信号与系统Chapter 3Answers (1).doc

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1、Chapter 33.1Solution:Fundamental period .3.2Solution: for, , , , , 3.3Solution:for the period ofis , the period ofis so the period ofis 6 , i.e. then, , , , 3.5Solution: (1). Because , then has the same period as ,that is , (2). 3.8Solution:while: is real and odd, then , , then and for so for3.13 So

2、lution:Fundamental period . Because So .3.15Solution:forif , it needs that isand is integer, so 3.22Solution: 3.34Solution:A periodic continous-signal has Fourier Series:. T is the fundamental period of .The output of LTI system with inputed is Its coefficients of Fourier Series: (a) .T=1, .(Note:If

3、 ,)So (b) .T=2, ,So ,(c) T=1, 3.35Solution: T= , .for,that is,and is integer, so .Let , it needs ,for .3.37Solution:A periodic sequence has Fourier Series:.N is the fundamental period of .The output of LTI system with inputed is .Its coefficients of Fourier Series: (a) .N=4, .So 3.40Solution:Accordi

4、ng to the property of fourier series: (a). (b).Because (c).Because (d). (e).first, the period of is then 3.43(a) Proof:(i)Because is odd harmonic,where for every non-zero even k.It is noticed that k is odd integers or k=0.That means (ii)Because of ,we get the coefficients of Fourier Series It is obvious that for every non-zero even k. So is odd harmonic,(b) Extra problems:, (1).Consider , when is (2).Consider , when isSolution:,(1).(for k can only has value 0)(2).(for k can only has value 1 and 1)

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