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2021年春季学期-信号与系统-第三次作业参考答案-第四道题

系统 2021 信号 作业 春季 参考答案 学期 第三次
2023-09-11 14:15:21 时间


本文是 2021年春季学期-信号与系统-第三次作业参考答案 中的参考答案。

 

▌第四道题

4. 求下列系统的单位冲激响应：

1)第一小题

2)第二小题

注：按照经典求解微分方程的方法进行求解，只是在确定0+时刻的起始条件的时候，需要使用到奇异函数匹配方法。

求解：

（1）第一小题

d 2 y ( t ) d t 2 + 8 d y ( t ) d t + 12 y ( t ) = 2 d x ( t ) d t {{d^2 y\left( t \right)} \over {dt^2 }} + 8{{dy\left( t \right)} \over {dt}} + 12y\left( t \right) = 2{{dx\left( t \right)} \over {dt}} dt2d2y(t)​+8dtdy(t)​+12y(t)=2dtdx(t)​

特征方程： λ 2 + 8 λ + 12 = 0 \lambda ^2 + 8\lambda + 12 = 0 λ2+8λ+12=0。特征根： λ 1 = − 2 ,    λ 2 = − 6 \lambda _1 = - 2,\,\,\lambda _2 = - 6 λ1​=−2,λ2​=−6。

系统的齐次解： y ( t ) = c 1 e − 2 t + c 2 e − 6 t y\left( t \right) = c_1 e^{ - 2t} + c_2 e^{ - 6t} y(t)=c1​e−2t+c2​e−6t

通过奇异函数匹配方法确定系统的初始条件。

当输入为 δ ( t ) \delta \left( t \right) δ(t)，可以知道方程右边最高的奇异函数导数为 δ ′ ( t ) \delta '\left( t \right) δ′(t)。近而可以确定方程左边最高导数想的奇异函数一般表达式为：
d 2 d t y ( t ) = a ⋅ δ ′ ( t ) + b ⋅ δ ( t ) + c ⋅ u ( t ) {{d^2 } \over {dt}}y\left( t \right) = a \cdot \delta '\left( t \right) + b \cdot \delta \left( t \right) + c \cdot u\left( t \right) dtd2​y(t)=a⋅δ′(t)+b⋅δ(t)+c⋅u(t)
近而： d d t y ( t ) = a ⋅ δ ( t ) + b ⋅ u ( t ) {d \over {dt}}y\left( t \right) = a \cdot \delta \left( t \right) + b \cdot u\left( t \right) dtd​y(t)=a⋅δ(t)+b⋅u(t)

y ( t ) = a ⋅ u ( t ) y\left( t \right) = a \cdot u\left( t \right) y(t)=a⋅u(t)

因此：

a ⋅ δ ′ ( t ) + b ⋅ δ ( t ) + c ⋅ u ( t ) + 8 ⋅ [ a ⋅ δ ( t ) + b ⋅ u ( t ) ] + 12 ⋅ a ⋅ u ( t ) = 2 ⋅ δ ′ ( t ) a \cdot \delta '\left( t \right) + b \cdot \delta \left( t \right) + c \cdot u\left( t \right) + 8 \cdot \left[ {a \cdot \delta \left( t \right) + b \cdot u\left( t \right)} \right] + 12 \cdot a \cdot u\left( t \right) = 2 \cdot \delta '\left( t \right) a⋅δ′(t)+b⋅δ(t)+c⋅u(t)+8⋅[a⋅δ(t)+b⋅u(t)]+12⋅a⋅u(t)=2⋅δ′(t)

a ⋅ δ ′ ( t ) + ( b + 8 a ) δ ( t ) + ( c + 8 b + 12 a ) ⋅ u ( t ) = 2 δ ′ ( t ) a \cdot \delta '\left( t \right) + \left( {b + 8a} \right)\delta \left( t \right) + \left( {c + 8b + 12a} \right) \cdot u\left( t \right) = 2\delta '\left( t \right) a⋅δ′(t)+(b+8a)δ(t)+(c+8b+12a)⋅u(t)=2δ′(t)
{ a = 2 b + 8 a = 0 c + 8 b + 12 a = 0 \left\{ \begin{matrix} {a = 2}\\{b + 8a = 0}\\{c + 8b + 12a = 0}\\\end{matrix} \right. ⎩⎨⎧​a=2b+8a=0c+8b+12a=0​

