本文是 2021年春季学期-信号与系统-第四次作业参考答案 的内容。

▌第一道题

1. 计算下列卷积积分

（1）第一小题

f(t)=t[u(t)−u(t−2)]∗δ(2−t)f\left( t \right) = t\left[ {u\left( t \right) - u\left( {t - 2} \right)} \right] * \delta \left( {2 - t} \right)f(t)=t[u(t)−u(t−2)]∗δ(2−t)

（2）第二小题

f(t)=u(t)∗e−3tu(t)f\left( t \right) = u\left( t \right) * e^{ - 3t} u\left( t \right)f(t)=u(t)∗e−3tu(t)

（3）第三小题

f(t)=[(t+2)u(t+2)−2t⋅u(t)+(t−2)⋅u(t−2)]∗[δ′(t+2)−δ′(t−2)]f\left( t \right) = \left[ {\left( {t + 2} \right)u\left( {t + 2} \right) - 2t \cdot u\left( t \right) + \left( {t - 2} \right) \cdot u\left( {t - 2} \right)} \right] * \left[ {\delta '\left( {t + 2} \right) - \delta '\left( {t - 2} \right)} \right]f(t)=[(t+2)u(t+2)−2t⋅u(t)+(t−2)⋅u(t−2)]∗[δ′(t+2)−δ′(t−2)]

（4）第四小题

f(t)=δ(t−1)∗δ(2−t)f\left( t \right) = \delta \left( {t - 1} \right) * \delta \left( {2 - t} \right)f(t)=δ(t−1)∗δ(2−t)

注：很抱歉，这个题目（1,3,4）需要用到奇异函数的卷积性质。可以直接根据卷积定义+delta(t)的性质求解。也可以参考课件2.4中的相关性质。

▌求解

第一小题

f(t)=t[u(t)−u(t−2)]∗δ(2−t)f\left( t \right) = t\left[ {u\left( t \right) - u\left( {t - 2} \right)} \right] * \delta \left( {2 - t} \right)f(t)=t[u(t)−u(t−2)]∗δ(2−t)

求解：

根据δ(t)\delta \left( t \right)δ(t)以下两个性质：

偶对称： δ(t)=δ(−t)\delta \left( t \right) = \delta \left( { - t} \right)δ(t)=δ(−t)
卷积特性：f(t)∗δ(t−t0)=f(t−t0)f\left( t \right) * \delta \left( {t - t_0 } \right) = f\left( {t - t_0 } \right)f(t)∗δ(t−t0)=f(t−t0)

可以进行如下化简：

f(t)=t[u(t)−u(t−2)]∗δ(2−t)f\left( t \right) = t\left[ {u\left( t \right) - u\left( {t - 2} \right)} \right] * \delta \left( {2 - t} \right)f(t)=t[u(t)−u(t−2)]∗δ(2−t)=t[u(t)−u(t−2)]∗δ(t−2)= t\left[ {u\left( t \right) - u\left( {t - 2} \right)} \right] * \delta \left( {t - 2} \right)=t[u(t)−u(t−2)]∗δ(t−2)=(t−2)⋅[u(t−2)−u(t−4)]= \left( {t - 2} \right) \cdot \left[ {u\left( {t - 2} \right) - u\left( {t - 4} \right)} \right]=(t−2)⋅[u(t−2)−u(t−4)]

信号的波形：

^{▲ 信号f(t)的波形}

第二小题

f(t)=u(t)∗e−3tu(t)f\left( t \right) = u\left( t \right) * e^{ - 3t} u\left( t \right)f(t)=u(t)∗e−3tu(t)

求解：

根据u(t)u\left( t \right)u(t)卷积特性：u(t)∗f(t)=∫−∞tf(τ)dτu\left( t \right) * f\left( t \right) = \int_{ - \infty }^t {f\left( \tau \right)d\tau }u(t)∗f(t)=∫−∞tf(τ)dτ。所以：

f(t)=u(t)∗e−3t⋅u(t)=∫0te−3τdτ=−13e−3t∣0t=13(1−e−3t)f\left( t \right) = u\left( t \right) * e^{ - 3t} \cdot u\left( t \right) = \int_0^t {e^{ - 3\tau } d\tau } = \left. {{{ - 1} \over 3}e^{ - 3t} } \right|_0^t = {1 \over 3}\left( {1 - e^{ - 3t} } \right)f(t)=u(t)∗e−3t⋅u(t)=∫0te−3τdτ=3−1e−3t∣∣∣∣0t=31(1−e−3t)

