奥本海默信号与系统第2版课后习题解答

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1、CHAPTER11.1to1.41-partoftext1.42(a)Periodic:Fundamentalperiod=0.5s(b)Nonperiodic(c)PeriodicFundamentalperiod=3s(d)PeriodicFundamentalperiod=2samples(e)Nonperiodic(f)Periodic:Fundamentalperiod=10samples(g)Nonperiodic(h)Nonperiodic(i)Periodic:Fundamentalperiod=1samplep

2、2l.43yt()=æö3cosæö200t+---èøèø62p=9cosæö200t+---èø69æöp=---cos400t+---12èø39(a)DCcomponent=---29æöp(b)Sinusoidalcomponent=---cos400t+---2èø39Amplitude=---21200Fundamentalfrequency=---------Hzp1.44TheRMSvalueofsinusoidalx(t)isA¤2.Hence,theaveragepowerofx(t)ina1-ohm22r

3、esistoris()A¤2=A/2.1.45LetNdenotethefundamentalperiodofx[N].whichisdefinedby2pN=------WTheaveragepowerofx[n]isthereforeN-112P=----åx[]nNn=0N-1122æö2pn=----åAcos----------+fNèøNn=02N-1A2æö2pn=------åcos----------+fNèøNn=01.46Theenergyoftheraisedcosinepulseispw¤12E=ò---

4、()cos()wt+1dt–pw¤41pw¤2=---ò()cos()wt+2cos()wt+1dt201pw¤æö11=---òèø---cos()2wt++---2cos()wt+1dt20221---æö3---æö---p-==3p¤4w2èø2èøw1.47Thesignalx(t)iseven;itstotalenergyistherefore52E=2òx()tdt024252=21ò()dt+25ò()–tdt041543=2[]t+2–---()5–tt=03t=4226==8+---------331.48(

5、a)Thedifferentiatoroutputisì1for–5<

6、2002025t31.51x(10t-5)1.0t00.10.50.91.01.52(a)x(t)1t-1123-1y(t-1)t-1123-1x(t)y(t-1)11t-123-141.52(b)x(t+1)x(t-1)11t-11234-1y(-t)y(-t)1t-2-11234-1x(t-1)y(-t)1t-2-11234-11.52(c)-2t-1123-1-2-1t1234x(t+1)y(t-2)-2-11234t51.52(d)x(t)1t-3-2-1123-1y(1/2t+1)6-5-4-3-2-1t1246-1.

7、0x(t-1)y(-t)1t-3-2-1123-11.52(e)x(t)1-4-3-2-1t123-1y(2-t)123t-4-3-2-1x(t)y(2-t)-1t123-161.52(f)x(t)1t-2-112-1y(t/2+1)1.0-5-3-2-1t-61123-1.0x(2t)y(1/2t+1)+1-0.5t-112-11.52(g)x(4-t)1t-7-6-5-4-3-2-1y(t)t-2-1124x(4-t)y(t)=0t-3-2-112371.53Wemayrepresentx(t)asthesuperposit

8、ionof4rectangularpulsesasfollows:g1(t)11234tg2(t)111234tg3(t)11234tg4(t)11234t0Togenerateg1(t)fromtheprescribedg(t),weletg()t=gatb(