Fourier series: Periodic signals and lti Systems ()=∑H(k k= ak一→H(ko)ak “g Soak-→|H(jkco)lkl H(7k)=1H(k0e∠B(ko) or powers of signals get modified through filter/system ncludes both amplitude phase akeJhwon

Fouriers derivation of the ct fourier transform x(t)-an aperiodic signal view it as the limit of a periodic signal as t→∞ For a periodic signal the harmonic components are spaced Oo=2π/ T apart. AsT→∞,Obo→>0, and harmonic components are space

2.0 引言 2.1 离散时间LTI系统——卷积和 2.2 连续时间LTI系统：卷积积分 2.3 LTI系统性质 2.4 用微分方程和差分方程描述的因果LTI系统 2.5 奇异函数

Motivation for the Laplace transform CT Fourier transform enables us to do a lot of things, e. g Analyze frequency response of lTi systems Sampling Modulation Why do we need yet another transform? One view of Laplace Transform is as an extension of the Fourier

Question: Are there sets of basic\ signals so that a) We can represent rich classes of signals as linear combinations of these building block signals b) The response of ltI Systems to these basic signals are both simp and insightful Fact: For LtI Systems(CT or dt) there are two natural choices for these building blocks Focus for now:

第一部分选择题(共30分) 一、单项选择题(本大题共10小题,每小题3分,共30分) 在每小题列出的四个选项中只有一个是符合题目要求的.请将其代码填在题后的括号内。错选或未选均无分。 串联谐振电路,固有谐振频率取决于 A电源电压幅值 B电源电压的初始相位 C.电源电压频率

3.1 X(eo)=2xnJe-jon where x[n] is a real sequence. Therefore X(e)=Rl∑xnlo/。 ∑xR(-mu)=∑ x[n]cos(on),and xmm)=m∑刈nm∑刈mc-m)=-2 xn] sin(oon) Since cos(on)and sin(on)are, respectively, even and odd functions of o, Xre(eJo) is an even function of o

REMINDER: Quiz 1 will be held from 7: 30 to 9: 30 p. m. Tuesday, October 14 The quiz will cover material in Chapters 1-3 of o&w Lectures and Recitations through september 26, Problem Sets 1-3 and that part of Problem Set 4 involving problems from Chapter 3 Reading assignments

PROBLEM SET 7 Issued: October 28. 2003 Due: November 5. 2003 REMINDER: Computer Lab 2 is also due on November 7 Reading Assignments Lectures #14-15 PS#7: Chapter 7(through Section 7. 4)and Chapter 8(through Section 8.4) of o&W Lectures #16-18 PS#8: Section 7.5 and Chapters 8 and 9(through Section 9.6)of O&W Exercise for home study(not to be turned in, although we will provide solutions)

The lowpass filter H(u) has a cutoff frequency wc=205T rad/ sec. Thus, c(t)is r(t) where all terms with frequency above we are removed by the lowpass filter. The terms which are kept have kwol 205T rad /sec k|< 10.25, so the output, ac(t),is r(t)= To obtain n, we sample c(t) every T=5 10-3 seconds with an impulse train The sampling frequency is 400T=2 x maximum frequency in c(t). Therefore