所以： { a = 2 b = − 16 c = 104 \left\{ \begin{matrix} {a = 2}\\{b = - 16}\\{c = 104}\\\end{matrix} \right. ⎩⎨⎧​a=2b=−16c=104​

由此可以得到：
y ′ ( 0 + ) − y ′ ( 0 − ) = b = − 16 y'\left( {0_ + } \right) - y'\left( {0_ - } \right) = b = - 16 y′(0+​)−y′(0−​)=b=−16 y ( 0 + ) − y ( 0 − ) = a = 2 y\left( {0_ + } \right) - y\left( {0_ - } \right) = a = 2 y(0+​)−y(0−​)=a=2

系统的起始条件：
y ′ ( 0 + ) = − 16 ,       y ( 0 + ) = 2 y'\left( {0_ + } \right) = - 16,\,\,\,\,\,y\left( {0_ + } \right) = 2 y′(0+​)=−16,y(0+​)=2

求完全解的待定系数：
{ c 1 + c 2 = 2 − 2 c 1 − 6 c 2 = − 16 \left\{ \begin{matrix} {c_1 + c_2 = 2}\\{ - 2c_1 - 6c_2 = - 16}\\\end{matrix} \right. {c1​+c2​=2−2c1​−6c2​=−16​

{ c 1 = − 1 c 2 = 3 \left\{ \begin{matrix} {c_1 = - 1}\\{c_2 = 3}\\\end{matrix} \right. {c1​=−1c2​=3​

系统的单位冲击响应：

y ( t ) = ( − e − 2 t + 3 e − 6 t ) ⋅ u ( t ) y\left( t \right) = \left( { - e^{ - 2t} + 3e^{ - 6t} } \right) \cdot u\left( t \right) y(t)=(−e−2t+3e−6t)⋅u(t)

（2）第二小题

d 2 d t 2 y ( t ) + 5 d d t y ( t ) + 4 y ( t ) = d d t x ( t ) + 2 x ( t ) {{d^2 } \over {dt^2 }}y\left( t \right) + 5{d \over {dt}}y\left( t \right) + 4y\left( t \right) = {d \over {dt}}x\left( t \right) + 2x\left( t \right) dt2d2​y(t)+5dtd​y(t)+4y(t)=dtd​x(t)+2x(t)

特征方程： λ 2 + 5 λ + 4 = 0 \lambda ^2 + 5\lambda + 4 = 0 λ2+5λ+4=0，求得特征根： λ 1 = − 1 ,    λ 2 = − 4 \lambda _1 = - 1,\,\,\lambda _2 = - 4 λ1​=−1,λ2​=−4。系统的齐次解： y ( t ) = c 1 e − t + c 2 e − 4 t y\left( t \right) = c_1 e^{ - t} + c_2 e^{ - 4t} y(t)=c1​e−t+c2​e−4t

由输入为， δ ( t ) \delta \left( t \right) δ(t)，可以知道方程左边最高导数项中的奇异函数为：

d 2 d t 2 y ( t ) = a ⋅ δ ′ ( t ) + b ⋅ δ ( t ) + c ⋅ u ( t ) {{d^2 } \over {dt^2 }}y\left( t \right) = a \cdot \delta '\left( t \right) + b \cdot \delta \left( t \right) + c \cdot u\left( t \right) dt2d2​y(t)=a⋅δ′(t)+b⋅δ(t)+c⋅u(t)

d d t y ( t ) = a ⋅ δ ( t ) + b ⋅ u ( t ) {d \over {dt}}y\left( t \right) = a \cdot \delta \left( t \right) + b \cdot u\left( t \right) dtd​y(t)=a⋅δ(t)+b⋅u(t)

y ( t ) = a ⋅ u ( t ) y\left( t \right) = a \cdot u\left( t \right) y(t)=a⋅u(t)

