【免费下载】信号与系统第二版课后习题解答3 4奥本海姆

x(t)Solution:

a42ej/3kfor, a01, a2ex[n]j/4k0k1Express x(t) in the form

2ej0t2ej0t4jej30t4jej30t34cos(t)8sin(t)44akejk(2/N)nSignal and System

, a2eChap 3

Fundamental period T8.02/8/43.1 A continuous-time periodic signal x(t) is real value and has a

fundamental period T=8. The nonzero Fourier series coefficients for x(t) are

3.2 A discrete-time periodic signal x[n] is real valued and has a fundamental period N=5.The nonzero Fourier series coefficients for x[n] are

4812cos(n)4cos(n)5453438512sin(n)4sin(n)5456*a1a12,a3a34j.

x(t)Akcos(ktk)*j/3a01,a2ej/4,a2ej/4,a4a42ex[n]A0Aksin(knk)a0a2ej(4/5)na2ej(4/5)na4ej(8/5)na4ej(8/5)nakej0kta1ej0ta1ej0ta3ej30ta3ej30t1ej/4ej(4/5)nej/4ej(4/5)n2ej/3ej(8/5)n2ej/3ej(8/5)n1

j/4Express x[n] in the form

kN,

a42ej/3,

T63.1.

Solution:

Solution:

period as x1(t),

for the period ofcos(coefficients ak such that

x(t)2cos(x(t)x(t)2cos(that is T2T1T, w2w1Determine the fundamental frequency

Signal and System

3.3 For the continuous-time periodic signal

kae25t)4sin(t)33so the period ofx(t)is 6, i.e. w02/6/3frequency 1 and Fourier coefficients ak. Given that

12cos(20t)4sin(50t)212(ej20tej20t)2j(ej50tej50t)21then, a02, a2a2, a52j, a52j2x2(t)x1(1t)x1(t1)25t)4sin(t)33and the coefficients ak You may use the properties listed in Table

(1). Because x2(t)x1(1t)x1(t1), then x2(t) has the same

How is the fundamental frequency 2 of x2(t) related to? Also, find

a relationship between the Fourier series coefficients bk of x2(t)

3.5 Let x1(t) be a continuous-time periodic signal with fundamental

25t)is T3, the period ofsin(t)is 332

kjk0t0 and the Fourier series

.

so

for

while:

(2). bk1TTa00, akak,.

and ak0 for k1 x(t)122 4 |x(t)|dt1.

20 3. ak0 for |k|1.

k a1(ejta12/2jkakejkw1akejkw1(akak)ejkw1T2, then 02/2x(t)2sin(t)Signal and System

H(j)h(t)ejtdtx(t)3.13 Consider a continuous-time LTI system whose frequency response is

If the input to this system is a periodic signal

ejt)2a1sin(t)Specify two different signals that satisfy these conditions.Solution:

3.8 Suppose given the following information about a signal x(t):1. x(t) is real and odd.

1x2(t)ejkw2tdt(x1(1t)x1(t1))ejkw1tdtTTT1x1(1t)ejkw1tdtx1(t1)ejkw1tdtTTakej0kta0a1ej0ta1ej0t 2. x(t) is periodic with period T=2 and has Fourier coefficients ak.

x(t) is real and odd, then ak is purely imaginary and odd，

1222222x(t)dtaaa2a10111203

akej0ktsin(4)SSolution:

H(jk0) So y(t)0.

1,0t4x(t)1,4t8x(t)y(t)x(t).

y(t)kfor

x(t)x(t)y(t)y(t)kkkk即对于所有的k，H(jk0)1Signal and System

Fundamental period T8.02/8/4sin(4k0)4,.......k0k00,.......k01,.......100H(j)0,.......100For what values of k is it guaranteed that ak0?

akej0ktakej0ktakH(jk0)ejkw0t4a0

11418 Because a0x(t)dt1dt(1)dt0TT8084With period T=8,determine the corresponding system output y(t).

Solution:

另：x(t)为实奇信号，则ak为纯虚奇函数，也可以得到a0为0。

When the input to this filter is a signal x(t) with fundamental period

3.15 Consider a continuous-time ideal lowpass filter S whose frequency response is

T/6and Fourier series coefficientsak, it is found that

akH(jk0)ejk0takH(jk0)ejk0t4

Solution:

H(j)for

for

output y(t) is identical to x(t).

T= /7, 02/T14.

that isk0250,........kk18..or..k17.

x(t)1,||2500,otherwisey(t)kbkakH(jkw0)1,.......k17H(jkw0)0,.......otherwisek1,.......100H(j)0,.......100Signal and System

即12k<100，k<=8，故当k>8时，ak＝0。

也就是说k0100, T/6012For what values of k is it guaranteed that ak0?

1,.......w250,H(jw)0,.......otherwise3.35.Consider a continuous-time LTI system S whose frequency response is

Let y(t)x(t),bkak, it needs ak0,for k18..or..k17.

