TEST BANK COMMUNICATION SYSTEMS CARLSO QUESTIONS WITH CORRECT ANSWERS 100% VERIFIED A+

TEST BANK se se




to accompany
se




CommunicationSystem se




s
An Introduction to Signals and Noise inElec
se se se se se se se




trical Communication
se




Fourth Edition se




A. Bruce Carlson
se se


Rensselaer Polytechnic Institute
se se




Paul B. Crilly se se


University of Tennessee se se




Janet C. Rutledge se se


University of Maryland at Baltimore
se se se se

,SolutionsseManualsetoseaccompany
COMMUNICATIONseSYSTEMS:seANseINTRODUCTIONseTOseSIGNALSseANDseNOISEseINseELECTRICALseCOMMUNICATION,seF
OURTHseEDITION
A.seBRUCEseCARLSON,sePAULseB.seCRILLY,seANDseJANETseC.seRUTLEDGE

PublishedsebyseMcGraw-HillseHigherseEducation,seanseimprintseofseTheseMcGraw-
HillseCompanies,seInc.,se1221seAvenueseofsetheseAmericas,seNewseYork,seNYse10020.seCopyrightse©seTheseMcGraw-
HillseCompanies,seInc.,se2002,se1986,se1975,se1968.s e Allserightssereserved.

Thesecontents,seorsepartssethereof,semaysebesereproducedseinseprintseformsesolelyseforseclassroomseusesewithseCOMMUNICATIONseSYSTEMS
:seANseINTRODUCTIONseTOseSIGNALSseANDseNOISEseINseELEseCTRICALseCOMMUNICATION,seprovidedse suchse reproductionsse bears
e copyrightsenotice,sebutsemaysenotsebesereproducedseinseanyseotherseformseorseforseanyseothersepurposesewithoutsethesepriorsewrittenseconsentseofse

TheseMcGraw-
HillseCompanies,seInc.,seincluding,sebutsenotselimitedseto,seinseanysenetworkseorseotherseelectronicsestorageseorsetransmission,seorsebroadcastsefor
sedistanceselearning.


www.mhhe.com

,Chapterse2

2.1-1
jf Aesejf nse=m
cns e =Ae dts e = sinc(mse−sen)se=
+eAe
T0s/2see
j2pse (sem−nse)fse 0ts e
se jf
se
T0
−seT0s e /
0 otherwise
2se


2.1-2




c0s e v(t)s e =0
T0se/se4s e 2psent T0s e /2 se 2psents e 2A se psen
cs e =se se+ Acosse
2 dt (−seA)cosse dts e = sinse

Tses e  
n T psen 2
0
T Tsese/
4se
0
0 0 0


n 0 1 2 3 4 5 6 7
cn 0 2seAse/sep 0 2seAse/se3p 0 2seAse/se5p 0 2seAse/se7p
argsecn 0 180 0 180


2.1-3




c0s e =s e v(t)s e =A se/2 se
cs e =se T se/2se
2 2seAtse 2psent A A

0


n  Ase− T0 cos dts e = sinpsense−se 2s e (cospsense−1)
T0 0   T0 psense (psen)
n 0 1 2 3 4 5 6
cn 0.5A 0.2A 0 0.02A 0 0.01A 0
argsecn 0 0 0 0


2.1-4




2
cses e =se T0se /2 s e
Acosse
2psets e (cont.)
=0

2-1

, Tsese
0
0 T
0 0




2-2