1、Chapter 1 Answers1.1 Converting from polar to Cartesian coordinates:1cos22je11cos()22je()injjjinjjj52jj 4(s)(4j941jj91jje1.2 converting from Cartesian to polar coordinates:, , 05j 2je23je, , 213j41jj, 4()jj 12j1.3. (a) = , =0, because E0tdePE(b) , .Therefore, = = = ,(2)4jtx12()dtx=P21limliTTtdtlim1(
2、c) =cos(t). Therefore, = = = ,2()t 232cost=2()lilicos()TTCOStt(d) , . Therefore, = 1nux4nuE20413nx=0，because 7.(b) The signal xn is shifted by 4 to the left. The shifted signal will be zero for n0.(c) The signal xn is flipped signal will be zero for n2.(d) The signal xn is flipped and the flipped si
3、gnal is shifted by 2 to the right. The new Signal will be zero for n4.(e) The signal xn is flipped and the flipped and the flipped signal is shifted by 2 to the left. This new signal will be zero for n0.1.5. (a) x(1-t) is obtained by flipping x(t) and shifting the flipped signal by 1 to the right.Th
4、erefore, x (1-t) will be zero for t-2.(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 bezero for t3.v x(b) Since x1(t) is an
5、 odd signal, is zero for all values of t.2v(c) 1332v nnnu Therefore, is zero when 3 and when .3v (d) 1 5541()()(2)()v tttttxeTherefore, is zero only when .v1.8. (a) 01()2cos()tt(b) 02)32s(cos(3)4t(c) 3()in(in)t ttxe(d) 2 24s10)s(1s(1)t t tte1.9. (a) is a periodic complex exponential.1()t102()jtjttx(
6、b) is a complex exponential multiplied by a decaying exponential. Therefore, 2tx() is not periodic.（c） is a periodic signal. = = .3n3n7jejnis a complex exponential with a fundamental period of .2(d) is a periodic signal. The fundamental period is given by N=m( )4 3/5= By choosing m=3. We obtain the
7、fundamental period to be 10.10().3m(e) is not periodic. is a complex exponential with =3/5. We cannot find any integer m such 5nx5nx0wthat m( ) is also an integer. Therefore, is not periodic.02w5nx1.10. x(t)=2cos(10t1)-sin(4t-1)Period of first term in the RHS = .210Period of first term in the RHS =
8、.4Therefore, the overall signal is periodic with a period which the least common -3-141-10-4111-1n5x3n20-1-2-3 1 2 3XnnFigure S 1.1210-1210-1 t1-2g(t)2-3 -3tFigure S 1.14x(t)multiple of the periods of the first and second terms. This is equal to .1.11. xn = 1+ 74jne25jPeriod of first term in the RHS
9、 =1.Period of second term in the RHS = =7 （when m=2 ）7/Period of second term in the RHS = =5 (when m=1)2Therefore, the overall signal xn is periodic with a period which is the least commonMultiple of the periods of the three terms inn xn.This is equal to 35.1.12. The signal xn is as shown in figure
10、S1.12. xn can be obtained by flipping un and thenShifting the flipped signal by 3 to the right. Therefore, xn=u-n+3. This implies thatM=-1 and no=-3.1.13y(t)= = =tdx)(dtt )()2(,tTherefore 24dtE1.14 The signal x(t) and its derivative g(t) are shown in Figure S1.14.Therefore )kktttg12(3)2(3)(This impl
11、ies that A =3, t =0, A =-3, and t =1.121.15 (a) The signal x n, which is the input to S , is the same as y n.Therefore ,2 21y n= x n-2+ x n-32= y n-2+ y n-311=2x n-2 +4x n-3 + ( 2x n-3+ 4x n-4)11=2x n-2+ 5x n-3 + 2x n-411The input-output relationship for S is 3yn=2xn-2+ 5x n-3 + 2x n-4(b) The input-
12、output relationship does not change if the order in which S and S are connected series 12reversed. . We can easily prove this assuming that S follows S . In this case , the signal x n, which is the 12 1input to S is the same as y n.12Therefore y n =2x n+ 4x n-11= 2y n+4 y n-122=2( x n-2+ x n-3 )+4(x
13、 n-3+ x n-4)12212=2 x n-2+5x n-3+ 2 x n-42The input-output relationship for S is once again yn=2xn-2+ 5x n-3 + 2x n-41.16 (a)The system is not memory less because yn depends on past values of xn.(b)The output of the system will be yn= =0n(c)From the result of part (b), we may conclude that the syste
14、m output is always zero for inputs of the form , k . Therefore , the system is not invertible .n1.17 (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 x (t)and x (t).12x (t) y (t)= x (sin
