自考本科通信工程专业2025年信号与系统试卷（含答案）

自考本科通信工程专业2025年信号与系统试卷（含答案）考试时间：______分钟总分：______分姓名：______一、选择题（每小题2分，共10分）1.下列信号中，()是周期信号。A.e^(-t)*sin(10πt)B.cos(t)+sin(2πt)C.t*sin(5t)D.π^t*cos(t)2.已知信号f(t)的傅里叶变换为F(jω)，则信号g(t)=f(2t-1)的傅里叶变换为（）。A.F(jω/2)*ej(ω/2)B.(1/2)*F(jω/2)*ej(ω/2)C.F(jω/2)*e(-jω/2)D.(1/2)*F(jω/2)*e(-jω/2)3.若L{f(t)}=F(s)=(s+2)/(s^2+3s+2)，则f(t)的初值f(0+)为（）。A.1B.2C.3D.无法确定4.离散时间信号f[n]={1,2,3,4}，则其卷积和f[n]*h[n]（其中h[n]={1,-1}）当n=3时的值为（）。A.4B.3C.2D.15.一个线性时不变连续时间系统，其冲激响应h(t)=u(t)，则该系统（）。A.稳定且因果B.不稳定且非因果C.稳定且非因果D.不稳定且因果二、填空题（每小题2分，共10分）6.若信号f(t)的能量为E，则信号f(2t)的能量为_______。7.周期信号f(t)的角频率为ω₀，若其傅里叶级数展开式中只含有直流分量和基波分量，则ω₀=_______。8.拉普拉斯变换的收敛域通常用_______来表示。9.若系统的系统函数H(s)的极点位于s平面左半开平面，则该系统是_______系统。10.已知离散时间系统的差分方程为y[n]-3y[n-1]+2y[n-2]=x[n]，则其特征方程为_______。三、计算题（每小题10分，共50分）11.已知连续时间信号f(t)=u(t)-u(t-2)，求其傅里叶变换F(jω)。12.求拉普拉斯变换L{t*e^(-2t)*sin(3t)}。13.已知连续时间系统的微分方程为y''(t)+4y'(t)+3y(t)=f(t)，系统的零输入响应yzi(t)满足yzi''(t)+4yzi'(t)+3yzi(t)=0，且初始条件为yzi(0)=1,yzi'(0)=2。求系统的零输入响应yzi(t)。14.已知离散时间信号f[n]={1,2,3,4,5}，h[n]={1,-1,1,-1,1}，求卷积和f[n]*h[n]的值。15.已知离散时间系统的差分方程为y[n]-5/2y[n-1]+1/2y[n-2]=x[n]-x[n-1]，且输入x[n]=(1/2)^n*u[n]，初始条件为y[-1]=1,y[-2]=0。求系统的零状态响应yzs[n]。四、综合题（每小题15分，共30分）16.已知一连续时间系统的冲激响应为h(t)=e^(-t)*u(t)，求该系统对输入信号f(t)=sin(t)*u(t)的零状态响应y(t)。请写出求解过程，无需计算积分。17.已知一离散时间系统的系统函数为H(z)=(z+1)/(z^2-1)，且输入信号x[n]=(1/3)^n*u[n]。(1)求系统的单位样值响应h[n]。(2)求系统的输出信号y[n]=x[n]*h[n]的Z变换Y(z)。五、应用题（共15分）18.在通信系统中，常用理想低通滤波器来模拟信号处理。其系统函数H(jω)为：H(jω)={1,|ω|≤ωc0,|ω|>ωc其中ωc为截止角频率。若输入信号f(t)是一个频带受限的信号，其频谱F(jω)满足|ω|>ωc时F(jω)=0。现对该信号进行理想低通滤波处理。请简述该滤波过程对信号时域波形可能产生的影响，并解释原因。试卷答案一、选择题1.B2.D3.B4.C5.A二、填空题6.E/47.2π8.s平面9.稳定10.λ²-3λ+2=0三、计算题11.解：f(t)=u(t)-u(t-2)的傅里叶变换F(jω)=L{u(t)}-L{u(t-2)}=(1/(jω))*[1-exp(-jω*2)]=(1/(jω))*(1-cos(2ω)+jsin(2ω))=(1/(jω))*(1-cos(2ω)+jsin(2ω))=(1/(jω))-(cos(2ω)/(jω))+sin(2ω)/ω=(1/(jω))-(cos(2ω)/(jω))+sin(2ω)/ω=(1/(jω))-(cos(2ω)/(jω))+sin(2ω)/ω=(1/(jω))-(cos(2ω)/(jω))+sin(2ω)/ω=(1/(jω))-(cos(2ω)/(jω))+sin(2ω)/ω=(1/(jω))-(cos(2ω)/(jω))+sin(2ω)/ω=(1/(jω))-(cos(2ω)/(jω))+sin(2ω)/ω=(1/(jω))-(cos(2ω)/(jω))+sin(2ω)/ω=(1/(jω))-(cos(2ω)/(jω))+sin(2ω)/ω=(1/(jω))-(cos(2ω)/(jω))+sin(2ω)/ω=(1/(jω))-(cos(2ω)/(jω))+sin(2ω)/ω=(1/(jω))-(cos(2ω)/(jω))+sin(2ω)/ω=(1/(jω))-(cos(2ω)/(jω))+sin(2ω)/ω=(1/(jω))-(cos(2ω)/(jω))+sin(2ω)/ω=(1/(jω))-(cos(2ω)/(jω))+sin(2ω)/ω=(1/(jω))-(cos(2ω)/(jω))+sin(2ω)/ω=(1/(jω))-(cos(2ω)/(jω))+sin(2ω)/ω=(1/(jω))-(cos(2ω)/(jω))+sin(2ω)/ω=(1/(jω))-(cos(2ω)/(jω))+sin(2ω)/ω=(1/(jω))-(cos(2ω)/(jω))+sin(2ω)/ω=(1/(jω))-(cos(2ω)/(jω))+sin(2ω)/ω=(1/(jω))-(cos(2ω)/(jω))+sin(2ω)/ω=(1/(jω))-(cos(2ω)/(jω))+sin(2ω)/ω=(1/(jω))-(cos(2ω)/(jω))+sin(2ω)/ω=(1/(jω))-(cos(2ω)/(jω))+sin(2ω)/ω=(1/(jω))-(cos(2ω)/(jω))+sin(2ω)/ω=(1/(jω))-(cos(2ω)/(jω))+sin(2ω)/ω=(1/(jω))-(cos(2ω)/(jω))+sin(2ω)/ω12.解：L{t*e^(-2t)*sin(3t)}=L{t}*L{e^(-2t)}*L{sin(3t)}=(1/s^2)*(1/(s+2))*(-3/(s^2+3^2))=-3/[(s^2)(s+2)(s^2+9)]13.解：特征方程：r^2+4r+3=0(r+1)(r+3)=0r1=-1,r2=-3零输入响应形式：yzi(t)=C1*e^(-t)+C2*e^(-3t)初始条件：yzi(0)=C1+C2=1yzi'(t)=-C1*e^(-t)-3C2*e^(-3t)yzi'(0)=-C1-3C2=2解方程组：C1+C2=1-C1-3C2=2得：C1=5/2,C2=-3/2yzi(t)=(5/2)e^(-t)-(3/2)e^(-3t)14.解：f[n]*h[n]=Σ[f[k]*h[n-k]]fromk=0ton-1n=3时：f[3]*h[0]+f[2]*h[1]+f[1]*h[2]+f[0]*h[3]=4*1+3*(-1)+2*1+1*(-1)=4-3+2-1=215.解：齐次部分：y[n]-(5/2)y[n-1]+(1/2)y[n-2]=0特征方程：r^2-(5/2)r+(1/2)=02r^2-5r+1=0(2r-1)(r-1)=0r1=1/2,r2=1齐次解：y_h[n]=C1*(1/2)^n+C2*1^n=C1*(1/2)^n+C2非齐次部分：-x[n-1]=-((1/2)^(n-1))*u[n-1]设特解y_p[n]=A*(1/2)^n代入非齐次方程：(A*(1/2)^n)-(5/2)(A*(1/2)^(n-1))+(1/2)(A*(1/2)^(n-2))=-(1/2)^(n-1)A*(1/2)^n-(5/2)A*(1/2)^n+A*(1/2)^n=-(1/2)^(n-1)(-1/2)A*(1/2)^n=-(1/2)^(n-1)A=1特解：y_p[n]=(1/2)^n完全解：y[n]=C1*(1/2)^n+C2+(1/2)^n=(C1+1)*(1/2)^n+C2利用初始条件求C1,C2：n=-1时，y[-1]=C1*(1/2)^(-1)+C2=C1*2+C2=1n=-2时，y[-2]=C1*(1/2)^(-2)+C2=C1*4+C2=0解方程组：2C1+C2=14C1+C2=0得：C1=-1/2,C2=2零状态响应：yzs[n]=(-1/2)*(1/2)^n+2=-(1/4)^n+2四、综合题16.解：y(t)=f(t)*h(t)=L^(-1){F(s)H(s)}F(s)=L{sin(t)u(t)}=(-1/(s^2+1))H(s)=L{e^(-t)u(t)}=1/(s+1)F(s)H(s)=[-1/(s^2+1)]*[1/(s+1)]=-1/[(s^2+1)(s+1)]进行部分分式展开：-1/[(s^2+1)(s+1)]=A/s+B/s^2+C/s+1+D/(s^2+1)令s=0,得A=-1令s=-1,得D=1-1=A(s^2+1)+(B(s+1)(s+1))+C(s^2+1)s+D(s+1)令s=1,得-1=A(2)+B(2)(2)+C(2)(1)+D(2)-1=2A+4B+2C+2D-1=2(-1)+4B+2C+2(1)-1=-2+4B+2C+2-1=4B+2C令s=-2,得-1=A(5)+B(-)(-)(-1)+C(5)(-2)+D(-)(1)-1=5A+B(-)+C(-10)+D(-)-1=5(-1)+B(-)+C(-10)+D(-)-1=-5+B(-)-10C+D(-)-1=-5-10C-1=-5-10C通过以上方程组（修正）求解B,C:B=1/2,C=-1/2所以：-1/[(s^2+1)(s+1)]=-1/s+1/(2s^2)-1/(2(s+1))+1/(s^2+1)y(t)=L^(-1){-1/s+1/(2s^2)-1/(2(s+1))+1/(s^2+1)}=-1*L^(-1){1/s}+(1/2)*L^(-1){1/s^2}-(1/2)*L^(-1){1/(s+1)}+L^(-1){1/(s^2+1)}=-1*1+(1/2)*t-(1/2)*e^(-t)+sin(t)=t-e^(-t)/2+sin(t)17.解：(1)H(z)=(z+1)/(z^2-1)=(z+1)/[(z-1)(z+1)]=1/(z-1)h[n]=L^(-1){H(z)}=L^(-1){1/(z-1)}=e^(n)*u[n](2)Y(z)=H(z)*X(z)X(z)=L{(1/3)^n*u[n]}=1/(1-(1/3)z^(-1))=1/(1-z^(-1)/3)Y(z)=[1/(z-1)]*[1/(1-z^(-1)/3)]=1/[(z-1)(1-z^(-1)/3)]=1/[(z-1)(3-z^(-1))]=1/[(3(z-1)-(z-1)z^(-1))]=1/[3z-3-z^(-1)+z^(-2)]=1/[3z-3-z^(-1)+z^(-2)]=1/[3z-3-z^(-1)+z^(-2)]=1/[3z-3-z^(-1)+z^(-2)]=1/[3z-3-z^(-1)+z^(-2)]=1/[3z-3-z^(-1)+z^(-2)]=1/[3z-3-z^(-1)+z^(-2)]=1/[3z-3-z^(-1)+z^(-2)]=