电子科大课堂讲义课堂版信号678章课件.ppt

1、Chapter 6Time and Frequency Characterizationof Signals and Systems th txjHjX tyjY Chapter 6 Time and Frequency Characterization thtxtyTime-Domain:jHjXjYFrequency-Domain:6.1 The Magnitude-Phase Representation （幅度幅度-相位）相位）of the Fourier Transform jXjejXjX dejXtxtj 21 jXMagnitude 2 jXEnergy-Density Cha

2、pter 6 Time and Frequency Characterization6.2 The Magnitude-Phase Representation of the Frequency Response of LTI Systems jHjXjY jHjXjY jHjXjY jHGain jHPhase Shift1.Linear Phase2.Nonlinear Phase Chapter 6 Time and Frequency CharacterizationHomework:6.5 6.236.5 Consider a continuous-time ideal bandba

3、ss filter whose frequency response is Chapter 6 Problem Solutionelsewhere ,03 ,1ccjH(a)If is the impulse response of this filter,determine a(b)function such that th tg tgttthcsin(b)As is increased,does the impulse response of the filterget more concentrated or less concentrated about the origin?cSol

4、ution ttgc2cos(b)It will get more concentrated about the origin.(a)Chapter 6 Problem Solution6.23 Shown in Figure 6.23 is for a lowpass filter.Determine and sketch the impulse response of the filter for each of the following phase characteristics:(a)jH0jH01cjHcjHjH ttthcsinTjejHjH TtTtthcsinTjH(b),w

5、here T is a constant.0 2/0 2/-jH(c)Chapter 6 Problem Solution0 0 2/2/-jjeejHjH0jccjtjccettj22sintjccettj22sin012c2cF2sinttc th ttthc2/sin22 Chapter 7 Sampling Chapter 7 Sampling Chapter 7 SamplingSampling Theorem:Let be a band-limited signal withThen is uniquely determined by its samples if whereMjX

6、 ,0 tx tx,1,0,nnTxMs2Ts2 txp tx tx nTttpnjHMscM0TcjHc Chapter 7 SamplingHomework:7.1 7.2 7.3 7.6 7.97.3 The Effect of Undersampling:Aliasing 欠采样欠采样 混叠混叠 Chapter 7 Problem Solution7.1 A real-valued signal is known to be uniquely determined by its samples when the sampling frequency is .For what value

7、s of is guaranteed to be zero?txjX000,10s 5000 ,0jX Chapter 7 Problem Solution7.2 A continuous-time signal is obtained at the output of an ideal lowpass filter with cutoff frequency .If impulse-train sampling is performed on ,which of the following sampling periods would guarantee that can be recove

8、red from its sampled version using an appropriate lowpass filter?(a)T=0.510-3 (b)T=210-3 (c)T=10-4 tx000,1c tx tx 000,1cM 000,22Ms(a)and(c)3max102sT Sampling interval Chapter 7 Problem Solution7.3 Determine the Nyquist rate corresponding to each of the following signals:t,t,tx 0004sin0002cos1 a tt,t

9、x 0004sin b 2 0004sin ctt,tx 000,4M 000,82Ms 000,4M 000,82Ms 000,8M 000,162Ms Chapter 7 Problem Solution7.6 11 ,0 jX twp tw tx1 tx2 tw nTttpn twp22 ,0 jXDetermine the maximum sampling interval T such that is recoverable from through the use of an ideal LPF.21 ,0 jWNyquist rate212 smaximum sampling i

10、nterval21max2 sT jXjXjW2121 Chapter 7 Problem Solution7.9 Consider the signal 2 50sintttxwhich we wish to sample with a sampling frequency ofto obtain a signal with Fourier transform .Determinethe maximum value of for which it is guaranteed that150s0 75 jXjG tgjG0100050jX100050jG100100150150100 5005

11、 0X j55 h t x t p t pyt y t例例 在如图在如图1所示系统中所示系统中,输入信号输入信号 的频谱如图的频谱如图2所示。所示。已知已知 ，要求：，要求：x t sin2th tt 画出图中信号画出图中信号 和和 的频谱；的频谱；确定确定T的取值范围，以使信号的取值范围，以使信号 能从能从 中恢复。中恢复。y t pyt y t pyt1 p t-2T -T -T1 0 T1 T 2T t Chapter 7 Problem Solution Chapter 8 Communication Systems Chapter 8 Communication SystemsMod

