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

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

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

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

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

Can be continuous Trajectory of a space shuttle Mass density in a cross-section of a brain · Can be discrete dNa base sequence Digital image pixels Can be 1-D,2-D,·N-D For this course: Focus on a single(1-D) independent variable which we call“time

Problems to be handed in: In this lab, you will complete the Basic, Intermediate, and Advanced Exercises for the Echo Cancellation problem considered in Section 2.10 on pages 44-46 of Buck, Daniel, and Singer(BDS). For all of the exercises, please include

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

Exercise for home study O&W4.47 tea this problem examines the Fourier transform of a continuous-time LTI system with a real, causal impulse response, h(t) (a) To prove that H(w) is completely specified by eh(u) for a real and causal h(t) lore the even part of a function, he(t)

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