15、(t)1x (t) y (t)= x (sin(t)22Let x (t) be a linear combination of x (t) and x (t).That is , x (t)=a x (t)+b x (t)3 1 312Where a and b are arbitrary scalars .If x (t) is the input to the given system ,then the corresponding output 3y (t) is y (t)= x ( sin(t)3 3=a x (sin(t)+ x (sin(t)12=a y (t)+ by (t)
16、Therefore , the system is linear.1.18.(a) Consider two arbitrary inputs x nand x n.12x n y n =101kxnkx n y n =2202nkLet x n be a linear combination of x n and x n. That is :3 1x n= ax n+b x n312where a and b are arbitrary scalars. If x n is the input to the given system, then the corresponding outpu
17、t y n is y n= 3 303knk= =a +b)(210kbxank 01kxn02kxn= ay n+b y nTherefore the system is linear.(b) Consider an arbitrary input x n.Let 14y n =101kxnkbe the corresponding output .Consider a second input x n obtained by shifting x n in time:2 1x n= x n-n 1The output corresponding to this input is y n=
18、= = 202knkn10n01knkAlso note that y n- n = .101xnkTherefore , y n= y n- n 21This implies that the system is time-invariant.(c) If B, then yn (2 n +1)B.nx0Therefore ,C (2 n +1)B.01.19 (a) (i) Consider two arbitrary inputs x (t) and x (t). x (t) y (t)= t x (t-1)121121x (t) y (t)= t x (t-1)2Let x (t) b
19、e a linear combination of x (t) and x (t).That is x (t)=a x (t)+b x (t)3 32where a and b are arbitrary scalars. If x (t) is the input to the given system, then the corresponding output 3y (t) is y (t)= t x (t-1)3323= t (ax (t-1)+b x (t-1)12= ay (t)+b y (t)Therefore , the system is linear.(ii) Consid
20、er an arbitrary inputs x (t).Let y (t)= t x (t-1)1121be the corresponding output .Consider a second input x (t) obtained by shifting x (t) in time:1x (t)= x (t-t )20The output corresponding to this input is y (t)= t x (t-1)= t x (t- 1- t )2210Also note that y (t-t )= (t-t ) x (t- 1- t ) y (t)102Ther
21、efore the system is not time-invariant.(b) (i) Consider two arbitrary inputs x nand x n. x n y n = x n-21211x n y n = x n-2.222Let x (t) be a linear combination of x nand x n.That is x n= ax n+b x n3 3where a and b are arbitrary scalars. If x n is the input to the given system, then the correspondin
22、g output 3y n is y n = x n-232=(a x n-2 +b x n-2)12=a x n-2+b x n-2+2ab x n-2 x n-212ay n+b y nTherefore the system is not linear.(ii) Consider an arbitrary input x n. Let y n = x n-2 112be the corresponding output .Consider a second input x n obtained by shifting x n in time:1x n= x n- n 20The outp
23、ut corresponding to this input is y n = x n-2.= x n-2- n 125Also note that y n- n = x n-2- n 10120Therefore , y n= y n- n 2This implies that the system is time-invariant.(c) (i) Consider two arbitrary inputs x nand x n.1x n y n = x n+1- x n-111x n y n = x n+1 - x n -1222Let x n be a linear combinati
24、on of x n and x n. That is :3 1x n= ax n+b x n31where a and b are arbitrary scalars. If x n is the input to the given system, then the corresponding output y n is y n= x n+1- x n-13 3=a x n+1+b x n +1-a x n-1-b x n -11212=a(x n+1- x n-1)+b(x n +1- x n -1)12= ay n+b y nTherefore the system is linear.
25、(ii) Consider an arbitrary input x n.Let y n= x n+1- x n-1111be the corresponding output .Consider a second input x n obtained by shifting x n in time: x n= x n-2 121n 0The output corresponding to this input is y n= x n +1- x n -1= x n+1- n - x n-1- n 221010Also note that y n-n = x n+1- n - x n-1- n
26、 1010Therefore , y n= y n-n This implies that the system is time-invariant.(d) (i) Consider two arbitrary inputs x (t) and x (t).12x (t) y (t)= 1d(t) x1x (t) y (t)= 2Let x (t) be a linear combination of x (t) and x (t).That is x (t)=a x (t)+b x (t)3 12312where a and b are arbitrary scalars. If x (t)
27、 is the input to the given system, then the corresponding output 3y (t) is y (t)= 33d(t)= (t)b+a21=a +b = ay (t)+b y (t) d12Therefore the system is linear.(ii) Consider an arbitrary inputs x (t).Let1y (t)= =(t) x1)(x-t 1tbe the corresponding output .Consider a second input x (t) obtained by shifting
