信号与系统 奥本海姆第九章答案

Chapter 9

9.21 Solution:

(a). Q

)()()(32t u e t u e t x t t −−+= ∴

)3)(2(523121)(+++=+++=s s s s s s X , 2}Re{−>s (b). Q

)()5(sin )()(54t u t e t u e t x t t −−+= ∴

)55)(55)(4(70155)5(541)(222j s j s s s s s s s X −++++++=++++=, 4}Re{−>s

(i). Q

)()()(t u t t x +=δ

∴ s s s s X 111)(+=+=, 0}Re{>s

(f). Q

)3()3()(t u t t x +=δ ∴

s s s s X 333/13131)(+=⋅+=, 0}Re{>s

9.22 Solution:

(a). Q 1211/6/6()9(3)(3)33

j j X s s s j s j s j s j −===+++−+−,0}Re{>s ∴ 3311()()()(sin 3)()663j t j t j j x t e u t e u t t u t −=−

+=

(b). Q 221/21/2()9(3)(3)33s s X s s s j s j s j s j =

==+++−+−,0}Re{>s ∴ 33211()()()(cos3)()22

j t j t x t e u t e u t t u t −=+= (c) From the property of shifting in the time-domain and (b),we can get 22()(cos(3))()()9s x t t u t s −−=−−⇔

−+,Re{}0s < So 2()cos(3)()()9

s g t t u t G s s =−−⇔=+,Re{}0s < From the property of shifting in the s-domain,we can get 321()(1)(1)9

s X s G s s +==+++,Re{}1s <− and 3()()(cos3)()t t x t e g t e t u t −−==−−

9.28. Solution:

(a). All possible ROCs:

),1(),1,1(),1,2(),2,(+∞−−−−−∞

(b). It’s obviouse to see:

)2,(−−∞ unstable & uncausal

)1,2(−− unstable & uncausal

)1,1(− stable & uncausal ),1(+∞ unstable & causal

9.31. Solution:

(a). Q )()(2)()(22t x t y dt t dy dt

t y d =−−

∴ )()(2)()(2s X s Y s sY s Y s =−− ∴

13/123/1)1)(2(121)()()(2+−+−=+−=−−==s s s s s s s X s Y s H

(b). 1. The system is stable.

∴ ROC: (-1,2)

∴ )()()(31231t u e t u e t h t t −−−−=

2. The system is causal.

∴ ROC: ),2(+∞

∴ )()()(31231t u e t u e t h t t −−=

3. The system is neither stable nor causal

∴ ROC: )1,(−−∞ ∴ )

()()(31231t u e t u e t h t t −+−−=− 9.32. Solution:

from (1)

Q t e t x 2)(=, for all t and x(t) is a eigen function

∴ t t s e e s H t y 2226

1|)()(=⋅== ∴ 61|)(2==s s H from (2)

Q

)()()()(2)(4t bu t u e t h dt t dh t +=+− ∴ s b s s h s sH ++=+41)(2)( ∴ )

2)(4()4()(++++=s s s s b s s H when 2=s , 6146262)2(=××+=

b h ∴ 862=+b ,1=b

∴ 0}Re{,........)

4(2)2)(4()2(2)(>+=+++=s s s s s s s s H

9.33. Solution:

Q )()()(||t u e t u e e t x t t t −+==−− ∴ )

1)(1(21111)(−+−=−−+=s s s s s X ∴ )()()(s H s X s Y =2

21)1)(1(22+++⋅−+−=

s s s s s )22)(1(22++−−=s s s )

22(5652)1(522++++−−=s s s s

1)1(541)1()1(52)1(521)1(54)1(52)1(52222+++++++−−=+++++−−=s s s s s s s ∴ )(sin 5

4)(cos 52)(52)(t tu e t tu e t u e t h t t t −−++−= 9.35. Solution:

According to the block-diagram, we will know (a) 126121611)(2222++−−=++−−=s s s s s

s s s s H , Re{s}>-1 ∴ )(6)()()()(2)(2222t x dt t dx dt

t x d t y dt t dy dt t y d −−=++ (b) It’s obviouse that this system is stable.