封面动图来自于: SHUTTERSTOCK网站
作业要求链接: 信号与系统 2022 春季学期第三次作业 https://zhuoqing.blog.csdn.net/article/details/123403423
§01 参考答案(1)
1.1 列写系统微分方程
(1)第一小题
求解: 根据题目所示电路图,列写网孔电路方程: 2 i 1 ( t ) + d i 1 ( t ) d t + ∫ − ∞ t i 1 ( τ ) d τ − ∫ − ∞ t i 2 ( τ ) d τ = e ( t ) 2i_1 \left( t \right) + {
{di_1 \left( t \right)} \over {dt}} + \int_{ - \infty }^t {i_1 \left( \tau \right)d\tau } - \int_{ - \infty }^t {i_2 \left( \tau \right)d\tau } = e\left( t \right) 2i1(t)+dtdi1(t)+∫−∞ti1(τ)dτ−∫−∞ti2(τ)dτ=e(t) ∫ − ∞ t [ i 2 ( τ ) − ∫ i 1 ( τ ) ] d τ + i 2 ( t ) = − v 0 ( t ) \int_{ - \infty }^t {\left[ {i_2 \left( \tau \right) - \int_{}^{} {i_1 \left( \tau \right)} } \right]d\tau } + i_2 \left( t \right) = - v_0 \left( t \right) ∫−∞t[i2(τ)−∫i1(τ)]dτ+i2(t)=−v0(t)
又 v 2 ( t ) = 2 d i 2 ( t ) d t v_2 \left( t \right) = 2{
{di_2 \left( t \right)} \over {dt}} v2(t)=2dtdi2(t) 经过校区 i 1 ( t ) , i 2 ( t ) i_1 \left( t \right),i_2 \left( t \right) i1(t),i2(t) 可以得到电路的输入输出方程为 2 d 3 d t 2 v 0 ( t ) + 5 d 2 d t 2 v 0 ( t ) + 5 d d t v 0 ( t ) + 3 v 0 ( t ) = 2 d d t e ( t ) 2{
{d^3 } \over {dt^2 }}v_0 \left( t \right) + 5{
{d^2 } \over {dt^2 }}v_0 \left( t \right) + 5{d \over {dt}}v_0 \left( t \right) + 3v_0 \left( t \right) = 2{d \over {dt}}e\left( t \right) 2dt2d3v0(t)+5dt2d2v0(t)+5dtdv0(t)+3v0(t)=2dtde(t)
(2)第二小题
求解: 这是一个带有互感的电路,根据原边回路典雅与副边回路电压列写回路电压方程:
1 C ∫ − ∞ t i 1 ( τ ) d τ + L d i 1 ( t ) d t + M d i 2 ( t ) d t + R i 1 ( t ) = e ( t ) {1 \over C}\int_{ - \infty }^t {i_1 \left( \tau \right)d\tau } + L{
{di_1 \left( t \right)} \over {dt}} + M{
{di_2 \left( t \right)} \over {dt}} + Ri_1 \left( t \right) = e\left( t \right) C1∫−∞ti1(τ)dτ+Ldtdi1(t)+Mdtdi2(t)+Ri1(t)=e(t) 1 C ∫ − ∞ t i 2 ( τ ) d τ + L d i 2 ( t ) d t + M d i 1 ( t ) d t + R i 2 ( t ) = 0 {1 \over C}\int_{ - \infty }^t {i_2 \left( \tau \right)d\tau } + L{
{di_2 \left( t \right)} \over {dt}} + M{
{di_1 \left( t \right)} \over {dt}} + Ri_2 \left( t \right) = 0 C1∫−∞ti2(τ)dτ+Ldtdi2(t)+Mdtdi1(t)+Ri2(t)=0
输出电压为 − R i 2 ( t ) = v 0 ( t ) - Ri_2 \left( t \right) = v_0 \left( t \right) −Ri2(t)=v0(t) 消去 i 1 ( t ) , i 2 ( t ) i_1 \left( t \right),i_2 \left( t \right) i1(t),i2(t) 可以得到电路输入输出电压方程 ( L 2 − M 2 ) d 4 d t 4 v 0 ( t ) + 2 R L d 3 d t 3 v 0 ( t ) + ( 2 L C + R 2 ) d 2 d t 2 v 0 ( t ) + 2 R C d d t v 0 ( t ) + 1 C 2 v 0 ( t ) = M R d 2 d t 2 e ( t ) \left( {L^2 - M^2 } \right){
{d^4 } \over {dt^4 }}v_0 \left( t \right) + 2RL{
{d^3 } \over {dt^3 }}v_0 \left( t \right) + \left( {
