设轮111转动的角度为θ 1 \theta_1θ1​，轮111的转矩方程为J 1d2θ1dt2+f 1dθ1dt+M 1= M J_1\frac{\text{d}^2\theta_1}{\text{d}t^2}+f_1\frac{\text{d}\theta_1}{\text{d}t}+M_1=MJ1​dt2d2θ1​​+f1​dtdθ1​​+M1​=M 轮2的转矩方程为J 2d2θdt2+f 2dθdt=M 2J_2\frac{\text{d}^2\theta}{\text{d}t^2}+f_2\frac{\text{d}\theta}{\text{d}t}=M_2J2​dt2d2θ​+f2​dtdθ​=M2​ 对上述两式取拉普拉斯变换，则 s (J 1s +f 1)θ 1( s ) +M 1( s ) = M ( s )s (J 2s +f 2) θ ( s ) =M 2( s ) s(J_1s+f_1)\theta_1(s)+M_1(s)=M(s) \\ s(J_2s+f_2)\theta(s)=M_2(s)s(J1​s+f1​)θ1​(s)+M1​(s)=M(s)s(J2​s+f2​)θ(s)=M2​(s) 轮111和轮222做工相等，则M 1 θ 1=M 2θ M 2=M 1θ1 θ=M 1r2r1M_1\theta_1=M_2\theta \\ M_2=M_1\frac{\theta_1}{\theta}=M_1\frac{r_2}{r_1}M1​θ1​=M2​θM2​=M1​θθ1​​=M1​r1​r2​​ 联立上述方程，消去中间变量M 1 (s)M_1(s)M1​(s)和M 2 (s)M_2(s)M2​(s)可得 s (J 1s +f 1) r2r1θ ( s ) + s (J 2s +f 2) r1r2θ ( s ) = M ( s ) s(J_1s+f_1)\frac{r_2}{r_1}\theta(s)+s(J_2s+f_2)\frac{r_1}{r_2}\theta(s)=M(s)s(J1​s+f1​)r1​r2​​θ(s)+s(J2​s+f2​)r2​r1​​θ(s)=M(s) 整理可得系统的传递函数 θ(s)M(s)= r1r2s(J1s+f1)+s(J2s+f2)(r 1 r 2 )2\frac{\theta(s)}{M(s)}=\frac{\cfrac{r_1}{r_2}}{s(J_1s+f_1)+s(J_2s+f_2)(\cfrac{r_1}{r_2})^2}M(s)θ(s)​=s(J1​s+f1​)+s(J2​s+f2​)(r2​r1​​)2r2​r1​​​

系统的开环传递函数 G ( s ) = 20K1s(s+20)(s+1+α)G(s)=\frac{20K_1}{s(s+20)(s+1+\alpha)}G(s)=s(s+20)(s+1+α)20K1​​ 系统的闭环特征方程 Δ ( s )= s ( s + 20 ) ( s + 1 + α ) + 20K 1 =s 3+ ( 21 + α )s 2+ 20 ( 1 + α ) s + 20K 1\begin{aligned} \Delta(s) &= s(s+20)(s+1+\alpha)+20K_1 \\ &=s^3+(21+\alpha)s^2+20(1+\alpha)s+20K_1 \end{aligned}Δ(s)​=s(s+20)(s+1+α)+20K1​=s3+(21+α)s2+20(1+α)s+20K1​​ 将s1 , 2=−3±j3s_{1,2}=-3\pm j3s1,2​=−3±j3代入闭环特征方程，得到方程组 { −3−60α+20K1=0 − 6 − 60 α + 20K1 = 0 − 264 + 42 α = 0\left\{\begin{array}{l} \xcancel{-3-60\alpha+20K_1=0} \\ -6-60\alpha+20K_1=0 \\ -264+42\alpha=0 \end{array}\right.⎩⎨⎧​−3−60α+20K1​=0 ​−6−60α+20K1​=0−264+42α=0​ 解得 {α =447 = 6.286K1 =2661140 = 19.007K1 =2661140 = 19.157\left\{\begin{array}{l} \alpha=\cfrac{44}{7}=6.286 \\ K_1=\cfrac{2661}{140}=19.007 \\ K_1=\cfrac{2661}{140}=19.157 \end{array}\right.⎩⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎧​α=744​=6.286K1​=1402661​=19.007K1​=1402661​=19.157​