因此：
a ⋅ δ ′ ( t ) + b ⋅ δ ( t ) + c ⋅ u ( t ) + 5 ⋅ [ a ⋅ δ ( t ) + b ⋅ u ( t ) ] + ⋅ a ⋅ u ( t ) = δ ′ ( t ) + 2 δ ( t ) a \cdot \delta '\left( t \right) + b \cdot \delta \left( t \right) + c \cdot u\left( t \right) + 5 \cdot \left[ {a \cdot \delta \left( t \right) + b \cdot u\left( t \right)} \right] + \cdot a \cdot u\left( t \right) = \delta '\left( t \right) + 2\delta \left( t \right) a⋅δ′(t)+b⋅δ(t)+c⋅u(t)+5⋅[a⋅δ(t)+b⋅u(t)]+⋅a⋅u(t)=δ′(t)+2δ(t)

a ⋅ δ ′ ( t ) + ( b + 5 a ) δ ( t ) + ( c + 5 b + 4 a ) ⋅ u ( t ) = δ ′ ( t ) + 2 δ ( t ) a \cdot \delta '\left( t \right) + \left( {b + 5a} \right)\delta \left( t \right) + \left( {c + 5b + 4a} \right) \cdot u\left( t \right) = \delta '\left( t \right) + 2\delta \left( t \right) a⋅δ′(t)+(b+5a)δ(t)+(c+5b+4a)⋅u(t)=δ′(t)+2δ(t)

{ a = 1 b + 5 a = 2 c + 5 b + 4 a = 0 \left\{ \begin{matrix} {a = 1}\\{b + 5a = 2}\\{c + 5b + 4a = 0}\\\end{matrix} \right. ⎩⎨⎧​a=1b+5a=2c+5b+4a=0​

{ a = 1 b = − 3 c = 11 \left\{ \begin{matrix} {a = 1}\\{b = - 3}\\{c = 11}\\\end{matrix} \right. ⎩⎨⎧​a=1b=−3c=11​

y ′ ( 0 + ) − y ′ ( 0 − ) = b = − 3 y'\left( {0_ + } \right) - y'\left( {0_ - } \right) = b = - 3 y′(0+​)−y′(0−​)=b=−3 y ( 0 + ) − y ( 0 − ) = a = 1 y\left( {0_ + } \right) - y\left( {0_ - } \right) = a = 1 y(0+​)−y(0−​)=a=1

系统的起始条件：
y ′ ( 0 + ) = − 3 ,       y ( 0 + ) = 1 y'\left( {0_ + } \right) = - 3,\,\,\,\,\,y\left( {0_ + } \right) = 1 y′(0+​)=−3,y(0+​)=1

求完全解中的待定系数：
{ c 1 + c 2 = 1 − c 1 − 4 c 2 = − 3 \left\{ \begin{matrix} {c_1 + c_2 = 1}\\{ - c_1 - 4c_2 = - 3}\\\end{matrix} \right. {c1​+c2​=1−c1​−4c2​=−3​

{ c 1 = 1 3 c 2 = 2 3 \left\{ \begin{matrix} {c_1 = {1 \over 3}}\\{c_2 = {2 \over 3}}\\\end{matrix} \right. {c1​=31​c2​=32​​

系统的单位冲击响应：

y ( t ) = ( 1 3 e − t + 2 3 e − 4 t ) ⋅ u ( t ) y\left( t \right) = \left( {{1 \over 3}e^{ - t} + {2 \over 3}e^{ - 4t} } \right) \cdot u\left( t \right) y(t)=(31​e−t+32​e−4t)⋅u(t)

 

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