{
{2L} \over C} + R^2 } \right){
{d^2 } \over {dt^2 }}v_0 \left( t \right) + {
{2R} \over C}{d \over {dt}}v_0 \left( t \right) + {1 \over {C^2 }}v_0 \left( t \right) = MR{
{d^2 } \over {dt^2 }}e\left( t \right) (L2−M2)dt4d4v0(t)+2RLdt3d3v0(t)+(C2L+R2)dt2d2v0(t)+C2Rdtdv0(t)+C21v0(t)=MRdt2d2e(t)
(3)第三小题
求解: 设 m 1 m_1 m1的速度为 v 1 ( t ) v_1 \left( t \right) v1(t),火箭 m 1 、 载 荷 仓 m 2 m_1 、载荷仓m_2 m1、载荷仓m2受力情况如下图所示:
根据牛顿第三定律,火箭的质量 m 1 , m 2 m_1 ,m_2 m1,m2之间的作用力 F k F_k Fk, F k ′ F'_k Fk′大小相等,方向相反,大小等于:
k ∫ − ∞ t [ v 1 ( τ ) − v 2 ( τ ) ] d τ k\int_{ - \infty }^t {\left[ {v_1 \left( \tau \right) - v_2 \left( \tau \right)} \right]d\tau } k∫−∞t[v1(τ)−v2(τ)]dτ
对于 m 1 m_1 m1可以建立如下方程:
e ( t ) − f 1 v 1 ( t ) − k ∫ − ∞ t [ v 1 ( τ ) − v 2 ( τ ) ] d τ = m 1 d v 1 ( t ) d t e\left( t \right) - f_1 v_1 \left( t \right) - k\int_{ - \infty }^t {\left[ {v_1 \left( \tau \right) - v_2 \left( \tau \right)} \right]d\tau } = m_1 {
{dv_1 \left( t \right)} \over {dt}} e(t)−f1v1(t)−k∫−∞t[v1(τ)−v2(τ)]dτ=m1dtdv1(t)
对于 m 2 m_2 m2可以建立如下方程:
k ∫ − ∞ t [ v 1 ( τ ) − v 2 ( τ ) ] d τ − f 2 v 2 ( t ) = m 2 d v 2 ( t ) d t k\int_{ - \infty }^t {\left[ {v_1 \left( \tau \right) - v_2 \left( \tau \right)} \right]d\tau } - f_2 v_2 \left( t \right) = m_2 {
{dv_2 \left( t \right)} \over {dt}} k∫−∞t[v1(τ)−v2(τ)]dτ−f2v2(t)=m2dtdv2(t)
由上式可以得到:
v 1 ( t ) = v 2 ( t ) + f 2 k d d t v 2 ( t ) + m 2 k d 2 d t 2 v 2 ( t ) v_1 \left( t \right) = v_2 \left( t \right) + { {f_2 } \over k}{d \over {dt}}v_2 \left( t \right) + { {m_2 } \over k}{ {d^2 } \over {dt^2 }}v_2 \left( t \right) v1(t)=v2(t)+kf2dtdv2(t)+km2dt2d2v2(t)
代入 m 1 m_1 m1对应的方程,化简可得:
d 3 d t 3 v 2 ( t ) + m 1 f 2 + m 2 f 1 m 1 m 2 d 2 d t 2 v 2 ( t ) + f 1 f 2 + k ( m 1 + m 2 ) m 1 m 2 d d t v 2 ( t ) + ( f 1 + f 2 ) k m 1 m 2 v 2 ( t ) = k m 1 m 2 e ( t ) { {d^3 } \over {dt^3 }}v_2 \left( t \right) + { {m_1 f_2 + m_2 f_1 } \over {m_1 m_2 }}{ {d^2 } \over {dt^2 }}v_2 \left( t \right) + { {f_1 f_2 + k\left( {m_1 + m_2 } \right)} \over {m_1 m_2 }}{d \over {dt}}v_2 \left( t \right) + { {\left( {f_1 + f_2 } \right)k} \over {m_1 m_2 }}v_2 \left( t \right) = {k \over {m_1 m_2 }}e\left( t \right) dt3d3v2(t)+m1m2m1f2+m2f1dt2d2v2(t)+m1m2f1f2+k(m1+m2)dtdv2(t)+m1m2(f1+f2)kv2(t)=m1m2ke(t)
1.2 液体混合差分方程
(1)第一小题
(1) 如果y[n]表示第n个循环结束时,A在混合液中所占的比例,则y[n-1]表示在第n-1个循环结束是A在混合液中所占的比例,根据题意可得:
900 ⋅ y [ n − 1 ] + x [ n ] 1000 = y [ n ] { {900 \cdot y\left[ {n - 1} \right] + x\left[ n \right]} \over {1000}} = y\left[ n \right] 1000900⋅y[n−1]+x[n]=y[n]