系统的闭环传递函数为 Φ ( s ) =1s+2\varPhi(s)=\frac{1}{s+2}Φ(s)=s+21​ 系统的误差传递函数Φ e( s ) = 1 − Φ ( s ) = s+1s+2=1 2⋅ s+112s+1\varPhi_e(s)=1-\varPhi(s)=\frac{s+1}{s+2}=\cfrac{1}{2}\cdot\frac{s+1}{\cfrac{1}{2}s+1}Φe​(s)=1−Φ(s)=s+2s+1​=21​⋅21​s+1s+1​ 误差传递函数的频率特性 {∣Φe ( j ω ) ∣ =12 ⋅ω2 + 114ω2 + 1∠Φe(jω)=−arctan⁡ω+arctan⁡1 2 ω ∠Φe ( j ω ) = arctan ⁡ ω − arctan ⁡12 ω\left\{\begin{array}{l} |\varPhi_e(j\omega)|=\cfrac{1}{2}\cdot\cfrac{\sqrt{\omega^2+1}}{\sqrt{\cfrac{1}{4}\omega^2+1}} \\ \xcancel{\angle\varPhi_e(j\omega)=-\arctan \omega+\arctan\cfrac{1}{2}\omega} \\ \angle\varPhi_e(j\omega)=\arctan \omega-\arctan\cfrac{1}{2}\omega \end{array}\right.⎩⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎧​∣Φe​(jω)∣=21​⋅41​ω2+1 ​ω2+1 ​​∠Φe​(jω)=−arctanω+arctan21​ω ​∠Φe​(jω)=arctanω−arctan21​ω​ 在输入信号r 1 (t)=sin⁡(t+30°)r_1(t)=\sin(t+30°)r1​(t)=sin(t+30°)作用下 {∣Φe1 ( j ω ) ∣∣ω=1 =10 5∠Φe 1 (jω)∣ω = 1 =−18.435° ∠Φe1 ( j ω )∣ω=1 = 18.435 °\left\{\begin{array}{l} |\varPhi_{e_1}(j\omega)| \Big| _{\omega=1} =\cfrac{\sqrt{10}}{5} \\ \xcancel{\angle\varPhi_{e_1}(j\omega) \Big| _{\omega=1} =-18.435°} \\ \angle\varPhi_{e_1}(j\omega) \Big| _{\omega=1} =18.435° \end{array}\right.⎩⎪⎪⎪⎪⎨⎪⎪⎪⎪⎧​∣Φe1​​(jω)∣∣∣∣​ω=1​=510 ​​∠Φe1​​(jω)∣∣∣​ω=1​=−18.435° ​∠Φe1​​(jω)∣∣∣​ω=1​=18.435°​ 在输入信号r 2 (t)=2cos⁡(2t−45°)r_2(t)=2\cos(2t-45°)r2​(t)=2cos(2t−45°)作用下 {∣Φe2 ( j ω ) ∣∣ω=2 =10 4∠Φe 2 (jω)∣ω = 2 =−18.435° ∠Φe2 ( j ω )∣ω=2 = 18.435 °\left\{\begin{array}{l} |\varPhi_{e_2}(j\omega)| \Big| _{\omega=2} =\cfrac{\sqrt{10}}{4} \\ \xcancel{\angle\varPhi_{e_2}(j\omega) \Big| _{\omega=2} =-18.435°} \\ \angle\varPhi_{e_2}(j\omega) \Big| _{\omega=2} =18.435° \end{array}\right.