Amiinezrr

(a) x 1(t) = e 2tu(t) (b) x 2(t) = ej(2tˇ=4) (c) x 3(t) = cos(t) (d) x 1n = (1 2) nun (e) x 2n = ej( ˇ=2n 8) (f) x 3n = cos(ˇ 4 n) Solution (a) E 1= R 1 0 e 2tdt= 1 4 P 1= 0, because E 14 You can visualize these using an Argand diagram, which is just a plot of imaginary part vs real part of a complex number For example, z= 3 j4 = 5ej0927 is plotted at rectangular coordinates (3,4) and polar coordinates (5,0927), where 0927 is the angle in radians measured counterclockwise from the positive real@b/(/c ª k z c 6 Ñ @ * ;1/(/ g i a ) ) i = è 6 Ì Å ì Ç _ ª d # ² j ' & = k ¦ ^ * Ø _ c æ _ * ä q ¯1 q # j í = â b = ³ Ì Å g ) ¥ z c ) ( h k k 4 k & n k ä i Ø _ q j j * Ç À ) á a ' j ' & = ³ Ì Å h i Ê # " ç i ` * Ë " à ¼ k 6 x 6 l q j ­ M Qy O Y V Lw I Cu 7 T Y T7 Z ƒEƒ†ƒj‰–ŒÎ •ÇŽ† ƒXƒ}ƒz