Chapter 5 Answers
5.1 (a) let x[n]= (1/2)n?1u[n-1].Using the Fourier transform analysis equation (5.9).the Fourier transform
X(ejw) of this signal is
X(ejw)?n????x[n]e
?jwn? =?(1/2)n?1e?jwn
n?1? =
?(1/2)n?0?ne?jw(n?1)
=e?jw 1?jw(1?(1/2)e)jw(b) Let x[n]=(1/2)|n?1|.Using the Fourier transform analysis equation (5.9).The Fourier transform x(e)of signal is
x(ejw)?n?????x[n]e?jwn=
n????=
0(1/2)?(n?1)e?jwn??(1/2)n?1e?jwnn?1?
The second summation in the right—hand side of the above equation is exactly the same as result of part (a).Now ,
n????(1/2)jw0?(n?1)e?jwn?(1/2)n?0?n?1ejwn1=(1/2)
(1?(1/2)ejw)
Therefore
x(e)5.2 (a) let
1=(1/2)+
(1?(1/2)ejw)e?jw?jw0.75e1=
(1?(1/2)e?jw)(1.25?cosw)x[n]??[n?1]??[n?1]jw . Using the Fourier
transform analysis equation (5.9).the Fourier transform x(e)of this signal is
x(e)?=
jwn?????x[n]e?jwn
+=
(b) Letx[n]??[n?2]??[n?2] .using the Fourier transform analysis equation (5.9). the Fourier transformx(ejw) of this signal is
?x(ejw)??x[n]e?jwn
n???e?jwejw2cosw=e-e=2jsin(2w)
5.3 We note from section 5.2 that a periodic signal with Fourier series representation
x[n]=
k??N?2jw?2jw?akejk(2?/N)n
2?k )Nhas a Fourier transform
X(ejw)=
k????2?a?(w?k?( a ) Consider the signalx1[n]?sin(?n??) .We note that the fundamental period of the signal x1[n] is N=6.
34The signal may be written as
??x1[n]= 1ej(3n?4) ?
2j?2?2?n?)?jnjn1?j(?1?j?1j?46 ?4e34= ee6 ee2j2j2jForm this , we obtain the non-zero Fourier series coefficients akof x1[n] the range
?2???3 as a?1??(1/2j)e4
Therefore , in the range ???w?? ,we obtain
2?2?X(ejw)?2?a1?(w?)?2?a?1?(w?) a1?(1/2j)e4j??j?66 ?(?/j){ej?/4?(w?2?/6)?e?j?/4?(w?2?/6)}
8(b) consider the signal x[n]?2?cos(?n??).we note that the fundamental period of the signal x1[n] is
26N=12.the signal maybe written as
x1[n]?2?(1/2)e?2?(1/2)e2?jn12j(n?)68???(1/2)e?j(n?)68??
e?j?8?(1/2)e?2??jn?j812e Form this ,we obtain the non-zero Fourier series coefficients ak of x2[n] in the range ?5?k?6 as a?1?(1/2)e?j8
Therefore ,in the range ,we obtain
2?2?X(ejw)?2?a0?(w)?2?a1?(w?)?2?a?1?(w?)a1?(1/2)ej8a0?2
?1212
?4??(w)??{ej?/8?(w?)?e?j?/8?(w?)}66??5.4 (a)Using the Fourier transform synthesis equation (5.8)
x1[n]?(1/2?)?X1(ejw)ejwndw
????(1/2?)?[2??(w)???(w??/2)???(w??/2)]ejwndw
????ej0?(1/2)ej(?/2)n?(1/2)e?j(?/2)n
?1?cos(?n/2)
(b)Using the transform synthesis equation (5.8)
?x2[n]?(1/2?)?X2(ejw)ejwndw
????(1/2?)?2jejwndw?(1/2?)?2jejwndw
??00??(j/?)[?1?ejn?j?n?ej?n?1 ]jn??(4/(n?))sin2(n?/2)
5.5 From the given information
?x[n]?(1/2?)?x(ejw)ejwndw
??=(1/2?)?|X(ejw)ej<{X(e???jw)}ejwn|)dw
?(1/2?)???/4??/4e3?w2ejwndw
sin((n?3/2)) 4??(n?3/2)The signal x[n]is zero when
?4(n?3/2)?4(n?3/2) is a nonzero integer multiple of ? or when |n|?? .the value of
can never be such that it is a nonzero integer multiple of ? .Therefore .x[n]=0 only for n=??
5.6 Throughout this problem, we assume that
FTX[n] ???x1(ejw)
(a) Using the time reversal property (Sec.5.3.6),we have
FTX(e?jw) x[-n]???Using the time shift property (Sec.5.3.3) on this .we have FTFTx[-n+1] ??? e?jwn x(e?jw) and x[-n-1] ??? ejwn x(e?jw) Therefore
FTx1[n]?x[-n+1]+x[-n-1]???e?jwn X(e?jw)+ ejwn X(e?jw) ???2X(e?jw)cosw
(b) Using the time reversal property (Sec.5.3.6) ,we have
FTx[-n] ???X(e?jw)
Using the same conjugation property on this ,we have
FTx*[-n] ???X*(e?jw) Therefore
FTx2[n]=(1/2)(x*[-n]+x[n])???(1/2)X(ejw)?X*(ejw)
???Re{X(ejw)}
(c) Using the differentiation frequency property (Sec.5.3.8),we have
dX(ejw)
nx[n]???jdwFTFTFTUsing the same property second time ,
d2X(ejw)
nx[n]????dw22FTTherefore
d2X(ejw)d2X(ejw)x3[n]?nx[n]?2nx[n]?1?????2j?X(ejw) 22dwdw2FT5.7 (a) Consider the signal y1[n] with Fourier transform
Y1(ejw)??sin(kw)
k?110jwWe see that Y1(e) is real and odd .From Table 5.1 , we know that the Fourier transform of a real and odd
signal is purely imaginary and odd. Therefore ,we may say that the Fourier transform of a purely imaginary and odd signal is real and odd. Using this observation, we conclude that y1[n] is purely imaginary and odd Note now that
X1(ejw)?e?jwY1(ejw) Therefore , x1[n]?y1[n?1].therefore , is also purely imaginary .but x1[n] is neither even nor odd (b)We note that X2(ejw)is purely imaginary and odd. Therefore, x2[n] has to be real and odd.
