信号与系统第一章作业

第一章习题

14

\[\begin{aligned} \frac{ \ {\rm d} x(t)}{ \ {\rm d} t} & = 3 \delta(t - 2k) + (-3) \delta (t - 2k + 1) \\ & = 3 g(t) - 3 g(t - 1) \end{aligned} \\ \therefore \begin{cases} A_1 = 3, \\ A_2 = -3, \\ t_1 = 0, \\ t_2 = 1 \end{cases}\]

15

(a)

\[\begin{aligned} S: y[n] & = y_2[n] \\ & = y_1[n - 2] + \frac{1}{2} y_1[n - 3] \\ & = (2 x[n - 2] + 4 x[n - 3]) + \frac{1}{2} (2x[n - 3] + 4x[n - 4]) \\ & = 2 x[n - 2] + 5 x[n - 3] + 2 x[n - 4] \end{aligned}\]

(b)

\[\begin{aligned} S: y[n] & = y_1[n] \\ & = 2 y_2[n] + 4 y_2[n - 1] \\ & = 2 (x[n - 2] + \frac{1}{2} x[n - 3]) + 4 (x[n - 3] + \frac{1}{2} x[n - 4]) \\ & = 2 x[n - 2] + 5 x[n - 3] + 2 x[n - 4] \end{aligned}\]

没有变化

20

(a)

已知

\[1： \begin{cases} x_a(t) = e^{i2t} = {\rm cos} \ (2t) + i \ {\rm sin} \ (2t) \\ y_a(t) = e^{i3t} = {\rm cos} \ (3t) + i \ {\rm sin} \ (3t) \end{cases}\] \[2： \begin{cases} x_b(t) = e^{-i2t} = {\rm cos} \ (-2t) + i \ {\rm sin} \ (-2t) \\ y_b(t) = e^{-i3t} = {\rm cos} \ (-3t) + i \ {\rm sin} \ (-3t) \end{cases}\]

可以得到

\[\begin{cases} x_1(t) = {\rm cos} \ (2t) = (e^{i2t} + e^{-i2t}) / 2 \\ y_1(t) = (e^{i3t} + e^{-i3t}) / 2 = {\rm cos} \ (3t) \end{cases}\]

(b)

已知

\[\begin{aligned} x_2(t) & = {\rm cos} \ (2 (t - \frac{1}{2})) \\ & = {\rm cos} \ (2t - 1) \\ & = {\rm cos} \ (2t) \cdot {\rm cos} \ 1 + {\rm sin} \ (2t) \cdot {\rm sin} \ 1 \\ & = x_1(t) \cdot {\rm cos} \ 1 + (- i \ x_a(t) + i \ x_b(t)) / 2 \cdot {\rm sin} 1 \end{aligned}\]

所以

\[\begin{aligned} y_2(t) & = y_1(t) \cdot {\rm cos} \ 1 + (- i \ y_a(t) + i \ y_b(t)) / 2 \cdot {\rm sin} \ 1 \\ & = {\rm cos} \ (3t) \cdot {\rm cos} \ 1 + {\rm sin} \ (3t) \cdot {\rm sin} 1 \\ & = {\rm cos} \ (3t - 1) \end{aligned}\]

25

(a) 是，$N = \frac{\pi}{2}$

(b) 不是

26

(b) 不是

(d) 是，$N = 8$

31