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hw03su09_soln

# hw03su09_soln - GEORGIA INSTITUTE OF TECHNOLOGY ECE 2025...

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Unformatted text preview: GEORGIA INSTITUTE OF TECHNOLOGY ECE 2025 SOLUTIONS Problem Set # 3 Summer 2009 Problem 3.1 (a) Since f i ( t ) = 1 2 π dΨ( t ) d t , we find in [0 , 1]: 1 Ψ( t ) = Ψ(0) + Z t 2 πτ 2 d τ, All solutions are of the form x ( t ) = cos Ψ(0) + 2 πt 3 3 ¶ , | t | < with x ( t ) = 0 outside the interval. Since, at t = ± 1, we have x = cos ( Ψ(0) ± 2 π 3 ) , we may find a continuous solution if cos ( Ψ(0) ± 2 π 3 ) = 0. This is not possible as we would need simultaneously Ψ(0) = ± 2 π 3 . Hence all solutions lead to discontinuous signals. (b) If f i ( t ) = 3cos(2 πt + π/ 2), then Ψ( t ) = Ψ(0) + 2 π Z t 3cos(2 πτ + π/ 2)d τ == 3sin(2 πt + π/ 2) + Ψ(0) . Hence x ( t ) = cos(3sin(2 πt + π/ 2) + Ψ(0)) . Problem 3.2 (a) Periodic. Fundamental frequency is GCD(16 , 20) π = 4 π rad/sec. Besides a DC compo- nent only the 4-th and 5-th harmonics are present. x =- 3 x 4 = 12 / (2 j )e jπ/ 6 = 6e j ( π/ 6- π/ 2) = 6e- jπ/ 3 1 When denoting integrals as the one shown, it is important to write the infinitesimal d τ . 1 x- 4 =- 12 / (2 j )e- jπ/ 6 = 6e j (- π/ 6+ π/ 2) = 6e jπ/ 3 x 5 =- 6 / (2)e j 5 π/ 8 = 3e j ( π 5 / 8- π ) = 3e- jπ 3 / 8 x- 5 =- 6 / (2)e- j 5 π/ 8 = 3e j (- π 5 / 8+ π ) = 3e jπ 3 / 8 (b) Not periodic. The ratios of the frequencies of the two sinusoidal components is e, which is not a rational number. (c) Periodic. 8cos(- 200 πt + π/ 3)cos(10 πt ) = 4cos(- 200 πt + π/ 3 + 10 πt ) + 4cos(- 200 πt + π/ 3- 10 πt ) = 4cos(- 190 πt + π/ 3) + 4cos(- 210 πt + π/ 3) = 4cos(190 πt- π/ 3) + 4cos(210 πt- π/ 3) Its fundamental frequency is GCD(190 , 210) π = 10 π rad/sec. Only the 19-th and 21-th harmonics are present....
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hw03su09_soln - GEORGIA INSTITUTE OF TECHNOLOGY ECE 2025...

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