T/7 and Fourier series coefficients ak,it is found that the

Chap 4

akejw0kt250,and 14akH(jkw0)ejkw0t4.1 Use the Fourier transform analysis equation(4.9)to calculate the

When the input to this system is a signal x(t) with fundamental period

5

k is integer, so

(a).

(a).

(b).

(b).

(a)ee22t1jtee  2(t1) Fourier transforms of;

u(t1)

X(j)x(t)ejtdt2t2jt11e(2j)t|1e2e(2j)t|12j2jSketch and label the magnitude of each Fourier transform.Solution:

Sketch and label the magnitude of each Fourier transform.Solution:

X(j)x(t)ejtdtX(j)x(t)ejtdtX(j)x(t)ejtdt2|t1| ejej2cosSignal and System

e2(t1)u(t1)ejtdt[(t1)(t-1)]e-jtdt(t1)edt(2j)t(2j)ej2e2e ee2j12j2j4.2 Use the Fourier transform analysis equation(4.9) to calculate the

Fourier transforms of:

{(2t)(t2)}ejtdt(b)

d{u(2t)u(t2)}dtd{u(2t)u(t2)}ejtdtdt6

1(a)(t1)(t1)

ejej4ej2j2j42 1   1dtee--e(b)e2(t1)jte-jt1dte(2j)te2dtdte22tejtdt(t-1)e-jtdt121x(t)2133eThat is tin Table4.1.

3X(j)21ej(t3)23j(t)w23j(w)23  eX(jw)e3ejwtdw1ejwtj2(a)x1(t)x(1t)x(1t)ejk3,..........k0323j3(t)2|X(j)|2{u(3)u(3)}1dw22{u(w3)u(w3)}e(2t)eSignal and System

e3j(t)2the inverse Fourier transform of X(j)X(j)e332jsin3(t)2jsin3(t)12233j(t)(t)22ej22jsin233(t)k,.........k0,1,2,...2 If x(t)=0 then (t3)02Use your answer to determine the values of t for which x(t)=0.

Solution:

4.6 Given that x(t) has the Fourier transform X(j), express the

4.5 Use the Fourier transform synthesis equation(4.8) to determine

Fourier transforms of the signals listed below in the terms of

X(j).You may find useful the Fourier transform properties listed

7

33j(w)23j(t)w23ejt3j3(t)2dtX(jw)ejX(jw)ejwtdwejwtdwdw(t2)ejtdtjX(j),where

(c).

(a).

(b).

then

(b)x2(t)x(3t6)d2(c) x3(t)2x(t1)dtFT4.11 Given the relationships

FTx(t)  X(j)FTx(t)  X(j)Determine the values of A and B.Solution:

FTx(t)  X(j)d2FT2  x(t)(j)X(j)2dtto show that g(t) has the form

Signal and System

1jbx(atb)  X(j)eaaa

FTx1(t)x(1t)x(1t) FTjx(1t)  X(j)eX1(j)X(j)ejX(j)ej2X(j)cosSolution:

Accorrding to the properties of the Fourier transform, we’ll get:

y(t)x(t)h(t)，And g(t)x(3t)h(3t)FTjx(1t)  X(j)eFTx2(t)x(3t6)  X2(j)d2FTx3(t)2x(t1)X3(j)2X(j)ejdtAnd given that x(t) has Fourier transform X(j) and h(t) has

Fourier transform H(j),use Fourier transform properties

g(t)Ay(Bt)8

FTjx(t1)  X(j)e1X(j)ej2333.

for

2. F{(1j)X(j)}Aewe are given the following facts:1. x(t) is real and nonnegative.

But

1and

So

y(t)x(t)h(t)1Y(j)93That isX(j)FT then

|X(j)|d2.

From (3), we know:And we also knowAe2tFrom (2), we know : Ae1FTx(3t)  X(j)331FTh(3t)  H(j)33(1j)X(j)=

G(j)2t2tFTu(t) g(t)x(3t)h(3t)FT g(t)1A,B33Signal and System

Determine a closed-form expression for x(t).Solution:

From (1), we know x(t) is real and x(t)0;

x(t)Aetu(t)Ae2tu(t)4.14 Consider a signal x(t) with Fourier transform X(j).Suppose

AAA(j1)(j2)j1j2FTY(j)X(j)H(j)FT (1j)X(j)u(t) 9

Aj21y(3t)311X(j)H(j)3333u(t),where A is independent of t.

X(j)d2Aj222A2022X(j)d2While x(t)0,

(e2t2e3te4t)dtSignal and System

e2te3te4t2A(2)2340A122tx(t)dt22122212A()A23412222A=2,Sothat is A12, A1212u(t)12(ete2t)u(t)10

Then x(t)Aeu(t)Aet20A(ete2t)2dt2