Chapter 1 Answers
1.1 Converting from polar to Cartesian coordinates:
1j?11?cos??? e222 1
2e?j?j?11?cos(??)?? 22eej?2?cos()?jsin()?j
??225?j2?ej9?4j?2?j
?1?j
?cos()?jsin()??j
22?j?? 42e?2(cos()?jsin())?1?je?2??442e?2ej?42e?9?j4?2e??j4?1?j
2ej??4?1?j
?21.2 converting from Cartesian to polar coordinates:
5?5ej0, ?2?2ej?, ?3j?3e2?j
?2??j13?j4, , 21?j?2?j?ee22?1?j??2e1?j3?j
??j(1?j)?e4, 1?j?4 2?j2??12
ee?1?j. (a) E?=
?e0??4tdt??1, P?=0, because E??? 42 (b) x2(t)?ej(2t?4), x(t)?1.Therefore, E?= P?=lim1T???x2(t)????2dt=
?2????dt=?,
(t)2T?x2?TTT2dt?lim1T??2T?T?Tdt??T????lim1?1
2?? (c)
x(t)=cos(t). Therefore, E22TT???TT??n=
?x3(t)dt=?????cos(t)dt=?,
nP?=lim1?cos(t)dt?lim1?2T2T,
n211?COS(2t)1dt? ?T22 (d)
?1??1?x[n]???u[n]x1[n]???u[n]?4??2? P?=0，because E?
?n?. Therefore, E?=?????x1[n]2???n?0??1???4??4 32=?, (e) [n]=e?j(2?8), [n]2=1. therefore, E?=??x2x2?x2[n]??
P?=limN1?N??2N?1n??Nnx2[n]2N1?lim1?1. ?N??2N?1n??N222 (f) x[n]=???. Therefore, E?=??=???=???, [n]3?cos?xcos(n)?3cos(n)?????4???4??4
????. (a) The signal x[n] is shifted by 3 to the right. The shifted signal will be zero for n<1, And n>7.
(b) The signal x[n] is shifted by 4 to the left. The shifted signal will be zero for n<-6. And n>0.
(c) The signal x[n] is flipped signal will be zero for n<-1 and n>2.
(d) The signal x[n] is flipped and the flipped signal is shifted by 2 to the right. The new Signal will be zero for n<-2 and n>4.
(e) The signal x[n] is flipped and the flipped and the flipped signal is shifted by 2 to the left.
This new signal will be zero for n<-6 and n>0.
. (a) x(1-t) is obtained by flipping x(t) and shifting the flipped signal by 1 to the right. Therefore, x (1-t) will be zero for t>-2.
P?=limN??1cos??N??2N?1n??N4n1?cos(n)12)?1 lim(?N??2N?1n?22??NN? (b) From (a), we know that x(1-t) is zero for t>-2. Similarly, x(2-t) is zero for t>-1, Therefore, x (1-t) +x(2-t) will be zero for t>-2.
(c) x(3t) is obtained by linearly compression x(t) by a factor of 3. Therefore, x(3t) will be
zero for t<1.
(d) x(t/3) is obtained by linearly compression x(t) by a factor of 3. Therefore, x(3t) will be
zero for t<9.
(a) x1(t) is not periodic because it is zero for t<0.
(b) x2[n]=1 for all n. Therefore, it is periodic with a fundamental period of 1. (c) x3[n] is as shown in the Figure . 1 1 1 … x3[ …-3 5
-401n 4
-1 -1 -1 Therefore, it is periodic with a fundamental period of 4.
. (a)
11[n]=x[n]?x[?n]?(u[n]?u[n?4]?u[?n]?u[?n?4]) x?v11212 Therefore, x[n]is zero for x[n]>3.
?v1 (b) Since x1(t) is an odd signal, x[n] is zero for all values of t.