28、 x (t) in time:21x (t)= x (t-t )10The output corresponding to this input is y (t)= =2(t2d(- 2t= )-0101Also note that y (t-t )= y (t)10(x)tt2Therefore the system is not time-invariant.61.20 (a) Givenx = y(t)=)(tjte2tje3x = y(t)=jttjSince the system liner) =1/2( + )tjetx21(/)jt(1tytj3tjTherefore(t)= c
29、os(2t) =cos(3t)1 (1ty(b) we know that(t)=cos(2(t-1/2)= ( + )/22jet2jjte2Using the linearity property, we may once again write(t)= ( + ) =( + )= cos(3t-1)x12jetjte)1tjt3jt3Therefore,(t)=cos(2(t-1/2) =cos(3t-1)(y1.21.The signals are sketched in figure S1.21.Figure S1.211.22 The signals are sketched in
30、 figure S1.221.23 The even and odd parts are sketched in Figure S1.23t-10-11 2213x(t-1)a11/2-1/2-1n73210x3- n0.50.5t3/2-3/2t4-1 321012x(2-t)10-112tx(2t+1)x(4-t/2)t10 12618412 )()(uxt10211/2-1/2-1n73210xn-4(b)-111/2-1/2n210x3n(c)1-12n0x3n+1(d)2112n0xnun-3=xn(f)7t-1/2x0(t)1/2-1-2 0 21t1-1/2x0(t)1/2-1
31、-2 0 21-1/2x0(t)1/2-1-2 0 21 tFigure S1.220 11n2(h)-4-1-211/2n20x3- n/2 +(-1)nxn/2(g)7 n10-7xonx0(t)-t/20 tx0(t)t3t/2-3t/20(a)10-1/2-7 7-1/2nxn1(c)Figure S1.24-2 01/22 tx0(t)(a)(b)(c)Figure S1.231/2 n1/2-771n(n)xe3(b) 1/207 1/2-1nxon3/251-5nxen03/2-3/2-1/24n1/2xon81.24 The even and odd parts are ske
32、tched in Figure S1.241.25 (a) periodic period=2 /(4)= /2(b) periodic period=2 /(4)= 2(c) x(t)=1+cos(4t-2 /3)/2. periodic period=2 /(4)= /2(d) x(t)=cos(4 t)/2. periodic period=2 /(4)= 1/2(e) x(t)=sin(4 t)u(t)-sin(4 t)u(-t)/2. Not period.(f) Not period.1.26 (a) periodic, period=7.(b) Not period.(c) pe
33、riodic, period=8.(d) xn=(1/2)cos(3 n/4+cos( n/4). periodic, period=8.(e) periodic, period=16.1.27 (a) Linear, stable(b) Not period.(c) Linear(d) Linear, causal, stable(e) Time invariant, linear, causal, stable(f) Linear, stable(g) Time invariant, linear, causal1.28 (a) Linear, stable(b) Time invaria
34、nt, linear, causal, stable(c)Memoryless, linear, causal(d) Linear, stable(e) Linear, stable(f) Memoryless, linear, causal, stable(g) Linear, stable1.29 (a) Consider two inputs to the system such thatand 111.Sexnyxn 221.Seynx Now consider a third input n= n+ n. The corresponding system output321Will
35、be 3122eeexnytherefore, we may conclude that the system is additiveLet us now assume that inputs to the system such that/4111.Sjexyxn and /222.jen Now consider a third input x3 n= x2 n+ x1 n. The corresponding system outputWill be /43331122/4/41212cossin/ije mej jeeynxnxyntherefore, we may conclude
36、that the system is additive(b) (i) Consider two inputs to the system such that9and 2111Sdxtxtyt 2221Sdxtxtyt Now consider a third input t= t+ t. The corresponding system output321Will be 32112dxtytttytherefore, we may conclude that the system is not additiveNow consider a third input x4 t= a x1 t. T
37、he corresponding system outputWill be 241121dtytxatdytTherefore, the system is homogeneous.(ii) This system is not additive. Consider the fowling example .Let n=2 n+2+2 n+1+2 n and n= n+1+ 2 n+1+ 3 n. The corresponding outputs evaluated at n=0 arex2100/yandyNow consider a third input x3 n= x2 n+ x1
38、n.= 3 n+2+4 n+1+5 nThe corresponding outputs evaluated at n=0 is y30=15/4. Gnarly, y30 .This 021n442,0,xnyotherwis445 4,10,axayntiTherefore, the system is homogenous.1.30 (a) Invertible. Inverse system y(t)=x(t+4)(b)Non invertible. The signals x(t) and x1(t)=x(t)+2 give the same output(c) n and 2 n
39、give the same outputd) Invertible. Inverse system; y(t)=dx(t)/dt(e) Invertible. Inverse system y(n)=x(n+1) for n 0 and yn=xn for n0(f) Non invertible. x(n) and x(n) give the same result(g)Invertible. Inverse system y(n)=x(1-n)(h) Invertible. Inverse system y(t)=dx(t)/dt(i) Invertible. Inverse system
40、 y(n) = x(n)-(1/2)xn-1(j) Non invertible. If x(t) is any constant, then y(t)=0(k) n and 2 n result in yn=0(l) Invertible. Inverse system: y(t)=x(t/2)(m) Non invertible x1 n= n+ n-1and x2 n= n give yn= n(n) Invertible. Inverse system: yn=x2n1.31 (a) Note that x2t= x1 t- x1 t-2. Therefore, using linearity we get y2 (t)=y1 (t)- y1 (t-2).this is shown in Figure S1.31(b)Note that x3 (t)= x1 t+ x1 t+1. .Therefore, using linearity we get Y3 (t)= y1 (t)+ y1 (t+2). this is
