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# s1 - ECE301 HW1 Solutions 1(a x(t = 2t u(t 2 Real part(see...

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ECE301 HW1 Solutions 1. (a) x ( t ) = 2 t u ( t 2) Real part: (see Fig. 1) Imaginary part is zero. E = lim T →∞ T T | x ( t ) | 2 dt = lim T →∞ T T 2 2 t u ( t 2) dt = lim T →∞ T 2 2 2 t dt = lim T →∞ 1 2 ln 2 2 2 t T 2 = 1 32 ln 2 P = 0 because E is finite. (b) x ( t ) = 2 e j (2 t + π 4 ) Real part: (see Fig. 2) Imaginary part: (see Fig. 3) P = lim T →∞ 1 2 T T T | x ( t ) | 2 dt = lim T →∞ 1 2 T T T 4 e j (2 t + π 4 ) 2 dt = lim T →∞ 1 2 T 4 T T 1 dt = 4 E = because P > 0. 1

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2. (a) For plot, see Fig. 4. P = 1 N N 1 n =0 | x [ n ] | 2 = 1 8 7 n =0 sin 2 ( π 4 n ) = 1 2 E = because P > 0. (b) For plot, see Fig. 5. E = lim N →∞ N n = N | x [ n ] | 2 = lim N →∞ N n = N 1 2 2 n u ( n + 1) = lim N →∞ N n = 1 1 2 2 n = 4 + n =0 1 2 2 n = 4 + 1 1 1 4 = 16 3 P = 0 because E is finite. 3. For plots, see Figs. 6-9. 4. (a) Periodic; T = lcm ( 2 π 3 , 2 π 4 ) = 2 π ; Neither even nor odd. (b) Not periodic; Neither even nor odd. (c) Periodic; N = 8; Neither even nor odd. (d) Not periodic; Neither even nor odd. 5. (1.23) For plots, see Figs. 10-15. 2
0 1 2 3 4 5 6 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 Figure 1: 1.(a) Re { x ( t ) } = 2 t u ( t 2) -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 Figure 2: 1.(b) Re { x ( t ) } = 2 cos(2 t + π 4 ) -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 Figure 3: 1.(b) Im { x ( t ) } = 2 sin(2 t + π 4 ) 3

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-2 0 2 4 6 8 10 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 Figure 4: 2.(a) x

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