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Unformatted text preview: 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 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 t ) = A cos(2 πf t − π/ 2) . Applying cos( θ ) = 1 2 ( e jθ + e- jθ ) we have v ( t ) = A 2 e- jπ/ 2 e j 2 πf t + A 2 e jπ/ 2 e j 2 π (- f ) 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 ) + A 2 e jπ/ 2 δ ( f + f ) . 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 jθ − e- jθ ) . 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 t − A 2 j e j 2 π (- f ) t = A 2 1 j e j 2...
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This note was uploaded on 11/30/2011 for the course EEE 1001 taught by Professor Phoon during the Spring '11 term at National University of Singapore.
- Spring '11