2 π ( - f 0 ) t = A 2 1 j e j 2 πf 0 t + A 2 1 j e j 2 π ( - f 0 ) t . Now, since j = 1 , we have that 1 j = j j 2 = j = e - jπ/ 2 since e j ( θ π/ 2 ) = cos( θ π/ 2) + j sin( θ

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EE1003_ClassroomExercise2 - Classroom Exercise#2 Heres the...

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