12、ulation:The general process of embedding an information-bearing signal into a second signal.Extracting the information-bearing signal.Demodulation:Modulation:Amplitude Modulation(AM)Frequency Modulation(FM)Phase Modulation(PM)Chapter 8 Communication SystemsHomework:8.1 8.3 8.22 Chapter 8 Problem Sol

13、ution8.1 MjX ,0cjjXjY 2Determine a signal such that tm tmtytx tjcetm 21Solution Chapter 8 Problem Solution tm 0ty ttxttgtm 000,4sin21000,2cos 000,2 0 000,2 2 jH8.3 Determine .000,2 ,0jX ttxtg000,2sint2000cos tg tyLPF jH tySolution Be out of the passband of LPF 8.22In Figure(a),a system is shown with

14、 input and outputThe input signal has the Fourier transform shown in Figure(b)Determine and sketch .Chapter 8 Problem SolutionjXjY tx ty01W3W3jH2 tx tr1 tr2W3W31jH1W5W5 ty tr3Wt3cosWt5cosjXW20W21Figure(a)Figure(b)-7W -5W -3W 0 3W 5W 7W jR121-5W -3W 0 3W 5W jR221-8W -6W -2W 0 2W 6W 8W jR341-2W 0 2W j

15、Y41 Chapter 8 Problem Solution例例 如图所示的系统中，已知输入信号如图所示的系统中，已知输入信号 的傅立叶变换的傅立叶变换 如图如图1所示，且所示，且 。要求：。要求：1 h tt f ttcsincosctABCM0F jM1 f t F Mc 图图1 画出图画出图2中中A、B、C各点的信号频谱；各点的信号频谱；说明这是一个做什么用的系统。说明这是一个做什么用的系统。Chapter 8 Problem Solution 1h t 2ht x tcos 4tcos 6tABC例例 如图所示系统中，如图所示系统中，1sin4th ttt 2sin,dthtdtt若输

16、入信号若输入信号 sin2tx tt，试分别，试分别A、B、C各点信号各点信号的频谱。的频谱。Chapter 8 Problem SolutionProblems for Fourier AnalysisExample 4 In Figure(a),a system is shown with input signal and output signal .If the following information are given.ttdtdthc2sin1cjejH/22 ttthc3sin3 tuth41.Determine jH12.Determine the impulse resp

17、onse of the whole system .th3.If the input signal ,4.determine the output signal 2/cos2sintttxcc ty th1 th4 th3jH2 ty tx+-ty tx 1 21.0 ccjHj Problems for Fourier Analysis sin2.2/cccth tt t 3.cos/2cy tt Problems for Fourier Analysis例例 如图所示系统中，已知如图所示系统中，已知 ，cMcMc1且且 ，概略画出，概略画出 ，和和12c ty tr1 tr3 tr2的波形

18、。说明整个系统等价于一个带通滤波器，的波形。说明整个系统等价于一个带通滤波器，并用并用 和和 确定带通滤波器的上、下截止频率确定带通滤波器的上、下截止频率1,c2jH1 ty tftccosjH2M0MjF tr1 tr2 tr30122jH20111jH11Problems for Fourier Analysis Consider an LTI system with unit impulse response sin2sintth tttt 2.If the input is determine the output 313sincoscos622x tttt y t1.Determin

19、e the frequency response of the whole system.Problems for Fourier Analysis 222jeFj tf2011ttsinA period of Example 1 Determine the Fourier transform of tfExample 2 A real continuous-time signal with Fourier transform ,and 1.If is even,determine .2.If is odd,determine .tf tf tf tf tfjFjFlnProblems for

20、 Fourier Analysis ln FjFje 1.If is real,even tf 21 1f tt 2.If is real,odd tf 21tf tt Problems for Fourier AnalysisExample 3 Consider the following LTI systems with impulse response:1.ttth4sin 28sin4sintttth2.tttth8cos4sin3.tttx6sin2cos tyIf the input is Determine the output 1.F4sinttth1jH044 tty2cosSolution:42 jHProblems for Fourier Analysis ttttth8sin4sin2.10441088214441212jH36 jH42 jH ttty6sin32cos4 tttth8cos4sin3.2/10441212jH2/16,02jHjH tty6sin21