⎩⎪⎪⎪⎪⎨⎪⎪⎪⎪⎧​∣Φe2​​(jω)∣∣∣∣​ω=2​=410 ​​∠Φe2​​(jω)∣∣∣​ω=2​=−18.435° ​∠Φe2​​(jω)∣∣∣​ω=2​=18.435°​ 系统的稳态误差为 es s (∞)=∣Φe 1 (j1)∣sin⁡(t+30°+∠Φe 1 (j1))−2∣Φe 2 (j2)∣cos⁡(2t−45°+∠Φe 2 (j2)) =10 5sin⁡(t+11.565°)−10 4cos⁡(2t−63.435°)\xcancel{ \begin{aligned} e_{ss}(\infin) &= |\varPhi_{e1}(j1)|\sin (t+30°+\angle\varPhi_{e1}(j1))-2|\varPhi_{e2}(j2)|\cos(2t-45°+\angle\varPhi_{e2}(j2)) \\ &=\cfrac{\sqrt{10}}{5}\sin(t+11.565°)-\cfrac{\sqrt{10}}{4}\cos(2t-63.435°) \end{aligned} }ess​(∞)​=∣Φe1​(j1)∣sin(t+30°+∠Φe1​(j1))−2∣Φe2​(j2)∣cos(2t−45°+∠Φe2​(j2))=510 ​​sin(t+11.565°)−410 ​​cos(2t−63.435°)​​ess( ∞ )= ∣Φe1( j 1 ) ∣ sin ⁡ ( t + 30 ° + ∠Φe1( j 1 ) ) − 2 ∣Φe2( j 2 ) ∣ cos ⁡ ( 2 t − 45 ° + ∠Φe2( j 2 ) )= 10 5sin ⁡ ( t + 48.435 ° ) − 10 2cos ⁡ ( 2 t − 26.565 ° ) \begin{aligned} e_{ss}(\infin) &= |\varPhi_{e1}(j1)|\sin (t+30°+\angle\varPhi_{e1}(j1))-2|\varPhi_{e2}(j2)|\cos(2t-45°+\angle\varPhi_{e2}(j2)) \\ &=\cfrac{\sqrt{10}}{5}\sin(t+48.435°)-\cfrac{\sqrt{10}}{2}\cos(2t-26.565°) \end{aligned}ess​(∞)​=∣Φe1​(j1)∣sin(t+30°+∠Φe1​(j1))−2∣Φe2​(j2)∣cos(2t−45°+∠Φe2​(j2))=510​​sin(t+48.435°)−210​​cos(2t−26.565°)​

负反馈系统根轨迹如左图，反馈通道零极点分布图如右图，当反馈通道增益 KH=5\sout{K_H=5}KH​=5​时，系统存在重根，求有重根时系统的闭环传递函数。 已知闭环系统的根轨迹和反馈通道零极点分布如下图所示，试求当反馈通道根轨迹增益K H =5K_H=5KH​=5时，并且闭环系统存在重极点的闭环传递函数。（图中 − 5.256 -5.256−5.256改为 − 5.68 -5.68−5.68） 由图可知，系统的根轨迹为180°180°180°根轨迹，系统的根轨迹方程为K ∗s+3s(s+5)(s+6)(s2+2s+2)= − 1 K^*\cfrac{s+3}{s(s+5)(s+6)(s^2+2s+2)}=-1K∗s(s+5)(s+6)(s2+2s+2)s+3​=−1 系统的反馈通道传递函数为 H ( s ) = KH(s+3)s2+2s+2H(s)=\cfrac{K_H(s+3)}{s^2+2s+2}H(s)=s2+2s+2KH​(s+3)​ 令K ∗ =K G K H K^*=K_GK_HK∗=KG​KH​，系统的闭环特征方程为 Δ ( s ) = s ( s + 5 ) ( s + 6 ) (s 2+ 2 s + 2 ) +K G K H( s + 3 ) = 0 \Delta(s)=s(s+5)(s+6)(s^2+2s+2)+K_GK_H(s+3)=0Δ(s)=s(s+5)(s+6)(s2+2s+2)+KG​KH​(s+3)=0 当K H =5K_H=5KH​=5，根轨迹存在重根 −5.265\sout{-5.265}−5.265​， 