???????1?v?2? (c)
1[n]???x3?v2n?n?? 1??1??1??x1[n]?x1[?n]?2???u[n?3]???u[?n?3]??2?????2???
Therefore, v1?v?x[n]? is zero when n<3 and when n3?5t5t44??.
(d)
11(t)??(x(t)?x(?t))??e?x?22?u(t?2)?eu(?t?2)?
?Therefore,
?v?x(t)?is zero only when t4??.
. (a) ?l{x1(t)}??2?2ecos(0t??)
(b) ??l{x(t)}??2cos(?)cos(3t?2?)?cos(3t)?e0tcos(3t?0)
24??0t (c) ??l{x(t)}??e?tsin(3??t)?e?tsin(3t??)
32 (d) ??l{x(t)}???e?2tsin(100t)?e?tsin(100t??)?e?2tcos(100t?)
42. (a) (t) is a periodic complex exponential.
?x1
2
x(t)?je1j10t?e???j?10t??2??
(b) x(t) is a complex exponential multiplied by a decaying exponential. Therefore,
is not periodic.
（c）x3[n] is a periodic signal.
x[n] is a complex exponential with a fundamental period of 2???.
32x3[n]=
ej7?n=ej?n.
(d) x[n] is a periodic signal. The fundamental period is given by N=m(
32?)
43?/5
=m(10). By choosing m=3. We obtain the fundamental period to be 10.
5(e) x[n] is not periodic. x[n] is a complex exponential with w=3/5. We cannot find any
50
integer m such that m(2? ) is also an integer. Therefore,
w.
x[n]5 is not periodic.
0x(t)=2cos(10t＋1)-sin(4t-1)
Period of first term in the RHS =2???.
1045Period of first term in the RHS =2??? .
2Therefore, the overall signal is periodic with a period which the least common multiple of the periods of the first and second terms. This is equal to? . x[n] = 1+ej4?n?ej5?n
2.
7Period of first term in the RHS =1.
Period of second term in the RHS =?2??=7 （when m=2）
???4/7?Period of second term in the RHS =?2?????2?/5?=5 (when m=1)
Therefore, the overall signal x[n] is periodic with a period which is the least common Multiple of the periods of the three terms inn x[n].This is equal to 35.
. The signal x[n] is as shown in figure . x[n] can be obtained by flipping u[n] and then
Shifting the flipped signal by 3 to the right. Therefore, x[n]=u[-n+3]. This implies that
M=-1 and no=-3. X[n] ---0 1 2 3 n
3 2 Figure S 1
?0,t??2 y(t)= x(?)dt =(?(??2)??(??2))dt=,?1,?2?t?2
???????0,t?2?t?tTherefore E??2
?dt?4?2 The signal x(t) and its derivative g(t) are shown in Figure .
x(t)
g(t) 1
-1 1
-1 0 1 2 0 t
-3 -3
-2
Figure S Therefore g(t)?3??(t?2k)?3??(t?2k?1)
k???k?????2 t This implies that A1=3, t1=0, A2=-3, and t2=1.
(a) The signal x2[n], which is the input to S2, is the same as y1[n].Therefore ,
1 x2[n-3] 21 = y1[n-2]+ y1[n-3]
2 y2[n]= x2[n-2]+
=2x1[n-2] +4x1[n-3] +
1( 2x1[n-3]+ 4x1[n-4]) 2 =2x1[n-2]+ 5x1[n-3] + 2x1[n-4] The input-output relationship for S is
y[n]=2x[n-2]+ 5x [n-3] + 2x [n-4]
(b) The input-output relationship does not change if the order in which S1and S2 are connected series reversed. . We can easily prove this assuming that S1 follows S2. In this case , the signal x1[n], which is the input to S1 is the same as y2[n]. Therefore y1[n] =2x1[n]+ 4x1[n-1]
= 2y2[n]+4 y2[n-1]
=2( x2[n-2]+
11 x2[n-3] )+4(x2[n-3]+ x2[n-4]) 22 =2 x2[n-2]+5x2[n-3]+ 2 x2[n-4]
The input-output relationship for S is once again
y[n]=2x[n-2]+ 5x [n-3] + 2x [n-4]
(a)The system is not memory less because y[n] depends on past values of x[n].
