HW11f09_soln

# HW11f09_soln - SOLUTIONS TO ECE 2025 FALL 2009 PROBLEM...

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1 S OLUTIONS TO ECE 2025 F ALL 2009 P ROBLEM S ET #11 PROBLEM 11.1*: Approach: We will not use the Fourier transform integral. Instead, we will use the table of Fourier transform pairs (Table 11-2) and the table of Fourier transform properties (Table 11-3). (a) From the trig identity for cos x we have x ( t ) = 3.5 + 3.5cos(26 π t ). Combining the linearity property (first row of Table 11-3) and the Fourier transform pairs for the constant signal 1 and the sinusoidal signal cos( ω 0 t ) (rows 8 and 11 of Table 11-2), we get: X ( j ω ) = 7 πδ ( ω ) + 3.5 πδ ( ω 26 π ) + 3.5 πδ ( ω + 26 π ). (b) We can write x ( t ) = y ( t 3) , where y ( t ) = δ ( t ) 2 w ( t ) , and where w ( t ) = e t u ( t ) . The Fourier transform of w ( t ) is given in the first row of Table 11-2. Therefore, from the linearity property we have Y ( j ω ) = 1 = . Finally, since x ( t ) = y ( t 3), from the delay property we have: X ( j ω ) = e 3 j ω Y ( j ω ) = e 3 j ω . (c) We can write x ( t ) = cos( ω 0 t ) y ( t ) , where ω 0 = 25 π /3, and y ( t ) = w ( t 9), where w ( t ) is a rectangular pulse of unit height and width T = 4, centered at time zero. From row 3 of Table 11-2 we have W ( j ω ) = 2sin(2 ω )/ ω . Therefore, since y ( t ) = w ( t 9), the delay property yields: Y ( j ω ) = e 9 j ω W ( j ω ) = 2 e 9 j ω. Finally, from the modulation property we have: X ( j ω ) = Y ( j ( ω ω 0 )) + Y ( j ( ω + ω 0 )) = e 9 j ( ω 25 π /3) + e 9 j ( ω +25 π /3) = e 9 j ω + . 2 5 j ω + --------------- 3 j ω + 5 j ω + 3 j ω + 5 j ω + 2 ω () sin ω -------------------- 1 2 -- 1 2 2 ω 25 π /3 sin ω 25 π ----------------------------------------------- 2 ω 25 π + sin ω 25 π + 2 ω 25 π sin ω 25 π ----------------------------------------------- 2 ω 25 π + sin ω 25 π +

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HW11f09_soln - SOLUTIONS TO ECE 2025 FALL 2009 PROBLEM...

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