Solution Manual for Principles of Communications , 7th Edition

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APPENDIX A: PHYSICAL NOISE SOURCES AND NOISE CALCULATIONS1Problem A.1All parts of the problem are solved using the relationVrm s=p4kT RBwherek=1:381023J/KB=30MHz= 3107Hza. ForR= 10;000ohms andT=T0= 290KVrm s=p4 (1:381023) (290) (104) (3107)=6:93105V rms=69:3V rmsb.Vrm sis smaller than the result in part (a) by a factor ofp10 = 3:16:ThusVrm s= 21:9V rmsc.Vrm sis smaller than the result in part (a) by a factor ofp100 = 10:ThusVrm s= 6:93V rmsd. Each answer becomes smaller by factors of2;p10 = 3:16;and10, respectively.Problem A.2UseI=IsexpeVkT1We wantI >20IsorexpeVkT1>20.a. AtT= 290K,ekT=1:610191:381023290= 40, so we haveexp (40V)>21givingV>ln (21)40= 0:0761voltsi2rm s=2eIB'2eBIsexpeVkTori2rm sB=2eIsexpeVkT=21:61019 1:5105exp (400:0761)=1:00751022A2/Hzb. IfT= 90K, thenekT= 129, and forI >20Is, we need exp(129V)>21orV >ln (21)129= 2:36102volts

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APPENDIX A: PHYSICAL NOISE SOURCES AND NOISE CALCULATIONS2Thusi2rm sB=21:61019 1:5105exp (1290:0236)=1:00791022A2/Hzapproximately as before.Problem A.3a. Use Nyquistís formula to get the equivalent circuit ofReqin parallel withRL, whereReqisgiven byReq=R3(R1+R2)R1+R2+R3The noise equivalent circuit consists of a rms noise voltage,Veq, in series withReqand a rms noisevoltage,VL, in series with withRLwith these series combinations being in parallel.The equivalentnoise voltages areVeq=p4kT ReqBVL=p4kT RLBThe rms noise voltages across the parallel combination ofReqandRL, by using superposition andvoltage division, areV01=Ve qRLReq+RLandV02=VLRe qReq+RLAdding noise powers to getV20we obtainV20=V2eqR2L(Req+RL)2+V2LR2eq(Req+RL)2=(4kT B)RLReqReq+RL=4kT BRLR3(R1+R2)R1R3+R2R3+R1RL+R2RL+R3RLNote that we could have considered the parallel combination ofR3andRLas an equivalent loadresistor and found the Thevenin equivalent. LetRjj=R3RLR3+RLThe Thevenin equivalent resistance of the whole circuit is thenReq2=Rjj(R1+R2)Rjj+R1+R2=R3RLR3+RL(R1+R2)R3RLR3+RL+R1+R2=RLR3(R1+R2)R1R3+R2R3+R1RL+R2RL+R3RL

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APPENDIX A: PHYSICAL NOISE SOURCES AND NOISE CALCULATIONS3and the mean-square output noise voltage is nowV20= 4kT BReq2which is the same result as obtained before.b. WithR1= 2000; R2=RL= 300;andR3= 500, we haveV20B=4kT BRLR3(R1+R2)R1R3+R2R3+R1RL+R2RL+R3RL=41:381023(290) (300) (500) (2000 + 300)2000 (500) + 300 (500) + 2000 (300) + 300 (300) + 500 (300)=2:7751018V2/HzProblem A.4Find the equivalent resistance for theR1; R2; R3combination and setRLequal to this to getRL=R3(R1+R2)R1+R2+R3Problem A.5Using Nyquistís formula, we Önd the equivalent resistance looking back into the terminals withVrm sacross them.It isReq=50kk20kk(5k+ 10k+ 5k)=50kk20kk20k=50kk10k=(50k) (10k)50k+ 10k=8;333ThusV2rm s=4kT ReqB=41:381023(400) (8333)2106=3:681010V2which givesVrm s= 19:18V rms