(b)The output of the system will be y[n]= ?[n]?[n?2]=0
(c)From the result of part (b), we may conclude that the system output is always zero for inputs of the form ?[n?k], k? ?. Therefore , the system is not invertible .
(a) The system is not causal because the output y(t) at some time may depend on future values of x(t). For instance , y(-?)=x(0).
(b) Consider two arbitrary inputs x1(t)and x2(t).
x1(t) ?y1(t)= x1(sin(t)) x2(t) ? y2(t)= x2(sin(t))
Let x3(t) be a linear combination of x1(t) and x2(t).That is , x3(t)=a x1(t)+b x2(t) Where a and b are arbitrary scalars .If x3(t) is the input to the given system ,then the corresponding output y3(t) is y3(t)= x3( sin(t))
=a x1(sin(t))+ x2(sin(t))
=a y1(t)+ by2(t) Therefore , the system is linear.
.(a) Consider two arbitrary inputs x1[n]and x2[n].
x1[n] ? y1[n] =x2[n ] ? y2[n] =
n?n0k?n?n0n?n0?x[k]
1k?n?n0?x[k]
2Let x3[n] be a linear combination of x1[n] and x2[n]. That is :
x3[n]= ax1[n]+b x2[n]
where a and b are arbitrary scalars. If x3[n] is the input to the given system, then the
n?n0corresponding output y3[n] is y3[n]=
n?n0k?n?n0?x[k]
3n?n0n?n012k?n?n0k?n?n0 =
k?n?n0?(ax[k]?bx[k])=a?x[k]+b?x[k]
12 = ay1[n]+b y2[n]
Therefore the system is linear.
(b) Consider an arbitrary input x1[n].Let
n?n0y1[n] =
k?n?n0?x[k]
1be the corresponding output .Consider a second input x2[n] obtained by shifting x1[n] in time: x2[n]= x1[n-n1]
The output corresponding to this input is
n?n0y2[n]=
k?n?n0?x[k]= ?x[k?n ]= ?x[k]
211k?n?n01k?n?n1?n0n?n1?n01k?n?n1?n0n?n0n?n1?n0Also note that y1[n- n1]=
?x[k].
Therefore , y2[n]= y1[n- n1] This implies that the system is time-invariant. (c) If x[n] (a) (i) Consider two arbitrary inputs x1(t) and x2(t). x1(t) ? y1(t)= tx1(t-1) x2(t) ? y2(t)= tx2(t-1) Let x3(t) be a linear combination of x1(t) and x2(t).That is x3(t)=a x1(t)+b x2(t) where a and b are arbitrary scalars. If x3(t) is the input to the given system, then the corresponding output y3(t) is y3(t)= tx3 (t-1) = t(ax1(t-1)+b x2(t-1)) = ay1(t)+b y2(t) Therefore , the system is linear. (ii) Consider an arbitrary inputs x1(t).Let y1(t)= tx1(t-1) be the corresponding output .Consider a second input x2(t) obtained by shifting x1(t) in time: x2(t)= x1(t-t0) The output corresponding to this input is y2(t)= tx2(t-1)= tx1(t- 1- t0) Also note that y1(t-t0)= (t-t0)x1(t- 1- t0)? y2(t) Therefore the system is not time-invariant. (b) (i) Consider two arbitrary inputs x1[n]and x2[n]. x1[n] ? y1[n] = x12[n-2] x2[n ] ? y2[n] = x22[n-2]. Let x3(t) be a linear combination of x1[n]and x2[n].That is x3[n]= ax1[n]+b x2[n] 22222222
信号与系统奥本海姆英文版课后答案
