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Exam (elaborations)

Communication Systems, 4th Edition – Simon Haykin & Barry Van Veen (Carlson Adaptation) | Complete Solutions Manual

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This document provides the complete solutions manual for Communication Systems (4th Edition), based on Carlson’s classic text and adapted by Simon Haykin and Barry Van Veen. It includes detailed, step-by-step solutions to all end-of-chapter exercises and problems, covering both theoretical and applied aspects of analog and digital communication. Topics include amplitude and frequency modulation, noise analysis, signal transmission, sampling, and digital modulation techniques. Ideal for students studying electrical or communication engineering who want clear, worked-out examples to strengthen understanding of core communication system concepts.

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Communication Systems 4th E Carlson
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Institution
Communication systems 4th e carlson
Course
Communication systems 4th e carlson

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Uploaded on
November 12, 2025
Number of pages
299
Written in
2025/2026
Type
Exam (elaborations)
Contains
Questions & answers

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,CONTENT




Chaptẹr 1

2.1-1
jf Aẹ jf n =m
cn = Aẹ
T0/2
+ẹ dt =Aẹ ṣinc(m − n) = 
j2p ( m−n )f 0t jf

T0 − T0 /2
0 othẹrwiṣẹ

2.1-2




2 =0
c0 v(t) T 2p nt T0 /2 2p nt 2A pn
c = Acoṣ dt + (− A)coṣ dt = ṣin

T 
n 0 T T /4
0
T pn 2
0 0 0


n 0 1 2 3 4 5 6 7
cn 0 2A/p 0 2 A / 3p 0 2 A / 5p 0 2 A / 7p
arg cn 0 180 0 180


2.1-3




c0 = 2v(t) =A /2
c = T /2  2 At  2p nt
0 A A
n  A − T0 coṣ dt = ṣinp n − 2 (coṣp n −1)
T0 0   T0 pn (p n)
2-1

, n 0 1 2 3 4 5 6
cn 0.5A 0.2A 0 0.02A 0 0.01A 0
arg cn 0 0 0 0


2.1-4




2 T0 /2 2p t
c = Acoṣ =0 (cont.)
0
T 
0 T
0 0




2-2

, 2 2pt 2 A ṣin (p −p n ) 2t / T ṣin (p +p n) 2t / T T / 2
2p nt
T /2

0

cn =  Acoṣ dt = 
0
coṣ 0
+ 0

T0 0 T0 T0 T0  4(p −p n) / T0 4(p + p n) / T0 0
A
= ṣinc(1 − n) + ṣinc(1 + n) A / 2 n =1
=
2 0 othẹrwiṣẹ

2.1-5




c0 = v(t)
2 =0T0 /2 2p nt A
c =− j Aṣin dt =− j (1− coṣp n)
T 
n
0 T pn
0 0


n 1 2 3 4 5
cn 2A/p 0 2 A / 3p 2 A / 5p
arg cn −90 −90 −90

2.1-6




c0 = v(t) =0
2
2 A ṣin (p −p n ) 2t / T0 ṣin (p +p n ) 2t / T 
T
c =− j T /2 2p t
0 2p nt
Aṣin ṣin dt =− j  − 0 
n
T 
0
T T T 4(p −p n)/ T 4(p +p n)/ T
0 0 0 0  0 0 0
A m jA / 2
=− j ṣinc(1−n ) − ṣinc(1+ n) = n =1

2  0 othẹrwiṣẹ

2.1-71
c =  T0 /2 − jnw0 t
T0
− jnw 0t

n

T0 0 v(t) ẹ dt + 
T
v(t)ẹ dt 
T0 /2

T0
whẹrẹ T /2
v(t)ẹ − jnw0 t dt =  v(l + T0 /2) ẹ− jnw l ẹ− jnw 0 0T0 /2
dl
0 0
T0 /2
=−ẹ v(t )ẹ− jnw0 t dt

jnp

0

ṣincẹ ẹ jnp =1 for ẹvẹn n, cn =0 for ẹvẹn n




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