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APPENDIX A: PHYSICAL NOISE SOURCES AND NOISE CALCULATIONS4Problem A.6To Önd the noise Ögure, we Örst determine the noise power due to a source at the output, thendue to the source and the network, and take the ratio of the latter to the former.Initally assumeunmatched conditions.The results areV20due toRS, only=R2kRLRS+R1+R2kRL2(4kT RSB)V20due toR1andR2=R2kRLRS+R1+R2kRL2(4kT R1B)+RLk(R1+RS)R2+ (R1+RS)kRL2(4kT R2B)V20due toRS; R1andR2=R2kRLRS+R1+R2kRL2[4kT(RS+R1)B]+RLk(R1+RS)R2+ (R1+RS)kRL2(4kT R2B)The noise Ögure is (after some simpliÖcation)F= 1 +R1RS+RLk(R1+RS)R2+ (R1+RS)kRL2RS+R1+R2kRLR2kRL2R2RSIn the above,RakRb=RaRbRa+RbNote that the noise due toRLhas been excluded because it belongs to the next stage. Since this isa matching circuit, we want the input matched to the source and the output matched to the load.Matching at the input requires thatRS=Rin=R1+R2kRL=R1+R2RLR2+RLand matching at the output requires thatRL=Rout=R2k(R1+RS) =R2(R1+RS)R1+R2+RSNext, these expressions are substituted back into the expression forF.After some simpliÖcation,this givesF= 1 +R1RS+2R2LRS(R1+R2+RS)=(RSR1)R22(R1+RS+RL) +R2L(R1+R2+RS)2R2RSNote that ifR1>> R2we then have matched conditions ofRL=R2andRS=R1:Then, thenoise Ögure simpliÖes toF= 2 + 16R1R2

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APPENDIX A: PHYSICAL NOISE SOURCES AND NOISE CALCULATIONS5Note that the simple resistive pad matching circuit is very poor from the standpoint of noise.Theequivalent noise temperature is found by usingTe=T0(F1)=T01 + 16R1R2Problem A.7a. The important relationships areFl= 1 +TelT0Tel=T0(Fl1)Te0=Te1+Te2Ga1+Te3Ga1Ga2Completion of the table givesAmpl. No.FTeiGai1not needed300 K10dB= 1026 dB864.5 K30dB= 1000311 dB3360.9 K30dB= 1000Therefore,Te0=300 + 864:510+3360:9(10) (1000)=386:8KHence,F0=1 +Te0T0=2:33 = 3:68dBb. With ampliÖers 1 and 2 interchangedTe0=864:5 + 30010 +3360:9(10) (1000)=865:14KThis gives a noise Ögure ofF0=1 + 865:14290=3:98 = 6dB

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APPENDIX A: PHYSICAL NOISE SOURCES AND NOISE CALCULATIONS6c. See part (a) for the noise temperatures.d. ForB= 50kHz,TS= 1000K, and an overall gain ofGa= 107, we have, for the conÖgurationof part (a)Pna;out=Gak(T0+Te0)B=1071:381023(1000 + 386:8)5104=9:57109wattsWe desirePsa;outPna;out= 104=Psa;out9:57109which givesPsa;out= 9:57105wattsFor part (b), we havePna;out=1071:381023(1000 + 865:14)5104=1:29108wattsOnce again, we desirePsa;outPna;out= 104=Psa;out1:29108which givesPsa;out= 1:29104wattsandPsa;in=Psa;outGa= 1:291011watts

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APPENDIX A: PHYSICAL NOISE SOURCES AND NOISE CALCULATIONS7Problem A.8a. The noise Ögure of the cascade isFoverall=F1+F21Ga1=L+F1(1=L) =LFb. For two identical attenuator-ampliÖer stagesFoverall=L+F1(1=L) +L1(1=L)L+F1(1=L)L(1=L) = 2LF12LF;L >>1c. Generalizing, forNstages we haveFoverallN F LProblem A.9a. The data for this problem isStageFiGi1 (preamp)2 dB = 1.58G12 (mixer)8 dB = 6.311.5 dB = 1.413 (ampliÖer)5 dB = 3.1630 dB = 1000The overall noise Ögure isF=F1+F21G1+F31G1G2which gives5dB= 3:16 = 1:58 + 6:311G1+ 3:1611:41G1or3:161:58=6:311G1+ 3:1611:41G1orG1=5:311:58 +2:16(1:41) (1:58) = 4:33 = 6:37dBb. First, assume thatG1= 15dB= 31:62:ThenF=1:58 + 6:31131:62+3:161(1:41) (31:62)=1:8 = 2:54dB

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APPENDIX A: PHYSICAL NOISE SOURCES AND NOISE CALCULATIONS8ThenTes=T0(F1)=290 (1:81)=230:95KandToverall=Tes+Ta=230:95 + 300=530:95KNow useG1as found in part (a):F=3:16Tes=290 (3:161) = 626:4KToverall=300 + 626:4 = 926:4Kc. ForG1= 15dB= 31:62; Ga= (31:62) (1:41) (1000) = 4:46104:ThusPna;out=GakToverallB=4:46104 1:381023(530:95)107=3:27106wattsForG1= 6:37dB= 4:33; Ga= (4:33) (1:41) (1000) = 6:11103:ThusPna;out=6:11103 1:381023(926:4)107=7:81107wattsNote that for the second case, we get less noise power out even wth a largerToverall.This is dueto the lower gain of stage 1, which more than compensates for the larger input noise power.d. A transmission line with lossL= 2dB connects the antenna to the preamp.We Örst ÖndTSfor the transmission line/preamp/mixer/amp chain:FS=FTL+F11GTL+F21GTLG1+F31GTLG1G2;whereGTL= 1=L= 102=10= 0:631andFTL=L= 102=10= 1:58Assume two cases forG1:15 dB and 6.37 dB.First, forG1= 15dB= 31:6, we haveFS=1:58 + 1:5810:631+6:311(0:631) (31:6) +3:161(0:631) (31:6) (1:41)=2:84