(d) ?Consider a signal y3[n]whose magnitude of the Fourier transform is |Y3(ejw)|?A(w) and whose phase
of the Fourier transform is<{Y3(ejw)}??(3/2)w .since|Y3(ejw)|?|Y3(e?jw)| and ,we may conclude that the
signal y3[n]is real (see Table 5.1,property5.3.4).
Now, consider the signal x3[n] with Fourier transform X3(ejw)?Y3(ejw)ej??Y3(jw).Using the result from previous paragraph and the linearity property of the Fourier transform .we may conclude that has to
jwreal .since the Fourier transform ,we may conclude that has to real . since the Fourier transform X3(e)is neither purely imaginary nor purely real .the signalx3[n] is neither even nor odd 5.8 Consider the signal
|n|?1x1[n]?{1,0, |n|?1 From the table 5.2, we know that
sin(3w/2) FTx1[n]???X1(ejw)?sin(w/2)Using the accumulation property (Table 5.1，Property 5.3.5),we have
?1jwj0x1[k]???X1(e)??X1(e)??(w?2?k) ??jw1?ek???k???FTnTherefore , in the range ???w?? ,
k????x[k]???1?eFT1n1?jwX1(ejw)?3??(w)
Also, in the range ???w??,
FT1???2??(w)
Therefore , in the range ???w?? ,
x[n]?1?k????x[k]???1?eFT1nn1?jwX1(ejw)?5??(w)
The signal x[n] has the desired Fourier transform .We may express x[n] mathematically as
1n??2= x[n]?1??x1[k]n?3?1?n?1k???4n?25.9 From property 5.3.4 in Table 5.1 , we know that for a real signal x[n], FTOd{x[n]}???jIm{X1(ejw)} From the given information
jImX1(ejw)?jsinw?jsin2w
?(1/2)(ejw?e?jw?e2jw?e?2jw) Therefore,
Od{x[n]}?IFT{jImX1(ejw)}?(1/2)(?[n?1]??[n?1]??[n?2]??[n?2]) We also know that Od{x[n]}?x[n]?x[?n]
2And that x[n]=0 for n>0. therefore
x[n]?2Od{x[n]}??[n?1]??[n?2] for n<0
Now we only have to find x[0] .Using Parseval’s relation ,we have
?21?jw|X(e)|dw??|x[n]|2
2????n???Form the given information, we can write
?13?(x[n])2?|x[n]|2?(x[n])2?2
n???? This gives x[0]=1.but since we are given that x[0]>0.we conclude that x[0]=1 Therefore
x[n]??[n]??[n?1]??[n?2]
5.10 From table 5.2 we know that
1n1FT()u[n]???121?e?jw2Using property 5.3.8 in table 5.1,
1?jwe 1nd1FTjw2x[n]?n()u[n]???X(e)?j{}?12dw1?1e?jw(1?e?jw)222??Therefore , x[n]?n(1)n?x[n]?X(ej0)?2
??2n?0??5.11 We know from the time expansion property (Table 5.1, Property 5.3.7) that
FT g[n]?x(2)[n]???G(ej?)?X(ej2?)
Therefore, G(ej?) is obtained by compressing X(ej?) by a factor of 2. Since we know that X(ej?) is periodic with a periodic of 2?, we my conclude that G(ej?) has a periodic which is (1/2)2?=?. Therefore,
G(ej?)?G(ej9???))anda??.
5.12 Consider the signal
?? ??sin4x1[n]????n??n????? For Table 5.2, we obtain the Fourier transform of x1[n] to be
?1,0?|?|??/4
X1(ej?)???0,?/4?|?|?? The plot of X1(ej?) is as shown in Figure S5.12. Now consider the signal x1[n]=( x2[n])2. Using the
multiplication property (Table 5.1, Property 5.5), we obtain the Fourier transform of x2[n] to be
X2(ej?)?(1/2?)[X1(ej?)?X1(ej?)]
This is plotted in Figure S5.12.
From Figure S5.12. It is clear that X2(ej?) is zero for ???/2. By using the convolution property (Table 5.1, Property 5.4), we know that
?sin()?cn? Y(ej?)?X(ej?)FT1???n??
The plot of
?sin?cn?FT????n? is shown in Figure S5.12. It is clear that of then ?/2????.
5.13 When two LTI systems connected in parallel, the impulse response of the overall system is the sum of the impulse response of the individual. Therefore, h[n]= h1[n]+ h2[n]
using the linearity property (Table 5.1, Property 5.3.2)
Given that h1[n]=(1/2)nu[n], we obtain
Therefore,
Taking the inverse Fourier transform, h2[n]=-2(1/4)nu[n].
5.14 From the given information, we have the Fourier transformG(ej?)of g[n] to be
Also, when the input to the system is x[n]=(1/4)nu[n], the output is g[n]. Therefore,
For Table 5.2, we obtain
Therefore,
Clearly, h[n] is a three point sequence. We have and We see that H(ej?)?H(ej(???)) only if h[1]=0.
We also have
j?/2Since we are also given that H(e)?1, we have h[0]-h[2]=1 (S5.14-1) Now not that
信号与系统奥本海姆英文版课后答案chapter5