化简可以得到:
1000 ⋅ y [ n ] − 900 ⋅ y h [ n − 1 ] = x [ n ] 1000 \cdot y\left[ n \right] - 900 \cdot yh\left[ {n - 1} \right] = x\left[ n \right] 1000⋅y[n]−900⋅yh[n−1]=x[n]
(2) 通过求解特征方程,得到特征根为0.9,所以差分方程的齐次解为: y h [ n ] = C ⋅ 0. 9 n y_h \left[ n \right] = C \cdot 0.9^n yh[n]=C⋅0.9n
设特解为: y p [ n ] = D y_p \left[ n \right] = D yp[n]=D。代入差分方程,求得: D = 1 2 D = {1 \over 2} D=21。
因而差分方程的完全解为:
y [ n ] = C ⋅ 0. 9 n + 0.5 y\left[ n \right] = C \cdot 0.9^n + 0.5 y[n]=C⋅0.9n+0.5
将初始条件 y [ 0 ] = 0 y\left[ 0 \right] = 0 y[0]=0代入完全解,求出待定系数C:
C = − 0.5 C = - 0.5 C=−0.5
所以差分方程的完全解为:
y [ n ] = − 0.5 × 0. 9 n + 0.5 y\left[ n \right] = - 0.5 \times 0.9^n + 0.5 y[n]=−0.5×0.9n+0.5
(3) 自由分量: − 0.5 × 0. 9 n - 0.5 \times 0.9^n −0.5×0.9n。强迫响应为: 0.5。
(4) 当 n → + ∞ n \to + \infty n→+∞, y [ ∞ ] = 0.5 y\left[ \infty \right] = 0.5 y[∞]=0.5。
由于每次导入的都是A,B各占50%的混合液,因此不管原先容器内的900升的混合液是怎样的,经过无限次的倒入、混匀、倒出的过程,A所占的比例最终趋近于50%。
1.3 微分、差分方程求解
1.3.1 必做题
(1)微分方程求解
Ⅰ.第一小问
求解:
- 求齐次解:
系统的特征方程为: λ 2 + 7 λ + 10 = 0 \lambda ^2 + 7\lambda + 10 = 0 λ2+7λ+10=0;
特征根: λ 1 = − 2 ; λ 2 = − 5 \lambda _1 = - 2;\,\,\,\,\lambda _2 = - 5 λ1=−2;λ2=−5;
系统齐次解: y h ( t ) = c 1 e − 2 t + c 2 e − 5 t y_h \left( t \right) = c_1 e^{ - 2t} + c_2 e^{ - 5t} yh(t)=c1e−2t+c2e−5t
- 求特解:
假设系统的特解: y s ( t ) = B y_s \left( t \right) = B ys(t)=B。代入方程,可得: 10 ⋅ B = 1 10 \cdot B = 1 10⋅B=1,因此: B = 0.1 B = 0.1 B=0.1。
- 完全解:
y ( t ) = y h ( t ) + y s ( t ) = c 1 e − 2 t + c 2 e − 5 t + 0.1 y\left( t \right) = y_h \left( t \right) + y_s \left( t \right) = c_1 e^{ - 2t} + c_2 e^{ - 5t} + 0.1 y(t)=yh(t)+ys(t)=c1e−2t+c2e−5t+0.1
利用奇异函数方法求系统的起始条件( 0 + 0_ + 0+时刻的状态):
d 2 d t 2 r ( t ) = a ⋅ δ ( t ) + b ⋅ u ( t ) {
{d^2 } \over {dt^2 }}r\left( t \right) = a \cdot \delta \left( t \right) + b \cdot u\left( t \right) dt2d2r(t)=a⋅δ(t)+b⋅u(t) d d t r ( t ) = a ⋅ u ( t ) {d \over {dt}}r\left( t \right) = a \cdot u\left( t \right) dtdr(t)=a⋅u(t) r ( t ) = 0 r\left( t \right) = 0 r(t)=0
代入方程左边: a ⋅ δ ( t ) + b ⋅ u ( t ) + 7 a ⋅ u ( t ) = δ ( t ) + u ( t ) a \cdot \delta \left( t \right) + b \cdot u\left( t \right) + 7a \cdot u\left( t \right) = \delta \left( t \right) + u\left( t \right) a⋅δ(t)+b⋅u(t)+7a⋅u(t)=δ(t)+u(t)
通过奇异函数系数匹配法可以得到:
{ a = 1 b + 7 a = 1 \left\{ \begin{matrix} {a = 1}\\{b + 7a = 1}\\\end{matrix} \right. {
a=1b+7a=1
{ a = 1 b = − 6 \left\{ \begin{matrix} {a = 1}\\{b = - 6}\\\end{matrix} \right. { a=1b=−6