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APPENDIX A: PHYSICAL NOISE SOURCES AND NOISE CALCULATIONS9This givesTS= 290 (2:841) = 534KandToverall= 534 + 300 = 834KNow, forG1= 6:37dB= 4:33, we haveFS=1:58 + 1:5810:631+6:311(0:631) (4:33) +3:161(0:631) (4:33) (1:41)=5:00This givesTS= 290 (5:001) = 1160KandToverall= 1160 + 300 = 1460KProblem A.10a. (a) UsingPna;out=GakTSB=108 1:381023(1800)3106with the given values yieldsPna;out= 7:45105wattsb. We wantPsa;outPna;out= 105orPsa;out=105 7:45105= 7:45wattsThis givesPsa;in=Psa;outGa= 7:45108= 7:45108watts=71:28dBW=41:28dBm

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APPENDIX A: PHYSICAL NOISE SOURCES AND NOISE CALCULATIONS10Problem A.11a. ForA= 1dB,Y= 1:259and the e§ective noise temperature isTe= 600(1:259) (300)1:2591= 858:3KForA= 1:5dB,Y= 1:413and the e§ective noise temperature isTe= 600(1:413) (300)1:4131= 426:4KForA= 2dB,Y= 1:585and the e§ective noise temperature isTe= 600(1:585) (300)1:5851= 212:8Kb. These values can be converted to noise Ögure usingF= 1 +TeT0WithT0= 290K, we get the following values:(1) ForA= 1dB;F= 5:98dB; (2) ForA= 1:5dB; F= 3:938dB; (3) ForA= 2dB; F= 2:39dB.Problem A.12a. Using the data given, we can determine the following:=0:039m4d2=202:4dBGT=39:2dBPTGT=74:2dBWThis givesPS=202:4 + 74:2 + 65 =127:2dBWb. UsingPn=kTeBfor the noise power, we getPn=10 log10kT0TeT0B=10 log10[kT0] + 10 log10TeT0+ 10 log10(B)=174 + 10 log101000290+ 10 log10106=108:6dBm=138:6dBW

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APPENDIX A: PHYSICAL NOISE SOURCES AND NOISE CALCULATIONS11c.PSPndB=127:2(138:6)=11:4dB=101:14= 13:8ratiod. Assuming the SNR=z=Eb=N0= 13:8;we get the results for various digital signalingtechniques given in the table below:Modulation typeError probabilityBPSKQp2z= 7:4108DPSK12ez= 5:06107Noncoh. FSK12ez=2= 5:03104QPSKSame as BPSK

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APPENDIX D: ZERO-CROSSING AND ORIGIN ENCIRCLEMENT STATISTICS1Problem D.1a. Considerz(t)=Acos (!0+!d)t+n(t);!0= 2f0; !d= 2fd=Acos (!0+!d)t+nc(t) cos!0tns(t) sin!0twith the PSD ofn(t)given bySn(t) =N0=2forjff0j B=2and0otherwise.Notealso that the PSD of the quadrature noise components isSnc(f) =Sns(f) =N0;jfj B=20;otherwiseAlso, the quadrature noise components are uncorrelated.It follows thatn0c(t) cos (!0+!d)tn0s(t) sin (!0+!d)t=n0c(t) [cos!0tcos!dtsin!0tsin!dt]n0s(t) [sin!0tcos!dt+ cos!0tsin!dt]=n0c(t) cos!dtn0s(t) sin!dtcos!0tn0c(t) sin!dt+n0s(t) cos!dtsin!0tnc(t) cos!0tns(t) sin!0tThereforenc(t)=n0c(t) cos!dtn0s(t) sin!dtns(t)=n0c(t) sin!dt+n0s(t) cos!dtThese two equations can be solved forn0c(t)andn0s(t).The results aren0c(t)=nc(t) cos!dt+ns(t) sin!dtn0s(t)=nc(t) sin!dt+ns(t) cos!dtTo Önd the PSDs ofn0c(t)andn0s(t)we Örst Önd their autocorrelation functions.BydeÖnition,Rn0c()=E[nc(t) cos!dt+ns(t) sin!dt][nc(t+) cos!d(t+) +ns(t+) sin!d(t+)]=Rc() cos!dtcos!d(t+) +Rs() sin!dtsin!d(t+)Noting thatRc() =Rs()and using a trig identity, we Önd thatRn0c() =Rc() cos!d

