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6.003 Signals and Systems DUE: 9/12/2003Bryan RussellProblem Set 1Problem 1(a)(√3 + j)5e−jπ3= (2ejπ6)5e−jπ3= 2ej5π6e−jπ3= 2ejπ2.|2ejπ2| = 2θ =π2See Figure 3Problem 2(a)x1 −t3: Shift left 1, flip, expand by 3.See Figure 4(b)x (t − 2) [δ(t −12) + u(3 − t)]x(t − 2): Shift right 2δ(t −12): Shift right12u(3 − t): Shift left 3, flipSee Figure 5Problem 3(a)x[2 − n]: Shift left 2, flip.See Figure 6(b)x[2n + 1]: Shift left 1, compress by 2.See Figure 7P1-1Problem 4Even part:See Figure 8Odd part:See Figure 9Problem 5(a)x(t) = (sin(4t − 1))2= (x1(t))2We see that x1(t) oscillates equally above and below the x-axis. Since we sqaure x1(t), weflip the negative values above the x-axis. So the period will be cut in half. Now, calculatethe period of x1(t).x1(t + T ) = sin(4(t + T ) − 1) = sin(4t − 1 + 4T )For this to be periodic, we need:4T = 2πkfor k = 0, ±1, ±2, . . .. Using our above analysis that the period will be cut in half, T0=12π2=π4.(b)Let us use the complex exponential:x1[n] = ej(4n+π4).So,x1[n + N] = ej(4(n+N )+π4)= ej4Nej(4n+π4).This is periodic if:ej4N= 1 ⇔ 4N = 2πk ⇔ N =π2kfor k = 1, ±1, ±2, . . .. However, notice that N can never be an integer. So, x[n] cannot beperiodic.(c)Let us use complex exponentials:x1[n] = (−1)nej2πn7= (−1)nx2[n].Let us first look at x2[n]:x2[n + N] = ej2π(n+N )7= ej2πN7ej2πn7.This is periodic if:ej2π(n+N )7= 1 ⇔2πN7= 2πk ⇔ N = 7kfor k = 1, ±1, ±2, . . .. We can easily observe that (−1)nhas period of 2. So, N0= 14.Problem 6(a)1. Memoryless? No—y depends on future and past values of x.P1-22. TI? Nox2(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? Yesx3(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] = δ[n] → y1[n] = δ[n]x2[n] = δ[n + 1] → y2[n] = −δ[n + 1]3. Linear? No. Let:x1[n] = δ[n]x2[n] = (−1)x1[n] = −δ[n] → y2[n] = −2δ[n] 6= (−1)y1[n] = −δ[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].2. TI? Yes. x2[n] = x1[n−n0] → y2[n] =P∞k=nx2[k] =P∞k=nx1[k−n0] =P∞k=n−n0x1[k]y1[n − n0] =P∞k=n−n0x1[k].3. Linear? Yes. x3[n] = ax1[n] + bx2[n] → y3[n] =P∞k=nx3[k] =P∞k=nax1[k] + bx2[k] =aP∞k=nx1[k] + bP∞k=nx2[k] = ay1[n] + by2[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 7x2(t) = −x1(t − 1) − 2x1(t − 2)y2(t) = −y1(t − 1) − 2y1(t − 2)See Figure 10P1-3Problem 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’);P1-4−3 −2 −1 0 1 2 3 4 5 6 7−1−0.500.511.522.53xFigure 1: Figure for part 8(a).P1-5−5 0 5 10−10123y1−5 0 5 10−10123y2−10 −5 0 5−10123y3−10 −5 0 5−10123y4Figure 2: Figure for part


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MIT 6 003 - Problem Set 1

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