信号与系统(习题课).ppt

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1、习题课SignalsandSystems第2章信号的时域分析第3章系统的时域分析第4章周期信号的频域分析第5章非周期信号的频域分析第6章系统的频域分析《信号与系统》习题课2-1定性绘出下列信号的波形(1)f(t)=u(t)-2u(t-1)-101t1-1-2u(t)-2u(t-1)f(t)1-1-101t2-1定性绘出下列信号的波形(2)f(t)=u(t+1)-2u(t)+u(t-1)1-1-101tf(t)1-1-101tu(t+1)-2u(t)u(t-1)f(t)1-12-1定性绘出下列信号的波形(2)f(t)=u(t+1)-2u(t)+u(t-1)另一种思路：f(

2、t)=u(t+1)-u(t)-[u(t)-u(t-1)]u(t+1)-u(t)=?-101tu(t)-u(t-1)=?-[u(t)-u(t-1)]=?2-1定性绘出下列信号的波形(4)f(t)=d(t-1)-2d(t-2)+d(t-3)f(t)1-12-2-3-2-10123t(1)(-2)(1)2-2定性绘出下列信号的波形(1)f(t)=u(t)u(3-t)f(t)1-1-3-2-10123tu(t)u(3-t)=u[-(t-3)]u(t-3)2-2定性绘出下列信号的波形(1)f(t)=u(t)u(3-t)f(t)1-1-3-2-10123t2-2定性绘出下列信号的波

3、形(3)f(t)=e-2tsin(2t)u(t)e-2tsin(2t)0p2p3pt1-1e-2tsin(2t)u(t)2-2定性绘出下列信号的波形(3)f(t)=e-0.5tsin(2t)u(t)0p2p3pt1-1e-0.5te-0.5tsin(2t)u(t)2-2定性绘出下列信号的波形(5)f(t)=(t-2)u(t)f(t)1-12-2-3-2-10123tt-2t(t-2)u(t)2-4利用单位阶跃信号u(t)表示下列信号-202tf(t)2f(t)=(t+2)u(t+2)u(-t)+2u(t)u(2-t)=(t+2)u(t+2)-tu(t)-2u(t-2)=

4、(t+2)[u(t+2)-u(t)]+2[u(t)-u(t-2)](a)2-4利用单位阶跃信号u(t)表示下列信号(b)f(t)213-3-2-10123tu(t+3)u(3-t)u(t+2)u(2-t)u(t+1)u(1-t)f(t)=u(t+3)u(3-t)+u(t+2)u(2-t)+u(t+1)u(1-t)=u(t+3)-u(t-3)+u(t+2)-u(t-2)+u(t+1)-u(t-1)2-4利用单位阶跃信号u(t)表示下列信号(c)01234tf(t)2-1f(t)=2u(t-1)u(2-t)-u(t-2)u(3-t)+u(t-3)u(4-t)=2[u(t-1

5、)-u(t-2)]-[u(t-2)-u(t-3)]+u(t-3)-u(t-4)=2u(t-1)-3u(t-2)+2u(t-3)-u(t-4)2-5写出下列信号的时域表达式(a)f(t)1-1-101tf(t)=t[u(t)-u(t-1)]+u(t-1)或者f(t)=tu(t)u(1-t)+u(t-1)2-5写出下列信号的时域表达式(c)f(t)1-1-101tf(t)=-t[u(t+1)-u(t)]+t[u(t)-u(t-1)]=-tu(t+1)u(-t)+tu(t)u(1-t)2-5写出下列信号的时域表达式(e)f(t)1-1-101tf(t)=u(t+1)-u(t)

6、+(1-2t)[u(t)-u(t-1)]-u(t-1)f(t)=u(t+1)u(-t)+(1-2t)u(t)u(1-t)-u(t-1)2-10已知信号波形,绘出下列信号波形f(t)12-3-2-10123t12-3-2-10123tf(-t)12-3-2-10123tf(t+2)12-3-2-10123tf(-3t)2-10已知信号波形,绘出下列信号波形f(t)12-3-2-10123t12-3-2-10123tf(-t)12-3-2-10123tf(5-3t)12-3-2-10123tf(-3t)连续LTI系统的响应经典时域分析方法全解＝齐次解＋特解卷积法完全响应＝零

7、输入响应＋零状态响应齐次解中0-时刻对应的分量卷积积分固有响应强迫响应例题：简单RC电路+－f(t)1W1F+－uc(t)已知f(t)=(1+e－3t)u(t)初始条件uC(0-)=1V求uC(t)。解：根据电容电流iC(t)=CduC(t)/dt得微分方程uC’(t)+uC(t)=f(t)特征方程s+1=0得特征根s1=－1(1)零输入响应（与齐次解形式相同）uCx(t)=K1e－t根据初始条件uC(0-)=1V得到K1=1，即零输入响应uCx(t)=e－t(2)冲激响应h(t)=Ae－tu(t)代入原微分方程－Ae－tu(t)+Ae－