信号的波形：

^{▲ 信号f(t)的波形}

第三小题

求解：

原来信号可以看成两个信号的卷积：f(t)=f1(t)∗f2(t)f\left( t \right) = f_1 \left( t \right) * f_2 \left( t \right)f(t)=f1(t)∗f2(t)：

f1(t)=[(t+2)u(t+2)−2t⋅u(t)+(t−2)⋅u(t−2)]f_1 \left( t \right) = \left[ {\left( {t + 2} \right)u\left( {t + 2} \right) - 2t \cdot u\left( t \right) + \left( {t - 2} \right) \cdot u\left( {t - 2} \right)} \right]f1(t)=[(t+2)u(t+2)−2t⋅u(t)+(t−2)⋅u(t−2)]
f2(t)=[δ′(t+2)−δ′(t−2)]f_2 \left( t \right) = \left[ {\delta '\left( {t + 2} \right) - \delta '\left( {t - 2} \right)} \right]f2(t)=[δ′(t+2)−δ′(t−2)]

f1(t)f_1 \left( t \right)f1(t)可以分解成两端组成：
f1(t)=(t+2)⋅u(t+2)−[(t+2)+(t−2)]⋅u(t)+(t−2)⋅u(t−2)f_1 \left( t \right) = \left( {t + 2} \right) \cdot u\left( {t + 2} \right) - \left[ {\left( {t + 2} \right) + \left( {t - 2} \right)} \right] \cdot u\left( t \right) + \left( {t - 2} \right) \cdot u\left( {t - 2} \right)f1(t)=(t+2)⋅u(t+2)−[(t+2)+(t−2)]⋅u(t)+(t−2)⋅u(t−2)=(t+2)⋅[u(t+2)−u(t)]−(t−2)⋅[u(t)−u(t−2)]= \left( {t + 2} \right) \cdot \left[ {u\left( {t + 2} \right) - u\left( t \right)} \right] - \left( {t - 2} \right) \cdot \left[ {u\left( t \right) - u\left( {t - 2} \right)} \right]=(t+2)⋅[u(t+2)−u(t)]−(t−2)⋅[u(t)−u(t−2)]

它实际上是一个中心位于0点的对称等腰三角形：

^{▲ f1(t)的波形}

它的导数为：

f1′(t)=[u(t+2)−u(t)]−[u(t)−u(t−2)]f_1 '\left( t \right) = \left[ {u\left( {t + 2} \right) - u\left( t \right)} \right] - \left[ {u\left( t \right) - u\left( {t - 2} \right)} \right]f1′(t)=[u(t+2)−u(t)]−[u(t)−u(t−2)]=u(t+2)−2u(t)+u(t−2)= u\left( {t + 2} \right) - 2u\left( t \right) + u\left( {t - 2} \right)=u(t+2)−2u(t)+u(t−2)

^{▲ f1'(t)的波形}

根据

δ′(t)\delta '\left( t \right)δ′(t)的卷积特性：f(t)∗δ′(t)=f′(t)f\left( t \right) * \delta '\left( t \right) = f'\left( t \right)f(t)∗δ′(t)=f′(t)，
卷积延迟特性：f(t)∗δ′(t−t0)=f′(t−t0)f\left( t \right) * \delta '\left( {t - t_0 } \right) = f'\left( {t - t_0 } \right)f(t)∗δ′(t−t0)=f′(t−t0)

所以计算f1(t)∗f2(t)f_1 \left( t \right) * f_2 \left( t \right)f1(t)∗f2(t)的结果就等于：
f(t)=f1′(t+2)−f1′(t−2)f\left( t \right) = f_1 '\left( {t + 2} \right) - f_1 '\left( {t - 2} \right)f(t)=f1′(t+2)−f1′(t−2)=u(t+4)−2u(t+2)+2u(t)−2u(t−2)+u(t−4)= u\left( {t + 4} \right) - 2u\left( {t + 2} \right) + 2u\left( t \right) - 2u\left( {t - 2} \right) + u\left( {t - 4} \right)=u(t+4)−2u(t+2)+2u(t)−2u(t−2)+u(t−4)

^{▲ f(t)的波形}

同学提问:

老师~我化完多出来后面那一串东西（如下）：

回复：你的推导是对的。针对你推导的表达式后面的几项，你根据下面表达式，看看是否它们最终都是0？
t⋅δ(t)=0⋅δ(t)=0t \cdot \delta \left( t \right) = 0 \cdot \delta \left( t \right) = 0t⋅δ(t)=0⋅δ(t)=0

喔喔喔，是不是说delta（t）只在t＝0的时候有取值，而这个时候t＝0，因此t*deltat就等于0？

对的。

第四小题

f(t)=δ(t−1)∗δ(2−t)f\left( t \right) = \delta \left( {t - 1} \right) * \delta \left( {2 - t} \right)f(t)=δ(t−1)∗δ(2−t)

求解：

根据δ(t)\delta \left( t \right)δ(t)的对称性以及卷积延时特性可知：

f(t)=δ(t−1)∗δ(2−t)=δ(t−3)f\left( t \right) = \delta \left( {t - 1} \right) * \delta \left( {2 - t} \right) = \delta \left( {t - 3} \right)f(t)=δ(t−1)∗δ(2−t)=δ(t−3)