Solution Manual for Principles of Communications , 7th Edition

Solution Manual for Principles of Communications, 7th Edition breaks down difficult textbook problems into simple solutions, making your study time more effective.

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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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