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APPENDIX D: ZERO-CROSSING AND ORIGIN ENCIRCLEMENT STATISTICS2Similarly, it can be shown thatRn0s() =Rs() cos!dThe PSDs follow by applying the modulation theorem:Sn0c(f)=Sn0s(f) = 12Snc(f+fd) + 12Snc(ffd)=8<:N0;jfj B=2fdN0=2; B=2fd<jfj B=2 +fd0;otherwiseb. To Önd the cross-spectral density ofn0c(t)andn0s(t), we Örst Önd their cross-correlationfunction:Rn0cn0s()=En0c(t)n0s(t+)=E[nc(t) cos!dt+ns(t) sin!dt][nc(t+) sin!d(t+) +ns(t+) cos!d(t+)]=Rc() cos!dtsin!d(t+) +Rs() sin!dtcos!d(t+)Again noting thatRc() =Rs()and using a trig identity, we Önd thatRn0cn0s() =Rs() sin!d=Rs()exp (j!d)exp (j!d)2jApplication of the frequency translation theorem results in the following for the cross-spectral density:Sn0cn0s(f)=j2 [Sns(ffd)Sns(f+fd)]=8<:jN0=2;B=2fdfB=2 +fdjN0=2;(B=2 +fd)f (B=2fd)0;otherwiseForfd6= 0this cross-PSD is not identically zero as a sketch will show.Thusn0c(t)andn0s(t)are not uncorrelated. However, they are when sampled at the same instant becauseRn0cn0s(0) = 0.Therefore, they are statistically independent when sampled at the sameinstant since they are jointly Gaussian.Problem D.2The derivation here follows S. O. Rice, "Noise in FM Receivers,"Proc.of the Symp.ofTime Series Analysis, M. Rosenblatt, ed., New York: Wiley (1963), pp. 395-422.

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APPENDIX D: ZERO-CROSSING AND ORIGIN ENCIRCLEMENT STATISTICS3a. We need the additional clicks due to modulation.Thus, considerz(t)=Acos (!0+!d)t+nc(t) cos!0tns(t) sin!0t=Acos (!0+!d)t+n0c(t) cos (!0+!d)tn0s(t) sin (!0+!d)t=A+n0c(t)cos (!0+!d)tn0s(t) sin (!0+!d)t,R(t) cos [(!0+!d)t+ (t)]whereR(t) =q[A+n0c(t)]2+ [n0s(t)]2andtan (t) =n0s(t)A+n0c(t)It follows from the cross-correlation function found in Problem D.1 thatRn0cn0s(0) =En0c(t)n0s(t)= 0From an appropriate sketch, it follows that the probability of a counter-clockwise originencirclement in a small time intervalisPcc= PrA+n0c(t)<0andn0s(t)undergoes a + tozero crossing in[0;]The condition forn0s(t)to undergo a + tozero crossing in[0;], from an appropriatesketch, is thatn0s(0)>0and thatn0s(0) +dn0s(t)dtt=0<0for alldn0s(t)dtt=0<0. ThereforePcc= PrA+n0c(0)<0; n0s(0)>0; n0s(0) +dn0s(t)dtt=0<0;dn0s(t)dtt=0<0Thus we need the joint pdf ofX,n0c(0); Y,n0s(0);andZ,dn0s(t)dtt=0These are jointly Gaussian random variables (the derivative is a linear operation).The

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APPENDIX D: ZERO-CROSSING AND ORIGIN ENCIRCLEMENT STATISTICS4following expectations hold:E[X]=E[Y] =E[Z] = 0EX2=EY2=Z11Sn0c(f)df=Z11Sn0s(f)df=N0B,aEZ2=Z11jj2fj2Sn0s(f)df=132N0B3+ 42f2dN0B,b+a!2dE[XY]=E[Y Z] = 0E[XZ]=dRn0cn0s()d=0=d(Rs() sin!d)d=0=a!dE[ZjX]=Edn0s(t)dtt=0jn0c(0)=E!dnc(0) +dns(t)dtt=0jnc(0)=!dE[nc(0)jnc(0)] =!dXTherefore, the joint pdf ofX; Y; ZisfXY Z(x; y; z)=fZjXY(zjx; y)fX(x)fY(y)=fZjX(zjx)fX(x)fY(y)=1(2)3=2apbexp("x2+y22a+ (z+a!d)22b#)

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