r ′ ( 0 + ) − r ′ ( 0 − ) = a = 1 r'\left( {0_ + } \right) - r'\left( {0_ - } \right) = a = 1 r′(0+)−r′(0−)=a=1
r ( 0 + ) − r ( 0 − ) = 0 r\left( {0_ + } \right) - r\left( {0_ - } \right) = 0 r(0+)−r(0−)=0
所以,系统的起始条件为:
r ′ ( 0 + ) = 4 , r ( 0 + ) = 2 r'\left( {0_ + } \right) = 4,\,\,\,\,r\left( {0_ + } \right) = 2 r′(0+)=4,r(0+)=2
代入系统完全解,求解待定系数:
{ − 2 c 1 − 5 c 2 = 4 c 1 + c 2 + 0.1 = 2 \left\{ \begin{matrix} { - 2c_1 - 5c_2 = 4}\\{c_1 + c_2 + 0.1 = 2}\\\end{matrix} \right. {
−2c1−5c2=4c1+c2+0.1=2
{ c 1 = 4.5 c 2 = − 2.6 \left\{ \begin{matrix} {c_1 = 4.5}\\{c_2 = - 2.6}\\\end{matrix} \right. {
c1=4.5c2=−2.6
系统的完全响应为:
y ( t ) = ( 4.5 e − 2 t − 2.6 e − 5 t + 0.1 ) ⋅ u ( t ) y\left( t \right) = \left( {4.5e^{ - 2t} - 2.6e^{ - 5t} + 0.1} \right) \cdot u\left( t \right) y(t)=(4.5e−2t−2.6e−5t+0.1)⋅u(t)
Ⅱ.第二小问
- 零状态相应、零输入响应
当系统为零状态的时候,对应的起始条件为:
r ′ ( 0 + ) = 1 , r ( 0 + ) = 0 r'\left( {0_ + } \right) = 1,\,\,\,\,r\left( {0_ + } \right) = 0 r′(0+)=1,r(0+)=0
{ − 2 c 1 − 5 c 2 = 1 c 1 + c 2 + 0.1 = 0 \left\{ \begin{matrix} { - 2c_1 - 5c_2 = 1}\\{c_1 + c_2 + 0.1 = 0}\\\end{matrix} \right. { −2c1−5c2=1c1+c2+0.1=0
{ c 1 = 1 6 c 2 = − 4 15 \left\{ \begin{matrix} {c_1 = {1 \over 6}}\\{c_2 = - {4 \over {15}}}\\\end{matrix} \right. { c1=61c2=−154
零状态响应为:
y z s ( t ) = 1 6 e − 2 t − 4 15 e − 5 t + 0.1 , t ≥ 0 y_{zs} \left( t \right) = {1 \over 6}e^{ - 2t} - {4 \over {15}}e^{ - 5t} + 0.1,\,\,\,\,t \ge 0 yzs(t)=61e−2t−154e−5t+0.1,t≥0
零输入响应为:
y z i ( t ) = y ( t ) − y z s ( t ) = 4 1 3 e − 2 t − 2 1 3 e − 5 t , t ≥ 0 y_{zi} \left( t \right) = y\left( t \right) - y_{zs} \left( t \right) = 4{1 \over 3}e^{ - 2t} - 2{1 \over 3}e^{ - 5t} ,\,\,\,\,t \ge 0 yzi(t)=y(t)−yzs(t)=431e−2t−231e−5t,t≥0
- 系统自由响应与强迫响应
自由响应为:
y h ( t ) = 4.5 e − 2 t − 2.6 e − 5 t , t ≥ 0 y_h \left( t \right) = 4.5e^{ - 2t} - 2.6e^{ - 5t} ,\,\,\,\,t \ge 0 yh(t)=4.5e−2t−2.6e−5t,t≥0
强迫响应为:
y f ( t ) = 0.1 , t ≥ 0 y_f \left( t \right) = 0.1,\,\,\,\,t \ge 0 yf(t)=0.1,t≥0
(2)差分方程求解
Ⅰ.第一小题
y [ n ] − 1 4 y [ n − 1 ] = ( − 1 ) n u [ n ] y\left[ n \right] - {1 \over 4}y\left[ {n - 1} \right] = \left( { - 1} \right)^n u\left[ n \right] y[n]−41y[n−1]=(−1)nu[n]
特征方程: λ − 1 4 = 0 \lambda - {1 \over 4} = 0 λ−41=0,求得特征根: λ 1 = 1 4 \lambda _1 = {1 \over 4} λ1=41。系统的齐次解: y h [ n ] = c 1 ( 1 4 ) n , n ≥ 0 y_h \left[ n \right] = c_1 \left( { {1 \over 4}} \right)^n ,\,\,n \ge 0 yh[n]=c1(41)n,n≥0
