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)}G j FT h t t ωπ=

))

-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()

().

()

1M

j k j k j k j k

k k b e Y e H e X e a e

ωω

ωω-=-==

=

-∑∑ 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)M

k j k

j k

j k N j k j k

k k b e

Y e H e X e a e ωωωωω-=-=-==

--∑∑

This may also be written as

()()0()1

()

()().

()

1M

j k

j k j j k N j j k

k k b e

Y e H e H e X e a e ωπωωωπωωπ---=--===

=-∑∑

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()()

5j

j j Y e H e X e j ωω

ωω

=

=

+ Taking the inverse Fourier transform, we obtain the impulse response to be 5()2().t h t e u t -=

Using the result derived in section 6.5.1,we have the step response of the system

128

52

()()*()[1]().5

t s t h t u t e u t -==-

The final value of the step response is 2().5

s ∞=

We also have

052()[1].5

t s e -∞=- 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).40

H j j j ωωω=++

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 (),0,14032,40H j ωωωωω-⎧⎪=⎨⎪⎩

(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

()10

1020,0.220log 20log (),0.250

28,50H j ωωωωω<<⎧

⎪

≈-<<<<⎨⎪->>⎩

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

2250250()()50.525(0.5)(50)

H j j j j j ωωωωω==

++++. 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

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