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Solutions Manual | Communication Systems | Carlson, Crilly, Rutledge | 4E

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The complete Solutions Manual for Communication Systems: An Introduction to Signals and Noise in Electrical Communication, Fourth Edition by A. Bruce Carlson, Paul B. Crilly, and Janet C. Rutledge. Includes detailed solutions for problems covering signals, noise, analog modulation, digital communication, and more. Essential for electrical engineering students!

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Uploaded on
December 11, 2025
Number of pages
407
Written in
2025/2026
Type
Exam (elaborations)
Contains
Questions & answers

Subjects

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Solutions Manual C




to accompany
C




CommunicationSystem C




s
An Introduction to Signals and Noise inEle
C C C C C C C




ctrical Communication
C




Fourth Edition C




A. Bruce Carlson
C C


Rensselaer Polytechnic Institute
C C




Paul B. Crilly C C


University of Tennessee C C




Janet C. Rutledge C C


University of Maryland at Baltimore
C C C C

,SolutionsCManualCtoCaccompany
COMMUNICATIONCSYSTEMS:CANCINTRODUCTIONCTOCSIGNALSCANDCNOISECINCELECTRICALCCOMMUNICATION,CFO
URTHCEDITION
A.CBRUCECCARLSON,CPAULCB.CCRILLY,CANDCJANETCC.CRUTLEDGE

PublishedCbyCMcGraw-HillCHigherCEducation,CanCimprintCofCTheCMcGraw-
HillCCompanies,CInc.,C1221CAvenueCofCtheCAmericas,CNewCYork,CNYC10020.CCopyrightC©CTheCMcGraw-
HillCCompanies,CInc.,C2002,C1986,C1975,C1968.C AllCrightsCreserved.

TheCcontents,CorCpartsCthereof,CmayCbeCreproducedCinCprintCformCsolelyCforCclassroomCuseCwithCCOMMUNICATIONCSYSTEMS:CA
NCINTRODUCTIONCTOCSIGNALSCANDCNOISECINCELECCTRICALCCOMMUNICATION,CprovidedC suchC reproductionsCbearCcopyr
ightCnotice,CbutCmayCnotCbeCreproducedCinCanyCotherCformCorCforCanyCotherCpurposeCwithoutCtheCpriorCwrittenCconsentCofCTheCMcGra
w-
HillCCompanies,CInc.,Cincluding,CbutCnotClimitedCto,CinCanyCnetworkCorCotherCelectronicCstorageCorCtransmission,CorCbroadcastCforCdist
anceClearning.

www.mhhe.com

,ChapterC2

2.1-1
jf AeCjf nC=m
cnC =Ae
T0C/2C
+ee
j2pC (Cm−nC)fC 0tC
Cjf dtC =A sinc(mC−Cn)C=
C
T0 0 otherwise
−CT0C /
2C


2.1-2




2 C =0
c0C v(t) T0C/C4C 2pCnt T0C /2 C 2pCntC 2A C pCn
cC =C + AcosC dtC (−CA)cosC dtC = sinC

TCC  
n 0 T TCC/ 4
T pCn 2
C
0
0 0 0


n 0 1 2 3 4 5 6 7
cn 0 2CAC/Cp 0 2CAC/C3p 0 2CAC/C5p 0 2CAC/C7p
argCcn 0 180 0 180


2.1-3




c0C =C v(t)C =A C/2 C
2
cC =C T C/2C 2CAtC 2pCnt A A

0


n  AC− T0 cos dtC = sinpCnC−C 2C (cospCnC−1)
T0 0   T0 pCnC (pCn)
n 0 1 2 3 4 5 6
cn 0.5A 0.2A 0 0.02A 0 0.01A 0
argCcn 0 0 0 0


2.1-4




2 2pCtC
cCC =C T0C /2 C
AcosC = (cont.)
0

2-1

, TCC
0
0 T
0 0




2-2
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