系统的特解: y s [ n ] = B ⋅ ( − 1 ) n y_s \left[ n \right] = B \cdot \left( { - 1} \right)^n ys[n]=B⋅(−1)n。代入方程左右: B ⋅ ( − 1 ) n − 1 4 ( − 1 ) ⋅ ( − 1 ) n = ( − 1 ) n B \cdot \left( { - 1} \right)^n - {1 \over 4}\left( { - 1} \right) \cdot \left( { - 1} \right)^n = \left( { - 1} \right)^n B⋅(−1)n−41(−1)⋅(−1)n=(−1)n
B ⋅ ( − 1 ) n − 1 4 ( − 1 ) ⋅ B ⋅ ( − 1 ) n = ( − 1 ) n B \cdot \left( { - 1} \right)^n - {1 \over 4}\left( { - 1} \right) \cdot B \cdot \left( { - 1} \right)^n = \left( { - 1} \right)^n B⋅(−1)n−41(−1)⋅B⋅(−1)n=(−1)n
所以: 5 4 ⋅ B = 1 {5 \over 4} \cdot B = 1 45⋅B=1,求得: B = 4 5 B = {4 \over 5} B=54。
系统的完全解: y [ n ] = c 1 ( 1 4 ) n + 4 5 ( − 1 ) n y\left[ n \right] = c_1 \left( { {1 \over 4}} \right)^n + {4 \over 5}\left( { - 1} \right)^n y[n]=c1(41)n+54(−1)n
有系统为零状态,根据差分方程可以计算出系统的初始条件为: y [ 0 ] = 1 y\left[ 0 \right] = 1 y[0]=1。求解完全解中的待定系数:, c 1 + 4 5 = 1 c_1 + {4 \over 5} = 1 c1+54=1,所以 c 1 = 1 5 c_1 = {1 \over 5} c1=51。
系统的零状态解为:
y [ n ] = 1 5 ( 1 4 ) n + 4 5 ( − 1 ) n , n ≥ 0 y\left[ n \right] = {1 \over 5}\left( {
{1 \over 4}} \right)^n + {4 \over 5}\left( { - 1} \right)^n ,\,\,\,\,n \ge 0 y[n]=51(41)n+54(−1)n,n≥0
Ⅱ.第二小题
y [ n ] − 1 4 y [ n − 1 ] = cos ( π 2 n ) u [ n ] y\begin{bmatrix} n \end{bmatrix} - {1 \over 4}y\begin{bmatrix} {n - 1} \end{bmatrix} = \cos \left( { {\pi \over 2}n} \right)u\begin{bmatrix} n \end{bmatrix} y[n]−41y[n−1]=cos(2πn)u[n]
由特征方程: λ − 1 4 = 0 \lambda - {1 \over 4} = 0 λ−41=0,求的特征根: λ 1 = 1 4 \lambda _1 = {1 \over 4} λ1=41。系统的齐次解: y h [ n ] = c 1 ( 1 4 ) n , n ≥ 0 y_h \left[ n \right] = c_1 \left( { {1 \over 4}} \right)^n ,\,\,\,\,n \ge 0 yh[n]=c1(41)n,n≥0
系统的特解: y s ( t ) = a ⋅ sin ( π ⋅ n 2 ) + b ⋅ cos ( π ⋅ n 2 ) y_s \left( t \right) = a \cdot \sin \left( { { {\pi \cdot n} \over 2}} \right) + b \cdot \cos \left( { { {\pi \cdot n} \over 2}} \right) ys(t)=a⋅sin(2π⋅n)+b⋅cos(2π⋅n)
代入系统方程左边: a ⋅ sin ( π ⋅ n 2 ) + b ⋅ cos ( π ⋅ n 2 ) − 1 4 [ a ⋅ sin ( π ⋅ n 2 − π 2 ) + b ⋅ cos ( π ⋅ n 2 − π 2 ) ] a \cdot \sin \left( { { {\pi \cdot n} \over 2}} \right) + b \cdot \cos \left( { { {\pi \cdot n} \over 2}} \right) - {1 \over 4}\left[ {a \cdot \sin \left( { { {\pi \cdot n} \over 2} - {\pi \over 2}} \right) + b \cdot \cos \left( { { {\pi \cdot n} \over 2} - {\pi \over 2}} \right)} \right] a⋅sin(2π⋅n)+b⋅cos(2π⋅n)−41[a⋅sin(2π⋅n−2π)+b⋅cos(2π⋅n−2π)] = ( a − 1 4 b ) sin ( π ⋅ n 2 ) + ( 1 4 a + b ) cos ( π ⋅ n 2 ) = cos ( π ⋅ n 2 ) = \left( {a - {1 \over 4}b} \right)\sin \left( { { {\pi \cdot n} \over 2}} \right) + \left( { {1 \over 4}a + b} \right)\cos \left( { { {\pi \cdot n} \over 2}} \right) = \cos \left( { { {\pi \cdot n} \over 2}} \right) =(a−41b)sin(2π⋅n)+(41a+b)cos(2π⋅n)=cos(2π⋅n)
