6 003 Signals and Systems DUE 9 12 2003 Bryan Russell Problem Set 1 Problem 1 a 5 3 j 5 e j 3 2ej 6 5 e j 3 2ej 6 e j 3 2ej 2 2ej 2 2 2 See Figure 3 Problem 2 a x 1 3t Shift left 1 flip expand by 3 See Figure 4 b x t 2 t 12 u 3 t x t 2 Shift right 2 t 12 Shift right 21 u 3 t Shift left 3 flip See Figure 5 Problem 3 a x 2 n Shift left 2 flip See Figure 6 b x 2n 1 Shift left 1 compress by 2 See Figure 7 P1 1 Problem 4 Even part See Figure 8 Odd part See Figure 9 Problem 5 a x t sin 4t 1 2 x1 t 2 We see that x1 t oscillates equally above and below the x axis Since we sqaure x1 t we flip the negative values above the x axis So the period will be cut in half Now calculate the period of x1 t x1 t T sin 4 t T 1 sin 4t 1 4T For this to be periodic we need 4T 2 k for k 0 1 2 Using our above analysis that the period will be cut in half T 0 1 2 2 4 b Let us use the complex exponential x1 n ej 4n 4 So x1 n N ej 4 n N 4 ej4N ej 4n 4 This is periodic if ej4N 1 4N 2 k N 2 k for k 1 1 2 However notice that N can never be an integer So x n cannot be periodic c Let us use complex exponentials 2 n x1 n 1 n ej 7 1 n x2 n Let us first look at x2 n 2 n N 2 N 2 n x2 n N ej 7 e j 7 ej 7 This is periodic if 2 n N ej 7 1 2 N 7 2 k N 7k for k 1 1 2 We can easily observe that 1 n has period of 2 So N0 14 Problem 6 a 1 Memoryless No y depends on future and past values of x P1 2 2 TI No x2 t x1 t t0 y2 t x2 t 3 x2 1 t x1 t 3 t0 x1 1 t t0 y1 t t0 x1 t t0 3 x1 1 t t0 3 Linear Yes x3 t ax1 t bx2 t y3 t x3 t 3 x3 1 t ax1 t 3 bx2 t 3 ax1 1 t bx2 1 t a x1 t 3 x1 1 t b x2 t 3 x2 1 t ay1 t by2 t 4 Causal No For t 1 y 1 depends on x 2 5 Stable Yes Suppose x t B t Then y t 2B b 1 Memoryless Yes since y depends only on current values of x 2 TI No Let x1 n n y1 n n x2 n n 1 y2 n n 1 3 Linear No Let x1 n n x2 n 1 x1 n n y2 n 2 n 6 1 y1 n n 4 Causal Yes since it is memoryless 5 Stable Yes If x n B y n 2B c 1 Memoryless No since y n depends on future values of x n P P P 2 TI Yes x2 n P x1 n n0 y2 n k n n0 x1 k k n x1 k n0 k n x2 k x k y1 n n0 1 k n n0 P P 3 Linear ax1 n bx2 n y3 n k n ax1 k bx2 k k n x3 k P P Yes x3 n x k ay n by n a k n x1 k b 2 1 2 k n 4 Causal No see memoryless answer 5 Stable No because we sum all future samples of x which imples that y is unbounded Problem 7 x2 t x1 t 1 2x1 t 2 y2 t y1 t 1 2y1 t 2 See Figure 10 P1 3 Problem 8 a nx 3 7 x 0 0 0 2 0 1 1 3 0 0 0 figure 1 stem nx x title x b y1 x y2 x y3 x y4 x ny1 1 9 ny2 4 6 ny3 3 1 7 ny4 4 1 6 c y1 Delayed by 2 y2 Advanced by 1 y3 Flipped y4 Flipped and then delayed by 1 figure 2 subplot 2 2 1 stem ny1 y1 title y1 subplot 2 2 2 stem ny2 y2 title y2 subplot 2 2 3 stem ny3 y3 title y3 subplot 2 2 4 stem ny4 y4 title y4 P1 4 x 3 2 5 2 1 5 1 0 5 0 0 5 1 3 2 1 0 1 2 3 Figure 1 Figure for part 8 a P1 5 4 5 6 7 y1 y2 3 3 2 2 1 1 0 0 1 5 0 5 1 5 10 0 y3 3 2 2 1 1 0 0 5 10 0 5 y4 3 1 10 5 0 1 10 5 Figure 2 Figure for part 8 c P1 6 5
View Full Document
Unlocking...