根轨迹存在重根−5.68-5.68−5.68，将其代入上式，解得 KG=1.696K G= 2.112 \sout{K_G=1.696} \\ K_G=2.112KG​=1.696​KG​=2.112 系统的前向通道传递函数为 G(s)=1.696s ( s + 5 ) ( s + 6 ) G ( s ) =2.112s(s+5)(s+6)\sout{G(s)=\cfrac{\sout{1.696}}{\sout{s(s+5)(s+6)}}} \\ G(s)=\cfrac{2.112}{s(s+5)(s+6)}G(s)=s(s+5)(s+6)​1.696​​G(s)=s(s+5)(s+6)2.112​ 系统的闭环传递函数为 Φ ( s )= G(s)1+G(s)H(s) = 1.696 (s2 + 2 s + 2 ) s ( s + 5 ) ( s + 6 ) (s2 + 2 s + 2 ) + 8.48 ( s + 3 ) = 2.112(s2+2s+2)s(s+5)(s+6)(s2+2s+2)+10.56(s+3)\begin{aligned} \varPhi(s) &= \cfrac{G(s)}{1+G(s)H(s)} \\ &\sout{= \cfrac{\sout{1.696(s^2+2s+2)}}{\sout{s(s+5)(s+6)(s^2+2s+2)+8.48(s+3)}}} \\ &= \cfrac{2.112(s^2+2s+2)}{s(s+5)(s+6)(s^2+2s+2)+10.56(s+3)} \end{aligned}Φ(s)​=1+G(s)H(s)G(s)​=s(s+5)(s+6)(s2+2s+2)+8.48(s+3)​1.696(s2+2s+2)​​​=s(s+5)(s+6)(s2+2s+2)+10.56(s+3)2.112(s2+2s+2)​​

单位负反馈系统 G(s)= 6 ( s + 1 )s2 ( s − 1 )\sout{G(s)=\cfrac{6(s+1)}{s^2(s-1)}}G(s)=s2(s−1)6(s+1)​​，画出奈奎斯图，判断稳定性，右半平面有几个极点，画出对数频率特性曲线。 单位负反馈系统开环传递函数G(s)=6(s+1)s2(s−1)G(s)=\cfrac{6(s+1)}{s^2(s-1)}G(s)=s2(s−1)6(s+1)​ （1）画出开环系统的Nyquist图，判断闭环系统稳定性，若系统不稳定，求处右半平面闭环极点个数。 （2）画出开环系统的Bode图（幅频特性和相频特性）。

（1） 系统的传递函数 G ( s ) = − 6(s+1)s2(−s+1)G(s)=-\frac{6(s+1)}{s^2(-s+1)}G(s)=−s2(−s+1)6(s+1)​ 系统的频率特性 {Φ ( j ω ) =6 ( 1 −ω 2)ω 2( 1 +ω 2)+ j12ω ( 1 +ω 2)∠ Φ ( j ω ) = − 180 ° − 180 ° + arctan ⁡ ω + arctan ⁡ ω\left\{\begin{array}{l} \varPhi(j\omega)=\cfrac{6(1-\omega^2)}{\omega^2(1+\omega^2)}+j\cfrac{12}{\omega(1+\omega^2)} \\ \angle\varPhi(j\omega)=-180°-180°+\arctan \omega+\arctan\omega \end{array}\right.⎩⎨⎧​Φ(jω)=ω2(1+ω2)6(1−ω2)​+jω(1+ω2)12​∠Φ(jω)=−180°−180°+arctanω+arctanω​ 则有{Φ(jω)∣ω →0 +=+∞∠Φ(jω)∣ω →0 +=−360° {Φ(jω)∣ω → + ∞ =0∠Φ(jω)∣ω → + ∞ =−180°\left\{\begin{array}{l} \varPhi(j\omega) \Big| _{\omega\rightarrow0^+} =+\infin \\ \angle\varPhi(j\omega) \Big| _{\omega\rightarrow0^+} =-360° \end{array}\right. \left\{\begin{array}{l} \varPhi(j\omega) \Big| _{\omega\rightarrow+\infin} =0 \\ \angle\varPhi(j\omega) \Big| _{\omega\rightarrow+\infin} =-180° \end{array}\right.