{ a − 1 4 b = 0 1 4 a + b = 1 \left\{ \begin{matrix} {a - {1 \over 4}b = 0}\\{ {1 \over 4}a + b = 1}\\\end{matrix} \right. { a−41b=041a+b=1
{ a = 4 17 b = 16 17 \left\{ \begin{matrix} {a = {4 \over {17}}}\\{b = { {16} \over {17}}}\\\end{matrix} \right. { a=174b=1716
系统的の完全解: y [ n ] = c 1 ( 1 4 ) n + 4 17 sin ( π ⋅ n 2 ) + 16 17 cos ( π ⋅ n 2 ) y\left[ n \right] = c_1 \left( { {1 \over 4}} \right)^n + {4 \over {17}}\sin \left( { { {\pi \cdot n} \over 2}} \right) + { {16} \over {17}}\cos \left( { { {\pi \cdot n} \over 2}} \right) y[n]=c1(41)n+174sin(2π⋅n)+1716cos(2π⋅n)
由系统为零状态,根据差分方程可以计算出系统的初始条件为: y [ 0 ] = 1 y\left[ 0 \right] = 1 y[0]=1。求解完全解中的待定系数:
c 1 + 16 17 = 1 c_1 + { {16} \over {17}} = 1 c1+1716=1
所以: c 1 = 1 17 c_1 = {1 \over {17}} c1=171。
系统的零状态解为:
y [ n ] = 1 17 ( 1 4 ) n + 4 17 sin ( π ⋅ n 2 ) + 16 17 cos ( π ⋅ n 2 ) , n ≥ 0 y\left[ n \right] = {1 \over {17}}\left( {
{1 \over 4}} \right)^n + {4 \over {17}}\sin \left( {
{
{\pi \cdot n} \over 2}} \right) + {
{16} \over {17}}\cos \left( {
{
{\pi \cdot n} \over 2}} \right),\,\,\,\,n \ge 0 y[n]=171(41)n+174sin(2π⋅n)+1716cos(2π⋅n),n≥0
Ⅲ.第三小题
特征方程为: α + 5 = 0 \alpha + 5 = 0 α+5=0
求得特征根: α = − 5 \alpha = - 5 α=−5
于是齐次解为: y h [ n ] = C ⋅ ( − 5 ) n y_h \left[ n \right] = C \cdot \left( { - 5} \right)^n yh[n]=C⋅(−5)n
由于输入信号是 n n n,所以令特解为: y p [ n ] = D 1 n + D 2 y_p \left[ n \right] = D_1 n + D_2 yp[n]=D1n+D2
将特解代入方程,有: D 1 n + D 2 + 5 { D 1 [ n − 1 ] + D 2 } = n D_1 n + D_2 + 5\left\{ {D_1 \left[ {n - 1} \right] + D_2 } \right\} = n D1n+D2+5{ D1[n−1]+D2}=n
比较上式左右同类项系数可得: D 1 = 1 6 , D 2 = 5 36 D_1 = {1 \over 6},\,\,\,D_2 = {5 \over {36}} D1=61,D2=365
则差分方程的完全解为: y [ n ] = y h [ n ] + y p [ n ] = C ( − 5 ) n + 1 6 n + 5 36 y\left[ n \right] = y_h \left[ n \right] + y_p \left[ n \right] = C\left( { - 5} \right)^n + {1 \over 6}n + {5 \over {36}} y[n]=yh[n]+yp[n]=C(−5)n+61n+365
将 y [ − 1 ] = 0 y\left[ { - 1} \right] = 0 y[−1]=0代入上式,可得: C = − 5 36 C = - {5 \over {36}} C=−365
所以: y [ n ] = 1 36 [ ( − 5 ) n + 1 + 6 n + 5 ] y\left[ n \right] = {1 \over {36}}\left[ {\left( { - 5} \right)^{n + 1} + 6n + 5} \right] y[n]=361[(−5)n+1+6n+5]
1.3.2 选做题
(1)微分方程求解
求解:
Ⅰ.第一小问
- 齐次解:
求解齐次解的过程与第一小题相同。
- 特解:
下面从求解系统的特解开始:
假设系统的特解为: y s ( t ) = B ⋅ e − 3 t y_s \left( t \right) = B \cdot e^{ - 3t} ys(t)=B⋅e−3t,代入方程左边: 9 B e − 3 t − 7 ⋅ 3 e − 3 t + 10 B ⋅ e − 3 t = − 3 e − 3 t + e − 3 t 9Be^{ - 3t} - 7 \cdot 3e^{ - 3t} + 10B \cdot e^{ - 3t} = - 3e^{ - 3t} + e^{ - 3t} 9Be−3t−7⋅3e−3t+10B⋅e−3t=−3e−3t+e−3t
化简可得方程: − 2 B = − 2 - 2B = - 2 −2B=−2,所以 B = 1 B = 1 B=1 B = 1 B = 1 B=1。
- 完全解:
系统的完全响应为:
y ( t ) = y h ( t ) + y s ( t ) = c 1 e − 2 t + c 2 e − 5 t + e − 3 t y\left( t \right) = y_h \left( t \right) + y_s \left( t \right) = c_1 e^{ - 2t} + c_2 e^{ - 5t} + e^{ - 3t} y(t)=yh(t)+ys(t)=c1e−2t+c2e−5t+e−3t
使用奇异函数匹配方法确定系统初始状态,将输入信号 e ( t ) = e − 3 t ⋅ u ( t ) e\left( t \right) = e^{ - 3t} \cdot u\left( t \right) e(t)=e−3t⋅u(t)代入系统方程左边,得到 δ ( t ) \delta \left( t \right) δ(t)的最高求导次数为1。
d 2 d t 2 r ( t ) = a ⋅ δ ( t ) + b ⋅ u ( t ) {
{d^2 } \over {dt^2 }}r\left( t \right) = a \cdot \delta \left( t \right) + b \cdot u\left( t \right) dt2d2r(t)=a⋅δ(t)+b⋅u(t)
d d t r ( t ) = a ⋅ u ( t ) {d \over {dt}}r\left( t \right) = a \cdot u\left( t \right) dtdr(t)=a⋅u(t)
r ( t ) = 0 r\left( t \right) = 0 r(t)=0
a ⋅ δ ( t ) + ( b + 7 a ) ⋅ u ( t ) = δ ( t ) − 2 ⋅ u ( t ) a \cdot \delta \left( t \right) + \left( {b + 7a} \right) \cdot u\left( t \right) = \delta \left( t \right) - 2 \cdot u\left( t \right) a⋅δ(t)+(b+7a)⋅u(t)=δ(t)−2⋅u(t)
平衡奇异函数方程西树,得到方程组: { a = 1 b + 7 a = − 2 \left\{ \begin{matrix} {a = 1}\\{b + 7a = - 2}\\\end{matrix} \right. { a=1b+7a=−2
{ a = 1 b = − 9 \left\{ \begin{matrix} {a = 1}\\{b = - 9}\\\end{matrix} \right. { a=1b=−9
因此可以确定系统的输出在0时刻的跳跃:
r ′ ( 0 + ) − r ′ ( 0 − ) = 1 r ( 0 + ) − r ( 0 − ) = 0 r'\left( {0_ + } \right) - r'\left( {0_{\rm{ - }} } \right){\rm{ = }}1\,\,\,\,r\left( {0_ + } \right) - r\left( {0_ - } \right) = 0 r′(0+)−r′(0−)=1r(0+)−r(0−)=0
系统的初始条件为: r ′ ( 0 + ) = 2 , r ( 0 + ) = 2 r'\left( {0_ + } \right) = 2,\,\,\,\,r\left( {0_ + } \right) = 2 r′(0+)=2,r(0+)=2
根据初试条件求解系统的待定系数:
{ − 2 c 1 − 5 c 2 − 3 = 3 c 1 + c 2 + 1 = 1 \left\{ \begin{matrix} { - 2c_1 - 5c_2 - 3 = 3}\\{c_1 + c_2 + 1 = 1}\\\end{matrix} \right. {
−2c1−5c2−3=3c1+c2+1=1
{ c 1 = 2 c 2 = − 2 \left\{ \begin{matrix} {c_1 = 2}\\{c_2 = - 2}\\\end{matrix} \right. { c1=2c2=−2
系统的完全响应为:
y ( t ) = 2 e − 2 t − 2 e − 5 t + e − 3 t , t ≥ 0 y\left( t \right) = 2e^{ - 2t} - 2e^{ - 5t} + e^{ - 3t} ,\,\,\,\,t \ge 0 y(t)=2e−2t−2e−5t+e−3t,t≥0
Ⅱ.第二小问
- 自由响应与强迫响应:
求解系统的零状态响应,根据完全解求解过程,系统的零状态的初始条件为:
r ′ ( 0 + ) = 1 , r ( 0 + ) = 0 r'\left( {0_ + } \right) = 1,\,\,\,\,r\left( {0_ + } \right) = 0 r′(0+)=1,r(0+)=0
求对应系统的待定系数:
{ − 2 c 1 − 5 c 2 − 3 = 1 c 1 + c 2 + 1 = 0 \left\{ \begin{matrix} { - 2c_1 - 5c_2 - 3 = 1}\\{c_1 + c_2 + 1 = 0}\\\end{matrix} \right. {
−2c1−5c2−3=1c1+c2+1=0
{ 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=−31c2=−32
因此对应的零状态响应为:
y z s ( t ) = − 1 3 e − 2 t − 2 3 e − 5 t + e − 3 t , t ≥ 0 y_{zs} \left( t \right) = - {1 \over 3}e^{ - 2t} - {2 \over 3}e^{ - 5t} + e^{ - 3t} ,\,\,\,\,t \ge 0 yzs(t)=−31e−2t−32e−5t+e−3t,t≥0
系统的零输入响应为:
y z i ( t ) = y ( t ) − y z s ( t ) = 2 1 3 e − 2 t − 1 1 3 e − 5 t , t ≥ 0 y_{zi} \left( t \right) = y\left( t \right) - y_{zs} \left( t \right) = 2{1 \over 3}e^{ - 2t} - 1{1 \over 3}e^{ - 5t} ,\,\,\,\,t \ge 0 yzi(t)=y(t)−yzs(t)=231e−2t−131e−5t,t≥0
- 自由响应与强迫响应:
系统的自由响应为:
y h ( t ) = 2 e − 2 t − 2 e − 5 t , t ≥ 0 y_h \left( t \right) = 2e^{ - 2t} - 2e^{ - 5t} ,\,\,\,\,t \ge 0 yh(t)=2e−2t−2e−5t,t≥0
系统的强迫响应为:
y f ( t ) = e − 3 t , t ≥ 0 y_f \left( t \right) = e^{ - 3t} ,\,\,\,\,t \ge 0 yf(t)=e−3t,t≥0
(2)差分方程求解
Ⅰ.第一小题
Ⅱ.第二小题
特征方程为: α 2 + 2 α + 1 = 0 \alpha ^2 + 2\alpha + 1 = 0 α2+2α+1=0
求得特征根: α 1 = α 2 = − 1 \alpha _1 = \alpha _2 = - 1 α1=α2=−1
于是齐次解为: y h [ n ] = ( C 1 n + C 2 ) ( − 1 ) n y_h \left[ n \right] = \left( {C_1 n + C_2 } \right)\left( { - 1} \right)^n yh[n]=(C1n+C2)(−1)n
由于输入信号为指数序列 3 n 3^n 3n,令特解为: y p [ n ] = D 1 ⋅ 3 n y_p \left[ n \right] = D_1 \cdot 3^n yp[n]=D1⋅3n
将特解代入方程,有: D 1 ⋅ 3 n + 2 D 1 ⋅ 3 n − 1 + D 1 ⋅ 3 n − 2 = 3 n D_1 \cdot 3^n + 2D_1 \cdot 3^{n - 1} + D_1 \cdot 3^{n - 2} = 3^n D1⋅3n+2D1⋅3n−1+D1⋅3n−2=3n
比较上式左右同类型系数,可以求得: D 1 = 9 16 D_1 {\rm{ = }}{9 \over {16}} D1=169
则差分方程的完全解为: y [ n ] = y h [ n ] + y p [ n ] = ( C 1 n + C 2 ) ( − 1 ) n + 9 16 ⋅ 3 n y\left[ n \right] = y_h \left[ n \right] + y_p \left[ n \right] = \left( {C_1 n + C_2 } \right)\left( { - 1} \right)^n + {9 \over {16}} \cdot 3^n y[n]=yh[n]+yp[n]=(C1n+C2)(−1)n+169⋅3n
将 y [ − 1 ] = 0 , y [ 0 ] = 0 y\left[ { - 1} \right] = 0,y\left[ 0 \right] = 0 y[−1]=0,y[0]=0代入上式,可得:
( − C 1 + C 2 ) × ( − 1 ) + 9 16 × 1 3 = 0 \left( { - C_1 + C_2 } \right) \times \left( { - 1} \right) + {9 \over {16}} \times {1 \over 3} = 0 (−C1+C2)×(−1)+169×31=0
C 2 + 9 16 = 0 C_2 + {9 \over {16}} = 0 C2+169=0
求解得到完全解中的待定系数: C 1 = − 3 4 , C 2 = − 9 16 C_1 = - {3 \over 4},\,\,\,C_2 = - {9 \over {16}} C1=−43,C2=−169
所以: y [ n ] = ( − 3 4 n − 9 16 ) ( − 1 ) n + 9 16 3 n y\left[ n \right] = \left( { - {3 \over 4}n - {9 \over {16}}} \right)\left( { - 1} \right)^n + {9 \over {16}}3^n y[n]=(−43n−169)(−1)n+1693n
■ 相关文献链接: