信号与系统奥本海姆英文版课后答案chapter6

127 Chapter 6 Answers

6.6 (b) the impulse response h1[n] is as shown in figure s6.6,as was increase ,it is clear that the significant

central lobe of h1[n] becomes more concentrated around the origin. consequently. h[n]=h1[n](-1)^n also

becomes more concentrated about the origin.

6.7 the frequency response magnitude |H(jw)| is as shown in figure s6.7.the frequency response of the

bandpass filter G(jw) will be given by

(){2()cos(4000)}GjFThtt

((4000))((4000))HjHj

This is as shown in figure s6.7

-6000 -4000 -2000 0 2000 4000 6000

Figure S6.7

(a) from the figure ,it is obvious that the passband edges are at 2000∏rad/sec and 6000∏rad/sec. this

translates to 1000HZ and 3000Hz,respectively.

(b) (b)from the figure ,it is obvious that the stopband edges are at 1600∏ rad/sec.this translates to 800Hz and

3200 Hz, respectively.

6.8 taking the Fourier transform of both sides of the first difference equation and simplifying, we obtain the

frequency response H(e^jw)of the first filter.

01()().()1MjkjkjkNjjkkkbeYeHeXeae

Taking the Fourier transform of both sides of the second difference equation and simplifying ,we obtain the

frequency response H1(e^jw) of the second filter.

01(1)()().()1(1)MkjkjkjkNjkjkkkbeYeHeXeae

This may also be written as

()()0()1()()().()1MjkjkjjkNjjkkkbeYeHeHeXeae

Therefore .the frequency response of the second filter is obtained bu shifting the frequency response of the first

filter by ∏.although the first fitter has its passband between-wp and wp. Therefore, the second filter will have

its passband between ∏-wp and ∏+wp.

6.9 taking the Fourier transform of the given differential equation and simplifying .we obtain the frequency of

the LTI system to be

()2()()5jjjYeHeXej

Taking the inverse Fourier transform, we obtain the impulse response to be

5()2().thteut

Using the result derived in section 6.5.1,we have the step response of the system -4000-2000-10001000 2000 4000 H(jG（j） 128 52()()*()[1]().5tsthtuteut

The final value of the step response is

2().5s

We also have

052()[1].5tse

Substituting s(t0)=(2/5)[1-1/e^2],in the above equation ,we obtain t0=2/5 sec

(a) we may rewrite H1(jw)to be

1()()(0.1).40Hjjj

we may then treat each of the two factors as individual first order systems and draw their bode magnitude

plots .the final bode will then be asum of these two bode plots .this is shown in the figures6.10

mathematically. the straight-line approximation of the bode magnitude plot is

101020,0.120log|()|20log(),0,14032,40Hj

Figure S6.10

(b) Using a similar approach as in part (a),we obtain the Bode plot to be as shown in

Figure S6.10.

Mathematically, the straight-line approximation of the BODE magnitude plot is

101020,0.220log20log(),0.25028,50Hj

6.10. (a) We may rewrite the given frequency response 1H(j) as

12250250()()50.525(0.5)(50)Hjjjjj.

We may then use an approach similar to the one used in Example 6.5 and in Problem

6.11 to obtain the Bode magnitude plot(with straight line approximations) shown in

Figure S6.11.

Mathematically, the straight-line approximation of the Bode magnitude plot is 10 1 

-28 0 20

0.1 10log|()|xj

100 50

0.1 1 40 0

-20 32 40

20 10log|()|xj



50

0 1 100 0.1

-68 -20

-40 20 40

0.5 

50 100 10

0.1 0.5 0 20



-60 10 10log|()|xj 10log|()|xj 129 10101020,0.520log20log()14,0.55040log()48,50Hj

(b)We may rewrite the frequency response 2 H( j) as

2H()j=20.02(50)()0.21jjj.

Again using an approach similar to the one used in Example 6.5,we may draw the

Bode magnitude plot by treating the first and second order factors separately. This

Givens us a Bode magnitude plot (using straight line) approximations as shown below:

Mathematically , the straight-line approximation of the Bode magnitude plot is

1010100,120log40log,15020log()34,50Hj

6.12. Using the Bode magnitude plot, specified in Figure P6.12(a). we may obtain an expression

For 1H(j). The figure shows that 1H(j) has the break frequencies1=1, 2=8,And

3=40. The frequency response rises as 20dB/decade after 1. At 2,this rise is canceled by a -20

dB/decade contribution. Finally, at 3 ,an additional -20 dB/decade. Contribution results in the

subsequent decay at the rate of -20 dB/decade, therefore, we may conclude that

1H()j=123()()()Ajjj.

We now need to find A. Note that when =0, 20101log(0)Hj=2.Therefore. 1H(0)j=0.05. From eq .