⎩⎨⎧​Φ(jω)∣∣∣​ω→0+​=+∞∠Φ(jω)∣∣∣​ω→0+​=−360°​⎩⎨⎧​Φ(jω)∣∣∣​ω→+∞​=0∠Φ(jω)∣∣∣​ω→+∞​=−180°​ 令1−ω 2 =01-\omega^2=01−ω2=0，奈奎斯曲线与虚轴交点为j6j6j6，系统完整的奈奎斯特图为 由系统的传递函数可知P=1P=1P=1，在(−1,j0)(-1,j0)(−1,j0)点左侧存在半次负穿越，则 Z = P − 2 (N +−N −) = 2 Z=P-2(N^+-N^-)=2Z=P−2(N+−N−)=2 系统在sss平面的右半平面存在222个极点，系统不稳定。 （2） 系统的截至频率为ωc = 6 r a d / s\xcancel{\omega_c=6rad/s}ωc​=6rad/s ​ω c =6 rad/s\quad\omega_c=\sqrt6rad/sωc​=6​rad/s，相角裕度 γ = − 198.924 °\xcancel{\gamma=-198.924°}γ=−198.924° ​γ=−44.415°\quad\gamma=-44.415°γ=−44.415°，系统的对数频率特性曲线。

单位负反馈系统 G(s)=250s ( s + 25 )\sout{G(s)=\cfrac{250}{s(s+25)}}G(s)=s(s+25)250​​，设计校正装置，要求静态误差系数 K=100\sout{K=100}K=100， γ⩾45°\sout{\gamma\geqslant45°}γ⩾45°​， ωc>65rad/s\sout{\omega_c>65rad/s}ωc​>65rad/s​，画出校正后渐近幅频特性曲线。 单位负反馈系统的开环传递函数为G 0 (s)= 250s(s+25)G_0(s)=\cfrac{250}{s(s+25)}G0​(s)=s(s+25)250​，要求系统静态速度误差系数为100，相角裕度γ⩾45°\gamma\geqslant45°γ⩾45°，截止频率ω c \omega_cωc​不低于65rad/s65rad/s65rad/s，设计一个合适的串联校正装置，求校正环节传递函数，并画出校正后开环系统渐进幅频特性曲线。 将系统传递函数化为 G ( s ) =10s(0.04s+1)G(s)=\frac{10}{s(0.04s+1)}G(s)=s(0.04s+1)10​ 根据题目的要求，设校正装置增益为K c =10K_c=10Kc​=10，则G 1( s ) =100s(0.04s+1)G_1(s)=\frac{100}{s(0.04s+1)}G1​(s)=s(0.04s+1)100​ 截止频率为ωc1= 46.978 r a d / s \omega_{c1}=46.978rad/sωc1​=46.978rad/s 相角裕度为γ 1= 180 ° − 90 ° − arctan ⁡ 0.04ωc1= 28.02 ° \gamma_{1}=180°-90°-\arctan0.04\omega_{c1}=28.02°γ1​=180°−90°−arctan0.04ωc1​=28.02° 故可使用超前校正，设校正装置为G c( s ) = Kc(τs+1)ατs+1G_c(s)=\frac{K_c(\tau s+1)}{\alpha\tau s+1}Gc​(s)=ατs+1Kc​(τs+1)​ 取校正后系统的截至频率为ω c =68rad/s\omega_c=68rad/sωc​=68rad/s，则系统的相角裕度为γ 2= 180 ° − 90 ° − arctan ⁡ 0.04ωc2= 20.185 ° \gamma_2=180°-90°-\arctan0.04\omega_{c2}=20.185°γ2​=180°−90°−arctan0.04ωc2​=20.185° 由20lg⁡∣G 1 (jω)∣=10lg⁡α20\lg|G_1(j\omega)|=10\lg\alpha20lg∣G1​(jω)∣=10lgα可得 α = 0.257 \alpha=0.257α=0.257 由ϕ m =arcsin⁡1−α1+α\phi_m=\arcsin\cfrac{1-\alpha}{1+\alpha}ϕm​=arcsin1+α1−α​可得ϕ m= 36.234 ° \phi_m=36.234°ϕm​=36.234° 校正后系统的相角裕度为 γ =γ 2+ϕ m= 20.185 ° + 36.123 ° = 56.308 ° \gamma=\gamma_2+\phi_m=20.185°+36.123°=56.308°γ=γ2​+ϕm​=20.185°+36.123°=56.308° 满足题目要求。由ω c = 1ατ\omega_c=\cfrac{1}{\sqrt{\alpha}\tau}ωc​=α​τ1​可得 τ = 0.029 \tau=0.029τ=0.029 校正装置为G c( s ) = 10(0.029s+1)0.00745s+1G_c(s)=\frac{10(0.029s+1)}{0.00745s+1}Gc​(s)=0.00745s+110(0.029s+1)​ 图略

对信号 p(t)=3cos⁡2πt+1.5cos⁡4πt+cos⁡8πt\sout{p(t)=3\cos2\pi t+1.5\cos4\pi t+\cos8\pi t}p(t)=3cos2πt+1.5cos4πt+cos8πt​采样，采样周期 T=0.1s\sout{T=0.1s}T=0.1s，是否满足采样定理，画出采样频谱，写出采样信号表达式。 设连续信号p(t)=3cos⁡2πt+1.5cos⁡4πt+cos⁡8πtp(t)=3\cos2\pi t+1.5\cos4\pi t+\cos8\pi tp(t)=3cos2πt+1.5cos4πt+cos8πt，系统采样周期T=0.1sT=0.1sT=0.1s （1）分析系统是否满足香农采样定理。 （2）在该采样周期下，p(t)p(t)p(t)的采样信号为p ∗ (t)p^*(t)p∗(t)的频谱图。

（1）ω s= 2π T= 20 π ,  ωmax= 8 π \omega_s=\cfrac{2\pi}{T}=20\pi,\ \omega_{max}=8\piωs​=T2π​=20π, ωmax​=8π 因为ω s >2ωm a x\omega_s>2\omega_{max}ωs​>2ωmax​，满足香农采样定理，采样信号能无失真还原。 （2） 采样信号表达式为p ∗( t )= p ( t ) ∑k=0∞ δ ( t − k T ) = ∑k=0∞ p ( k T ) δ ( t − k T ) = ∑k=0∞ [ 3 cos ⁡ ( 2 π ⋅ 0.1 ⋅ k ) + 1.5 cos ⁡ ( 4 π ⋅ 0.1 ⋅ k ) + cos ⁡ ( 8 π ⋅ 0.1 ⋅ k ) ] δ ( t − k T ) = ∑k=0∞ [ 3 cos ⁡ ( 0.2 π k ) + 1.5 cos ⁡ ( 0.4 π k ) + cos ⁡ ( 0.8 π k ) ] δ ( t − k T )\begin{aligned} p^*(t) &= p(t)\displaystyle\sum_{k=0}^{\infin}\delta(t-kT) \\ &= \displaystyle\sum_{k=0}^{\infin}p(kT)\delta(t-kT) \\ &= \displaystyle\sum_{k=0}^{\infin}[3\cos(2\pi\cdot0.1\cdot k)+1.5\cos(4\pi\cdot0.1\cdot k)+\cos(8\pi\cdot0.1\cdot k)]\delta(t-kT) \\ &= \displaystyle\sum_{k=0}^{\infin}[3\cos(0.2\pi k)+1.5\cos(0.4\pi k)+\cos(0.8\pi k)]\delta(t-kT) \end{aligned}p∗(t)​=p(t)k=0∑∞​δ(t−kT)=k=0∑∞​p(kT)δ(t−kT)=k=0∑∞​[3cos(2π⋅0.1⋅k)+1.5cos(4π⋅0.1⋅k)+cos(8π⋅0.1⋅k)]δ(t−kT)=k=0∑∞​[3cos(0.2πk)+1.5cos(0.4πk)+cos(0.8πk)]δ(t−kT)​ 