信号与系统奥本海姆英文版课后答案chapter3

1、Chapter 3 Answers3.1 Using the Fourier series synthesis eq. (3.38) 3.2 Using the Fourier series synthesis eq. (3.95) 3.3 The given signal is Form this we may conclude that the fundamental frequency of x(t) is . The non-zero Fourier series coefficients of x(t) are , , 3.4 Since , , Therefore, Now , and for 3.5 Both and are periodic with fundamental period , Since y(t) is a linear combination of and ,it is also periodic with fundamental period , Therefore . Since .using the results in Table 3.1 we

2、 have Therefore 3.6 (a) Comparing with the Fourier series synthesis eq. (3.38) , we obtain the Fourier series coefficients to be Form Table 3.1 we know that if is real ,then has to be conjugate-symmetric,i.e, Since this is not true for , the signal is not real valued .Similarly , the Fourier series coefficients of are Form Table 3.1 we know that if is real ,then has to be conjugate-symmetric,i.e, Since this is not true for , the signal is real valued .Similarly , the Fourier series coefficients

3、of are Form Table 3.1 we know that if is real ,then has to be conjugate-symmetric,i.e, Since this is not true for , the signal is real valued .(b) For a signal to be even , its Fourier series coefficients must be even . This is true only for .3.7 Given that we have Therefore , When using given information Therefore , 3.8 Since x(t) is real and odd(clue 1), its Fourier series coefficients are purely imaginary and odd (See Table 3.1) Therefore , and ,Also since it is given that for , the only unkn

4、own Fourier series coefficients are and .Using Parsevals relation for the given signal we have Using the information given in clue (4) along with the above equation , Therefore or The two possible signals which satisfy the given information are and 3.9 The period of the given signal is 4 .Therefore , This gives , 3.10. Since the Fourier series coefficients repeat every N , we have , and Furthermore ,since the signal is real and odd ,the Fourier series coefficients willbe purely imaginary and odd

5、 . Therefore , and Finally 3.11 Since the Fourier series coefficients repeat every N=10,we have Furthermore ,since xn is real and even , is also real and even .Therefore We are also given that Using Parsevals relation , Therefore for k=2 ,.8, Now using the synthesis eq. (3.94) , we have 3.12. Using the multiplication property (see Table 3.2 ) , we have Since is 1 for all values of k, it is clear that +2+3will be for all values of k, Therefore, for all k,3.13 Let us first evaluate the Fourier ser

6、ies coefficients of .Clearly ,since is eal and odd,is purely imaginary and odd Therefore, =0. Now, = = -=Clearly, the above expression evaluates to zero or all even values of k Therefore.=When is passed through an LTI system with frequency response , the output is given by (see Section 3.8)Where , Since is non zero only or odd values of k, we need to evaluate the above summation only or odd k, Furthermore ,note thatis always zero or odd values of k, Therefore,3.14 The signal is periodic with per

7、iod N=4, Its Fourier series coefficients are for all kFrom the results presented in Section 3.8 , we know that the output is given by = +From the given information , we know that is = = = = Comparing this with eq. (S3.14-1), we have And and 3.15 From the results o Section 3.8,Where , Since is zero for , the largest value of |k| or which is nonzero should be such that This implies that |k|8, Therefore , for |k|8, is guaranteed to be zero.3.16 (a) The given signal is Therefore, is periodic with period N=2 and its Fourier series coefficients in the range are and Using the results derived in Section 3.8 , the output is given by =0 + = 0,(b) The signal is periodic with period N= 16 The signal may be written as = = Therefore, the non-zero Fourier series coefficients of in the range are , Using the results derived in Section 3.8,the output is given by = = (c) The s

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