Chapter 2 answers (a) We have know that (S-1)1223401-1nx[n]The signals x[n] and h[n] are as how in Figure Sn201-12h[n]2Figure SFrom this figure we can easily see that the above convolution sum reduce
Chapter 2 answers (a) We have know that (S-1)1223401-1nx[n]The signals x[n] and h[n] are as how in Figure Sn201-12h[n]2Figure SFrom this figure we can easily see that the above convolution sum reduce
Chapter 4 (a)Let then the Fourier transform is : is as shown in figure .(b) Let then the Fourier transform is :12002(a)(b) is as shown in figure figure (a) Let then the Fourier transform is : is
Chapter 7 Answers
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归纳记忆:1F 变换对2 常用函数 F变换对:δ(t)ε(t)e -?t ε(t) gτ(t)sgn (t)e –?|t| 1 12πδ(ω)44 非周期信号的频谱傅里叶变换45傅里叶变换的性质一、线性(LinearProperty)若f1(t) ←→F1(jω), f2(t) ←→F2(jω)则证: F[a f1(t) + b f2(t)]= [a F1(jω) + b F2(jω) ][a f
41信号分解为正交函数42傅里叶级数43周期信号的频谱44非周期信号的频谱傅里叶变换45傅里叶变换的性质46周期信号的傅里叶变换47LTI系统的频域分析第四章 连续信号与系统的频域分析时域分析,以冲激函数为基本信号,任意输入信号可分解为一系列冲激函数之和;yf(t) = h(t)*f(t) 本章将以正弦信号和虚指数信号ejωt为基本信号,任意输入信号可分解为一系列不同频率的正弦信号或虚指数信号之和
第五章 连续系统的复频域分析51拉普拉斯变换52拉普拉斯变换的性质53拉普拉斯变换逆变换54复频域分析频域分析以虚指数信号ejωt为基本信号,任意信号可分解为众多不同频率的虚指数分量之和。使响应的求解得到简化,物理意义清楚,但也有不足:(1)有些重要信号不存在傅里叶变换,如e?t ?(t) ;(2)对于给定初始状态的系统难以利用频域分析,全响应困难在这一章将通过把频域中的傅里叶变换推广到复频域来解
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2-1: (7)12 3 513-22:求零输入响应:代入初始条件有:完全响应:AT0T02利用欧拉公式:4-9(4):tu(t)(2)作Z反变换有:7-15反Z变换7-21(1):练习题: 7-27-157-8(2) 7-21(1)6-1(5) 6-2(3) 6-12(1) 6-146-254-5 4-9(3)(4) 4-13(6) 4-303-14(4) 3-2
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