对上式进行傅里叶变换（cosωt→π[δ(ω−ω 0 )+δ(ω+ω 0 )]cos\omega t\rightarrow\pi[\delta(\omega-\omega_0)+\delta(\omega+\omega_0)]cosωt→π[δ(ω−ω0​)+δ(ω+ω0​)]）P ∗( j ω )=1 T∑k=−∞∞ P ( j ω − j kωs ) =π T∑k=−∞∞ { 3 [ δ ( ω −ω1 − kωs ) + δ ( ω +ω1 − kωs ) ] + 1.5 [ δ ( ω −ω2 − kωs ) + δ ( ω +ω2 − kωs ) ] + [ δ ( ω −ω3 − kωs ) + δ ( ω +ω3 − kωs ) ] } =π T∑k=−∞∞ { 3 [ δ ( ω − 2 π − 20 π k ) + δ ( ω + 2 π − 20 π k ) ] + 1.5 [ δ ( ω − 4 π − 20 π k ) + δ ( ω + 4 π − 20 π k ) ] + [ δ ( ω − 8 π − 20 π k ) + δ ( ω + 8 π − 20 π k ] }\begin{aligned} P^*(j\omega) &= \cfrac{1}{T}\displaystyle\sum_{k=-\infin}^{\infin}P(j\omega-jk\omega_s) \\ &= \cfrac{\pi}{T}\displaystyle\sum_{k=-\infin}^{\infin}\{ 3[\delta(\omega-\omega_1-k\omega_s)+\delta(\omega+\omega_1-k\omega_s)]+1.5[\delta(\omega-\omega_2-k\omega_s)+\delta(\omega+\omega_2-k\omega_s)]+[\delta(\omega-\omega_3-k\omega_s)+\delta(\omega+\omega_3-k\omega_s)]\} \\ &= \cfrac{\pi}{T}\displaystyle\sum_{k=-\infin}^{\infin}\{ 3[\delta(\omega-2\pi-20\pi k)+\delta(\omega+2\pi-20\pi k)]+1.5[\delta(\omega-4\pi-20\pi k)+\delta(\omega+4\pi-20\pi k)]+[\delta(\omega-8\pi-20\pi k)+\delta(\omega+8\pi-20\pi k]\} \end{aligned}P∗(jω)​=T1​k=−∞∑∞​P(jω−jkωs​)=Tπ​k=−∞∑∞​{3[δ(ω−ω1​−kωs​)+δ(ω+ω1​−kωs​)]+1.5[δ(ω−ω2​−kωs​)+δ(ω+ω2​−kωs​)]+[δ(ω−ω3​−kωs​)+δ(ω+ω3​−kωs​)]}=Tπ​k=−∞∑∞​{3[δ(ω−2π−20πk)+δ(ω+2π−20πk)]+1.5[δ(ω−4π−20πk)+δ(ω+4π−20πk)]+[δ(ω−8π−20πk)+δ(ω+8π−20πk]}​ 采样信号频谱为

r(t)=1(t)\sout{r(t)=1(t)}r(t)=1(t)​， c(0)=c˙(0)=0\sout{c(0)=\dot{c}(0)=0}c(0)=c˙(0)=0​，画出系统误差相轨迹，系统是否稳定，稳定时是否有稳态误差，稳态误差是多少。 设c(0)=c ˙ (0)=0c(0)=\dot{c}(0)=0c(0)=c˙(0)=0，输入为单位阶跃信号，求 （1）e−e ˙ e-\dot{e}e−e˙平面的相轨迹图 （2）判断系统的稳定性，并求出最大稳态误差。 G ( s ) =30s(2s+1)G(s)=\frac{30}{s(2s+1)}G(s)=s(2s+1)30​

（1） 非线性环节是死区继电特性 u ={ Mx⩾0.5 0∣x∣<0.5−Mx⩽−0.5u= \begin{cases} M & x\geqslant0.5 \\ 0 & |x|